JEE Periodic Table & Periodicity Questions

The elements given are Li, Na, Be, Mg, B, and Al. We need to find which has the least atomic radius.

Atomic radii (approximate, in pm):

Li: 152, Na: 186, Be: 112, Mg: 160, B: 87, Al: 143

Boron (B) has the smallest atomic radius among these elements (approximately 87 pm).

Boron has atomic number 5 with electronic configuration $$1s^2 2s^2 2p^1$$. It is a metalloid in Group 13.

The oxide of boron is $$B_2O_3$$ (boron trioxide), which has the formula $$A_2O_3$$ where A = B.

$$B_2O_3$$ is an acidic oxide.

The answer is Option B: $$A_2O_3$$.

Find the number of valence electrons given that the lowest oxidation number of an atom in compound $$A_2B$$ is $$-2$$.

In the compound $$A_2B$$, the atom with oxidation number $$-2$$ is $$B$$, since it appears once and having the negative charge makes chemical sense for a non-metal. An atom that achieves an oxidation state of $$-2$$ gains 2 electrons to complete its octet, meaning it already has 6 electrons in its valence shell and needs 2 more to reach 8. $$ \text{Valence electrons} = 8 - |-2| = 8 - 2 = 6 $$. Elements with 6 valence electrons belong to Group 16 (e.g., O, S). In compounds like $$Na_2O$$ or $$H_2S$$, oxygen and sulfur have an oxidation state of $$-2$$, consistent with having 6 valence electrons. Also, if $$B$$ is $$-2$$, then in $$A_2B$$: $$2x + (-2) = 0 \implies x = +1$$ for each $$A$$, consistent with alkali metals (1 valence electron) and confirming that $$B$$ has 6 valence electrons.

The answer is 6.

Metal ions characterized by flame test:

$$Sr^{2+}$$: crimson red — YES

$$Ba^{2+}$$: apple green — YES

$$Ca^{2+}$$: brick red — YES

$$Cu^{2+}$$: blue-green — YES

$$Zn^{2+}$$: no characteristic flame color — NO

$$Co^{2+}$$: no characteristic flame color — NO

$$Fe^{2+}$$: no characteristic flame color — NO

Total = 4.

Therefore, the answer is $$\boxed{4}$$.

The electronegativity values on the Pauling scale are C (Carbon) = 2.55, P (Phosphorus) = 2.19, Br (Bromine) = 2.96, and At (Astatine) ≈ 2.2.

Since electronegativity increases from left to right across a period and from bottom to top in a group, the elements can be ranked by comparing these values.

Ordering from highest to lowest gives $$ Br (2.96) > C (2.55) > At (2.2) > P (2.19) $$. Therefore, the correct order is Br > C > At > P.

Electron gain enthalpy, $$\Delta_{eg}H$$, is the enthalpy change associated with the addition of an electron to an isolated gaseous atom.

For most non-metals, energy is released when an electron is added. Therefore, electron gain enthalpy values are generally negative.

As we move down a group, electron gain enthalpy usually becomes less negative because the incoming electron is added farther from the nucleus and experiences greater shielding.

An important exception occurs in the second period. Due to their very small size, elements such as fluorine and oxygen experience significant interelectronic repulsions, making their electron gain enthalpies less negative than expected.

The approximate electron gain enthalpy values are

$$F=-328\ \text{kJ mol}^{-1}$$

$$Cl=-349\ \text{kJ mol}^{-1}$$

$$I=-295\ \text{kJ mol}^{-1}$$

$$At\approx -270\ \text{kJ mol}^{-1}$$

$$S=-200\ \text{kJ mol}^{-1}$$

$$Se=-195\ \text{kJ mol}^{-1}$$

$$Te=-190\ \text{kJ mol}^{-1}$$

$$Po\approx -174\ \text{kJ mol}^{-1}$$

Checking each statement algebraically,

For Option A,

$$\Delta_{eg}H(Cl)<\Delta_{eg}H(F)$$

$$-349<-328$$

This statement is correct.

For Option B,

$$\Delta_{eg}H(Se)<\Delta_{eg}H(S)$$

$$-195<-200$$

This statement is incorrect because

$$-195>-200$$

For Option C,

$$\Delta_{eg}H(I)<\Delta_{eg}H(At)$$

$$-295<-270$$

This statement is correct.

For Option D,

$$\Delta_{eg}H(Te)<\Delta_{eg}H(Po)$$

$$-190<-174$$

This statement is correct.

Therefore, the only incorrect statement is

$$\Delta_{eg}H(Se)<\Delta_{eg}H(S)$$

Hence, the correct answer is

$$\text{Option B}$$

All the given ions are isoelectronic, i.e. each of them possesses $$18$$ electrons:

$$S^{2-}\;(Z = 16,\; 16 + 2 = 18\ \text{electrons})$$
$$Cl^- \;(Z = 17,\; 17 + 1 = 18\ \text{electrons})$$
$$K^+ \;(Z = 19,\; 19 - 1 = 18\ \text{electrons})$$
$$Ca^{2+}\;(Z = 20,\; 20 - 2 = 18\ \text{electrons})$$

For an isoelectronic series the principal quantum number $$n$$ is the same, so the ionic radius depends mainly on the effective nuclear charge $$Z_{\text{eff}}$$ experienced by the electrons.

Rule: Within an isoelectronic series, ionic radius decreases as the atomic number $$Z$$ (and hence $$Z_{\text{eff}}$$) increases, because the nucleus pulls the same number of electrons more strongly.

Ordering the ions by increasing atomic number:

$$Z = 16 \;(S^{2-}) \lt 17 \;(Cl^-) \lt 19 \;(K^+) \lt 20 \;(Ca^{2+})$$

Therefore the radii follow the opposite trend:

$$\text{smallest radius} \; Ca^{2+} \lt K^+ \lt Cl^- \lt S^{2-} \; \text{largest radius}$$

Writing the ions in increasing order of ionic radii:

$$Ca^{2+} \lt K^+ \lt Cl^- \lt S^{2-}$$

This matches Option D.

Final Answer: Option D

We need to evaluate both the Assertion and the Reason.

Assertion A: The energy required to form $$Mg^{2+}$$ from $$Mg$$ is much higher than that required to produce $$Mg^{+}$$.

To form $$Mg^{+}$$ from $$Mg$$, we need the first ionization energy ($$IE_1$$):

$$ Mg \rightarrow Mg^{+} + e^{-} \quad (IE_1 = 737.7 \text{ kJ/mol}) $$

To form $$Mg^{2+}$$ from $$Mg$$, we need both the first and second ionization energies ($$IE_1 + IE_2$$):

$$ Mg \rightarrow Mg^{2+} + 2e^{-} \quad (IE_1 + IE_2 = 737.7 + 1450.7 = 2188.4 \text{ kJ/mol}) $$

Since $$IE_1 + IE_2$$ is much greater than $$IE_1$$ alone, the energy required to form $$Mg^{2+}$$ from $$Mg$$ is indeed much higher than that required to form $$Mg^{+}$$. Assertion A is true.

Reason R: $$Mg^{2+}$$ is a smaller ion and carries more charge than $$Mg^{+}$$.

After removing the first electron, $$Mg^{+}$$ has the electronic configuration $$1s^2 2s^2 2p^6 3s^1$$. Removing the second electron from $$Mg^{+}$$ to form $$Mg^{2+}$$ requires more energy because:

• $$Mg^{+}$$ already has a net positive charge, so the remaining electrons experience a greater effective nuclear charge ($$Z_{eff}$$).
• $$Mg^{2+}$$ (with configuration $$1s^2 2s^2 2p^6$$) is smaller in size than $$Mg^{+}$$, meaning the electron being removed is held more tightly by the nucleus.

The fact that $$Mg^{2+}$$ is smaller and carries more charge than $$Mg^{+}$$ directly explains why the second ionization energy is higher, and therefore why forming $$Mg^{2+}$$ requires much more total energy. Reason R is true and is the correct explanation of A.

The correct answer is Option D: Both A and R are true and R is the correct explanation of A.

Assertion (A): The first ionization enthalpy of 3d series elements is more than that of Group 2 metals.

Reason (R): In 3d series, successive filling of d-orbitals takes place.

Analysis of Assertion (A):

The first ionization enthalpies (in kJ/mol) of the 3d series elements range from approximately 631 (Sc) to 906 (Zn). When compared with the Group 2 element of the same period (Ca, IE₁ = 590 kJ/mol), all 3d elements have higher ionization enthalpies. This is because the nuclear charge increases across the period while d-electrons provide poor shielding.

Assertion (A) is True.

Analysis of Reason (R):

In the 3d series (Sc to Zn), d-orbitals are progressively filled from $$3d^1$$ to $$3d^{10}$$.

Reason (R) is True.

Connection: The successive filling of d-orbitals results in poor shielding of nuclear charge, leading to higher effective nuclear charge and thus higher ionization enthalpy compared to the preceding Group 2 element. R correctly explains A.

The correct answer is Option A: Both (A) and (R) are true and (R) is the correct explanation of (A).

We need to match elements with their flame test colours.

The characteristic flame colours of alkali and alkaline earth metals are:

• K (Potassium) — Violet
• Ca (Calcium) — Brick Red
• Sr (Strontium) — Crimson Red (or Apple Red)
• Ba (Barium) — Apple Green (or Yellow-Green)

A. K → II (Violet)

B. Ca → I (Brick Red)

C. Sr → IV (Crimson Red)

D. Ba → III (Apple Green)

The correct matching is A-II, B-I, C-IV, D-III.

The correct answer is Option C: A-II, B-I, C-IV, D-III.

We need to identify which compound does not exist among the given options.

Option A: NaO$$_2$$ (Sodium superoxide)

Alkali metals react with oxygen to form different products. Lithium forms Li$$_2$$O (oxide), sodium forms Na$$_2$$O$$_2$$ (peroxide), and potassium/rubidium/cesium form superoxides (KO$$_2$$, RbO$$_2$$, CsO$$_2$$). Sodium does not readily form a stable superoxide NaO$$_2$$ under normal conditions. This compound is considered non-existent or extremely unstable.

Option B: BeH$$_2$$ - Beryllium hydride exists as a polymeric solid.

Option C: PbEt$$_4$$ - Tetraethyl lead exists and was historically used as an anti-knock agent in gasoline.

Option D: NH$$_4$$BeF$$_4$$ - Ammonium tetrafluoroberyllate is a known compound.

The correct answer is Option A: NaO$$_2$$.

We need to identify the least stable hydride among BeH$$_2$$, HF, NH$$_3$$, and LiH.

We start by noting that

The stability of a hydride depends on the strength of the bond between the element and hydrogen. Key factors include:

- Higher electronegativity of the bonded atom leads to stronger, more polar bonds with H, increasing stability.

- Higher bond energy means greater stability.

- Electron-deficient hydrides (like those of Group 2 and 13) tend to be less stable.

Next,

HF: Fluorine has the highest electronegativity (3.98). The H-F bond is very strong (bond energy ~568 kJ/mol). HF is very stable.

Nitrogen has high electronegativity (3.04). The N-H bond is strong (bond energy ~391 kJ/mol). NH$$_3$$ is stable.

LiH: This is an ionic hydride (Li$$^+$$ H$$^-$$). Despite the low electronegativity of Li, the ionic bond provides reasonable stability. LiH is moderately stable.

Beryllium has low electronegativity (1.57) and forms an electron-deficient covalent hydride. Be has only 2 valence electrons and forms only 2 bonds with H, leaving empty orbitals. BeH$$_2$$ is a polymeric hydride in solid state (with 3-centre 2-electron bonds) that readily decomposes. It is the least stable.

The least stable hydride is BeH$$_2$$.

The correct answer is Option 1: BeH$$_2$$.

In alkali metals, the melting point decreases as we go down the group due to increasing atomic size and weakening of metallic bonding.

Order: Li > Na > K > Rb > Cs

Caesium (Cs) has the least melting point (28.4°C) among alkali metals.

Ionisation enthalpy (IE) is the energy required to remove the most loosely bound electron from a gaseous atom.

General periodic trend: Across a period (left → right) the nuclear charge increases while the principal quantum number remains the same, so atomic size decreases and the IE normally increases.
However, two well-known deviations appear in every short period:

Case 1: Be vs B - Be has the configuration $$1s^22s^2$$ (completely filled $$2s$$ subshell) while B is $$1s^22s^22p^1$$. A completely filled subshell is extra-stable, so more energy is required to remove an electron from Be than from B. Therefore $$\text{IE(Be)} \gt \text{IE(B)}$$, opposite to the simple left→right trend.

Case 2: N vs O - N is $$1s^22s^22p^3$$ (exactly half-filled $$2p$$ subshell); O is $$1s^22s^22p^4$$. The half-filled $$2p$$ subshell of N is more stable than the partly filled subshell of O, so $$\text{IE(N)} \gt \text{IE(O)}$$, again reversing the straightforward trend.

Placing the seven given elements in Period 2 and applying these two exceptions:

1. Start at the far left: $$\text{IE(Li)}$$ is the least.
2. Between B and Be, the exception “Be > B” applies, so $$\text{IE(B)} \lt \text{IE(Be)}$$.
3. C follows the normal order, coming after Be.
4. Between N and O, the exception “N > O” applies, so $$\text{IE(O)} \lt \text{IE(N)}$$.
5. F, at the extreme right, has the maximum IE in the period.

Arranging from the smallest to the largest ionisation enthalpy:

$$\text{Li} \lt \text{B} \lt \text{Be} \lt \text{C} \lt \text{O} \lt \text{N} \lt \text{F}$$

This matches Option D.

Answer: Option D

The assertion is incorrect because alkali metals and their salts impart characteristic colours in an oxidizing flame, not in a reducing flame. The high temperature of the oxidizing flame excites the valence electrons of the metal atoms, and when these electrons return to their ground state, they emit characteristic colours.

The reason is correct because flame tests are commonly used for the qualitative detection of alkali metals based on the characteristic colours they produce.

Hence, the assertion is incorrect but the reason is correct. Therefore, the correct answer is $$\boxed{\text{Option C}}$$.

Statement I: Lithium and Magnesium do not form superoxide.

Statement II: The ionic radius of Li⁺ is larger than ionic radius of Mg²⁺.

Analysis of Statement I:

Lithium, due to its small size and high charge density, strongly polarizes the larger superoxide ion (O₂⁻), destabilizing it. Li forms oxide (Li₂O) and peroxide (Li₂O₂) but not a stable superoxide. Similarly, Mg forms MgO but not a stable superoxide.

Statement I is correct.

Analysis of Statement II:

According to NCERT data:

Li⁺ ionic radius ≈ 76 pm, Mg²⁺ ionic radius ≈ 72 pm (Shannon radii)

However, using the values commonly cited in NCERT textbooks: Li⁺ ≈ 60 pm, Mg²⁺ ≈ 65 pm.

Using NCERT values, Li⁺ is actually smaller than Mg²⁺.

Statement II is incorrect (in the NCERT context).

Since Statement I is correct and Statement II is incorrect, the answer is Option B: Statement I is correct but Statement II is incorrect.

We need to identify the correct statements about alkali metals.

Statement A: The order of standard reduction potential (M$$^+$$|M) for alkali metal ions is Na > Rb > Li. The standard reduction potentials are approximately: Li at $$-3.04$$ V, Na at $$-2.71$$ V, K at $$-2.93$$ V, Rb at $$-2.98$$ V, and Cs at $$-3.03$$ V. Although Li has the highest ionization energy, its exceptionally high hydration energy (due to the very small Li$$^+$$ ionic radius) makes its reduction potential the most negative. The order Na ($$-2.71$$) > Rb ($$-2.98$$) > Li ($$-3.04$$) is correct.

Statement B: CsI is claimed to be highly soluble in water. CsI consists of two large ions with low lattice energy but also low hydration energies. In the context of JEE, this claim is not considered a standard correct statement. This is incorrect.

Statement C: Lithium carbonate is claimed to be highly stable to heat. However, Li$$_2$$CO$$_3$$ decomposes on heating:

$$\text{Li}_2\text{CO}_3 \xrightarrow{\Delta} \text{Li}_2\text{O} + \text{CO}_2$$

The small, highly charged Li$$^+$$ cation strongly polarizes the large carbonate ion, destabilizing it. So Li$$_2$$CO$$_3$$ is the least stable alkali metal carbonate to heat. This is incorrect.

Statement D: Potassium dissolved in concentrated liquid ammonia is claimed to be blue and paramagnetic. However, dilute solutions of alkali metals in ammonia are blue and paramagnetic, while concentrated solutions are bronze and diamagnetic due to pairing of solvated electrons. This is incorrect.

Statement E: All alkali metal hydrides are ionic solids. Indeed, all alkali metals form ionic hydrides MH with M$$^+$$ cations and H$$^-$$ anions, adopting the NaCl-type crystal structure. This is correct.

The correct statements are A and E only. Hence, the correct answer is Option 3.

Metallic character increases down a group and from right to left in a period.

K (Group 1, Period 4) > Ca (Group 2, Period 4) > Be (Group 2, Period 2).

The correct answer is Option 4: K > Ca > Be.

Assertion (A) is correct.

The standard reduction potential of beryllium is less negative than that of the other alkaline earth metals.

$$E^\circ(\mathrm{Be^{2+}/Be})=-1.97\ \mathrm{V}$$

whereas for the heavier Group 2 metals the values are more negative.

The reduction potential depends on the balance of atomisation enthalpy, ionisation enthalpy and hydration enthalpy involved in the process

$$M_{(s)} \rightarrow M^{2+}_{(aq)} + 2e^-$$

For beryllium,

- Atomisation enthalpy is very high because of its strong metallic bonding.

- First and second ionisation enthalpies are exceptionally high due to its small atomic size.

- Hydration enthalpy of $$\mathrm{Be^{2+}}$$ is also very large because of its high charge density.

Although hydration releases a large amount of energy, it does not completely compensate for the very high atomisation and ionisation enthalpies.

As a result, beryllium has a lower tendency to oxidise than the other alkaline earth metals, leading to a less negative standard reduction potential.

Hence, Reason (R) is also correct and it correctly explains Assertion (A).

Therefore,

$$\boxed{\text{Both A and R are true, and R is the correct explanation of A}}$$

We need to identify compound 'A' formed when BeO reacts with ammonia and hydrogen fluoride, which on thermal decomposition gives $$BeF_2$$ and $$NH_4F$$.

Analyse the decomposition products.

The thermal decomposition gives $$BeF_2$$ and $$NH_4F$$. Let us work backwards to identify 'A'.

If compound 'A' decomposes to give $$BeF_2 + 2NH_4F$$, then 'A' must contain Be, N, H, and F atoms in the combined ratio. Adding the atoms:

$$BeF_2 + 2NH_4F = Be + 2N + 8H + 4F$$

This corresponds to the formula $$(NH_4)_2BeF_4$$ (ammonium tetrafluoroberyllate).

Verify the formation reaction.

The reaction of BeO with $$NH_3$$ and HF can be written as:

$$BeO + 2NH_3 + 4HF \rightarrow (NH_4)_2BeF_4 + H_2O$$

This is consistent because:

- BeO provides the beryllium

- HF provides the fluorine

- $$NH_3$$ and HF together form $$NH_4^+$$ ions

- The product $$(NH_4)_2BeF_4$$ contains the tetrafluoroberyllate ion $$[BeF_4]^{2-}$$ with two ammonium counterions

Verify the decomposition.

$$(NH_4)_2BeF_4 \xrightarrow{\Delta} BeF_2 + 2NH_4F$$

Atoms balance: Be: 1=1, N: 2=2, H: 8=8, F: 4=2+2=4. Correct.

The correct answer is Option (2): $$(NH_4)_2BeF_4$$.

We need to evaluate both the Assertion and the Reason.

Assertion A: Physical properties of isotopes of hydrogen are different.

The three isotopes of hydrogen are:

• Protium ($$^{1}H$$): mass number = 1
• Deuterium ($$^{2}H$$ or D): mass number = 2
• Tritium ($$^{3}H$$ or T): mass number = 3

Physical properties such as boiling point, melting point, and density differ among these isotopes. For example:

• Boiling point of $$H_2O$$ = 100°C, while boiling point of $$D_2O$$ = 101.4°C
• Density of $$H_2O$$ = 1.00 g/mL, while density of $$D_2O$$ = 1.107 g/mL

Assertion A is true.

Reason R: The mass difference between isotopes of hydrogen is very large.

The relative mass differences between hydrogen isotopes are the largest among all elements:

• Deuterium is 2 times as heavy as Protium (ratio = 2:1)
• Tritium is 3 times as heavy as Protium (ratio = 3:1)

For comparison, carbon isotopes $$^{12}C$$ and $$^{13}C$$ have a mass ratio of only 13:12, which is a much smaller relative difference. The exceptionally large relative mass difference in hydrogen isotopes is unique among all elements. Reason R is true.

Explanation of the link: Physical properties depend on the mass of atoms/molecules. Since hydrogen isotopes have the largest relative mass differences, the differences in their physical properties (such as boiling point, melting point, bond energy, and density) are pronounced and measurable. The large mass difference is the direct cause of the significant variation in physical properties.

Therefore, both A and R are true, and R is the correct explanation of A.

The correct answer is Option A: Both A and R are true and R is the correct explanation of A.

Hence, Option A is correct.

Let us match each item from List-I with List-II:

A. K (Potassium): Potassium ions play a crucial role in the sodium-potassium pump in biological systems, which maintains the electrochemical gradient across cell membranes. → III. Sodium potassium pump

B. KCl (Potassium chloride): KCl is widely used as a potash fertilizer in agriculture. → II. Fertilizer

C. KOH (Potassium hydroxide): KOH is used as an absorbent of CO$$_2$$ in various applications. → IV. Absorbent of CO$$_2$$

D. Li (Lithium): Lithium is used in thermonuclear reactions (hydrogen bombs), where lithium deuteride serves as the fusion fuel. → I. Thermonuclear reactions

The correct matching is: A(III), B(II), C(IV), D(I).

The metallic character of an element depends on its tendency to lose electrons. Metallic character decreases from left to right across a period due to a decrease in atomic size and an increase in ionization energy, while it increases down a group because of an increase in atomic size and a decrease in ionization energy.

Among the given elements, potassium ((K)) belongs to Group 1 and Period 4, making it the most metallic because it loses its valence electron most readily. Magnesium ((Mg)), a Group 2 element in Period 3, is less metallic than potassium but more metallic than beryllium. Beryllium ((Be)), which lies above magnesium in the same group, has a smaller atomic size and higher ionization energy, making it less metallic than magnesium. Silicon ((Si)) is a metalloid located in Group 14 and is the least metallic among the given elements.

Therefore, the correct order of metallic character is

$$Si < Be < Mg < K.$$

Hence, the correct answer is $$\boxed{\text{Option A}}$$.

