JEE Periodic Table & Periodicity Questions

Ionisation enthalpy (IE) of an ion is the energy required to remove one electron from that ion and convert it into the next higher-charged ion.
Two main points decide the magnitude of an IE:

(i) Higher positive charge on the ion → stronger attraction for the remaining electrons → larger IE.
(ii) Extra stability of half-filled $$d^{5}$$ or completely filled $$d^{10}$$ subshells → larger IE, because removing an electron disturbs that stable configuration.

Let us write the ground-state electronic configurations (after removing the outer $$4s$$ electrons first) and analyse every pair.

Case A:

$$Mn^{+} : [Ar]\;3d^{5}\,4s^{1}$$
$$Cr^{+} : [Ar]\;3d^{5}$$
To ionise $$Mn^{+}$$ we remove the remaining $$4s^{1}$$ electron, reaching the very stable $$[Ar]\;3d^{5}$$ state. This is easy.
To ionise $$Cr^{+}$$ we would have to disturb the already stable $$3d^{5}$$ set. This is difficult.
Hence $$IE(Mn^{+}) \lt IE(Cr^{+})$$, the given inequality is correct.

Case B:

Successive ionisation energies always rise because the nuclear charge experienced per electron increases after each removal. Therefore $$IE(Mn^{2+})$$ (third IE of Mn) is definitely greater than $$IE(Mn^{+})$$ (second IE of Mn).
So $$IE(Mn^{+}) \lt IE(Mn^{2+})$$, the given inequality is correct.

Case C:

$$Fe^{2+} : [Ar]\;3d^{6}$$
$$Fe^{3+} : [Ar]\;3d^{5}$$ (half-filled, extra stable)
Removing one electron from $$Fe^{2+}$$ produces the very stable half-filled configuration of $$Fe^{3+}$$, so $$IE(Fe^{2+})$$ is comparatively low.
Removing one more electron from $$Fe^{3+}$$ would break that half-filled stability, so $$IE(Fe^{3+})$$ is very high.
Therefore $$IE(Fe^{2+}) \lt IE(Fe^{3+})$$, the given inequality is correct.

Case D:

$$Mn^{2+} : [Ar]\;3d^{5}$$ (half-filled, highly stable)
$$Fe^{2+} : [Ar]\;3d^{6}$$
To ionise $$Mn^{2+}$$ we would have to disturb the stable $$3d^{5}$$ arrangement, so $$IE(Mn^{2+})$$ is very large.
To ionise $$Fe^{2+}$$ we move from $$3d^{6} \rightarrow 3d^{5}$$, attaining the half-filled stability; this process needs less energy.
Hence $$IE(Mn^{2+}) \gt IE(Fe^{2+})$$.
The statement given in Option D, $$Mn^{2+} \lt Fe^{2+}$$, is opposite to the actual order and is therefore incorrect.

Among the four options, only Option D shows a wrong relationship of ionisation enthalpies.

Final answer: Option D (4).

Group-17 elements in their normal sequence are $$F, Cl, Br, I$$. For each periodic property we compare the numerical values of these four elements and see whether the trend is strictly monotonic (regular) or has an exception (irregular).

Case 1: Covalent radius

General rule along a group: covalent radius increases because each successive element has an extra electron shell.

Experimental values (pm): $$F = 71,\; Cl = 99,\; Br = 114,\; I = 133$$. The values rise smoothly with no reversal. Hence the covalent radius shows a regular trend, not an irregularity.

Case 2: Electron affinity

General rule down a group: electron affinity becomes less exothermic (numerical value decreases) as atomic size increases.

Experimental values (kJ mol$$^{-1}$$): $$F = -328,\; Cl = -349,\; Br = -325,\; I = -295$$. Notice the first step $$F \rightarrow Cl$$ is opposite to the expected trend because $$|EA_{Cl}| \gt |EA_{F}|$$. This reversal is due to very small size of $$F$$ causing greater inter-electronic repulsion in the compact $$2p$$ subshell.

Since the trend is not monotonic, electron affinity shows an irregularity in the first four halogens.