Assertion A: Sodium is about 30 times as abundant as potassium in the oceans.

This statement is true. The concentration of sodium in seawater is approximately 10,800 ppm, while potassium is only about 380 ppm, making sodium roughly 28-30 times more abundant.

Reason R: Potassium is bigger in size than sodium.

This statement is true. Potassium (atomic radius $$\approx 227$$ pm) is larger than sodium (atomic radius $$\approx 186$$ pm), since potassium lies one period below sodium in Group 1.

Connecting A and R:

The larger size of potassium ions ($$K^+$$) allows them to be more readily adsorbed and trapped by clay minerals and silicate structures in rocks and soil. This means potassium is preferentially retained on land rather than being washed into the oceans. Sodium ions ($$Na^+$$), being smaller, are less strongly held by soil particles and are more easily leached into rivers and eventually the oceans.

Thus, R correctly explains A -- the larger size of potassium is the key reason it is less abundant in seawater.

The correct answer is Option A: Both A and R are true and R is the correct explanation of A.

Electron gain enthalpy (EGE) is the energy change when an electron is added to a neutral gaseous atom.

Noble gases (Ne, Ar) have completely filled electron shells, so they have very positive (highly unfavorable) electron gain enthalpies. They strongly resist gaining an electron.

Halogens (F, Cl) have the most negative (highly favorable) electron gain enthalpies. Among halogens, Cl has the most negative EGE (more negative than F, due to the small size of F causing electron-electron repulsion).

Among noble gases, Ne has a more positive EGE than Ar (smaller size, more repulsion).

Therefore, the maximum difference in electron gain enthalpies would be between the species with the most positive EGE (Ne) and the most negative EGE (Cl).

The correct answer is Ne and Cl.

Noble (inert) gases have positive electron gain enthalpy. They possess completely filled and highly stable electron configurations: He (1s$$^2$$), Ne (2s$$^2$$2p$$^6$$), Ar (3s$$^2$$3p$$^6$$), Kr (4s$$^2$$4p$$^6$$), Xe (5s$$^2$$5p$$^6$$). Adding an extra electron would place it in the next higher principal shell, which is energetically very unfavourable; therefore, the process of electron addition is endothermic (positive electron gain enthalpy).

As we move down the group from He to Xe, the atomic size increases, so the next available shell is farther from the nucleus; the shielding effect increases, making it progressively more difficult to stabilise the incoming electron; and the incoming electron must enter an increasingly higher energy level (n = 2 for He, n = 3 for Ne, n = 4 for Ar, etc.), all of which make the electron gain enthalpy more positive. Therefore, the positive electron gain enthalpy increases as we go down the group.

In increasing order of positive electron gain enthalpy: $$\text{He} < \text{Ne} < \text{Kr} < \text{Xe}$$. He has the least positive electron gain enthalpy and Xe has the most positive.

Answer: Option B

Isoelectronic species have the same number of electrons.

Counting electrons for each species:

• O²⁻: 8 + 2 = 10 electrons
• F⁻: 9 + 1 = 10 electrons
• Al: 13 electrons
• Mg²⁺: 12 - 2 = 10 electrons
• Na⁺: 11 - 1 = 10 electrons
• O⁺: 8 - 1 = 7 electrons
• Mg: 12 electrons
• Al³⁺: 13 - 3 = 10 electrons
• F: 9 electrons

Species with 10 electrons: O²⁻, F⁻, Mg²⁺, Na⁺, Al³⁺ → 5 isoelectronic species

The total number of isoelectronic species is $$\mathbf{5}$$.

We need to find the valence shell electron configuration of the element just above "E" in group 16. Element "E" is in Period 4, Group 16, which corresponds to Selenium (Se). Therefore, the element just above it in Period 3 is Sulphur (S).

Sulphur (S) has atomic number 16 and an electronic configuration of $$1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^4$$; hence the valence shell (n = 3) configuration is $$3s^2\, 3p^4$$. Thus, the correct answer is Option C: $$3s^2\, 3p^4$$.

We need to evaluate the Assertion-Reason statement about $$O^{2-}$$ and $$Mg^{2+}$$.

First, we check whether they are isoelectronic. $$O^{2-}$$ has 10 electrons (8 + 2), and $$Mg^{2+}$$ also has 10 electrons (12 - 2), so the Reason is true.

Next, we examine whether their ionic radii are the same. Since both species have 10 electrons but different nuclear charges (Z = 8 for $$O^{2-}$$ and Z = 12 for $$Mg^{2+}$$), the one with the greater nuclear charge pulls the electron cloud more tightly and has a smaller ionic radius. Therefore, $$r(O^{2-}) > r(Mg^{2+})$$, and the Assertion is false.

In conclusion, the Assertion is false but the Reason is true.

The correct answer is Option D.

We need to evaluate two statements about atomic and ionic radii.

Statement I: In Cl₂ molecule, the covalent radius is double the atomic radius of chlorine.

This statement is incorrect. In fact, the covalent radius is half the internuclear distance in a Cl₂ molecule. The atomic radius (van der Waals radius) of chlorine is always greater than its covalent radius. The covalent radius of Cl is approximately 99 pm, while its van der Waals radius is about 175 pm. So the covalent radius is NOT double the atomic radius; rather, the covalent radius is less than the atomic radius.

Statement II: Radius of anionic species is always greater than their parent atomic radius.

This statement is correct. When an atom gains electrons to form an anion, the increased electron-electron repulsion causes the electron cloud to expand. Additionally, the same nuclear charge now holds more electrons, leading to a decreased effective nuclear charge per electron. Therefore, anions are always larger than their parent atoms.

For example:

• Cl atom: 99 pm (covalent radius)
• Cl⁻ ion: 181 pm
• O atom: 73 pm
• O²⁻ ion: 140 pm

Since Statement I is incorrect and Statement II is correct, the answer is Option D.

Identify the species and their isoelectronic nature.

All the given ions are isoelectronic with 10 electrons each:

$$N^{3-}$$ (7 + 3 = 10 e$$^-$$), $$O^{2-}$$ (8 + 2 = 10 e$$^-$$), $$F^-$$ (9 + 1 = 10 e$$^-$$), $$Na^+$$ (11 - 1 = 10 e$$^-$$), $$Mg^{2+}$$ (12 - 2 = 10 e$$^-$$).

Apply the rule for isoelectronic species.

For isoelectronic species, the ionic radius decreases as the nuclear charge (atomic number) increases. This is because a higher nuclear charge pulls the same number of electrons closer to the nucleus.

Arrange in order of increasing ionic size.

Nuclear charges: $$Mg^{2+}$$ (Z=12) > $$Na^+$$ (Z=11) > $$F^-$$ (Z=9) > $$O^{2-}$$ (Z=8) > $$N^{3-}$$ (Z=7)

Therefore, ionic size increases as:

$$Mg^{2+} < Na^+ < F^- < O^{2-} < N^{3-}$$

This matches Option A.

The answer is $$\boxed{\text{Option A}}$$.

First Ionization Energy Values:

$$\mathrm{Be = 899\ kJ\ mol^{-1}}$$

$$\mathrm{B = 801\ kJ\ mol^{-1}}$$

$$\mathrm{N = 1402\ kJ\ mol^{-1}}$$

$$\mathrm{O = 1314\ kJ\ mol^{-1}}$$

Generally, ionization energy increases across a period due to increasing effective nuclear charge.

However, exceptions occur because completely filled and half-filled subshells possess extra stability.

1. Beryllium vs Boron

Electronic configurations:

$$\mathrm{Be : 1s^2\ 2s^2}$$

$$\mathrm{B : 1s^2\ 2s^2\ 2p^1}$$

Beryllium has a completely filled $$\mathrm{2s}$$ subshell, which is highly stable.

In boron, the outermost electron enters the higher-energy $$\mathrm{2p}$$ orbital.

It is easier to remove the single $$\mathrm{2p}$$ electron from boron than to disturb the stable filled $$\mathrm{2s}$$ subshell of beryllium.

Therefore:

$$\mathrm{IE(Be) > IE(B)}$$

2. Nitrogen vs Oxygen

Electronic configurations:

$$\mathrm{N : 1s^2\ 2s^2\ 2p^3}$$

$$\mathrm{O : 1s^2\ 2s^2\ 2p^4}$$

Nitrogen has an exactly half-filled $$\mathrm{2p}$$ subshell, giving extra stability due to symmetry and exchange energy.

In oxygen, one $$\mathrm{2p}$$ orbital contains paired electrons, producing inter-electronic repulsion.

Hence, it is easier to remove one electron from oxygen than to break the stable half-filled configuration of nitrogen.

Therefore:

$$\mathrm{IE(N) > IE(O)}$$

We need to identify which pairs of elements have nearly the same or identical electron gain enthalpies.

Pair (A): Rb and Cs

Both rubidium and caesium are alkali metals in the same group. As we go down the group, the electron gain enthalpy becomes less negative but for heavier alkali metals (Rb and Cs), the values become very similar due to the shielding effect and large atomic size. The electron gain enthalpies of Rb and Cs are nearly identical (both around -47 to -46 kJ/mol).

Pair (B): Na and K

Sodium and potassium are also alkali metals, but their electron gain enthalpies differ noticeably. Na has an electron gain enthalpy of about -53 kJ/mol while K has about -48 kJ/mol. These are not nearly the same.

Pair (C): Ar and Kr

Both argon and krypton are noble gases with completely filled electron shells. Noble gases have near-zero (positive) electron gain enthalpies because they resist gaining an extra electron. The electron gain enthalpies of Ar and Kr are nearly identical (both are positive and very close to zero).

Pair (D): I and At

Iodine and astatine are halogens. Their electron gain enthalpies are different — iodine has a much more negative value compared to astatine, which is a larger and more metallic element.

Therefore, the pairs with nearly identical electron gain enthalpies are A (Rb and Cs) and C (Ar and Kr).

The correct answer is Option C: A & C only.

Elements of the third period have the principal quantum number $$n = 3$$ for their valence-shell electrons. The four given configurations therefore correspond to

$$A : 3s^{2}\; \Longrightarrow\;$$ magnesium (Mg)
$$B : 3s^{2}\,3p^{1}\; \Longrightarrow\;$$ aluminium (Al)
$$C : 3s^{2}\,3p^{3}\; \Longrightarrow\;$$ phosphorus (P)
$$D : 3s^{2}\,3p^{4}\; \Longrightarrow\;$$ sulphur (S)

In a period, the first ionization enthalpy (I.E.) generally increases from left to right because the nuclear charge increases while the valence shell remains the same. However, two well-known exceptions disturb the smooth rise:

Case 1: The jump from a filled $$s$$ subshell ($$ns^{2}$$) to the first $$p$$ electron ($$ns^{2}np^{1}$$).
Removing an electron from the new, higher-energy $$p$$ subshell needs less energy than removing it from the lower-energy, fully-filled $$s$$ subshell. Hence
$$\text{I.E.}(Al) \lt \text{I.E.}(Mg)$$

Case 2: Half-filled and fully-filled subshell stability.
A half-filled $$p^{3}$$ subshell (as in P) is extra stable. When we move from $$p^{3}$$ to $$p^{4}$$ (S), the added electron must pair up in an already occupied orbital, introducing inter-electronic repulsion and slightly lowering the I.E. Therefore
$$\text{I.E.}(S) \lt \text{I.E.}(P)$$

Combining the normal trend with the two exceptions gives the overall order

$$\text{Al} \lt \text{Mg} \lt \text{S} \lt \text{P}$$

Replacing element symbols with the given letters:

$$B \lt A \lt D \lt C$$

Hence the correct order of first ionization enthalpy is described by Option B.

Answer → Option B \bigl(B < A < D < C\bigr)

We need to find the correct decreasing order of metallic character for the elements Na, Mg, Be, Si, and P.

Metallic character depends on how easily an atom can lose electrons. It increases as we go down a group (larger atomic size, easier to lose outer electrons) and decreases as we go from left to right across a period (increasing nuclear charge holds electrons more tightly).

Let us identify the positions of these elements in the periodic table. Na (sodium) is in Group 1, Period 3. Mg (magnesium) is in Group 2, Period 3. Be (beryllium) is in Group 2, Period 2. Si (silicon) is in Group 14, Period 3. P (phosphorus) is in Group 15, Period 3.

Among the Period 3 elements, the metallic character decreases from left to right: $$Na > Mg > Si > P$$. Now, Be is in the same group as Mg but in a higher period (Period 2), so Be is less metallic than Mg (since metallic character increases going down the group).

Comparing Be with Si and P: Be is in Group 2 (a metal), while Si is a metalloid and P is a non-metal. So Be is more metallic than both Si and P.

Putting it all together, the decreasing order of metallic character is $$Na > Mg > Be > Si > P$$.

Hence, the correct answer is Option A.

We are given the first ionization enthalpies: Na = 496 kJ/mol, Mg = 737 kJ/mol, and we need to find the first ionization enthalpy of Al. The question also mentions 786 kJ/mol, but the standard known values are Na = 496, Mg = 737, and Al = 577 kJ/mol.

In the periodic table, as we move from left to right across a period, the ionization enthalpy generally increases due to increasing nuclear charge. However, Al (electronic configuration: $$[Ne] 3s^2 3p^1$$) has a lower first ionization enthalpy than Mg ($$[Ne] 3s^2$$). This is because the outermost electron of Al is in the 3p orbital, which is higher in energy and better shielded by the 3s electrons compared to Mg's 3s electron. The 3p electron is easier to remove than a paired 3s electron.

Among the given options, 577 kJ/mol correctly reflects this trend — it is lower than Mg's 737 kJ/mol but higher than Na's 496 kJ/mol, consistent with the anomalous dip at Al in the ionization energy trend across Period 3.

Hence, the correct answer is Option C.

We need to identify a metal with a very low melting point whose position in the periodic table is closer to a metalloid.

Analyzing the options:

Option A: Al (Aluminium) - Melting point is $$660°C$$, which is relatively high. It is close to metalloids but does not have a very low melting point.

Option B: Ga (Gallium) - Gallium has an exceptionally low melting point of $$29.76°C$$ (it melts in your hand). In the periodic table, Gallium is in Group 13, Period 4, and is adjacent to the metalloid Germanium (Ge). This makes it close to the metal-metalloid boundary.

Option C: Se (Selenium) - Selenium is itself a non-metal/metalloid, not a metal.

Option D: In (Indium) - Indium has a melting point of $$156.6°C$$, which is low but not as remarkably low as Gallium. It is also farther from the metalloid boundary compared to Gallium.

Gallium perfectly fits both criteria: it has a very low melting point and its periodic position is close to metalloids.

The correct answer is Option B.

We need to determine the correct order of electron gain enthalpy (in terms of magnitude, more negative = more exothermic = higher tendency to gain electron) for Cl, F, Te, and Po.

Among halogens, chlorine has the most negative electron gain enthalpy (not fluorine). This is because fluorine is very small and has high electron-electron repulsion in its compact 2p orbitals, making it slightly less favorable to add an electron compared to chlorine.

So: $$Cl > F$$ (in terms of magnitude of electron gain enthalpy)

Te and Po belong to Group 16. As we go down the group, the electron gain enthalpy generally decreases in magnitude due to increasing atomic size and shielding effects.

So: $$Te > Po$$

Halogens (Group 17) generally have more negative electron gain enthalpies than chalcogens (Group 16) of similar periods because halogens need just one electron to complete their octet.

So: $$Cl > F > Te > Po$$

Hence, the correct answer is Option B.

We need to identify which of the given elements is a metalloid.

The commonly recognized metalloids are: Boron (B), Silicon (Si), Germanium (Ge), Arsenic (As), Antimony (Sb), Tellurium (Te), and Polonium (Po).

Option A: Sc (Scandium) — This is a transition metal.

Option B: Pb (Lead) — This is a metal (post-transition metal in Group 14).

Option C: Bi (Bismuth) — This is a metal (post-transition metal in Group 15).

Option D: Te (Tellurium) — This is a metalloid (Group 16).

The correct answer is Option D: Te.

We need to evaluate the Assertion-Reason statement about ionization enthalpies of oxygen and nitrogen. The first ionization enthalpy of oxygen (1314 kJ/mol) is indeed lower than that of nitrogen (1402 kJ/mol), which is a well-known exception in the periodic trend; thus Assertion A is correct.

Nitrogen has the electronic configuration 1s² 2s² 2p³, corresponding to half-filled 2p orbitals that confer extra stability, whereas oxygen has 1s² 2s² 2p⁴. In oxygen, the fourth electron in the 2p subshell must pair with one of the existing electrons, leading to increased electron-electron repulsion. This makes it easier to remove an electron from oxygen compared to nitrogen; therefore, Reason R is correct.

The reason directly explains why oxygen has a lower ionization enthalpy—the extra electron-electron repulsion due to pairing in the 2p orbital makes the electron easier to remove. Hence, R is the correct explanation of A.

Option A: Both A and R are correct and R is the correct explanation of A.

Silver is a transition metal. Transition metals generally have high melting points because their atomic structure allows for strong metallic bonding. They have a large number of delocalized electrons (from both s and d orbitals) that act like a strong "glue" holding the crystal lattice together. Breaking these strong bonds requires a significant amount of heat energy.

In contrast, the other three options are famous exceptions in the metal family due to their unusually weak metallic bonds:

• Mercury (Hg) is a liquid at room temperature.
• Gallium (Ga) and Cesium (Cs) have such low melting points that they will actually melt from the ambient heat of your hand.

We need to identify the incorrect statement about first ionization enthalpies.

Option A: The first ionization enthalpy of K is less than that of Na and Li

In Group 1 (alkali metals), ionization enthalpy decreases down the group: Li > Na > K. So the ionization enthalpy of K is indeed less than both Na and Li. This statement is correct.

Option B: Xe does not have the lowest first ionization enthalpy in its group

In Group 18 (noble gases), the ionization enthalpy generally decreases down the group, but Radon (Rn) is below Xe and has a lower ionization enthalpy. However, due to relativistic effects and the stability of the filled shells, the trend holds that Rn has the lowest. So Xe does NOT have the lowest ionization enthalpy in Group 18. This statement is correct.

Option C: The first ionization enthalpy of element with atomic number 37 is lower than that of element with atomic number 38

Element 37 is Rubidium (Rb, Group 1) and element 38 is Strontium (Sr, Group 2). As we move from Rb to Sr, the nuclear charge increases and the electron is being removed from the same shell. Group 2 elements generally have higher ionization enthalpy than Group 1 elements in the same period. So IE of Rb < IE of Sr. This statement is correct.

Option D: The first ionization enthalpy of Ga is higher than that of the d-block element with atomic number 30

Element with atomic number 30 is Zinc (Zn). Gallium (Ga, Z = 31) has a lower ionization enthalpy than Zinc. This is because Zn has a completely filled $$3d^{10}4s^2$$ configuration which is extra stable, and its ionization enthalpy (906 kJ/mol) is higher than that of Ga (579 kJ/mol). So Ga does NOT have higher ionization enthalpy than Zn.

This statement is incorrect.

The correct answer is Option D.

The electronic configuration is $$[Rn] 5f^{14} 6d^1 7s^2$$.

Radon (Rn) has atomic number $$Z = 86$$.

The configuration contributes $$5f^{14}$$ (14) + $$6d^1$$ (1) + $$7s^2$$ (2) = 17 electrons, giving $$Z = 86 + 17 = 103$$, which corresponds to the element Lawrencium (Lr).

For element 103, the IUPAC systematic name uses the roots for each digit:

- 1 → un

- 0 → nil

- 3 → tri

Adding the suffix "-ium": Unniltrium

Therefore, the correct answer is Option C: Unniltrium.

Ionic mobility in aqueous solution depends on the hydrated ionic radius, with a smaller hydrated radius corresponding to higher ionic mobility. Among alkaline earth metal ions, the smaller bare ions (like $$Be^{2+}$$ and $$Mg^{2+}$$) have higher charge density, attracting more water molecules and thus forming a larger hydrated radius. In contrast, larger bare ions (like $$Sr^{2+}$$) have lower charge density, attract fewer water molecules, and therefore exhibit a smaller hydrated radius.

The order of hydrated ionic radii is $$Be^{2+} > Mg^{2+} > Ca^{2+} > Sr^{2+}$$. Since ionic mobility is inversely related to the hydrated ionic radius, the mobility order becomes $$\text{Ionic mobility: } Sr^{2+} > Ca^{2+} > Mg^{2+} > Be^{2+}$$. Hence, $$Sr^{2+}$$ has the highest ionic mobility in aqueous solution due to its smallest hydrated radius among the given options. The correct answer is Option A.

Since element A belongs to Group 1 and shows similarity to element B in Group 2, and given that A has the maximum hydration enthalpy in Group 1, we first identify the member of Group 1 with the highest hydration enthalpy.

In Group 1 (alkali metals: Li, Na, K, Rb, Cs), hydration enthalpy decreases down the group because the ionic size increases. Therefore, Lithium (Li), having the smallest ionic radius, exhibits the maximum hydration enthalpy.

Furthermore, elements in the periodic table often display diagonal relationships, whereby an element resembles the one diagonally below and to the right. Consequently, Lithium (Group 1, Period 2) shows a diagonal relationship with Magnesium (Group 2, Period 3).

This similarity between Li and Mg arises from their comparable charge densities (charge/size ratio), electronegativity values, and polarizing powers.

Hence, element B is Mg, and the correct answer is Option A: Mg.

We need to determine the correct order of melting points for the alkaline earth metals Be, Mg, Ca, and Sr.

The melting points of these metals do not follow a simple trend because of differences in crystal structure. Beryllium adopts a hexagonal close-packed structure and melts at 1287°C, magnesium also has a hexagonal close-packed structure with a melting point of 650°C, calcium crystallizes in a face-centred cubic lattice and melts at 842°C, and strontium forms a face-centred cubic arrangement and melts at 777°C.

Since beryllium has the highest melting point (1287°C), followed by calcium (842°C), then strontium (777°C), and finally magnesium (650°C), arranging them in decreasing order gives $$\text{Be}(1287°C) > \text{Ca}(842°C) > \text{Sr}(777°C) > \text{Mg}(650°C)$$.

This sequence corresponds to Option C, so the correct answer is $$Be > Ca > Sr > Mg$$.

This question is about the diagonal relationship between Lithium (Li) and Magnesium (Mg). Let us analyze each statement.

Statement (A): Both LiCl and MgCl$$_2$$ are soluble in ethanol.

Due to the diagonal relationship, both Li and Mg form compounds with significant covalent character. LiCl and MgCl$$_2$$ are both covalent enough to be soluble in organic solvents like ethanol. This is correct.

Statement (B): The oxides Li$$_2$$O and MgO combine with excess of oxygen to give superoxide.