Case 3: Ionic radius

For isoelectronic $$X^-$$ ions, radius increases smoothly with increasing nuclear charge because shielding outweighs attraction. Measured radii (pm): $$F^- = 133,\; Cl^- = 181,\; Br^- = 196,\; I^- = 220$$, a regular rise. Thus ionic radius does not show any irregularity.

Case 4: First ionization energy

General rule down a group: first ionization energy decreases because outer electrons are farther from the nucleus and better shielded.

Values (kJ mol$$^{-1}$$): $$F = 1681,\; Cl = 1251,\; Br = 1140,\; I = 1008$$, a smooth decrease with no reversal. Therefore the trend is regular.

Only electron affinity (property B) displays an irregular pattern for the first four Group-17 elements.

Correct choice: Option C  -  B only.

Group-wise names used in the periodic table:

• Group 15 elements are called pnicogens (pnictogens).
• Group 16 elements are called chalcogens.
• Group 17 elements are called halogens.
• Group 18 elements are called noble gases.

Now locate each super-heavy element given in LIST-II:

Case 1: $$\mathbf{Mc}$$ (Moscovium) has atomic number $$115$$ and lies in Group $$15$$. Hence it belongs to the pnicogen family.

Case 2: $$\mathbf{Lv}$$ (Livermorium) has atomic number $$116$$ and lies in Group $$16$$. Hence it belongs to the chalcogen family.

Case 3: $$\mathbf{Ts}$$ (Tennessine) has atomic number $$117$$ and lies in Group $$17$$. Hence it belongs to the halogen family.

Case 4: $$\mathbf{Og}$$ (Oganesson) has atomic number $$118$$ and lies in Group $$18$$. Hence it belongs to the noble-gas family.

Therefore, the correct matching is:
A. Pnicogen   → Mc (IV)
B. Chalcogen → Lv (III)
C. Halogen   → Ts (I)
D. Noble gas → Og (II)

Thus the required option is:
Option B: A-IV, B-III, C-I, D-II

Let us identify the periods of each element:

- Iridium (Ir): Atomic number 77, Period 6 (6th row, 5d transition metal)

- Platinum (Pt): Atomic number 78, Period 6 (6th row, 5d transition metal)

- Osmium (Os): Atomic number 76, Period 6 (6th row, 5d transition metal)

- Palladium (Pd): Atomic number 46, Period 5 (5th row, 4d transition metal)

Iridium, Platinum, and Osmium all belong to Period 6, while Palladium belongs to Period 5.

The correct answer is Option 4: Palladium.

We need to find which electronegativity order is incorrect.

Key electronegativity values (Pauling scale):

• Mg = 1.31, Be = 1.57, B = 2.04, N = 3.04
• Al = 1.61, Si = 1.90, C = 2.55
• S = 2.58, Cl = 3.16, O = 3.44, F = 3.98

Option A: Mg < Be < B < N. 1.31 < 1.57 < 2.04 < 3.04 ✓ (Correct order)

Option B: S < Cl < O < F. 2.58 < 3.16 < 3.44 < 3.98 ✓ (Correct order)

Option C: Al < Si < C < N. 1.61 < 1.90 < 2.55 < 3.04 ✓ (Correct order)

Option D: Al < Mg < B < N. Claimed: Al (1.61) < Mg (1.31)? But 1.61 > 1.31, so Mg < Al, not Al < Mg. This order is INCORRECT. ✗

The incorrect electronegativity order is Option D: Al < Mg < B < N. The correct order should be Mg < Al < B < N. Therefore, the answer is Option D.

Atomic radius decreases from left to right across a period because nuclear charge increases while the principal quantum number $$n$$ remains the same.
Atomic radius increases down a group because a new shell is added, so the outer electrons are farther from the nucleus despite the higher nuclear charge.

We now test each option one by one.

Case A: $$r_{Br} \lt r_{K}$$
Both Br and K lie in the 4th period. K is in Group 1 and Br in Group 17. Moving from Group 1 to Group 17 in the same period, atomic radius must decrease. Hence $$r_{Br} \lt r_{K}$$ is a correct trend.

Case B: $$r_{Mg} \lt r_{Al}$$
Mg and Al are neighbours in the 3rd period (Mg: Group 2, Al: Group 13). Across a period, radius should decrease, so $$r_{Mg} \gt r_{Al}$$ in reality. The given inequality reverses this order, so it is incorrect.

Case C: $$r_{Rb} \lt r_{Cs}$$
Rb and Cs belong to Group 1. Going down the group from Rb (5th period) to Cs (6th period), radius increases, i.e. $$r_{Rb} \lt r_{Cs}$$. This statement is correct.

Case D: $$r_{Al} \lt r_{Cs}$$
Al is in the 3rd period, Cs in the 6th period. Down a group and also across many periods toward the left, Cs becomes much larger than Al, therefore $$r_{Al} \lt r_{Cs}$$ is also correct.

Only Case B shows an inequality opposite to the actual periodic trend, so Case B (Option B) is the incorrect trend.

Final answer: Option B.

The first-ionisation enthalpy is the energy required to remove the most loosely bound electron from a gaseous atom.

Periodic trend:
  • Along a period (left → right) the ionisation enthalpy generally increases because nuclear charge increases while the atomic radius decreases.
  • Down a group (top → bottom) the ionisation enthalpy generally decreases because atomic size increases and the outer electron is farther from the nucleus with greater shielding.

Therefore, the elements that lie farthest down the periodic table and belong to the extreme left (Group 1) will exhibit the lowest first-ionisation enthalpy.

Identify the elements corresponding to the given atomic numbers:
  • $$32$$ → Ge (Group 14, Period 4)
  • $$35$$ → Br (Group 17, Period 4)
  • $$87$$ → Fr (Group 1, Period 7)
  • $$19$$ → K (Group 1, Period 4)

Among these, both K and Fr are Group 1 elements, so they already have a lower ionisation enthalpy than Ge and Br.
Between K and Fr, Francium (Fr) lies lower in the same group (Period 7 vs. Period 4), so its atomic size is larger and the nuclear attraction on the outermost electron is weakest.

Hence the element with the lowest first-ionisation enthalpy is Francium, atomic number $$87$$.

Option C (atomic number 87) is correct.

The periodic properties involved in the five statements are electron affinity, ionisation enthalpy, electronegativity and electropositivity. We examine each one separately.

Case A: “The process of the addition of an electron to a neutral gaseous atom is always exothermic.”
Adding an electron to some atoms (for example $$Be,\,Mg,\,N,\,P$$ and the noble gases) is endothermic because the incoming electron must be placed either in a higher-energy subshell or into a completely new shell that is already shielded. Hence electron-gain enthalpy is not always negative.
Statement A is false.

Case B: “The process of removing an electron from an isolated gaseous atom is always endothermic.”
Ionisation energy (first ionisation enthalpy) is the energy required to pull an electron out of the attraction of the nucleus in the gaseous state. Work must always be done on the system, so the process is always endothermic (positive enthalpy change).
Statement B is true.

Case C: “The 1st ionization energy of boron is less than that of beryllium.”
Electronic configurations:
$$Be: 1s^2\,2s^2$$  ($$2s$$ subshell is fully filled and comparatively stable)
$$B : 1s^2\,2s^2\,2p^1$$  (the outermost electron is in a higher-energy $$2p$$ subshell)
It is easier to remove the lone $$2p$$ electron of boron than a $$2s$$ electron of beryllium, so $$IE_1(B) \lt IE_1(Be)$$.
Statement C is true.

Case D: “The electronegativity of C is 2.5 in $$CH_4$$ and $$CCl_4$$.”
Pauling electronegativity of carbon is quoted as 2.5 (more precisely 2.55). However, electronegativity is an empirical property that can vary slightly with the molecular environment, oxidation state and hybridisation. In $$CCl_4$$, carbon is strongly electron-deficient due to four $$Cl$$ atoms pulling electron density away, so its effective electronegativity is not the same numerical value that is assigned to carbon in non-polar $$CH_4$$. Thus the given fixed value “2.5” for both molecules is not strictly correct.
Statement D is false.