Lithium forms only the normal oxide ($$Li_2O$$) when burned in excess oxygen — it does not form peroxide or superoxide due to the small size of the $$Li^+$$ ion, which cannot stabilize the large superoxide ion $$O_2^-$$. Similarly, MgO does not form superoxides. Only heavier alkali metals like K, Rb, and Cs form superoxides. This is incorrect.

Statement (C): LiF is less soluble in water than other alkali metal fluorides.

LiF has a very high lattice energy due to the small sizes of both $$Li^+$$ and $$F^-$$ ions. The high lattice energy makes it difficult to dissolve, so LiF is the least soluble among alkali metal fluorides. This is correct.

Statement (D): Li$$_2$$O is more soluble in water than other alkali metal oxides.

Actually, alkali metal oxides generally react with water rather than simply dissolving. Among alkali metal oxides, $$Li_2O$$ reacts less vigorously with water compared to $$Na_2O$$, $$K_2O$$, etc. The solubility of alkali metal oxides increases down the group. So $$Li_2O$$ is NOT more soluble than other alkali metal oxides. This is incorrect.

The correct statements are (A) and (C) only.

The correct answer is Option A: (A) and (C) only.

The densities of alkali metals are:

- Lithium (Li): $$0.534 \text{ g cm}^{-3}$$

- Sodium (Na): $$0.971 \text{ g cm}^{-3}$$

- Potassium (K): $$0.862 \text{ g cm}^{-3}$$

- Rubidium (Rb): $$1.532 \text{ g cm}^{-3}$$

- Caesium (Cs): $$1.873 \text{ g cm}^{-3}$$

Arranging in increasing order of density:

$$Li \lt K \lt Na \lt Rb \lt Cs$$

$$0.534 \lt 0.862 \lt 0.971 \lt 1.532 \lt 1.873$$

Note: The general trend in alkali metals is that density increases down the group, but potassium is an anomaly — it is less dense than sodium. This is because the increase in atomic volume of potassium is disproportionately larger compared to its increase in atomic mass.

Therefore, the correct answer is Option A: $$Li \lt K \lt Na \lt Rb \lt Cs$$.

Analyzing Assertion A:

LiF is indeed sparingly soluble in water. This is because both $$Li^+$$ and $$F^-$$ are very small ions, resulting in a very high lattice energy. The high lattice energy makes it difficult for water molecules to overcome the strong electrostatic attraction between the ions, leading to low solubility. This statement is true.

Analyzing Reason R:

The reason states that $$Li^+$$ has the smallest ionic radius among its group members and hence has the least hydration enthalpy. This is false. According to the relationship between ionic size and hydration enthalpy, a smaller ion has a higher charge density, which attracts water molecules more strongly. Therefore, $$Li^+$$ actually has the highest (most negative) hydration enthalpy among the alkali metal cations, not the least.

The order of hydration enthalpy (magnitude) is:

$$Li^+ > Na^+ > K^+ > Rb^+ > Cs^+$$

The actual reason LiF is sparingly soluble is that its very high lattice energy (due to both small ions) outweighs the hydration enthalpy, making dissolution energetically unfavorable.

Since Assertion A is true but Reason R is false, the correct answer is Option C.

We need to determine the thermal decomposition products of lithium nitrate and sodium nitrate when heated separately.

Lithium, being the smallest alkali metal, has a very high charge density. Due to its strong polarising power, lithium nitrate decomposes differently from the nitrates of other alkali metals. When heated, $$LiNO_3$$ decomposes completely into the oxide:

$$4LiNO_3 \xrightarrow{\Delta} 2Li_2O + 4NO_2 + O_2$$

This happens because the small $$Li^+$$ ion strongly polarises the large nitrate ion, destabilising it and causing it to break down entirely into the oxide rather than just the nitrite.

Now, sodium nitrate behaves like a typical alkali metal nitrate (Na, K, Rb, Cs). These metals have lower polarising power, so their nitrates decompose only partially — losing one oxygen to form the nitrite:

$$2NaNO_3 \xrightarrow{\Delta} 2NaNO_2 + O_2$$

Hence, lithium nitrate gives $$Li_2O$$ and sodium nitrate gives $$NaNO_2$$ upon heating.

Hence, the correct answer is Option C: $$Li_2O$$ and $$NaNO_2$$.

We need to identify which s-block element cannot be qualitatively confirmed by the flame test.

Option A: Li — Gives a crimson red flame.

Option B: Na — Gives a golden yellow flame.

Option C: Rb — Gives a red/violet flame.

Option D: Be — Does not give a characteristic flame color.

Beryllium and magnesium have very high ionization energies compared to other s-block elements. The energy of the flame is insufficient to excite their electrons to higher energy levels, so they do not produce a characteristic flame color. Therefore, Be cannot be confirmed by the flame test.

The correct answer is Option D: Be.

We need to determine the correct order of density among the alkaline earth metals Be, Mg, Ca, and Sr.

Recall that the densities of alkaline earth metals (Group 2) are:

$$Be = 1.85 \text{ g/cm}^3$$

$$Mg = 1.74 \text{ g/cm}^3$$

$$Ca = 1.55 \text{ g/cm}^3$$

$$Sr = 2.63 \text{ g/cm}^3$$

Arranging these values in decreasing order gives:

$$Sr (2.63) > Be (1.85) > Mg (1.74) > Ca (1.55)$$

Although atomic mass increases down the group from Be to Ca, the atomic radius increases more rapidly, causing the density to decrease. However, Sr has a significantly higher atomic mass while the increase in volume is not as large, so its density is the highest among these four.

Hence, the correct order of density is $$Sr > Be > Mg > Ca$$, which corresponds to Option C.

Let us analyze each statement:

Option A: "Low solubility of LiF in water is due to its small hydration enthalpy."

This statement is incorrect. LiF has low solubility in water despite having a very high hydration enthalpy (since both Li$$^+$$ and F$$^-$$ are small ions with high charge density). The actual reason for its low solubility is its very high lattice enthalpy, which exceeds the hydration enthalpy. So the statement incorrectly attributes the low solubility to small hydration enthalpy, when in fact LiF has a high hydration enthalpy — its lattice enthalpy is simply even higher.

Option B: "$$KO_2$$ is paramagnetic."

This is correct. Potassium superoxide ($$KO_2$$) contains the superoxide ion $$O_2^-$$, which has one unpaired electron, making it paramagnetic.

Option C: "Solution of sodium in liquid ammonia is conducting in nature."

This is correct. When sodium dissolves in liquid ammonia, it forms solvated electrons ($$Na^+$$ and $$e^-_{(amm)}$$), which are responsible for the electrical conductivity and the characteristic blue colour of the solution.

Option D: "Sodium metal has higher density than potassium metal."

This is correct. Sodium has a density of 0.97 g/cm$$^3$$ while potassium has a density of 0.86 g/cm$$^3$$. As we go down the alkali metal group, despite increasing atomic mass, the atomic volume increases more rapidly, causing potassium to have a lower density than sodium.

Therefore, the incorrect statement is Option A.

Let us classify each oxide:

$$SO_3$$ (Sulfur trioxide): Sulfur is a non-metal. $$SO_3$$ reacts with water to form sulfuric acid: $$SO_3 + H_2O \rightarrow H_2SO_4$$. Therefore, $$SO_3$$ is an acidic oxide.

$$SiO_2$$ (Silicon dioxide): Silicon is a metalloid. $$SiO_2$$ reacts with bases (e.g., $$NaOH$$) but not with acids. Therefore, $$SiO_2$$ is an acidic oxide.

$$CaO$$ (Calcium oxide): Calcium is an alkaline earth metal. $$CaO$$ reacts with water to form calcium hydroxide: $$CaO + H_2O \rightarrow Ca(OH)_2$$, which is a strong base. Therefore, $$CaO$$ is a basic oxide.

$$Al_2O_3$$ (Aluminium oxide) reacts with both acids and bases. Therefore, $$Al_2O_3$$ is an amphoteric oxide.

Hence, the correct answer is Option C (CaO).

We need to match the alkali metals with their characteristic flame emission wavelengths.

Known flame emission wavelengths of alkali metals:

• Li (Lithium): Crimson red flame, wavelength approximately 670.8 nm
• Na (Sodium): Golden yellow flame, wavelength approximately 589.2 nm
• Rb (Rubidium): Red/violet flame, wavelength approximately 780.0 nm
• Cs (Caesium): Blue/violet flame, wavelength approximately 455.5 nm

Matching:

• (A) Li \(\rightarrow\) (I) 670.8 nm
• (B) Na \(\rightarrow\) (II) 589.2 nm
• (C) Rb \(\rightarrow\) (III) 780.0 nm
• (D) Cs \(\rightarrow\) (IV) 455.5 nm

Hence, the correct answer is Option A ((A)-(I), (B)-(II), (C)-(III), (D)-(IV)).

We need to evaluate both statements about lithium-magnesium alloy and magnesium ions.

Statement I: "An alloy of lithium and magnesium is used to make aircraft plates."

This statement is actually about lithium-aluminium alloys, not lithium-magnesium alloys. Lithium-aluminium alloys (Al-Li alloys) are widely used in aerospace applications because they are lightweight and have high strength. While lithium-magnesium alloys do exist, they are not the standard alloys used for aircraft body/plates. So Statement I is false.

Statement II: "The magnesium ions are important for cell-membrane integrity."

Magnesium ions ($$Mg^{2+}$$) are not primarily associated with cell-membrane integrity. It is calcium ions ($$Ca^{2+}$$) that play a key role in maintaining cell-membrane integrity and stability. Magnesium ions are mainly important as cofactors in enzymatic reactions, ATP metabolism, and nucleic acid chemistry. So Statement II is also false.

Since both statements are incorrect, the correct answer is that both Statement I and Statement II are false.

Hence, the correct answer is Option B.

In the flame test, different metal cations produce characteristic flame colours. We are told that a green flame with blue centre was observed.

The characteristic flame colours of the given cations are:

Option A: Calcium ($$Ca^{2+}$$) - Produces a brick-red (orange-red) flame.

Option B: Copper ($$Cu^{2+}$$) - Produces a green flame with a blue centre. This is the characteristic flame colour of copper salts, especially copper(II) chloride.

Option C: Barium ($$Ba^{2+}$$) - Produces an apple green or yellowish-green flame (without a blue centre).

Option D: Strontium ($$Sr^{2+}$$) - Produces a crimson red flame.

The description of a green flame with a blue centre specifically matches copper.

The correct answer is Option B: Copper.

We need to find the oxidation state of Be in compound [A], which is formed when BeO reacts with HF in the presence of ammonia. When BeO reacts with HF in the presence of $$NH_3$$, it forms ammonium tetrafluoroberyllate: $$BeO + 4HF + 2NH_3 \rightarrow (NH_4)_2[BeF_4] + H_2O$$ So compound [A] is $$(NH_4)_2[BeF_4]$$.

On thermal decomposition, compound [A] gives compound [B] and ammonium fluoride: $$(NH_4)_2[BeF_4] \xrightarrow{\Delta} BeF_2 + 2NH_4F$$ So compound [B] is $$BeF_2$$. This is consistent with the information given.

In $$(NH_4)_2[BeF_4]$$, $$NH_4^+$$ has a charge of +1 each (two of them contribute +2) and $$[BeF_4]^{2-}$$ has a charge of -2. In the $$[BeF_4]^{2-}$$ ion, each F has an oxidation state of -1, so 4 fluorine atoms contribute $$4 \times (-1) = -4$$. Let the oxidation state of Be be $$x$$; then $$x + (-4) = -2$$, giving $$x = +2$$.

The oxidation state of Be in compound [A] is +2.

We need to arrange Mg, Al, S, P, and Si in increasing order of their first ionisation enthalpy.

These elements belong to the third period. The general trend is that ionisation enthalpy increases across a period from left to right. However, there are two well-known exceptions in the third period.

First, Al (group 13) has a lower ionisation enthalpy than Mg (group 2). This is because Mg has a fully filled 3s orbital ($$3s^2$$), which is extra stable. Al loses an electron from the 3p orbital, which is easier to remove.

Second, S (group 16) has a lower ionisation enthalpy than P (group 15). This is because P has a half-filled 3p orbital ($$3p^3$$), which is extra stable. S has one paired electron in a 3p orbital, and the electron-electron repulsion makes it easier to remove.

So the correct increasing order is: $$Al < Mg < Si < S < P$$.

Hence, the correct answer is Option 1.

We have to judge each statement separately and then see whether the Reason really explains the Assertion.

First let us look at the trend of metallic and non-metallic character within one period of the modern periodic table. On moving from the extreme left (alkali metals) to the extreme right (halogens and noble gases) the effective nuclear charge $$Z_{\text{eff}}$$ experienced by the outer electrons keeps on increasing because the atomic number increases but the additional electrons enter the same valence shell. Due to this higher $$Z_{\text{eff}}$$ the attraction between the nucleus and the valence electrons becomes stronger. A stronger attraction naturally makes it harder for an atom to lose electrons; therefore the metallic character, which is the tendency to lose electrons and form positive ions, decreases. Simultaneously, the stronger nuclear pull favours the gain of electrons; thus the non-metallic character, which is the tendency to gain electrons and form negative ions, increases. Hence the statement

$$\text{Metallic character} \downarrow \quad\text{and}\quad \text{Non-metallic character} \uparrow\quad \text{from left to right in a period}$$

is correct. So, Assertion (A) is true.

Now we examine the Reason (R), which cites two energetic quantities:

(i) Ionisation enthalpy $$\Delta_{\text{i}}H$$ : the energy required to remove one mole of electrons from one mole of gaseous atoms.

(ii) Electron gain enthalpy (electron affinity) $$\Delta_{\text{eg}}H$$ : the enthalpy change when one mole of electrons is added to one mole of gaseous atoms.

Across a period, because $$Z_{\text{eff}}$$ increases, the outer electrons are held more tightly, so

$$\Delta_{\text{i}}H \uparrow \quad \bigl(\text{ionisation enthalpy increases}\bigr).$$

For electron gain enthalpy the convention is that a more negative value indicates a greater tendency to accept an electron. Across a period the attraction for an extra electron normally becomes stronger; thus the magnitude of $$\Delta_{\text{eg}}H$$ increases, making the value more negative. In words, electron gain enthalpy becomes more negative, or its magnitude increases; it does not decrease. Therefore the statement in the Reason

“... and decrease in electron gain enthalpy”

is incorrect because the trend is actually the opposite. So, Reason (R) is false.

Since the Assertion is correct but the Reason is wrong, the Reason cannot possibly be the correct explanation of the Assertion.

Hence, the correct answer is Option D.

Element X has a 1st ionization energy of 495 kJ/mol and a very high 2nd ionization energy of 4563 kJ/mol. The huge jump between the 1st and 2nd ionization energies indicates that X has only 1 valence electron. This is characteristic of sodium (Na), which has the configuration $$[Ne]3s^1$$. After removing the single valence electron, the next electron must come from the stable noble gas core, requiring much more energy.

Element Y has a 1st ionization energy of 731 kJ/mol and a 2nd ionization energy of 1450 kJ/mol. The ratio of the 2nd to 1st IE is roughly 2, with no dramatic jump. This suggests Y has more than one valence electron. Magnesium (Mg) with configuration $$[Ne]3s^2$$ has two valence electrons and shows a gradual increase in successive ionization energies, consistent with these values.

Therefore, X = Na and Y = Mg, which corresponds to option (1).

Element E has outermost configuration $$4s^2 4p^1$$, which places it in group 13, period 4. This is Gallium (Ga).

The diagonal relationship in the periodic table occurs between elements that are one period down and one group to the right. The element diagonally related to Ga (period 4, group 13) in the p-block of period 5 would be in period 5, group 14 — this is Tin (Sn).

The electronic configuration of Tin (Sn, Z = 50) is [Kr] $$4d^{10}5s^2 5p^2$$.

This matches option 4: [Kr] $$4d^{10}5s^2 5p^2$$.

The electron gain enthalpy (electron affinity) of halogens measures the energy released when a neutral atom gains an electron. Among the halogens, the general trend is that electron gain enthalpy becomes less negative as we go down the group due to increasing atomic size.

However, fluorine is an exception. Despite being the smallest halogen, fluorine has a smaller electron gain enthalpy (in absolute value) than chlorine. This is because fluorine's very small size leads to strong electron-electron repulsion in the compact 2p orbitals, making it less favorable to add an electron compared to chlorine.

The correct order of absolute values of electron gain enthalpy is: $$Cl > F > Br > I$$. The approximate values are: Cl ($$-349$$ kJ/mol), F ($$-328$$ kJ/mol), Br ($$-325$$ kJ/mol), and I ($$-295$$ kJ/mol).

This matches option (C): $$Cl > F > Br > I$$.

We need to determine the correct order of electron gain enthalpy ($$\Delta_{eg}H$$) for the Group 16 elements: O, S, Se, and Te. Electron gain enthalpy is the energy change when an electron is added to a neutral gaseous atom to form a negative ion: $$X(g) + e^- \to X^-(g)$$. A more negative value indicates a greater tendency to accept an electron.

As we move down Group 16 from S to Se to Te, the atomic size increases progressively. The incoming electron enters orbitals that are farther from the nucleus, so the effective nuclear attraction on the added electron decreases. Consequently, the electron gain enthalpy becomes less negative in the order $$S > Se > Te$$.

Oxygen, despite being at the top of the group with the smallest atomic size, does not follow the expected trend. The 2p orbitals of oxygen are very compact, and adding an extra electron to the already electron-rich 2p subshell causes severe electron-electron repulsion. This repulsion makes it energetically less favorable for oxygen to accept an additional electron compared to sulfur. As a result, oxygen has a less negative electron gain enthalpy than sulfur.

The experimentally determined electron gain enthalpies are approximately: $$S \approx -200 \text{ kJ/mol}$$, $$O \approx -141 \text{ kJ/mol}$$, $$Se \approx -195 \text{ kJ/mol}$$, and $$Te \approx -190 \text{ kJ/mol}$$. This gives the order (from most negative to least negative): $$S > Se > Te > O$$.

Therefore, the correct order of electron gain enthalpy is $$S > Se > Te > O$$, which corresponds to Option (3).

First, let us note the atomic numbers of the parent atoms: $$\text P\,(Z=15),\; \text S\,(Z=16),\; \text{Cl}\,(Z=17),\; \text K\,(Z=19),\; \text{Ca}\,(Z=20).$$

Now we write the electronic population of each given ion. We subtract or add electrons according to the charge:

$$\begin{aligned} \text P^{3-}: &\; 15+3 = 18 \text{ electrons},\\ \text S^{2-}: &\; 16+2 = 18 \text{ electrons},\\ \text{Cl}^-: &\; 17+1 = 18 \text{ electrons},\\ \text K^+: &\; 19-1 = 18 \text{ electrons},\\ \text{Ca}^{2+}: &\; 20-2 = 18 \text{ electrons}. \end{aligned}$$

So every species possesses exactly $$18$$ electrons. Such a group is called an isoelectronic series.

For an isoelectronic series, we apply the rule: “When the number of electrons is the same, the ionic radius decreases as the nuclear charge $$Z$$ increases.” The reason is that a greater positive charge in the nucleus pulls the same electron cloud more strongly, producing a smaller radius. Symbolically we can write

$$\text{larger } Z \;\Longrightarrow\; \text{larger } Z_{\text{eff}} \;\Longrightarrow\; \text{smaller radius}.$$

Hence we simply arrange the ions in order of increasing atomic number (increasing $$Z$$) to obtain decreasing radius. Starting with the smallest atomic number 15 and moving upward:

$$Z=15 \;(\text P^{3-}) \gt Z=16 \;(\text S^{2-}) \gt Z=17 \;(\text{Cl}^-) \gt Z=19 \;(\text K^+) \gt Z=20 \;(\text{Ca}^{2+}).$$

Thus the descending order of ionic radii is

$$\text P^{3-} \gt \text S^{2-} \gt \text{Cl}^- \gt \text K^+ \gt \text{Ca}^{2+}.$$

When we match this sequence with the listed alternatives, we see that it coincides exactly with Option A.

Hence, the correct answer is Option A.

We have to arrange the ions $$\text{K}^+ ,\; \text{Na}^+ ,\; \text{Al}^{3+} ,\; \text{Mg}^{2+}$$ in increasing order of their ionic radii. The key idea from chemical periodicity is: for species having the same number of electrons (they are called isoelectronic), a larger nuclear charge ($$Z$$, the atomic number) pulls the electron cloud more strongly toward the nucleus, so the ionic radius becomes smaller. Mathematically one writes

$$Z_{\text{effective}} \propto Z$$

and

$$\text{Greater } Z_{\text{effective}} \; \Longrightarrow \; \text{Smaller radius}.$$

Let us first note the atomic numbers and the total electrons after ionisation:

$$ \begin{aligned} \text{Na}^+ &: Z = 11,\; \text{electrons} = 11 - 1 = 10, \\ \text{Mg}^{2+} &: Z = 12,\; \text{electrons} = 12 - 2 = 10, \\ \text{Al}^{3+} &: Z = 13,\; \text{electrons} = 13 - 3 = 10, \\ \text{K}^+ &: Z = 19,\; \text{electrons} = 19 - 1 = 18. \\ \end{aligned} $$

We see that $$\text{Na}^+ ,\; \text{Mg}^{2+},\; \text{Al}^{3+}$$ all possess $$10$$ electrons, hence they form an isoelectronic series. For this trio the order of $$Z$$ is

$$Z(\text{Al}^{3+}) = 13 \; > \; Z(\text{Mg}^{2+}) = 12 \; > \; Z(\text{Na}^+) = 11.$$

Therefore, applying the earlier rule, their radii follow the reverse order:

$$r(\text{Al}^{3+}) \; < \; r(\text{Mg}^{2+}) \; < \; r(\text{Na}^+).$$

Now let us compare $$\text{Na}^+$$ with $$\text{K}^+$$. These two ions are not isoelectronic: $$\text{Na}^+$$ has $$10$$ electrons (configuration $$1s^2 2s^2 2p^6$$), while $$\text{K}^+$$ has $$18$$ electrons (configuration $$1s^2 2s^2 2p^6 3s^2 3p^6$$). Because $$\text{K}^+$$ possesses electrons in the third shell ($$n = 3$$), its electron cloud is farther from the nucleus than that of $$\text{Na}^+$$ whose electrons reside only up to the second shell ($$n = 2$$). Hence

$$r(\text{K}^+) \; > \; r(\text{Na}^+).$$

Combining both deductions we obtain the full ascending sequence of radii:

$$r(\text{Al}^{3+}) \; < \; r(\text{Mg}^{2+}) \; < \; r(\text{Na}^+) \; < \; r(\text{K}^+).$$

Expressed simply, the order is

$$\text{Al}^{3+} \; < \; \text{Mg}^{2+} \; < \; \text{Na}^+ \; < \; \text{K}^+.$$

Comparing this with the given options, we find that Option C exactly matches this order.

Hence, the correct answer is Option C.

Let us examine each statement about Mendeleev.

Statement (1): He authored the textbook "Principles of Chemistry" — this is correct. Mendeleev did write this famous textbook in which he developed the periodic law.

Statement (2): "At the time he proposed the Periodic Table, the structure of atom was known" — this is incorrect. Mendeleev proposed his Periodic Table in 1869. The internal structure of the atom (discovery of the electron by J.J. Thomson) was not established until 1897, and Rutherford's nuclear model came in 1911. Therefore, atomic structure was completely unknown when Mendeleev proposed his table.

Statement (3): Element with atomic number 101 is named after him (Mendelevium) — this is correct.

Statement (4): He invented the accurate barometer — this is incorrect as a factual matter (the barometer was invented by Torricelli), but this is a distractor. The question asks for a statement about Mendeleev that is "incorrect," and statement (2) is the most clearly incorrect claim.

The incorrect statement is option (2): atomic structure was not known at the time Mendeleev proposed the Periodic Table.

We begin by recalling the usual definitions.

• A hydroxide that produces $$\text{OH}^-$$ ions readily in water and neutralises acids is called basic.
• A hydroxide that donates $$\text{H}^+$$ (or accepts $$\text{OH}^-$$) and neutralises bases is called acidic.
• A hydroxide that can behave both as an acid and as a base is called amphoteric.

Now we examine each hydroxide one by one, using periodic trends and well-known reactions to justify its behaviour.

For sodium hydroxide, $$\text{NaOH}$$:
Sodium belongs to Group 1. Alkali-metal hydroxides are strongly basic because they dissociate completely:
$$\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-$$
The liberation of $$\text{OH}^-$$ makes the solution strongly basic. Hence $$\text{NaOH}$$ → Basic ⇒ (ii).

For beryllium hydroxide, $$\text{Be(OH)}_2$$:
Beryllium sits at the top of Group 2 and possesses a very high charge density. Owing to this high polarising power its hydroxide neither dissociates completely as a base nor remains strictly acidic; instead it can react with both acids and bases:
$$\text{Be(OH)}_2 + 2\,\text{HCl} \rightarrow \text{BeCl}_2 + 2\,\text{H}_2\text{O}$$
$$\text{Be(OH)}_2 + 2\,\text{OH}^- \rightarrow [\text{Be(OH)}_4]^{2-}$$
Because it shows dual behaviour, $$\text{Be(OH)}_2$$ is amphoteric ⇒ (iii).

For calcium hydroxide, $$\text{Ca(OH)}_2$$:
Calcium is lower down the same Group 2. The larger ionic size reduces polarising power, so its hydroxide acts predominantly as a base:
$$\text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2\,\text{OH}^-$$
Therefore $$\text{Ca(OH)}_2$$ is basic ⇒ (ii).

For boric acid written as $$\text{B(OH)}_3$$:
Boron is a typical non-metal. The molecule accepts $$\text{OH}^-$$ ions (acts as a Lewis acid) forming $$\text{[B(OH)}_4]^-$$. Experimentally its aqueous solution has $$\text{pH}<7$$. So $$\text{B(OH)}_3$$ behaves as an acid ⇒ (i).

For aluminium hydroxide, $$\text{Al(OH)}_3$$:
Like beryllium, aluminium has appreciable charge density and shows dual behaviour:
With acid: $$\text{Al(OH)}_3 + 3\,\text{HCl} \rightarrow \text{AlCl}_3 + 3\,\text{H}_2\text{O}$$
With base: $$\text{Al(OH)}_3 + \text{OH}^- \rightarrow [\text{Al(OH)}_4]^-$$
Thus $$\text{Al(OH)}_3$$ is amphoteric ⇒ (iii).

Collecting all the matches we have
$$(a)\;\text{NaOH}\; \rightarrow\; (ii)\quad\text{Basic}$$ $$(b)\;\text{Be(OH)}_2\; \rightarrow\; (iii)\quad\text{Amphoteric}$$ $$(c)\;\text{Ca(OH)}_2\; \rightarrow\; (ii)\quad\text{Basic}$$ $$(d)\;\text{B(OH)}_3\; \rightarrow\; (i)\quad\text{Acidic}$$ $$(e)\;\text{Al(OH)}_3\; \rightarrow\; (iii)\quad\text{Amphoteric}$$

Comparing with the given options, the sequence (ii, iii, ii, i, iii) corresponds to Option B.

Hence, the correct answer is Option B.

We identify the elements by their atomic numbers: X has atomic number 33, Y has atomic number 53, and Z has atomic number 83.

Element X (Z = 33) is Arsenic (As). Arsenic is located in Group 15, Period 4. It is classified as a metalloid, as it exhibits properties intermediate between metals and non-metals.

Element Y (Z = 53) is Iodine (I). Iodine is located in Group 17, Period 5. It is classified as a non-metal (a halogen).

Element Z (Z = 83) is Bismuth (Bi). Bismuth is located in Group 15, Period 6. It is classified as a metal, being the heaviest stable element in its group where metallic character increases going down the group.

Therefore, X is a metalloid, Y is a non-metal, and Z is a metal, which corresponds to option (2).

We recall the general rule that the first ionisation enthalpy, symbolised as $$IE_1$$, increases from left to right across a period because the nuclear charge increases while the principal quantum number $$n$$ remains the same. This means the outermost electrons are held more tightly, so more energy is required to remove one electron.

Moving through the third period we therefore expect $$IE_1(\text{Na}) < IE_1(\text{Mg}) < IE_1(\text{Al}) < IE_1(\text{Si}) < IE_1(\text{P}) < IE_1(\text{S}) \dots$$, but two well-known exceptions disturb this smooth rise. These exceptions arise from extra stability associated with completely filled and half-filled subshells.

First we compare magnesium and aluminium.

We write their electronic configurations in the ground state:

$$\text{Mg : } 1s^22s^22p^63s^2$$

$$\text{Al : } 1s^22s^22p^63s^23p^1$$

Magnesium has a completely filled $$3s$$ subshell ($$3s^2$$). Aluminium’s outer electron is the first electron in the $$3p$$ subshell ($$3p^1$$). A completely filled subshell is especially stable because there is no other orbital of the same energy available to lower the electron-electron repulsion. Removing an electron from magnesium would destroy this stable configuration, so comparatively more energy is required. In contrast, removing the single $$3p$$ electron from aluminium is easier because it restores the stable $$3s^2$$ core. Thus

$$IE_1(\text{Al}) < IE_1(\text{Mg}).$$

Next we compare phosphorus and sulfur. Their configurations are

$$\text{P : } 1s^22s^22p^63s^23p^3$$

$$\text{S : } 1s^22s^22p^63s^23p^4$$

Phosphorus has a half-filled $$3p$$ subshell ($$3p^3$$). A half-filled set of degenerate $$p$$ orbitals is of extra stability because each electron occupies a separate orbital with parallel spins, minimising electron-electron repulsion. Sulfur, on the other hand, already has one of its $$3p$$ orbitals doubly occupied ($$3p^4$$), increasing repulsion. Therefore it is slightly easier to remove an electron from sulfur than from phosphorus. Consequently

$$IE_1(\text{S}) < IE_1(\text{P}).$$

Combining these two anomalies with the usual left-to-right increase gives the order (from lowest to highest) as

$$IE_1(\text{Al}) < IE_1(\text{Mg}) < IE_1(\text{S}) < IE_1(\text{P}).$$

Translating this inequality back into the required arrangement we obtain

Al < Mg < S < P.

This sequence exactly matches Option C in the list provided.

Hence, the correct answer is Option C.

The first ionization energy of magnesium (Mg) is 737.7 kJ/mol. We need elements X and Y with higher ionization energies and element Z with a lower ionization energy than Mg.

The relevant ionization energies are approximately: Ar (1520.6 kJ/mol), Cl (1251.2 kJ/mol), Ne (2080.7 kJ/mol), Li (520.2 kJ/mol), and Na (495.8 kJ/mol). Both Ar and Cl have higher IE than Mg, while Na has a lower IE.

Option (3) gives X = Ar, Y = Cl, Z = Na. Both Ar and Cl have ionization energies greater than Mg, and Na has a lower ionization energy. This satisfies all the given conditions.

The correct answer is option (3): argon, chlorine and sodium.

We have to compare the ionic radius of the nitride ion $$\mathrm{N^{3-}}$$ with the already-given radii of $$\mathrm{F^-}$$ and $$\mathrm{O^{2-}}$$ and with the covalent radius of neutral nitrogen.

First, notice that $$\mathrm{N^{3-}},\; \mathrm{O^{2-}},\; \mathrm{F^-},$$ and $$\mathrm{Ne}$$ all possess the same total number of electrons:

$$\begin{aligned} \mathrm{N^{3-}} &:& 7 + 3 = 10 \text{ electrons},\\ \mathrm{O^{2-}} &:& 8 + (-2) = 10 \text{ electrons},\\ \mathrm{F^-} &:& 9 + (-1) = 10 \text{ electrons},\\ \mathrm{Ne} &:& 10 \text{ electrons}. \end{aligned}$$

Species holding the same electron count form what is called an isoelectronic series. In such a series, all ions (or atoms) have an identical electronic cloud, but their nuclear charge $$Z$$ differs. The rule for an isoelectronic series is stated as follows:

Rule. Within an isoelectronic series, the ionic radius decreases as the nuclear charge $$Z$$ increases, because more‐positive nuclei pull the same electron cloud closer.

Now, list the atomic numbers (nuclear charges):

$$\begin{aligned} Z(\mathrm{N}) &= 7,\\ Z(\mathrm{O}) &= 8,\\ Z(\mathrm{F}) &= 9,\\ Z(\mathrm{Ne}) &= 10. \end{aligned}$$

Hence, for the isoelectronic sequence

$$\mathrm{N^{3-}} \; (\!Z=7) \; \lt \; \mathrm{O^{2-}} \; (\!Z=8) \; \lt \; \mathrm{F^-} \; (\!Z=9) \; \lt \; \mathrm{Ne} \; (\!Z=10)$$

the magnitude of the radii follows the opposite order:

$$r(\mathrm{N^{3-}}) \; \gt \; r(\mathrm{O^{2-}}) \; \gt \; r(\mathrm{F^-}) \; \gt \; r(\mathrm{Ne}).$$

We are given the actual experimental values

$$r(\mathrm{O^{2-}}) = 1.40\;\text{Å}, \qquad r(\mathrm{F^-}) = 1.33\;\text{Å},$$

so by the above ordering we must have

$$r(\mathrm{N^{3-}}) \; \gt \; 1.40\;\text{Å}.$$

Consequently $$r(\mathrm{N^{3-}})$$ is certainly larger than both $$r(\mathrm{O^{2-}})$$ and $$r(\mathrm{F^-})$$. In addition, the covalent radius of neutral nitrogen is only $$0.74\;\text{Å},$$ so $$r(\mathrm{N^{3-}})$$ is also larger than that.

Therefore, the correct qualitative statement is: “It is bigger than $$\mathrm{O^{2-}}$$ and $$\mathrm{F^-}$$.” This corresponds to Option B.

Hence, the correct answer is Option B.

We need to identify the pair of oxides that are both acidic in nature.

$$B_2O_3$$ (boron trioxide) is a well-known acidic oxide. It reacts with water to form boric acid and with bases to form borates.

$$SiO_2$$ (silicon dioxide) is also an acidic oxide. It reacts with strong bases to form silicates, though it does not dissolve in water readily.

CaO (calcium oxide) and BaO (barium oxide) are basic oxides, as they are oxides of Group 2 metals. $$N_2O$$ (nitrous oxide) is a neutral oxide.

Therefore, the pair of oxides that are both acidic is $$B_2O_3$$ and $$SiO_2$$.

The correct answer is Option (4).

We recall that deuterium, symbolised as $$^2\text{H}$$ or $$\text{D}$$, is an isotope of ordinary hydrogen, written as $$^1\text{H}$$. Both isotopes possess one proton in the nucleus, so their chemical properties, which largely depend on electronic configuration, are almost identical. However, there is an important difference: deuterium contains one neutron in addition to the single proton, whereas protium (ordinary hydrogen) does not.

This extra neutron doubles the mass of the nucleus. Consequently, the atomic mass of deuterium is $$2\,\text{u}$$ while that of protium is $$1\,\text{u}$$. The heavier mass influences the velocity of the atoms or molecules in a gas at a given temperature. According to the kinetic-theory relation $$\dfrac{1}{2}m v^2 = \dfrac{3}{2}kT,$$ where $$m$$ is the mass of a particle, $$v$$ its root-mean-square speed, $$k$$ Boltzmann’s constant and $$T$$ absolute temperature, we observe that for a fixed $$T$$ the speed $$v$$ is inversely proportional to the square root of the mass $$m$$:

$$v \propto \dfrac{1}{\sqrt{m}}.$$

Substituting $$m = 1\,\text{u}$$ for protium and $$m = 2\,\text{u}$$ for deuterium, we immediately see that

$$\dfrac{v_{\text{D}}}{v_{\text{H}}} = \sqrt{\dfrac{m_{\text{H}}}{m_{\text{D}}}} = \sqrt{\dfrac{1}{2}} = \dfrac{1}{\sqrt{2}} \lt 1.$$

Thus deuterium atoms move more slowly than ordinary hydrogen atoms at the same temperature. Reaction rates for gases depend on collision frequency, which in turn depends on molecular speed. Because $$v_{\text{D}}$$ is smaller, the number of effective collisions per unit time is reduced. Therefore reactions involving deuterium proceed more slowly than analogous reactions with protium. This phenomenon is known as the “isotope effect”.

Now let us match this reasoning with the given options:

A. “reacts vigorously than hydrogen” - contradicts the isotope effect.
B. “reacts just as hydrogen” - ignores the mass-dependent rate difference.
C. “emits $$\beta^+$$ particles” - refers to radioactive decay, but deuterium is stable and does not undergo $$\beta^+$$ emission.
D. “reacts slower than hydrogen” - exactly expresses the isotope effect we have explained.

Hence, the correct answer is Option D.

We begin by recalling that the hydroxides of the alkaline-earth metals have the general formula $$M(OH)_2,$$ where $$M$$ stands for $$\mathrm{Be,\,Mg,\,Ca,\,Sr,\,Ba}.$$

The first point concerns their behaviour toward a concentrated external base (alkali). A hydroxide will dissolve in an alkali only if it is amphoteric, i.e. if it can react further with hydroxide ion $$\mathrm{OH^-}$$ to give a complex anion. The relevant amphoteric equilibrium is stated here:

$$M(OH)_2(s)+2\,\mathrm{OH^-}(aq)\;\rightleftharpoons\;[M(OH)_4]^{2-}(aq).$$

Experimentally, only $$\mathrm{Be(OH)_2}$$ is sufficiently amphoteric to undergo this reaction. The other hydroxides - $$\mathrm{Mg(OH)_2},\,\mathrm{Ca(OH)_2},\,\mathrm{Sr(OH)_2},\,\mathrm{Ba(OH)_2}$$ - are purely basic; they do not form the above complex and therefore do not dissolve in excess alkali. Hence the blanket statement “None of the alkaline-earth metal hydroxides dissolve in alkali” is false because $$\mathrm{Be(OH)_2}$$ is a clear counter-example.

Next we discuss the solubility of these hydroxides in water. Lattice energy ($$U_L$$) decreases more rapidly than hydration energy ($$H_h$$) when we go down the group, because the cationic radius increases markedly while the anionic radius (that of $$\mathrm{OH^-}$$) remains constant. The condition for dissolution is stated as

$$\lvert H_h\rvert \; >\; U_L.$$

As $$U_L$$ falls down the group, this inequality becomes progressively easier to satisfy, causing solubility to rise in the order

$$\mathrm{Be(OH)_2}<\mathrm{Mg(OH)_2}<\mathrm{Ca(OH)_2}<\mathrm{Sr(OH)_2}<\mathrm{Ba(OH)_2}.$$

Therefore the statement “Solubility of alkaline earth metal hydroxides in water increases down the group” is true.

Summarising, Statement I is incorrect whereas Statement II is correct. This matches Option B in the given list.

Hence, the correct answer is Option B.

We match each element with its characteristic property.

(a) Ba (Barium): Barium belongs to Group 2 (alkaline earth metals), period 6. Its outer electronic configuration is $$6s^2$$ — matching (ii).

(b) Ca (Calcium): Calcium sulphate and calcium oxalate are both sparingly soluble in water. Specifically, calcium oxalate (CaC$$_2$$O$$_4$$) is insoluble in water and is used in the gravimetric estimation of calcium — matching (iii).

(c) Li (Lithium): Lithium, unlike other alkali metals, forms organic solvent-soluble organolithium compounds (e.g., butyllithium dissolves in hydrocarbons). This diagonal relationship with Mg means its salts (like LiCl) are often soluble in organic solvents — matching (i).

(d) Na (Sodium): Sodium hydroxide (NaOH) is a very strong monoacidic base. It is a strong base that fully dissociates in water and is monoacidic (one OH$$^-$$ per formula unit) — matching (iv).

The correct matching is (a)-(ii), (b)-(iii), (c)-(i) and (d)-(iv), which is option (1).

We have to examine each statement and decide which one does not agree with the accepted facts about dihydrogen.

Let us start with the fundamental equilibrium for the thermal dissociation of dihydrogen:

$$\mathrm{H_2(g)} \;\rightleftharpoons\; 2\,\mathrm{H(g)}$$

For this reaction the equilibrium constant in terms of pressure is

$$K_p \;=\;\frac{(p_{\mathrm H})^{2}}{p_{\mathrm{H_2}}}$$

where $$p_{\mathrm H}$$ and $$p_{\mathrm{H_2}}$$ are the partial pressures of the atom and the molecule respectively.

Suppose we begin with 1 mol of $$\mathrm{H_2}$$ at 1 bar. After reaching equilibrium at 2000 K, let the degree of dissociation be $$\alpha$$. Then

• moles of $$\mathrm{H_2}$$ become $$1-\alpha$$,
• moles of $$\mathrm{H}$$ become $$2\alpha$$,
• total moles $$=\;1+\alpha$$.

Hence the mole fractions are

$$x_{\mathrm{H_2}}=\frac{1-\alpha}{1+\alpha}, \qquad x_{\mathrm{H}}=\frac{2\alpha}{1+\alpha}.$$

Because the total pressure is 1 bar, the partial pressures equal these mole fractions, so

$$K_p \;=\;\frac{(x_{\mathrm H})^{2}}{x_{\mathrm{H_2}}}$$

Substituting the expressions for the mole fractions, we get

$$K_p=\frac{\left(\dfrac{2\alpha}{1+\alpha}\right)^2}{\dfrac{1-\alpha}{1+\alpha}} =\;4\alpha^{2}\,\frac{1+\alpha}{1-\alpha}.$$

At 2000 K the tabulated value is $$K_p\approx1.6\times10^{-4}$$. Because $$\alpha$$ is very small, we can safely put $$1+\alpha\approx1$$ and $$1-\alpha\approx1$$, giving

$$K_p\;\approx\;4\alpha^{2}$$

and hence

$$\alpha\;\approx\;\frac{1}{2}\sqrt{K_p} =\;\frac{1}{2}\sqrt{1.6\times10^{-4}} =\;\frac{1}{2}\times\,1.26\times10^{-2} \approx6.3\times10^{-3}.$$

This is

$$\alpha\times100\;\approx\;0.63\%,$$

that is, only about $$0.63\%$$ of the molecular hydrogen splits into atoms. The value quoted in the NCERT text is even smaller, $$0.081\%.$$ In either case the dissociation is far below $$8.1\%.$$ Therefore, the statement “At around 2000 K, the dissociation of dihydrogen into its atoms is nearly 8.1 %” is incorrect.

Now we briefly justify the other three statements:

• Atomic hydrogen can indeed be produced by subjecting hot $$\mathrm{H_2}$$ to ultraviolet radiation, so statement A is correct.

• The bond dissociation enthalpy of $$\mathrm{H_2}$$ (about 436 kJ mol-1) is the highest among all diatomic molecules having a single bond; hence statement C is correct.

• Zinc, being amphoteric, liberates $$\mathrm{H_2}$$ both with dilute acids such as $$\mathrm{HCl}$$ and with strong bases such as $$\mathrm{NaOH(aq)}$$, so statement D is also correct.

Since only statement B is wrong, it is the incorrect option.

Hence, the correct answer is Option B.

Assertion A states that hydrogen is the most abundant element in the Universe, but it is not the most abundant gas in the troposphere. This is true because while hydrogen constitutes about 75% of the elemental mass in the Universe (mostly in stars), the most abundant gas in Earth's troposphere is nitrogen ($$N_2$$, about 78%), followed by oxygen ($$O_2$$, about 21%). Hydrogen gas is present in only trace amounts in the troposphere.

Reason R states that hydrogen is the lightest element. This is true, as hydrogen has an atomic mass of approximately 1 u, making it the lightest of all elements.

The reason R correctly explains assertion A. Because hydrogen is the lightest element, hydrogen molecules have the highest average molecular speed at any given temperature. At Earth's temperatures, a significant fraction of hydrogen molecules can exceed the escape velocity of Earth's gravitational field. Over geological timescales, most of the free hydrogen has escaped from the troposphere into space. This is why, despite being the most abundant element in the Universe, hydrogen is not the most abundant gas in the troposphere.

Therefore, the correct answer is Option (2): Both A and R are true and R is the correct explanation of A.

We begin with the metal $$\text{Li}$$, lithium. Lithium is extraordinarily light and, when alloyed with other metals or when converted to lithium-based greases, it provides a low-friction, high-melting lubricant. Such greases are packed inside high-speed bearings of motor engines. Hence lithium corresponds to $$\text{bearings for motor engines}$$, i.e. entry $$v$$.

Now we consider $$\text{Na}$$, sodium. A very important property of liquid sodium is its exceptionally high thermal conductivity and its low neutron-absorption cross-section. Because of these two facts, liquid sodium is circulated through the core in many fast breeder nuclear reactors to remove heat. Therefore sodium must be matched with $$\text{coolant in fast breeder nuclear reactor}$$, i.e. entry $$iii$$.

Next comes $$\text{K}$$, potassium. Potassium hydroxide, $$\text{KOH}$$, readily reacts with carbon dioxide according to the equation $$\text{2 KOH + CO}_2 \longrightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O}.$$ Because of this ready absorption, potassium compounds are widely used in air-purifying canisters and other devices as an $$\text{absorbent of CO}_2$$. So potassium matches with entry $$ii$$.

Finally, for $$\text{Cs}$$, caesium, we recall its very low ionisation enthalpy. When a caesium-coated surface is exposed to light, electrons are emitted even under weak illumination. This makes caesium ideal for constructing $$\text{photoelectric cells}$$, which convert light directly into electric current. Thus caesium corresponds to entry $$i$$.

Collecting all the matches we have obtained:

$$\begin{aligned} \text{Li} &\rightarrow v \\ \text{Na} &\rightarrow iii \\ \text{K} &\rightarrow ii \\ \text{Cs} &\rightarrow i \end{aligned}$$

This sequence corresponds exactly to Option D in the given list.

Hence, the correct answer is Option D.

The first three options represent diagonal relationships in the periodic table: Li-Mg (Group 1, Period 2 with Group 2, Period 3), Be-Al (Group 2, Period 2 with Group 13, Period 3), and B-Si (Group 13, Period 2 with Group 14, Period 3). Diagonal relationships arise because moving diagonally across the periodic table results in similar charge-to-size ratios, leading to similar chemical properties.

However, Li-Na are both elements in Group 1 (alkali metals). This is not a diagonal relationship but rather a relationship within the same group. While elements in the same group share some properties, this is a fundamentally different type of relationship compared to the diagonal pairs listed in the other options.

Therefore, Li-Na differs in its mutual relationship from the other sets, which are all diagonal pairs.

We are told that an s-block element M reacts with oxygen to form an oxide of formula MO$$_2$$, which is pale yellow in colour and paramagnetic.

Among s-block elements, when reacted with excess oxygen: lithium forms Li$$_2$$O (normal oxide), sodium forms Na$$_2$$O$$_2$$ (peroxide), potassium forms KO$$_2$$ (superoxide), rubidium and caesium also form superoxides. Calcium, barium, and strontium can form peroxides CaO$$_2$$, BaO$$_2$$, etc., and magnesium forms MgO.

The formula MO$$_2$$ with M being an s-block element points to either a peroxide (like Na$$_2$$O$$_2$$, CaO$$_2$$) or a superoxide (KO$$_2$$, RbO$$_2$$). A superoxide KO$$_2$$ has the formula MO$$_2$$ where M is K (potassium, atomic mass ~39), and potassium is indeed an s-block element. Superoxides contain the O$$_2^-$$ ion, which has an unpaired electron, making it paramagnetic. KO$$_2$$ is also known to be pale yellow in colour.

In contrast, peroxides like Na$$_2$$O$$_2$$ have a different formula (M$$_2$$O$$_2$$, not MO$$_2$$) and CaO$$_2$$ is a peroxide that is white and diamagnetic. Superoxides contain O$$_2^-$$ with one unpaired electron, explaining the paramagnetism. The pale yellow colour is characteristic of KO$$_2$$.

Therefore, element M is potassium (K), which is option 4.

We have the Assertion (A): “Barium carbonate is insoluble in water and is highly stable.” To analyse this, we recall the general solubility trend of group-2 carbonates. As we move down the alkaline-earth metal column from $$\mathrm{Mg^{2+}}$$ to $$\mathrm{Ba^{2+}},$$ the lattice energy decreases more slowly than the hydration energy, causing the net energy change for dissolution to become more positive. Consequently, $$\mathrm{BaCO_3}$$ does not obtain sufficient hydration enthalpy to break its crystal lattice, so it remains practically insoluble in water. Furthermore, its large, low-charge-density cation $$\mathrm{Ba^{2+}}$$ exerts only weak polarising power on the carbonate ion $$\mathrm{CO_3^{2-}},$$ making the salt reluctant to decompose when heated. Hence, barium carbonate is indeed highly thermally stable. Therefore, statement (A) is true.

Now we examine the Reason (R): “The thermal stability of the carbonates increases with increasing cationic size.” The thermal decomposition of any group-2 carbonate may be represented as

$$\mathrm{MCO_3(s) \;\xrightarrow{\Delta}\; MO(s) + CO_2(g)}$$

where $$\mathrm{M^{2+}}$$ is the alkaline-earth metal ion. The enthalpy change for this reaction depends largely on the lattice energies of the carbonate and the oxide produced. As we descend the group, the ionic radius of $$\mathrm{M^{2+}}$$ increases. Because a larger cation has lower polarising power, it disturbs the electron cloud of the carbonate ion less, so the bond between $$\mathrm{C}$$ and $$\mathrm{O}$$ in $$\mathrm{CO_3^{2-}}$$ is less weakened. Consequently, more heat is required to break the carbonate lattice, and the decomposition temperature rises. Thus, the trend “larger cation → greater thermal stability” is correct. Therefore, statement (R) is also true.

We further notice that (R) directly provides the underlying principle that explains why $$\mathrm{BaCO_3}$$ in (A) is highly stable: barium has the largest cationic size among the common alkaline-earth metals, so its carbonate possesses the greatest thermal stability. Hence, the truth of (R) supports and rationalises the truth of (A).

Because both (A) and (R) are true, the most appropriate choice is Option B.

Hence, the correct answer is Option B.

We match each alkaline earth metal with its well-known application.

Beryllium (Be) is transparent to X-rays due to its low atomic number and low X-ray absorption. It is used to make windows of X-ray tubes. This matches (iv).

Magnesium (Mg) burns with a brilliant white light in air. This property makes it useful in incendiary bombs and signal flares. This matches (iii).

Calcium (Ca) is a strong reducing agent and is used in the extraction of metals through metallothermic reduction, for example in extracting uranium, vanadium, and chromium from their compounds. This matches (ii).

Radium (Ra) is radioactive and emits alpha, beta, and gamma radiation. It was historically used in the treatment of cancer via radiation therapy (brachytherapy). This matches (i).

The correct matching is a - iv, b - iii, c - ii, d - i, which corresponds to Option B.

Alkali metals and their salts produce characteristic flame colours when heated. The flame colour corresponds to specific wavelengths of light emitted when excited electrons return to lower energy levels. The key flame colours and their wavelengths for alkali metal chlorides are as follows.

Lithium chloride (LiCl) produces a crimson red flame, which corresponds to a wavelength of approximately $$670.8$$ nm. Sodium chloride (NaCl) produces the well-known intense yellow flame, corresponding to the sodium D-line at approximately $$589.2$$ nm. Rubidium chloride (RbCl) produces a red to red-violet flame at approximately $$780.0$$ nm (near the infrared boundary). Caesium chloride (CsCl) produces a blue-violet flame at approximately $$455.5$$ nm.

Matching the salts with their flame colour wavelengths: (a) LiCl matches with (ii) 670.8 nm, (b) NaCl matches with (iv) 589.2 nm, (c) RbCl matches with (iii) 780.0 nm, and (d) CsCl matches with (i) 455.5 nm.

Therefore, the correct answer is Option (4): (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i).

We have four alkali-metal elements in List I: $$\text{Li},\; \text{Na},\; \text{K},\; \text{Cs}$$ and four characteristic properties in List II that must be matched one-to-one.

First, recall two fundamental periodic facts.

1. Down the group, hydration energy falls much faster than lattice energy.    Consequently the heavier alkali-metal halides (especially those with large anions such as $$I^-$$) become progressively less soluble in water.

2. Only lithium compounds often show behaviour resembling the alkaline-earth metals because the small $$\text{Li}^+$$ ion has an abnormally high polarising power.    Hence compounds such as $$\text{Li}_2\text{CO}_3$$ decompose readily on heating, a trait not shared by the carbonates of the heavier group members.

Now we examine each property one by one and attach it to the most appropriate element.

Property (iv) “Carbonate salt decomposes easily on heating.” The thermal decomposition reaction is $$\text{M}_2\text{CO}_3 \;\overset{\Delta}{\rightarrow}\; \text{M}_2\text{O}+ \text{CO}_2.$$ Among the alkali metals, only $$\text{Li}_2\text{CO}_3$$ executes this reaction at a moderate temperature; $$\text{Na}_2\text{CO}_3,\,\text{K}_2\text{CO}_3$$ and $$\text{Cs}_2\text{CO}_3$$ are heat-stable. So property (iv) must be allotted to $$\text{Li}.$$

Property (iii) “Bicarbonate salt used in fire extinguisher.” The well-known soda-acid and dry-powder extinguishers employ sodium bicarbonate: $$\text{NaHCO}_3 + \text{H}^+ \longrightarrow \text{Na}^+ + \text{H}_2\text{O} + \text{CO}_2 \uparrow.$$ Thus property (iii) belongs to $$\text{Na}.$$

Property (ii) “Most abundant element in cell fluid.” Physiology tells us that the intracellular fluid of animals and plants is rich in $$\text{K}^+$$ ions; sodium predominates outside the cell. Therefore property (ii) must match $$\text{K}.$$

Property (i) “Poor water solubility of $$I^-$$ salt.” Because hydration energy is smallest for the largest alkali-metal ion, $$\text{Cs}^+$$, its iodide $$\text{CsI}$$ is the least soluble of the series (while still appreciably soluble, it is markedly less so than $$\text{LiI}, \text{NaI}, \text{KI}$$). Hence property (i) is assigned to $$\text{Cs}.$$

Summarising the matches we obtained:

$$\begin{aligned} \text{Li} &\rightarrow (iv)\\ \text{Na} &\rightarrow (iii)\\ \text{K} &\rightarrow (ii)\\ \text{Cs} &\rightarrow (i) \end{aligned}$$

Looking at the given options, the sequence $$(a)-(iv),\;(b)-(iii),\;(c)-(ii),\;(d)-(i)$$ corresponds exactly to Option A.

Hence, the correct answer is Option A.

We begin by recalling the common metallurgical processes and the furnaces or techniques associated with them. We shall compare every item of List I with the most appropriate item of List II.

First, we look at item (a) of List I: “Concentration of Ag ore”. In metallurgy, silver (Ag) is frequently extracted from its ores such as argentite, $$Ag_{2}S$$, by the Mac-Arthur-Forrest cyanide process. In that process the ore is leached with a dilute aqueous solution of sodium cyanide, $$NaCN$$, which selectively dissolves silver according to the reaction

$$Ag_{2}S + 4NaCN \rightarrow 2Na[Ag(CN)2] + Na_{2}S.$$

Thus, concentration of Ag ore is correctly matched with “Leaching with dilute NaCN solution”, i.e. List-II item (iii).

Now we consider item (b) of List I: “Blast furnace”. A blast furnace is the tall shaft furnace used for the smelting of iron oxide ores. Inside the furnace, iron oxide is reduced to molten iron that contains about $$4\%$$ carbon and many other impurities. This impure molten iron solidifies to give what metallurgists call pig iron. Hence, the blast furnace is associated with “Pig iron”, i.e. List-II item (ii).

Next we move to item (c) of List I: “Blister copper”. Blister copper (so called because of its blister-like surface caused by escaping $$SO_{2}$$) is produced by melting copper matte in a reverberatory furnace. In this furnace, the charge is heated by hot gases which “reverberate” from the roof onto the ore, aiding oxidation of $$FeS$$ and $$Cu_{2}S$$ to yield almost pure copper. Therefore, blister copper matches with “Reverberatory furnace”, i.e. List-II item (i).

Finally we study item (d) of List I: “Froth flotation method”. The principle of froth flotation is based on the preferential wettability of ore and gangue particles. It is widely used for the concentration of sulphide ores such as $$ZnS$$, $$PbS$$, etc. Hence, “Froth flotation method” corresponds to “Sulfide ores”, i.e. List-II item (iv).

Collecting all the established matches, we have

$$\begin{aligned} (a) &\rightarrow (iii),\\ (b) &\rightarrow (ii),\\ (c) &\rightarrow (i),\\ (d) &\rightarrow (iv). \end{aligned}$$

Comparing this sequence with the options provided, we find that Option A exactly reproduces the same set of pairings.

Hence, the correct answer is Option A.

Assertion (A): Aluminium is obtained industrially by the electrolysis of a molten mixture containing $$\text{Al}_2\text{O}_3$$ (alumina) dissolved in cryolite ($$\text{Na}_3\text{AlF}_6$$). This statement is TRUE because alumina alone has a very high melting point (about $$2050^\circ\text{C}$$). Adding cryolite lowers the melting point to nearly $$950^\circ\text{C}$$ and increases the electrical conductivity, making large-scale electrolysis feasible. Hence, (A) is correct.

Reason (R): In cryolite, $$\text{Na}_3\text{AlF}_6$$, let the oxidation state of Al be $$x$$.
Total charge on six fluoride ions $$=6(-1)=-6$$.
Total charge contributed by three sodium ions $$=3(+1)=+3$$.
The compound is electrically neutral, so
$$x + (+3) + (-6) = 0 \Longrightarrow x -3 = 0 \Longrightarrow x = +3$$.
Therefore the oxidation state of aluminium in cryolite is indeed $$+3$$. Hence, (R) is also correct.

Link between (A) and (R): The reason for introducing cryolite during electrolysis is to lower the melting point and improve conductivity, not because aluminium happens to be in the $$+3$$ oxidation state (Al is already in the $$+3$$ state in alumina as well). Thus, while (R) is a true statement, it does not explain why cryolite is mixed with alumina.

Therefore, both (A) and (R) are correct, but (R) is not the correct explanation of (A).

Option C is the most appropriate choice.

A photoelectric cell works on the principle of the photoelectric effect. For a metal to be used as an electrode in a photoelectric cell, it must have a sufficiently low work function (ionization energy) so that visible or near-ultraviolet light can eject electrons from its surface.

Among the alkali metals, the work functions decrease as we go down the group: Li (2.42 eV) $$>$$ Na (2.28 eV) $$>$$ Rb (2.09 eV) $$>$$ Cs (1.95 eV). While all alkali metals have relatively low ionization energies, cesium (Cs) has the lowest work function and is the most commonly used metal in photoelectric cells. It is sensitive to visible light and is the standard choice for practical photoelectric applications.

Lithium and sodium have work functions that are too high for efficient use in standard photoelectric cells operating with visible light. Rubidium, while having a low work function, is less commonly used compared to cesium. Among the given metals, only cesium is practically employed as an electrode in photoelectric cells.

Therefore, the number of metals that can be used as electrodes in a photoelectric cell is $$\boxed{1}$$.

We need to find how many of the given compounds are amphoteric.

An amphoteric compound is one that can react with both acids and bases.

(A) BeO — Beryllium oxide is amphoteric. It reacts with acids: $$BeO + 2HCl \to BeCl_2 + H_2O$$, and with bases: $$BeO + 2NaOH \to Na_2BeO_2 + H_2O$$.

(B) BaO — Barium oxide is a strongly basic oxide. It reacts with acids but not with bases. So it is not amphoteric.

(C) $$Be(OH)_2$$ — Beryllium hydroxide is amphoteric. It dissolves in both acids and bases: $$Be(OH)_2 + 2HCl \to BeCl_2 + 2H_2O$$ and $$Be(OH)_2 + 2NaOH \to Na_2[Be(OH)_4]$$.

(D) $$Sr(OH)_2$$ — Strontium hydroxide is a strong base. It is not amphoteric.

Among the alkaline earth metals, beryllium compounds show amphoteric behaviour due to the very small size and high polarising power of $$Be^{2+}$$. The oxides and hydroxides of Ba and Sr are strongly basic.

So, the number of amphoteric compounds is 2 (BeO and $$Be(OH)_2$$).

So, the answer is $$2$$.

We begin by recalling the IUPAC systematic nomenclature that was used for newly discovered elements before they received their permanent names. The rule states that each individual digit of the atomic number is replaced by a specific Latin-Greek prefix and, after writing the three (or more) prefixes in sequence, we add the common ending “ium.” The mapping of digits to prefixes is as follows: $$0 \rightarrow \text{nil}, \; 1 \rightarrow \text{un}, \; 2 \rightarrow \text{bi}, \; 3 \rightarrow \text{tri}, \; 4 \rightarrow \text{quad}, \; 5 \rightarrow \text{pent}, \; 6 \rightarrow \text{hex}, \; 7 \rightarrow \text{sept}, \; 8 \rightarrow \text{oct}, \; 9 \rightarrow \text{enn}.$$

Now we examine the given temporary name “unnilennium.” We separate it into its component prefixes:

“un” + “nil” + “enn” + “ium.”

Using the digit-prefix chart stated above, we translate each prefix back into its numerical value:

$$\text{un} \rightarrow 1, \quad \text{nil} \rightarrow 0, \quad \text{enn} \rightarrow 9.$$

Placing these digits in the same order in which their prefixes appeared, we form the atomic number:

$$1\,0\,9 \;=\; 109.$$

So, the element whose temporary systematic name was “unnilennium” possesses the atomic number $$109$$.

Looking at the options given, we see that $$109$$ corresponds to Option A.

Hence, the correct answer is Option A.

We have been given the first ionisation enthalpy $$I_1 = 496\;{\rm kJ\;mol^{-1}}$$ and the second ionisation enthalpy $$I_2 = 4560\;{\rm kJ\;mol^{-1}}$$ for an unknown metal M. A very large jump from $$I_1$$ to $$I_2$$ tells us that, after one electron has been removed to give $$\mathrm{M^+}$$, removal of the second electron is energetically very unfavourable. In simple terms, the metal prefers the oxidation state $$+1$$.

If the metal forms a hydroxide, that hydroxide must therefore contain the monovalent cation $$\mathrm{M^+}$$. Combining charge balance with the monovalent anion $$\mathrm{OH^-}$$, the formula of the hydroxide is

$$\mathrm{MOH}.$$

Now we look at its reaction with hydrochloric acid. Writing the neutralisation reaction, we obtain

$$\mathrm{MOH + HCl \longrightarrow MCl + H_2O}.$$

The coefficients are all unity, so 1 mole of $$\mathrm{MOH}$$ consumes exactly 1 mole of $$\mathrm{HCl}$$.

Next we consider sulphuric acid. The reaction of the monobasic hydroxide with the dibasic acid is

$$2\,\mathrm{MOH + H_2SO_4 \longrightarrow M_2SO_4 + 2\,H_2O}.$$

Here 2 moles of the base react with 1 mole of $$\mathrm{H_2SO_4}$$. Dividing through by 2, we see that 1 mole of $$\mathrm{MOH}$$ requires

$$\dfrac{1}{2}\;{\rm mol} = 0.5\;{\rm mol}$$

of $$\mathrm{H_2SO_4}$$ for complete neutralisation.

Collecting the two results, to react completely with 1 mole of the metal hydroxide we need

$$1\;{\rm mol\;HCl} \quad\text{and}\quad 0.5\;{\rm mol\;H_2SO_4}.$$

Hence, the correct answer is Option D.

We first write the ground-state electronic configurations. For beryllium we have $$\mathrm{Be}(Z = 4):\;1s^{2}\;2s^{2}.$$ For boron we have $$\mathrm{B}(Z = 5):\;1s^{2}\;2s^{2}\;2p^{1}.$$

The first-ionisation enthalpy is the energy required to remove the outer-most (valence) electron from a gaseous atom. According to the general periodic trend, ionisation enthalpy increases from left to right in a period because nuclear charge increases while the principal quantum number remains the same. However, boron shows an exception: its first-ionisation enthalpy is lower than that of beryllium. To understand why, we analyse the nature of the electrons that are removed.

Be loses a $$2s$$ electron ($$2s^{2}\rightarrow2s^{1}$$), whereas B loses a $$2p$$ electron ($$2p^{1}\rightarrow2p^{0}$$). Now we discuss the four given statements one by one in the light of shielding, penetration and atomic size.

Statement (I) says “it is easier to remove a $$2p$$ electron than a $$2s$$ electron.” A $$2p$$ electron is farther from the nucleus on average and is shielded more effectively by the inner $$1s^{2}$$ and $$2s^{2}$$ electrons than a $$2s$$ electron. Hence the attraction between the nucleus and a $$2p$$ electron is weaker, so less energy is needed to remove it. Therefore statement (I) is true.

Statement (II) says “the $$2p$$ electron of B is more shielded from the nucleus by the inner core of electrons than the $$2s$$ electrons of Be.” Inside boron the $$2p$$ electron experiences the repelling effect of the already present $$2s^{2}$$ electrons in the same shell as well as the $$1s^{2}$$ core. By contrast, each $$2s$$ electron in beryllium is only shielded by the $$1s^{2}$$ core. Thus the effective nuclear charge felt by the $$2p$$ electron in B is lower than that felt by a $$2s$$ electron in Be. So the shielding of the $$2p$$ electron in B is indeed greater. Hence statement (II) is true.

Statement (III) says “$$2s$$ electron has more penetration power than $$2p$$ electron.” Penetration refers to the probability of finding an electron close to the nucleus. For a given principal quantum number $$n=2$$, the order of penetration is $$2s > 2p.$$ Because an $$s$$ orbital possesses a finite probability density at the nucleus whereas a $$p$$ orbital has a node there, the $$2s$$ electron penetrates closer to the nucleus, feels a higher effective nuclear charge, and is held more tightly. Hence statement (III) is true.

Statement (IV) says “atomic radius of B is more than Be.” Moving from Be to B across the same period, the effective nuclear charge increases (because of the extra proton) and the principal quantum number remains $$n=2$$. Greater nuclear attraction pulls the electron cloud inward, so the atomic radius actually decreases. Therefore $$r_{\text{B}} < r_{\text{Be}}.$$ Statement (IV) is false.

Summarising, statements (I), (II) and (III) are correct, while statement (IV) is incorrect.

Hence, the correct answer is Option C.

Across a period in the periodic table (for example from lithium to neon in the second period) the nuclear charge increases because one proton is added to the nucleus for every step to the right. At the same time, electrons are being added to the same principal energy level, so the extra nuclear charge is not completely shielded. Because of this, the effective nuclear charge, usually written as $$Z_{\text{eff}}$$, steadily rises from left to right.

Now, let us recall how the principal periodic properties depend on $$Z_{\text{eff}}$$:

1. Ionization enthalpy (also called first ionization energy) is the energy required to remove the outer-most electron. The greater the attraction between the nucleus and that outer electron, the higher this energy will be. Therefore, as $$Z_{\text{eff}}$$ increases across a period, ionization enthalpy also increases.

2. Electronegativity is the tendency of an atom in a molecule to attract the shared pair of electrons. A higher effective nuclear charge pulls electrons more strongly, so electronegativity also increases from left to right in a period.

3. Electron gain enthalpy (or electron affinity) is the enthalpy change when an isolated neutral atom gains an electron to form an anion. A higher $$Z_{\text{eff}}$$ means the added electron is more strongly attracted, releasing more energy (the value becomes more negative). In magnitude, electron gain enthalpy therefore also increases across a period.

4. Atomic radius is half the distance between the nuclei of two identical bonded atoms. A rise in $$Z_{\text{eff}}$$ pulls the electron cloud closer to the nucleus. So, instead of increasing, the atomic radius actually decreases from left to right in a period.

We observe that ionization enthalpy, electronegativity and electron gain enthalpy all show an increasing trend across a period, while atomic radius shows a decreasing trend—exactly the opposite direction. Therefore, the property whose magnitude varies oppositely compared with the others is the atomic radius.

Hence, the correct answer is Option D.

First, let us recall the Aufbau principle, which states that electrons occupy the available orbitals in the order of increasing $$(n+l)$$ value, and if two orbitals have the same $$(n+l)$$ value, then the one with the lower $$n$$ (principal quantum number) is filled first.

Now, we list all the orbitals in the increasing order of $$(n+l)$$ up to the region where the sixth period ends (i.e., up to the noble-gas element radon, $$Z = 86$$):

$$1s,\, 2s,\, 2p,\, 3s$$, $$3p,\, 4s,\, 3d,\, 4p$$, $$5s,\, 4d,\, 5p,\, 6s$$, $$4f,\, 5d,\, 6p,\, 7s,\ldots$$

We see that the very first orbital whose principal quantum number is $$6$$ is $$6s$$. So, after finishing $$5p$$ (which completes the fifth period at xenon, $$Z = 54$$), the next orbital to receive electrons is $$6s$$.

After filling the $$6s$$ orbital with two electrons, we check the next entry in our ordered list of orbitals. That orbital is $$4f$$. Although its principal quantum number $$n = 4$$ is numerically less than 6, its $$(n+l)$$ value $$=4+3=7$$ is larger than that of $$6s$$ $$\big(n+l = 6+0=6\big)$$, so, according to the Aufbau ordering, $$4f$$ comes directly after $$6s$$. Hence, as we progress through the sixth period (elements $$Z = 57$$ to $$Z = 70$$, the lanthanoids), the $$4f$$ subshell gets filled.

Once $$4f$$ is completely filled with its $$14$$ electrons, the next orbital in the list is $$5d$$. Thus, elements $$Z = 71$$ to $$Z = 80$$ (the sixth-period transition elements) add electrons to $$5d$$.

When $$5d$$ reaches its capacity of $$10$$ electrons, the sequence finally moves on to $$6p$$. The $$6p$$ orbitals are filled across the last part of the sixth period, ending with radon, $$Z = 86$$.

Putting all these steps together, the orbitals that receive electrons in the sixth period are, in chronological order, $$6s \rightarrow 4f \rightarrow 5d \rightarrow 6p$$.

Comparing this list with the supplied options, we observe that Option A lists the orbitals exactly as $$6s,\,4f,\,5d,\,6p$$, which matches our derived sequence.

Hence, the correct answer is Option A.

We first note that the species $$\text{O}^{2-},\;\text{N}^{3-},\;\text{F}^- ,\;\text{Na}^+ ,\;\text{Mg}^{2+}$$ and $$\text{Al}^{3+}$$ all possess the same total number of electrons.

Counting one by one, we have:

$$\begin{aligned} \text{N}^{3-}: &\;7+3=10\text{ e}^- \\ \text{O}^{2-}: &\;8+2=10\text{ e}^- \\ \text{F}^- : &\;9+1=10\text{ e}^- \\ \text{Na}^+: &\;11-1=10\text{ e}^- \\ \text{Mg}^{2+}: &\;12-2=10\text{ e}^- \\ \text{Al}^{3+}: &\;13-3=10\text{ e}^- \end{aligned}$$

Since every ion contains $$10$$ electrons, they are termed isoelectronic.

For any series of isoelectronic ions, we use the fact that:

$$\text{Ionic radius} \propto \frac1Z$$

where $$Z$$ is the atomic number (nuclear charge). The mathematical statement above expresses that, with the electron cloud held constant, a larger positive nuclear charge pulls the same set of electrons closer, so the radius becomes smaller. Conversely, a smaller nuclear charge allows the cloud to expand, giving a larger radius.

We therefore list the atomic numbers:

$$\begin{aligned} Z(\text{N}) &= 7,\\ Z(\text{O}) &= 8,\\ Z(\text{F}) &= 9,\\ Z(\text{Na}) &= 11,\\ Z(\text{Mg}) &= 12,\\ Z(\text{Al}) &= 13. \end{aligned}$$

Now we arrange the ions from the highest $$Z$$ (smallest radius) to the lowest $$Z$$ (largest radius):

$$\text{Al}^{3+}\;(Z=13)\; < \;\text{Mg}^{2+}\;(Z=12)\; < \;\text{Na}^+\;(Z=11)\; < \;\text{F}^-\;(Z=9)\; < \;\text{O}^{2-}\;(Z=8)\; < \;\text{N}^{3-}\;(Z=7).$$

Comparing this derived order with the given alternatives, we find perfect agreement with Option C.

Hence, the correct answer is Option C.

We begin by identifying the element whose atomic number is $$101$$. Consulting the extended periodic table, we find that atomic number $$101$$ corresponds to the element Mendelevium whose symbol is $$\text{Md}$$. This element lies in the series of fourteen elements that follow actinium ($$Z=89$$) and extend from thorium ($$Z=90$$) to lawrencium ($$Z=103$$). This horizontal block is known as the actinoid (actinide) series. So, the element with $$Z=101$$ is an actinoid.

Next, we locate the element whose atomic number is $$104$$. Atomic number $$104$$ corresponds to Rutherfordium with symbol $$\text{Rf}$$. After the completion of the actinoid series at lawrencium ($$Z=103$$), the periodic table resumes its normal long-form layout. Rutherfordium therefore occupies the position directly below hafnium ($$Z=72$$) and zirconium ($$Z=40$$). All three elements belong to the vertical column that, in modern IUPAC numbering, is called Group 4 of the periodic table. Group 4 contains the general configuration $$[ \text{noble gas} ]\,n s^2 (n-1)d^2$$ and is part of the $$d$$-block transition metals.

Hence, the element with atomic number $$101$$ is an actinoid, while the element with atomic number $$104$$ lies in Group 4.

Comparing with the given options, we see that Option C states “Actinoids and Group 4,” which matches our deduction.

Hence, the correct answer is Option C.

We recall the periodic trend that as we move from left to right in a period, the nuclear charge increases while the added electrons enter the same principal shell. Because the effective nuclear charge ($$Z_{\text{eff}}$$) felt by the outermost electron increases, the first ionization energy—defined as the energy needed to remove one mole of electrons from one mole of gaseous atoms, $$M(g) \rightarrow M^{+}(g) + e^{-}$$—generally increases across a period.

However, there are characteristic small drops when the added electron begins to occupy a new subshell of higher energy (such as from $$s$$ to $$p$$) or when electron-electron repulsion in a doubly occupied orbital becomes significant. In Period 3 the relevant electronic configurations are:

$$\begin{aligned} \text{Na} &:& [\text{Ne}]\,3s^{1} \\ \text{Mg} &:& [\text{Ne}]\,3s^{2} \\ \text{Al} &:& [\text{Ne}]\,3s^{2}\,3p^{1} \\ \text{Si} &:& [\text{Ne}]\,3s^{2}\,3p^{2} \end{aligned}$$

Starting with sodium, the single $$3s$$ electron is weakly held, so the first ionization energy is the smallest, about $$496\;\text{kJ mol}^{-1}$$.

Moving to magnesium, the outermost electron is still in the $$3s$$ subshell but the nucleus now has one extra proton. The additional positive charge outweighs the slight increase in shielding, so the ionization energy rises to approximately $$737\;\text{kJ mol}^{-1}$$.

For aluminium, the outermost electron enters the higher-energy $$3p$$ subshell. Because a $$3p$$ electron is less tightly held than a $$3s$$ electron, there is an anomalous dip despite the increasing nuclear charge. Consequently, the first ionization energy falls to about $$577\;\text{kJ mol}^{-1}$$.

Finally, in silicon the added electron pairs up in the $$3p$$ subshell, and the steadily increasing $$Z_{\text{eff}}$$ now dominates again. Hence the ionization energy climbs to roughly $$786\;\text{kJ mol}^{-1}$$.

Collecting the values in order Na, Mg, Al, Si we obtain:

$$496,\;737,\;577,\;786\;(\text{kJ mol}^{-1})$$

This exact sequence matches Option A.

Hence, the correct answer is Option A.

We begin with the basic periodic trends for atomic (covalent) radius.

1. Along any period (that is, from left to right in the same row of the periodic table) the atomic number $$Z$$ increases but the electrons are being added to the same principal energy level. The effective nuclear charge $$Z_{\text{eff}}$$ therefore increases, pulling the electron cloud closer to the nucleus.
So, within a single period: $$\text{Atomic radius decreases from left to right.}$$

2. Down any group (that is, from top to bottom in the same column) a new principal energy level is added each time. Even though $$Z_{\text{eff}}$$ also rises a little, the increase in principal quantum number $$n$$ dominates, keeping the outer electrons farther from the nucleus.
So, within a group: $$\text{Atomic radius increases from top to bottom.}$$

Now we list the given elements with their periodic positions:

$$\begin{aligned} \text{C}&:& \text{Group 14, Period 2}\\ \text{O}&:& \text{Group 16, Period 2}\\ \text{F}&:& \text{Group 17, Period 2}\\ \text{Cl}&:& \text{Group 17, Period 3}\\ \text{Br}&:& \text{Group 17, Period 4} \end{aligned}$$

First, let us compare the three elements that lie in the same period (Period 2): $$\text{C, O, F}.$$ Moving from C to O to F we move left → right, so the radii must decrease:

$$r_{\text{F}} < r_{\text{O}} < r_{\text{C}}.$$

Next, consider the three halogens $$\text{F, Cl, Br}$$ which lie in the same group (Group 17). Going down the group the radius increases, hence

$$r_{\text{F}} < r_{\text{Cl}} < r_{\text{Br}}.$$

To place C among Cl and Br we recall that C is in Period 2 while Cl is in Period 3. Because an extra principal shell is present in Period 3, $$r_{\text{Cl}}$$ is larger than $$r_{\text{C}}.$$ Numerically, their covalent radii are approximately $$r_{\text{C}}\approx 77\,\text{pm},\; r_{\text{Cl}}\approx 99\,\text{pm}.$$ Thus

$$r_{\text{C}} < r_{\text{Cl}}.$$

Combining every inequality obtained, starting from the smallest value and moving upward:

$$r_{\text{F}} < r_{\text{O}} < r_{\text{C}} < r_{\text{Cl}} < r_{\text{Br}}.$$

Translating back to the element symbols given in the question: $$(c)\;F \; <\; (b)\;O \; <\; (a)\;C \; <\; (d)\;Cl \; <\; (e)\;Br.$$

This sequence matches exactly with Option C.

Hence, the correct answer is Option C.

First, let us recall the two kinds of thermochemical quantities involved here.

1. Ionisation enthalpy (also called ionisation energy) is the energy required to remove an electron from an isolated gaseous atom:

$$X_{(g)} \to X^+_{(g)} + e^-$$

Because energy has to be supplied, ionisation enthalpy is always positive; hence every ionisation process is endothermic.

2. Electron-gain enthalpy (also called electron affinity) is the energy change when an electron is added to an isolated gaseous species:

$$Y_{(g)} + e^- \to Y^-_{(g)}$$

If energy is released, the enthalpy change is negative and the process is exothermic. If energy has to be supplied (positive enthalpy change) the process is endothermic.

With this information we now examine the four options one by one.

Option A: $$Ar_{(g)} + e^- \to Ar^-_{(g)}$$

Argon is a noble-gas atom with a completely filled octet. Adding an extra electron would place it in a new, higher-energy shell, so energy must be supplied. Therefore the electron-gain enthalpy of Ar is positive; the process is endothermic.

Option B: $$H_{(g)} + e^- \to H^-_{(g)}$$

Hydrogen has the electronic configuration $$1s^1$$. On taking one more electron it attains the stable helium-like configuration $$1s^2$$. The system releases energy in doing so; hence the electron-gain enthalpy of hydrogen is negative. Because energy is released, this process is exothermic, i.e. it is not endothermic.

Option C: $$O^-_{(g)} + e^- \to O^{2-}_{(g)}$$

The anion $$O^-$$ already carries a negative charge. Introducing a second electron increases electron-electron repulsion markedly, so extra energy must be supplied. Thus the second electron-gain enthalpy of oxygen is positive; the process is endothermic.

Option D: $$Na_{(g)} \to Na^+_{(g)} + e^-$$

This is the first ionisation of sodium. As stated earlier, every ionisation requires energy input, so the process is endothermic.

Summarising, Options A, C and D are endothermic, whereas Option B releases energy and is not endothermic.

Hence, the correct answer is Option B.

In the modern periodic table, elements of the 3rd period appear in the horizontal sequence from sodium (Na, atomic number 11) to argon (Ar, atomic number 18). Moving from left to right across a period, the metallic character of the elements decreases, while the non-metallic character increases. This gradual change in character is directly reflected in the nature of their oxides.

We recall the general trend:

$$\text{Metallic character}\;\downarrow \quad\Longrightarrow\quad \text{Basic nature of oxide}\;\downarrow \quad\text{and}\quad \text{Acidic nature of oxide}\;\uparrow$$

So, as we progress from the metallic left side to the non-metallic right side of a period:

$$\text{Strongly basic oxides} \;\longrightarrow\; \text{Amphoteric oxides} \;\longrightarrow\; \text{Strongly acidic oxides}$$

Listing the 3rd-period elements with the nature of their important oxides:

$$ \begin{aligned} \text{Na (11)} & : \text{Na}_2\text{O} \;-\; \text{strongly basic} \\ \text{Mg (12)} & : \text{MgO} \;-\; \text{basic} \\ \text{Al (13)} & : \text{Al}_2\text{O}_3 \;-\; \text{amphoteric} \\ \text{Si (14)} & : \text{SiO}_2 \;-\; \text{weakly acidic} \\ \text{P (15)} & : \text{P}_4\text{O}_{10} \;-\; \text{acidic} \\ \text{S (16)} & : \text{SO}_3 \;-\; \text{acidic} \\ \text{Cl (17)} & : \text{Cl}_2\text{O}_7 \;-\; \text{strongly acidic} \\ \end{aligned} $$

From the question, the oxide of element X is basic, the oxide of Y is amphoteric, and the oxide of Z is acidic. Matching these descriptions with the list above:

• A basic oxide corresponds to an element like Na (11) or Mg (12).
• The only amphoteric oxide in this period is Al2O3, so Y must be Al (atomic number 13).
• Acidic oxides begin with SiO2 (Si, 14) and become more acidic as we go further right. Any of Si, P, S, or Cl satisfies “acidic”.

For the atomic numbers to satisfy X < Y < Z, we simply pick the first basic element to the left of Al and the first acidic element to the right of Al. The smallest such set is:

$$ X = \text{Mg}\;(12),\quad Y = \text{Al}\;(13),\quad Z = \text{Si}\;(14). $$

This clearly gives:

$$X < Y < Z.$$

Among the options, this inequality matches Option B.

Hence, the correct answer is Option B.

For any atom in the gaseous state, the quantity of energy released when one electron is added is called its electron-gain enthalpy (also called electron affinity). A larger release of energy means a more negative value of $$\Delta_{\text{eg}}H$$ and therefore a stronger tendency to accept an electron.

General periodic trends help us decide which element of each pair shows the larger (more negative) $$\Delta_{\text{eg}}H$$ :

• Along a period from left to right, nuclear charge increases and size decreases, so electron-gain enthalpy usually becomes more negative.

• Down a group, atomic size increases and the added electron feels less attraction, so $$\Delta_{\text{eg}}H$$ usually becomes less negative. However, very small atoms of the second period (such as F, O, N) suffer noticeably from inter-electronic repulsion in their compact $$2p$$ orbitals, producing some well-known exceptions.

Now we compare each given pair.

First pair: $$\text{F}$$ and $$\text{Cl}$$ lie in Group 17. Although fluorine is above chlorine, its tiny $$2p$$ subshell is already crowded, so the incoming electron experiences appreciable repulsion. Therefore the energy released on adding an electron is actually less for F than for Cl. In numbers, $$\Delta_{\text{eg}}H(\text{F}) \approx -328\;\text{kJ mol}^{-1}$$ whereas $$\Delta_{\text{eg}}H(\text{Cl}) \approx -349\;\text{kJ mol}^{-1}$$. So chlorine releases more energy than fluorine.

Second pair: $$\text{S}$$ and $$\text{Se}$$ belong to Group 16. Moving down from sulfur to selenium the atomic size grows, and there is no exceptional crowding because the comparison is between $$3p$$ (for S) and $$4p$$ (for Se) orbitals. Consequently $$\Delta_{\text{eg}}H$$ becomes less negative down the group. Hence $$\Delta_{\text{eg}}H(\text{S})$$ is more negative than $$\Delta_{\text{eg}}H(\text{Se})$$, so sulfur releases more energy than selenium.

Third pair: $$\text{Li}$$ and $$\text{Na}$$ are the first two members of Group 1. For the alkali metals the usual down-the-group decrease in electron-gain enthalpy operates without any exceptions, because both $$\text{Li}$$ and $$\text{Na}$$ would place the extra electron into an $$s$$ orbital, and the size increase dominates. Numerically, $$\Delta_{\text{eg}}H(\text{Li}) \approx -60\;\text{kJ mol}^{-1}$$ while $$\Delta_{\text{eg}}H(\text{Na}) \approx -53\;\text{kJ mol}^{-1}$$. Therefore lithium releases more energy than sodium.

Collecting the conclusions: • Between F and Cl, the element releasing more energy is $$\text{Cl}$$. • Between S and Se, the element releasing more energy is $$\text{S}$$. • Between Li and Na, the element releasing more energy is $$\text{Li}$$.

Thus the required set is $$\text{Cl},\; \text{S},\; \text{Li}$$.

Hence, the correct answer is Option B.

First, we recall the general rule about the nature of oxides. Oxides of highly electropositive metals (especially the alkali and alkaline-earth metals) are basic. Oxides of non-metals are usually acidic because the non-metal is electronegative and pulls electron density away from oxygen, making the O atom less able to donate a pair. Finally, certain metal oxides lying near the metal-non-metal border (for example $$Al,\,Zn,\,Sn,\,Pb$$) can behave both as acids and as bases; these are called amphoteric oxides.

We now analyse every oxide that appears in the four options one by one, classifying each as acidic, basic or amphoteric.

1. $$Na_2O$$
Sodium is an alkali metal. Alkali-metal oxides react with water to give a strong base: $$Na_2O + H_2O \rightarrow 2\,NaOH$$. Hence $$Na_2O$$ is basic.

2. $$SO_3$$
Sulphur is a non-metal, and its oxide reacts with water to give sulphuric acid: $$SO_3 + H_2O \rightarrow H_2SO_4$$. Therefore $$SO_3$$ is acidic.

3. $$Al_2O_3$$
Aluminium lies on the diagonal boundary between metals and non-metals. Its oxide dissolves in acids as well as in strong bases, showing dual behaviour:

$$Al_2O_3 + 6\,HCl \rightarrow 2\,AlCl_3 + 3\,H_2O$$ (acts as a base)

$$Al_2O_3 + 2\,NaOH + 3\,H_2O \rightarrow 2\,Na[Al(OH)_4]$$ (acts as an acid)

Hence $$Al_2O_3$$ is amphoteric.

4. $$Cl_2O$$
Chlorine is a non-metal; its oxide reacts with water to form hypochlorous acid: $$Cl_2O + H_2O \rightarrow 2\,HOCl$$. Thus $$Cl_2O$$ is acidic.

5. $$CaO$$
Calcium is an alkaline-earth metal. $$CaO + H_2O \rightarrow Ca(OH)_2$$, producing a strong base. So $$CaO$$ is basic.

6. $$P_4O_{10}$$
Phosphorus is a non-metal. $$P_4O_{10} + 6\,H_2O \rightarrow 4\,H_3PO_4$$. Hence $$P_4O_{10}$$ is acidic.

7. $$N_2O_3$$
Nitrogen is a non-metal, and the oxide reacts with water to give nitrous acid: $$N_2O_3 + H_2O \rightarrow 2\,HNO_2$$. Therefore $$N_2O_3$$ is acidic.

8. $$Li_2O$$
Lithium is an alkali metal. $$Li_2O + H_2O \rightarrow 2\,LiOH$$. Thus $$Li_2O$$ is basic.

9. $$MgO$$
Magnesium, an alkaline-earth metal, forms oxide that reacts with acids producing salt and water, so it is basic.

With all the classifications complete, we inspect every option to see whether the first oxide is acidic, the second one basic, and the third one amphoteric, exactly in that order.

Option A: $$Na_2O$$ (basic), $$SO_3$$ (acidic), $$Al_2O_3$$ (amphoteric) - the sequence is basic-acidic-amphoteric, not correct.

Option B: $$Cl_2O$$ (acidic), $$CaO$$ (basic), $$P_4O_{10}$$ (acidic) - the third oxide is not amphoteric, so the option is wrong.

Option C: $$N_2O_3$$ (acidic), $$Li_2O$$ (basic), $$Al_2O_3$$ (amphoteric). This exactly matches the required pattern acidic-basic-amphoteric.

Option D: $$MgO$$ (basic), $$Cl_2O$$ (acidic), $$Al_2O_3$$ (amphoteric) - again the order is basic-acidic-amphoteric, hence not correct.

Only Option C satisfies the given order.

Hence, the correct answer is Option C.

We recall the definition first. Electron-gain enthalpy (also called electron affinity) is the enthalpy change when one mole of gaseous atoms accepts one mole of electrons to form gaseous anions. The more negative the value, the easier it is for that atom to accept an electron.

Across a period, effective nuclear charge increases and size decreases, so electron gain enthalpy generally becomes more negative. Down a group, atomic size increases, shielding increases and the added electron feels less attraction; therefore, electron gain enthalpy should normally become less negative from top to bottom of a group.

However, in the halogen group (Group 17) there is a very well-known exception: chlorine shows a more negative electron-gain enthalpy than fluorine. This happens because:

$$\text{(i) Fluorine is very small,}$$

$$$\text{(ii) The 2p subshell into which the extra electron must enter is already crowded,}$$$

$$$\text{(iii) Electron-electron repulsion partially offsets the nuclear attraction.}$$$

Hence the accepted experimental order is

$$$\Delta_{\text{eg}}H(\text{Cl}) < \Delta_{\text{eg}}H(\text{F}) < \Delta_{\text{eg}}H(\text{Br}) < \Delta_{\text{eg}}H(\text{I})$$$

(Remember that “more negative” means “smaller” on the number line.)

Numerically, the widely quoted values are approximately

$$\Delta_{\text{eg}}H(\text{F}) = -333\;\text{kJ mol}^{-1},$$

$$\Delta_{\text{eg}}H(\text{Cl}) = -349\;\text{kJ mol}^{-1},$$

$$\Delta_{\text{eg}}H(\text{Br}) = -325\;\text{kJ mol}^{-1},$$

$$\Delta_{\text{eg}}H(\text{I}) = -296\;\text{kJ mol}^{-1}.$$

Now we examine the options given:

Option A: $$-296,\,-325,\,-333,\,-349$$

Option B: $$-349,\,-333,\,-325,\,-296$$

Option C: $$-333,\,-349,\,-325,\,-296$$

Option D: $$-333,\,-325,\,-349,\,-296$$

We compare each list with the correct sequence $$\text{F},\;\text{Cl},\;\text{Br},\;\text{I}$$ having the numerical values just stated.

Option C exactly reproduces $$$-333\;(\text{F}),\;-349\;(\text{Cl}),\;-325\;(\text{Br}),\;-296\;(\text{I}).$$$

None of the other options aligns with both the correct order of elements and their accepted magnitudes.

Hence, the correct answer is Option 3.

First of all, successive ionization enthalpies $$I_{1}, I_{2}, I_{3}, I_{4}, I_{5}, \ldots$$ represent the energies required to remove the first, second, third and so on electrons from an isolated gaseous atom of an element. In general, as more electrons are removed, the positive charge on the ion left behind increases, so the remaining electrons are held more strongly; therefore $$I_{2} > I_{1},\; I_{3} > I_{2}$$ and so forth.

However, when all the outer-shell (valence) electrons have been removed and we begin to take electrons out of the next inner shell, the required energy shows a very large sudden jump. This is because inner-shell electrons are much closer to the nucleus and feel a far stronger effective nuclear charge. Therefore, by locating the first big jump in the given data, we can identify how many electrons were present in the outermost shell, i.e. how many valence electrons the neutral atom had.

We are given the following five successive ionization enthalpies of the unknown element:

$$I_{1}=800\ \text{kJ mol}^{-1},$$ $$I_{2}=2427\ \text{kJ mol}^{-1},$$ $$I_{3}=3658\ \text{kJ mol}^{-1},$$ $$I_{4}=35024\ \text{kJ mol}^{-1},$$ $$I_{5}=32824\ \text{kJ mol}^{-1}.$$

Let us now examine the relative increases step by step.

First, we compare $$I_{2}$$ with $$I_{1}$$:

$$I_{2}-I_{1}=2427-800 = 1627\ \text{kJ mol}^{-1}.$$

Next, we compare $$I_{3}$$ with $$I_{2}$$:

$$I_{3}-I_{2}=3658-2427 = 1231\ \text{kJ mol}^{-1}.$$

Both of these increases (1627 and 1231) are of the same order of magnitude, showing the normal gradual rise expected while we are still removing electrons from the same outer shell.

Now we compare $$I_{4}$$ with $$I_{3}$$:

$$I_{4}-I_{3}=35024-3658 = 31366\ \text{kJ mol}^{-1}.$$

Here we see an enormous jump (over 30 000 kJ mol−1), far larger than the earlier differences. This indicates that after the removal of the first three electrons, we have exhausted the valence shell and are now attempting to ionize an inner-shell electron, which is much more tightly bound.

Therefore, the neutral atom must have had exactly three electrons in its outermost shell. In other words, the number of valence electrons is $$3$$.

Among the given options, “3” corresponds to Option C.

Hence, the correct answer is Option C.

We first observe that the ions $$O^{2-},\;F^- ,\;Na^+$$ and $$Mg^{2+}$$ all possess exactly ten electrons. This is because

$$O^{2-}: 8\text{ (protons)}+2\text{ (extra electrons)}=10\text{ e}^-$$
$$F^-: 9+1=10\text{ e}^-$$
$$Na^+: 11-1=10\text{ e}^-$$
$$Mg^{2+}: 12-2=10\text{ e}^-$$

Species having the same number of electrons are called isoelectronic. For an isoelectronic series the principal factor governing ionic radius is the magnitude of the nuclear charge $$Z$$: the larger the value of $$Z$$, the stronger the electrostatic attraction exerted on the electron cloud, and hence the smaller the ionic radius.

Let us therefore list the nuclear charges explicitly:

$$\begin{aligned} O^{2-}:&\; Z=8,\\[2pt] F^-:&\; Z=9,\\[2pt] Na^+:&\; Z=11,\\[2pt] Mg^{2+}:&\; Z=12. \end{aligned}$$

Because ionic radius $$r$$ varies inversely with $$Z$$ within an isoelectronic sequence, we write symbolically

$$r\propto\frac1Z.$$

Hence a smaller nuclear charge gives a larger radius. Sorting the ions from the smallest nuclear charge to the largest nuclear charge gives

$$8\;(O^{2-}) < 9\;(F^-) < 11\;(Na^+) < 12\;(Mg^{2+}).$$

Correspondingly, the ionic radii follow the reverse order:

$$r(O^{2-}) > r(F^-) > r(Na^+) > r(Mg^{2+}).$$

Comparing with the options, this sequence matches Option B.

Hence, the correct answer is Option B.

We have to compare the solubilities of the sulphates of two successive Group 2 (alkaline-earth) elements, namely $$\text{BeSO}_4$$ and $$\text{MgSO}_4$$, when they are placed in water. To do this we recall the general principle that, for an ionic solid to dissolve, the magnitude of its hydration enthalpy must at least compensate for its lattice enthalpy.

For any salt $$\text{M}^{2+}\text{SO}_4^{2-}$$ belonging to the alkaline-earth series, two energetic quantities are important:

1. Lattice enthalpy, written $$\Delta H_{\text{lattice}},$$ which arises from the electrostatic attraction inside the crystal. 2. Hydration enthalpy, written $$\Delta H_{\text{hyd}},$$ which is released when the gaseous ions $$\text{M}^{2+}$$ and $$\text{SO}_4^{2-}$$ are surrounded by water molecules.

The overall enthalpy change for dissolution is

$$\Delta H_{\text{solution}} = \Delta H_{\text{lattice}} + \Delta H_{\text{hyd}}.$$

If $$\Delta H_{\text{solution}}$$ is negative or only slightly positive, the salt is appreciably soluble; if it is strongly positive, the salt remains largely undissolved.

Now we examine the periodic-trend arguments step by step:

• As we move down the Group 2 column, the ionic radius of $$\text{M}^{2+}$$ increases. So $$r_{\text{Be}^{2+}} < r_{\text{Mg}^{2+}} < r_{\text{Ca}^{2+}} \ldots$$

• Because lattice enthalpy roughly varies as $$\dfrac{z^{+}z^{-}}{r^{+}+r^{-}},$$ a larger cation radius decreases lattice enthalpy. Hence $$\Delta H_{\text{lattice}}$$ becomes less negative (weaker in magnitude) from BeSO4 to MgSO4 to CaSO4, and so on.

• Hydration enthalpy varies approximately as $$\dfrac{z^{2}}{r},$$ so a smaller cation radius gives a more negative (more exothermic) $$\Delta H_{\text{hyd}}.$$ Therefore the hydration enthalpy decreases markedly as we move down the group.

Combining the two trends:

Be}^{2+}: \; \Delta H_{\text{lattice}} \text{ very large negative}, \; \Delta H_{\text{hyd}} \text{ even more negative

Because the hydration enthalpy dominates, $$\Delta H_{\text{solution}}$$ is sufficiently negative, so $$\text{BeSO}_4$$ dissolves readily — it is highly soluble.

Mg}^{2+}: \; \Delta H_{\text{lattice}} \text{ slightly smaller in magnitude}, \; \Delta H_{\text{hyd}} \text{ still large though less than Be}^{2+

The hydration enthalpy still outweighs the lattice enthalpy, and $$\Delta H_{\text{solution}}$$ remains near zero or slightly negative. Hence $$\text{MgSO}_4$$ is also highly soluble in water. (Indeed, the familiar Epsom salt is $$\text{MgSO}_4\cdot7\text{H}_2\text{O},$$ which dissolves easily.)

Starting with $$\text{CaSO}_4$$ and continuing to $$\text{BaSO}_4,$$ the hydration enthalpy drops below the lattice enthalpy, making those sulphates sparingly soluble; however, our question concerns only BeSO4 and MgSO4.

So the solubility pattern obtained is:

$$\text{BeSO}_4 : \text{ high}$$

$$\text{MgSO}_4 : \text{ high}$$

This matches Option C in the given list.

Hence, the correct answer is Option C.

When an alkali metal is burnt in excess of air, the nature of the oxide obtained depends strongly on the size of the metal cation $$M^+$$ and on the particular oxide anion that can be stabilised by that cation. The sequence of possible oxide anions is:

$$O^{2-}\;(oxide)\;,$$ $$O_2^{2-}\;(peroxide)\;,$$ $$O_2^{-}\;(superoxide)$$

As the cation becomes larger down the group, its ability to stabilise the bigger and less charge-dense anions increases. Mathematically, lattice energy $$U_L$$ is roughly proportional to $$\dfrac{Z^{+}\,Z^{-}}{r^{+}+r^{-}}$$, where $$r^{+}$$ and $$r^{-}$$ are the ionic radii and $$Z^{+},Z^{-}$$ are the ionic charges. A small $$M^+$$ gives a high lattice energy only with the small $$O^{2-}$$ ion; a larger $$M^+$$ gives sufficient lattice energy even with the bulkier $$O_2^{2-}$$ or $$O_2^{-}$$ ions.

We now examine each metal one by one.

For lithium we have the smallest cation $$Li^+$$. Its high polarising power favours the compact oxide ion $$O^{2-}$$. Thus the dominant reaction in excess air is

$$4\,Li\;+\;O_2\;\longrightarrow\;2\,Li_2O$$

so the principal oxide is $$Li_2O$$.

For sodium the cation $$Na^+$$ is larger. It can still form $$Na_2O$$ in limited oxygen, but in excess oxygen it preferentially forms the peroxide:

$$2\,Na\;+\;O_2\;\longrightarrow\;Na_2O_2$$

Therefore, the major product under the stated conditions is $$Na_2O_2$$.

For potassium the cation $$K^+$$ is even larger. It is able to stabilise the still larger superoxide ion $$O_2^{-}$$. In excess oxygen we have

$$K\;+\;O_2\;\longrightarrow\;KO_2$$

So the principal oxide here is $$KO_2$$.

Collecting the three results, the major oxides obtained are

$$Li_2O,\qquad Na_2O_2,\qquad KO_2.$$

Comparing with the given options, we see that Option B lists exactly this sequence.

Hence, the correct answer is Option B.

We begin by recalling a basic periodic-table idea known as the diagonal relationship. According to this concept, the first element of a period in the $$s$$-block often shows properties similar to the second element of the next period lying diagonally below it. Thus, $$\text{Li (Group 1, Period 2)}$$ is diagonally related to $$\text{Mg (Group 2, Period 3)}$$ and they exhibit many chemical similarities.

Now let us match the two special properties mentioned in the question with the behaviour of different $$s$$-block elements:

1. They do not form solid hydrogencarbonates.

   • For most alkali metals, a salt such as $$\text{NaHCO}_{3}$$ or $$\text{KHCO}_{3}$$ can be isolated in the solid state.
   • However, $$\text{LiHCO}_{3}$$ is not stable as a solid. When an attempt is made to isolate it, it decomposes: $$\text{2 LiHCO}_{3 \,(aq)} \;\longrightarrow\; \text{Li}_{2}\text{CO}_{3 \,(s)} + \text{CO}_{2} + \text{H}_{2}\text{O}.$$
   • A very similar situation exists for magnesium. The salt $$\text{Mg(HCO}_{3})_{2}$$ only survives in aqueous solution; on heating or isolation it converts to $$\text{MgCO}_{3}$$, releasing $$\text{CO}_{2}$$ and water.

2. They react with nitrogen to give nitrides.

   • Lithium burns in nitrogen to give the nitride $$\text{Li}_{3}\text{N}$$: $$6\text{Li} + \text{N}_{2} \rightarrow 2\text{Li}_{3}\text{N}.$$
   • Magnesium also combines directly with nitrogen to form $$\text{Mg}_{3}\text{N}_{2}$$: $$3\text{Mg} + \text{N}_{2} \rightarrow \text{Mg}_{3}\text{N}_{2}.$$

Both of these characteristic reactions are uncommon for the heavier alkali metals such as Na, K, Rb, Cs and for the heavier alkaline-earth metals like Ca, Sr, Ba. Thus, if two elements satisfy both conditions simultaneously, the only reasonable pair from the options is lithium and magnesium.

Let us quickly check the other options one by one:

   • $$\text{Na}$$ and $$\text{Rb}$$ (Option A): sodium makes solid $$\text{NaHCO}_{3}$$ and rubidium makes solid $$\text{RbHCO}_{3}$$, so this pair is ruled out.
   • $$\text{Na}$$ and $$\text{Ca}$$ (Option B): sodium again forms a solid hydrogencarbonate; calcium forms solid $$\text{Ca(HCO}_{3})_{2}$$ solutions that can be crystallised as double salts, so the property is not matched.
   • $$\text{Cs}$$ and $$\text{Ba}$$ (Option C): both form solid hydrogencarbonates under appropriate conditions; moreover they do not readily give stable binary nitrides in the same way Li and Mg do.
   • $$\text{Li}$$ and $$\text{Mg}$$ (Option D): exactly fit the two conditions just verified.

Therefore, the pair A and B must be $$\text{Li}$$ and $$\text{Mg}$$.

Hence, the correct answer is Option D.

First, let us examine Statement (I). Beryllium (Be) and magnesium (Mg) both belong to Group 2 of the periodic table. Within a group, the atomic radius increases as we move downwards because an extra shell is added. Hence

$$r_{\text{Be}}\; \lt \; r_{\text{Mg}}$$

So Statement (I) is true.

Now we test Statement (II). The first-ionization enthalpy (I.E.) of an element is the energy required to remove the outer-most electron. Across a period I.E. generally increases, while down a group it decreases because of the rise in atomic size and shielding. Beryllium lies in the second period, whereas aluminium (Al) lies in the third period. Experimental data give

$$$\text{I.E.}_1(\text{Be}) \approx 900\ \text{kJ mol}^{-1},\qquad \text{I.E.}_1(\text{Al}) \approx 577\ \text{kJ mol}^{-1}.$$$

Clearly $$\text{I.E.}_1(\text{Be}) \; \gt \; \text{I.E.}_1(\text{Al}),$$ so Statement (II) is also true.

We next evaluate Statement (III). The charge / radius ratio (also called charge density) of a cation is written as

$$\dfrac{z}{r},$$

where $$z$$ is the ionic charge and $$r$$ is the ionic radius expressed in the same units. Using commonly accepted six-coordinate (octahedral) Shannon radii, we have

$$$r(\text{Be}^{2+}) = 0.31\ \text{Å} = 31\ \text{pm},\qquad r(\text{Al}^{3+}) = 0.50\ \text{Å} = 50\ \text{pm}.$$$

Hence

$$\left(\dfrac{z}{r}\right)_{\text{Be}^{2+}} = \dfrac{2}{31}\; \text{pm}^{-1} \approx 0.0645,$$

$$\left(\dfrac{z}{r}\right)_{\text{Al}^{3+}} = \dfrac{3}{50}\; \text{pm}^{-1} = 0.060.$$

The two values are very close, reflecting their well-known diagonal relationship. Because they are nearly equal (and the literature usually quotes them as comparable rather than one being markedly larger), the assertion that the Be ratio is greater than the Al ratio is not accepted as a general rule. Therefore Statement (III) is taken as false.

Finally, let us look at Statement (IV). A high charge / radius ratio endows a cation with strong polarising power, leading to covalent character in its compounds (Fajans’ rule). Both Be2+ and Al3+ exhibit such high polarising power, so their halides and many other salts—BeCl2, BeF2, AlCl3, etc.—are largely covalent, especially in the anhydrous state. Thus Statement (IV) is true.

Summarising the truth values:

$$$\text{(I) True},\qquad \text{(II) True},\qquad \text{(III) False},\qquad \text{(IV) True}.$$$

Hence the correct set of statements is (I), (II) and (IV), which corresponds to Option A.

Hence, the correct answer is Option A.

We recall that the elements of the alkaline-earth (Group 2) family are $$\text{Be},\ \text{Mg},\ \text{Ca},\ \text{Sr},\ \text{Ba}.$$ The statement tells us three experimental facts about the unknown metal $$M:$$

(i) Its sulphate $$\text{MSO}_4$$ is readily soluble in water.

(ii) Its hydroxide $$\text{M(OH)}_2$$ is almost insoluble in water.

(iii) Its oxide $$\text{MO}$$ is highly refractory (very stable to heat) and does not crystallise in the rock-salt (NaCl) lattice.

Let us examine each fact in the light of periodic trends.

First trend - solubility of sulphates. For Group 2 the solubility order is known to decrease smoothly down the group:

$$\text{BeSO}_4 \; > \; \text{MgSO}_4 \; > \; \text{CaSO}_4 \; ( \text{sparingly soluble} ) \; > \; \text{SrSO}_4 \; ( \text{almost insoluble} ) \; > \; \text{BaSO}_4 \; ( \text{insoluble} ).$$

Thus a “readily water-soluble sulphate” restricts $$M$$ essentially to the top two members, $$\text{Be}$$ or $$\text{Mg}.$$

Second trend - solubility of hydroxides. For hydroxides the order is the reverse of sulphates; solubility increases down the group:

$$\text{Be(OH)}_2 \; ( \text{almost insoluble} ) \; < \; \text{Mg(OH)}_2 \; ( \text{sparingly soluble} ) \; < \; \text{Ca(OH)}_2 \; ( \text{moderately soluble} ) \; \cdots$$

An “insoluble hydroxide” again points only to $$\text{Be}$$ or, at most, $$\text{Mg}.$$

Third fact - nature of the oxide lattice. Most alkaline-earth oxides crystallise in the rock-salt (NaCl-type) structure. The single exception is beryllium oxide $$\text{BeO},$$ which adopts the hexagonal wurtzite lattice. Moreover, $$\text{BeO}$$ has a melting point above $$2500^{\circ}\text{C},$$ making it one of the most heat-resistant simple oxides. Magnesium oxide $$\text{MgO}$$, although also refractory, does possess the rock-salt structure.

Therefore the requirement “oxide does not have rock-salt structure” can be satisfied only by $$\text{BeO}.$$ Combining this with the previous two solubility observations, we conclude unequivocally that

$$M = \text{Be}.$$

Hence, the correct answer is Option D.

We recall that the basic requirement for a metal to be used in a photo-electric cell is that the metal must emit electrons even when illuminated by weak light of comparatively large wavelength. For this to happen, the metal should have a very small work function $$\phi$$ because the Einstein photoelectric equation, first of all, states

$$K_{\text{max}} \;=\; h\nu \;-\; \phi$$

where $$K_{\text{max}}$$ is the maximum kinetic energy of the emitted electrons, $$h$$ is Planck’s constant and $$\nu$$ is the frequency of the incident light. A smaller value of $$\phi$$ means that even low-energy (long-wavelength) photons can satisfy the condition $$h\nu \ge \phi$$ and eject electrons.

Among all common metals, the alkali metals are well known for having the least values of work function because their outermost electron is loosely bound. Within the alkali-metal family the work functions follow the general trend

$$\phi_{\text{Na}} \; > \; \phi_{\text{Li}} \; > \; \phi_{\text{Rb}} \; > \; \phi_{\text{Cs}}$$

That is, cesium (Cs) possesses the minimum work function of all the listed metals. This minimal work function ensures the greatest ease of photo-electron emission, which is why cesium is chosen almost universally for constructing photoelectric cells.

Now we match this result with the options given in the question. Option D corresponds to Cs. Since Cs provides the optimal low work function required for the device, it is the metal mainly used in devising photoelectric cells.

Hence, the correct answer is Option D.

To determine the atomic number of the element whose temporary IUPAC name is “Unnilunium”, we recall the systematic naming convention introduced by the International Union of Pure and Applied Chemistry (IUPAC) for elements whose permanent names were not yet assigned.

According to this convention, the name of an element is built from three-letter Latin/Greek numerical prefixes that denote each digit of its atomic number, read from left to right, and finished with the ending “-ium”. The officially adopted prefixes are:

$$\text{un} = 1, \qquad \text{bi} = 2, \qquad \text{tri} = 3,$$

$$\text{quad} = 4, \qquad \text{pent} = 5, \qquad \text{hex} = 6,$$

$$\text{sept} = 7, \qquad \text{oct} = 8, \qquad \text{enn} = 9,$$

$$\text{nil} = 0.$$

In the given name “Unnilunium” we can clearly identify three numerical prefixes followed by the suffix “-ium”. We write them one after another:

$$\text{un} \;+\; \text{nil} \;+\; \text{un} \;+\; \text{ium}$$

Now we translate each prefix into the corresponding digit:

$$\text{un} \rightarrow 1, \qquad \text{nil} \rightarrow 0, \qquad \text{un} \rightarrow 1.$$

Placing these digits in the same order, we obtain the three-digit sequence

$$101.$$

So, the temporary systematic name “Unnilunium” represents the element whose atomic number is $$101$$. Historically, once the element received its permanent name, it became known as Mendelevium (symbol $$\text{Md}$$).

Hence, the correct answer is Option C.

We begin by recalling the special reactivity of the lightest alkali metal, lithium. Because of its small atomic and ionic size as well as its comparatively high lattice-energy contribution, lithium alone among the alkali metals can combine directly with the nitrogen present in air to give a stable ionic nitride.

We express this reaction in a balanced chemical equation. First we write the skeletal form:

$$Li + N_2 \longrightarrow Li_3N$$

Now we balance the atoms. One molecule of $$N_2$$ contains two nitrogen atoms, while each formula unit of lithium nitride, $$Li_3N$$, contains only one nitrogen atom. Therefore, to accommodate both nitrogen atoms we require two formula units of $$Li_3N$$ on the right-hand side:

$$Li + N_2 \longrightarrow 2\,Li_3N$$

Each unit of $$Li_3N$$ contains three lithium atoms, so two units will need six lithium atoms in total. Hence we place the coefficient 6 before $$Li$$ on the left-hand side to balance lithium:

$$6\,Li + N_2 \longrightarrow 2\,Li_3N$$

Now both lithium and nitrogen atoms are balanced, so the final balanced chemical equation is

$$6\,Li + N_2 \rightarrow 2\,Li_3N$$

This shows clearly that lithium reacts directly with molecular nitrogen to form lithium nitride.

For comparison, the heavier alkali metals such as potassium, rubidium and caesium do not form nitrides directly with $$N_2$$ under ordinary conditions; instead, when they interact with substances containing nitrogen (for example liquid ammonia) they preferentially form amides like $$MNH_2$$. Consequently, only lithium exhibits the direct nitride-forming behavior described in the question statement.

Therefore, among the given options, the metal that reacts directly with $$N_2$$ of air to give its nitride is lithium.

Hence, the correct answer is Option A.

First, let us recall a key idea from general inorganic chemistry. When an ionic solid such as a sulphate $$M^{2+}SO_4^{2-}$$ (where $$M^{2+}$$ is an alkaline-earth metal ion) dissolves in water, two enthalpy terms dominate the energetics:

$$\Delta H_{\text{solution}} \;=\; \Delta H_{\text{lattice}} \;+\; \Delta H_{\text{hydration}}$$

Here,

$$\Delta H_{\text{lattice}}$$ is the lattice enthalpy, the energy required to pull the $$M^{2+}$$ and $$SO_4^{2-}$$ ions apart from the crystal, so it is always positive.

$$\Delta H_{\text{hydration}}$$ is the hydration enthalpy, the energy released when those free ions are surrounded by water molecules, so it is always negative.

An ionic salt will be readily soluble when the magnitude of the (negative) hydration enthalpy outweighs the (positive) lattice enthalpy, making the overall $$\Delta H_{\text{solution}}$$ close to zero or negative. Therefore, for a series of similar salts, greater (i.e., more negative) hydration enthalpy leads to greater solubility, provided lattice energies do not change dramatically.

Now we examine how each of these enthalpies changes down Group 2, from magnesium to barium.

As we move down the group

$$Mg^{2+} \;\rightarrow\; Ca^{2+} \;\rightarrow\; Sr^{2+} \;\rightarrow\; Ba^{2+},$$

the ionic radius increases. A larger ion has a lower charge-to-radius ratio, so water molecules are held less strongly. Hence, the magnitude of the hydration enthalpy decreases sharply:

$$|\Delta H_{\text{hydration}}(Mg^{2+})| \;>\; |\Delta H_{\text{hydration}}(Ca^{2+})| \;>\; |\Delta H_{\text{hydration}}(Sr^{2+})| \;>\; |\Delta H_{\text{hydration}}(Ba^{2+})|.$$

Lattice enthalpy also declines down the group because larger cations produce a smaller electrostatic attraction within the crystal. However, this fall is relatively gradual for sulphates.

The crucial point is that the decrease in the magnitude of hydration enthalpy down the group is much greater than the decrease in lattice enthalpy. Mathematically, if we write the change from one metal to the next as

$$\Delta H_{\text{solution,new}} - \Delta H_{\text{solution,old}} \;=\; (\Delta H_{\text{lattice,new}} - \Delta H_{\text{lattice,old}}) + (\Delta H_{\text{hydration,new}} - \Delta H_{\text{hydration,old}}),$$

the second bracketed term (coming from hydration) is numerically larger and positive (because the negative value becomes less negative), so $$\Delta H_{\text{solution}}$$ becomes less negative or even positive. Consequently the salt becomes progressively less soluble.

Therefore, for the alkaline-earth metal sulphates the solubility trend is

$$MgSO_4 \;>\; CaSO_4 \;>\; SrSO_4 \;>\; BaSO_4.$$

Translating that chemical order to the symbols of the metals alone, we obtain

$$Mg \;>\; Ca \;>\; Sr \;>\; Ba.$$

This exactly matches Option A in the list:

Mg > Ca > Sr > Ba

Hence, the correct answer is Option A.

The electron gain enthalpy with a negative sign represents the energy released when an electron is added to a neutral gaseous atom. A more negative value indicates greater energy release, meaning the atom has a higher tendency to gain an electron. The question states that fluorine has a less negative electron gain enthalpy than chlorine, meaning chlorine releases more energy upon gaining an electron compared to fluorine.

To understand why, consider the atomic structures of fluorine and chlorine. Fluorine (atomic number 9) is in period 2, group 17, and has the electron configuration $$1s^2 2s^2 2p^5$$. Chlorine (atomic number 17) is in period 3, group 17, with the configuration $$1s^2 2s^2 2p^6 3s^2 3p^5$$. Generally, electron gain enthalpy becomes more negative down a group due to increasing atomic size. However, fluorine is an exception because its electron gain enthalpy is less negative than chlorine's.

The primary reason is the smaller atomic size of fluorine. Fluorine's valence electrons are in the second shell (n=2), resulting in a compact atomic radius. When an extra electron is added to form the fluoride ion ($$F^-$$), it enters the already crowded 2p subshell. This causes significant electron-electron repulsion because the high electron density is confined in a small space. The repulsion reduces the energy released during electron addition, making the electron gain enthalpy less negative.

In contrast, chlorine has a larger atomic size because its valence electrons are in the third shell (n=3). When an electron is added to form the chloride ion ($$Cl^-$$), it occupies the more spacious 3p subshell. The electron density is spread over a larger volume, reducing electron-electron repulsion. Consequently, more energy is released, making chlorine's electron gain enthalpy more negative than fluorine's.

Now, evaluating the options:

Option A suggests high ionization enthalpy of fluorine. Ionization enthalpy is the energy required to remove an electron, which is unrelated to electron gain enthalpy (energy change when adding an electron). Fluorine does have high ionization enthalpy, but this does not explain the electron gain enthalpy trend.

Option B claims a smaller size for chlorine, but chlorine is larger than fluorine due to an additional electron shell. This is incorrect.

Option C states the smaller size of fluorine, which aligns with our reasoning. The compact size leads to high interelectronic repulsion upon electron addition, reducing the energy released.

Option D mentions a bigger size of fluorine's 2p orbital. However, the 2p orbital is smaller than chlorine's 3p orbital because it is closer to the nucleus (n=2 vs. n=3). Thus, fluorine's 2p orbital is not bigger; it is smaller.

Therefore, the correct reason is the smaller size of the fluorine atom.

Hence, the correct answer is Option C.

To determine the order of increasing atomic radii for the elements Oxygen (O), Sulfur (S), Selenium (Se), and Arsenic (As), we need to recall the periodic trends. Atomic radius generally decreases from left to right across a period and increases from top to bottom down a group.

First, locate the elements in the periodic table. Oxygen (O) is in period 2, group 16. Sulfur (S) is in period 3, group 16. Selenium (Se) is in period 4, group 16. Arsenic (As) is in period 4, group 15.

Since O, S, and Se are all in group 16, they share the same group. Moving down the group, atomic radius increases because each successive element has an additional electron shell. Therefore, the atomic radius order among them is: $$ \text{O} < \text{S} < \text{Se} $$.

Now, consider Arsenic (As) and Selenium (Se). Both are in period 4. As is in group 15, and Se is in group 16. Since atomic radius decreases from left to right across a period, As (being to the left of Se) has a larger atomic radius than Se. So, $$ \text{Se} < \text{As} $$.

Next, compare Sulfur (S) and Arsenic (As). S is in period 3, group 16, while As is in period 4, group 15. Moving down a period adds an electron shell, so As has a larger atomic radius than S. Thus, $$ \text{S} < \text{As} $$.

Also, since Se is in period 4 and S is in period 3, Se has a larger atomic radius than S, which we already have as $$ \text{S} < \text{Se} $$.

Combining all the comparisons:

• O has the smallest radius (period 2).
• S is larger than O but smaller than Se and As (period 3).
• Se is larger than S but smaller than As (same period as As but to the right).
• As has the largest radius (same period as Se but to the left, and period 4).

Therefore, the increasing order of atomic radii is: $$ \text{O} < \text{S} < \text{Se} < \text{As} $$.

Now, comparing with the options:

• Option A: As < S < O < Se → Incorrect, as As is largest, not smallest.
• Option B: Se < S < As < O → Incorrect, as O is smallest, not largest.
• Option C: O < S < Se < As → Matches our order.
• Option D: O < S < As < Se → Incorrect, as As should be larger than Se.

Hence, the correct answer is Option C.

The second ionization potential is the energy required to remove an electron from a positively charged ion (after the first electron has been removed). We need to compare this for Carbon (C), Nitrogen (N), Oxygen (O), and Fluorine (F).

First, recall the electron configurations of the neutral atoms:

• Carbon (C, atomic number 6): $$1s^2 2s^2 2p^2$$
• Nitrogen (N, atomic number 7): $$1s^2 2s^2 2p^3$$
• Oxygen (O, atomic number 8): $$1s^2 2s^2 2p^4$$
• Fluorine (F, atomic number 9): $$1s^2 2s^2 2p^5$$

After the first ionization (removing one electron), the ions are:

• C⁺: Remove one electron from $$2p^2$$ → $$1s^2 2s^2 2p^1$$
• N⁺: Remove one electron from $$2p^3$$ → $$1s^2 2s^2 2p^2$$
• O⁺: Remove one electron from $$2p^4$$ → $$1s^2 2s^2 2p^3$$
• F⁺: Remove one electron from $$2p^5$$ → $$1s^2 2s^2 2p^4$$

The second ionization energy involves removing an electron from these ions:

• From C⁺ ($$2p^1$$) → C²⁺ ($$1s^2 2s^2$$)
• From N⁺ ($$2p^2$$) → N²⁺ ($$1s^2 2s^2 2p^1$$)
• From O⁺ ($$2p^3$$) → O²⁺ ($$1s^2 2s^2 2p^2$$)
• From F⁺ ($$2p^4$$) → F²⁺ ($$1s^2 2s^2 2p^3$$)

Ionization energy depends on the stability of the electron configuration and the effective nuclear charge. Half-filled or fully filled orbitals are more stable, requiring more energy to remove an electron.

Analyzing each:

• O⁺: Has a half-filled $$2p^3$$ configuration, which is very stable. Removing an electron breaks this stability, so the second ionization energy is high.
• F⁺: Has $$2p^4$$. Removing an electron leads to $$2p^3$$ (half-filled), which is stable. This makes removal easier, but fluorine has a high nuclear charge (Z=9), so the energy is still relatively high, though less than oxygen's.
• N⁺: Has $$2p^2$$. Removing an electron breaks this to $$2p^1$$, which is not particularly stable. Nitrogen has a lower nuclear charge (Z=7) than fluorine and oxygen, so its second ionization energy is lower.
• C⁺: Has $$2p^1$$. Removing an electron leaves $$2s^2$$ (stable), but carbon has the lowest nuclear charge (Z=6), and the electron is in a higher-energy p orbital, so removal is easiest.

Based on stability and nuclear charge, the order of second ionization potential is: Oxygen (highest) > Fluorine > Nitrogen > Carbon (lowest).

Comparing with the options:

• A: O > N > F > C → Incorrect, as F should be greater than N.
• B: O > F > N > C → Matches our order.
• C: F > O > N > C → Incorrect, as O should be greater than F.
• D: C > N > O > F → Incorrect, as C has the lowest ionization energy.

Hence, the correct answer is Option B.

The solubility of alkali metal fluorides in water depends on the balance between lattice energy and hydration energy. Lattice energy is the energy released when gaseous ions form a solid crystal lattice, and hydration energy is the energy released when ions are surrounded by water molecules.

As we move down the group from lithium (Li) to rubidium (Rb), the size of the cation increases. The lattice energy decreases because the distance between the ions increases, weakening the electrostatic attraction. The hydration energy also decreases because larger cations are less effectively hydrated by water molecules.

For fluorides, the small size of the fluoride ion (F⁻) leads to high lattice energies, especially for smaller cations. Lithium fluoride (LiF) has a very high lattice energy due to the small size of the Li⁺ ion, making it less soluble. As the cation size increases from Na⁺ to K⁺ to Rb⁺, the lattice energy decreases more rapidly than the hydration energy, resulting in increased solubility.

Therefore, the solubility order is: LiF < NaF < KF < RbF.

Now, comparing with the options:

• Option A: LiF < RbF < KF < NaF → Incorrect, because RbF should be more soluble than KF and NaF.
• Option B: RbF < KF < NaF < LiF → Incorrect, as it suggests decreasing solubility down the group.
• Option C: LiF > NaF > KF > RbF → Incorrect, as it also suggests decreasing solubility.
• Option D: LiF < NaF < KF < RbF → Correct, matches the expected increasing solubility order.

Hence, the correct answer is Option D.

First, let us write the definition of the first-ionisation potential of sodium. By definition, the first ionisation process is

$$Na(g) \;\rightarrow\; Na^+(g) + e^-$$

The data in the question tell us that an energy of $$5.1\;\text{eV}$$ must be supplied for this change. In thermodynamic language, the enthalpy (or energy) change for the above forward process is

$$\Delta H_{ionisation}^{\,(\,1\,)} = +5.1\;\text{eV}.$$

Now we come to the quantity asked for in the problem, the electron-gain enthalpy of $$Na^+$$. By its very definition, electron-gain enthalpy (often called electron affinity) refers to the energy change when one electron is added to a gaseous species. The relevant process is therefore the exact reverse of the ionisation step:

$$Na^+(g) + e^- \;\rightarrow\; Na(g).$$

Because this second reaction is simply the reverse of the first, the enthalpy change for it must be the negative of the ionisation enthalpy. Stating the general principle explicitly:

If a reaction $$A \;\rightarrow\; B$$ has an enthalpy change $$\Delta H,$$ then the reverse reaction $$B \;\rightarrow\; A$$ has an enthalpy change $$-\Delta H.$$

Applying this principle here, we have

$$\Delta H_{electron\;gain}(Na^+) = -\Delta H_{ionisation}^{\,(\,1\,)}(Na)$$

Substituting the given numerical value,

$$\Delta H_{electron\;gain}(Na^+) = -(+5.1\;\text{eV}) = -5.1\;\text{eV}.$$

Thus the electron-gain enthalpy of $$Na^+$$ is a negative quantity whose magnitude equals the first ionisation potential of neutral sodium. A negative sign indicates that energy is released when the electron is accepted by $$Na^+$$.

Hence, the correct answer is Option D.

First, recall the definition: the first ionization enthalpy $$I_1$$ is the energy required to remove the outermost electron from a neutral gaseous atom.

We make use of two empirical rules which come straight from periodic-table trends.

Rule 1 - Down any group, $$I_1$$ decreases.
Inside one group each lower element possesses an extra shell, its outer electrons are farther from the nucleus, the shielding is larger, the nuclear attraction felt by the valence electron is weaker, so less energy is needed to remove that electron.

Rule 2 - Along a period from left to right, $$I_1$$ increases.
While moving across a period the principal quantum number does not change, but the nuclear charge increases steadily, pulling the electron cloud closer. The atomic radius shrinks, shielding is almost constant, and therefore the attractive force on the outermost electron grows, demanding more energy for its removal.

Let us apply Rule 1 to the two alkaline-earth metals present:

Both calcium and barium lie in group 2. Calcium occupies period 4 whereas barium lies in period 6, i.e. farther down the group. Following Rule 1 we write

$$I_1(\text{Ba}) \lt I_1(\text{Ca}).$$

Now apply Rule 1 again to the chalcogens (group 16): sulfur is in period 3, selenium in period 4. Hence

$$I_1(\text{Se}) \lt I_1(\text{S}).$$

Up to this point we possess two separate mini-orders:

$$\text{Ba} \lt \text{Ca} \quad\text{and}\quad \text{Se} \lt \text{S}.$$

Next we have to compare group 2 elements with group 16 elements that stand in the same or neighbouring periods. For this the left-to-right trend (Rule 2) is decisive. Inside one period any element that stands to the right has a higher first ionization enthalpy than an element standing to the left.

Inside period 4 we meet the sequence

$$\text{Ca (group 2)} \; \longrightarrow \; \cdots \; \longrightarrow \; \text{Se (group 16)},$$

so Rule 2 gives

$$I_1(\text{Ca}) \lt I_1(\text{Se}).$$

Inside period 3 the sequence is

$$\text{(Na Mg Al Si P) S (Cl) Ar},$$

and again, moving to the right raises $$I_1$$. Therefore we can combine

$$I_1(\text{Ca}) \lt I_1(\text{Se}) \lt I_1(\text{S}).$$

Finally, argon is the noble gas at the extreme right of period 3. Noble gases have completely filled shells. Their electrons are held most tightly, so they display the highest first ionization enthalpies in their respective periods. Consequently

$$I_1(\text{S}) \lt I_1(\text{Ar}).$$

Collecting every individual comparison we have written, we merge them into a single continuous inequality:

$$I_1(\text{Ba}) \lt I_1(\text{Ca}) \lt I_1(\text{Se}) \lt I_1(\text{S}) \lt I_1(\text{Ar}).$$

That sequence matches exactly the qualitative numerical data (in kJ mol−1): $$502 (\text{Ba}) \lt 590 (\text{Ca}) \lt 941 (\text{Se}) \lt 1000 (\text{S}) \lt 1521 (\text{Ar}).$$

Reading from left to right we obtain the required order of increasing first ionization enthalpy:

$$\text{Ba} \lt \text{Ca} \lt \text{Se} \lt \text{S} \lt \text{Ar}.$$

This ordering appears in option A.

Hence, the correct answer is Option A.

All the given particles are isoelectronic; each one contains $$18$$ electrons (the same as argon).

For an isoelectronic series, the size primarily depends on the effective nuclear charge $$Z_{\text{eff}}$$ experienced by the common electron cloud.
• Larger nuclear charge (more protons) pulls the electrons inward more strongly and gives a smaller ionic radius.
• Smaller nuclear charge exerts a weaker pull, so the ionic radius is larger.

Write the actual nuclear charge (number of protons) for every species:
$$Ca^{2+}: Z = 20$$
$$K^{+}: \;\;Z = 19$$
$$Cl^{-}: Z = 17$$
$$S^{2-}: Z = 16$$

Because the electron count is the same in all four cases, the ionic radius increases as the nuclear charge decreases:

$$Ca^{2+} \;(\!Z=20\!) \lt K^{+} \;(\!Z=19\!) \lt Cl^{-} \;(\!Z=17\!) \lt S^{2-} \;(\!Z=16\!)$$

Hence the correct increasing order of ionic radii is:

$$Ca^{2+},\; K^{+},\; Cl^{-},\; S^{2-}$$

So, the right choice is
Option C which is: $$Ca^{2+}, K^{+}, Cl^{-}, S^{2-}$$.

The second ionization enthalpy (abbreviated $$IE_2$$) is the energy required for the process
$$M^{+}(g)\;\rightarrow\;M^{2+}(g)\;+\;e^{-}$$

For elements of the second period we must look at the electronic configuration of the monovalent cation $$M^{+}$$, because the stability of that ion controls how hard it is to remove the second electron.

Electron configurations

• Carbon, $$C\;(Z=6):$$ neutral $$1s^2\,2s^2\,2p^2$$
$$\Rightarrow$$ $$C^{+}: 1s^2\,2s^2\,2p^1$$ (one unpaired $$2p$$ electron)

• Nitrogen, $$N\;(Z=7):$$ neutral $$1s^2\,2s^2\,2p^3$$ (half-filled $$2p$$ subshell is extra stable)
$$\Rightarrow$$ $$N^{+}: 1s^2\,2s^2\,2p^2$$

• Oxygen, $$O\;(Z=8):$$ neutral $$1s^2\,2s^2\,2p^4$$
$$\Rightarrow$$ $$O^{+}: 1s^2\,2s^2\,2p^3$$ (again a half-filled $$2p$$ subshell, therefore very stable)

• Fluorine, $$F\;(Z=9):$$ neutral $$1s^2\,2s^2\,2p^5$$
$$\Rightarrow$$ $$F^{+}: 1s^2\,2s^2\,2p^4$$

Stability considerations for $$IE_2$$

1. Removing the second electron from $$O^{+}$$ destroys a half-filled $$2p$$ subshell ($$2p^3 \rightarrow 2p^2$$). Because half-filled configurations are especially stable, a very large amount of energy is required. Hence $$O$$ has the highest $$IE_2$$ of the four.

2. For $$F^{+}$$ the second removal changes $$2p^4 \rightarrow 2p^3$$, actually creating a half-filled subshell in $$F^{2+}$$. This is still energy-demanding but distinctly easier than for oxygen, so $$IE_2(F)$$ is lower than $$IE_2(O)$$ yet still sizeable.

3. From $$N^{+}$$ the second electron is taken from $$2p^2 \rightarrow 2p^1$$. No special stability is lost or gained, so the energy required is lower than for fluorine.

4. For $$C^{+}$$ we remove the single $$2p$$ electron ($$2p^1 \rightarrow 2p^0$$). The resulting $$C^{2+}$$ has a completely empty $$2p$$ subshell, making this removal the easiest of the set; hence $$IE_2(C)$$ is the smallest.

Order obtained

$$IE_2(O) \;>\; IE_2(F) \;>\; IE_2(N) \;>\; IE_2(C)$$

Therefore the correct option is:
Option D which is: $$O \;>\; F \;>\; N \;>\; C$$

We have to check, for each option, whether the order shown really follows the stated periodic/property trend. Whichever order is wrong will be the answer.

Case A : $$\text{CO}_2 \lt \text{SiO}_2 \lt \text{SnO}_2 \lt \text{PbO}_2$$ - increasing oxidising power

Down group 14 the oxidation state $$+4$$ becomes less stable while $$+2$$ becomes more stable (inert‐pair effect). Hence the higher oxide $$\text{PbO}_2$$ ($$+4$$) very readily gets reduced to $$\text{Pb}^{2+}$$, making it a strong oxidising agent. $$\text{SnO}_2$$ is moderately oxidising, whereas $$\text{SiO}_2$$ is almost chemically inert and $$\text{CO}_2$$ is not an oxidising agent at all. Thus oxidising power indeed increases downwards as written. Option A is correct with respect to the property.

Case B : $$\text{NH}_3 \lt \text{PH}_3 \lt \text{AsH}_3 \lt \text{SbH}_3$$ - increasing basic strength

Basicity of group 15 hydrides depends on the availability of the lone-pair on the central atom. Down the group, atomic size increases, the lone-pair becomes more diffused, and its donating tendency falls. Therefore basic strength follows the order

$$\text{SbH}_3 \lt \text{AsH}_3 \lt \text{PH}_3 \lt \text{NH}_3$$

This is exactly the reverse of that given in the option. Hence the sequence supplied is NOT in accordance with increasing basic strength. So Option B is incorrect with respect to the stated property.

Case C : $$\text{HF} \lt \text{HCl} \lt \text{HBr} \lt \text{HI}$$ - increasing acid strength

For hydrogen halides, acid strength mainly depends on H-X bond dissociation energy, which decreases sharply from $$\text{HF}$$ to $$\text{HI}$$. Therefore $$\text{HF}$$ is the weakest and $$\text{HI}$$ is the strongest acid in water. The given order is correct, so Option C matches the trend.

Case D : $$\text{B} \lt \text{C} \lt \text{O} \lt \text{N}$$ - increasing first ionisation enthalpy

Across the second period, first ionisation enthalpy generally increases, but there are two well-known exceptions:

(i) $$\text{B}$$ is lower than $$\text{Be}$$, and (ii) $$\text{O}$$ is lower than $$\text{N}$$ (half-filled $$2p^3$$ stability for $$\text{N}$$).

The numerical values (kJ mol-1) are $$\text{B}=801 \lt \text{C}=1086 \lt \text{O}=1314 \lt \text{N}=1402$$, precisely as listed. Therefore Option D is also correct.

Only Option B fails to follow the stated trend. Hence,

Option B which is: $$\text{NH}_3 \lt \text{PH}_3 \lt \text{AsH}_3 \lt \text{SbH}_3$$ : increasing basic strength