Case E: “Li is the most electropositive among elements of group I.”
Electropositivity (tendency to lose an electron) increases down the group because atomic size grows and ionisation energy decreases. Therefore $$Cs$$ (and then $$Rb,\,K,\,Na$$) are more electropositive than $$Li$$.
Statement E is false.

Only statements B and C are correct. Hence the correct choice is:

Option A: B and C only

Find the most stable oxidation states of the elements with highest and lowest first ionisation enthalpies among In, Tl, Al, Pb, Sn, Ge.

The elements belong to Groups 13 (Al, In, Tl) and 14 (Ge, Sn, Pb). General trend: IE decreases down a group but with irregularities due to poor shielding by d and f electrons. Among these, Ge has the highest IE (smaller size, Group 14, Period 4) and In or Tl has the lowest (larger size, Group 13). Specifically, Tl has a relatively high IE due to the inert pair effect and poor shielding. In actually has the lowest first IE among these elements.

Ge (highest IE) has electronic config [Ar]3d¹⁰4s²4p². Its most stable oxidation state is +4 (using all 4 valence electrons). The inert pair effect is weak for lighter elements.

In (lowest IE) has config [Kr]4d¹⁰5s²5p¹. Its most stable oxidation state is +3 (using all 3 valence electrons). Unlike Tl, In does not show a strong inert pair effect.

The most stable oxidation states are +4 (highest IE element) and +3 (lowest IE element). The correct answer is Option C: +4 and +3.

The electronic configurations correspond to atomic numbers as follows:

$$A: 1s^2\,2s^2\,2p^3 \;\;\Rightarrow\;\; Z=7 \;(\text{Nitrogen, }N)$$
$$B: 1s^2\,2s^2\,2p^4 \;\;\Rightarrow\;\; Z=8 \;(\text{Oxygen, }O)$$
$$C: 1s^2\,2s^2\,2p^5 \;\;\Rightarrow\;\; Z=9 \;(\text{Fluorine, }F)$$
$$D: 1s^2\,2s^2\,2p^2 \;\;\Rightarrow\;\; Z=6 \;(\text{Carbon, }C)$$

All four elements lie in Period 2 of the periodic table in the order
$$C\;(Z=6) \;\lt\; N\;(Z=7) \;\lt\; O\;(Z=8) \;\lt\; F\;(Z=9)$$ from left to right.

Pauling’s electronegativity increases left → right across a period because nuclear charge increases while shielding remains nearly constant. Hence

$$\chi_{C} \;\lt\; \chi_{N} \;\lt\; \chi_{O} \;\lt\; \chi_{F}$$

Translating this trend back to the given symbols:

$$D\;(C) \;\lt\; A\;(N) \;\lt\; B\;(O) \;\lt\; C\;(F)$$

Correct order of increasing electronegativity: $$D \lt A \lt B \lt C$$

Therefore, the correct option is Option D.

Group 14 elements with the given atomic numbers: C(6), Si(14), Sn(50), Pb(82). Which has the lowest melting point?

Melting points of Group 14 elements:

C (diamond): ~3550°C, Si: 1414°C, Ge: 938°C, Sn: 232°C, Pb: 327°C

Among the given options, Sn (Z=50) has the lowest melting point at 232°C.

The correct answer is Option 4: 50 (Tin/Sn).

In periodic-property matching questions, first recall how each property varies and then find the description that matches this trend.

Case A: Second ionisation enthalpy (IE2)
• For an alkali-metal atom the first electron is removed from the $$ns^{1}$$ subshell, but the second electron must be removed from the stable noble-gas core $$ns^{0}$$ configuration, making IE2 extraordinarily high.
• Hence the correct description is “exceptionally high when removal produces a noble-gas configuration”.
⇒ (A) → (IV)

Case B: Electron-gain enthalpy (electron affinity)
• Within a group, added electrons enter the same type of subshell but the atomic size increases, lessening the effective nuclear attraction.
• Therefore the energy released (numerically the negativity of the value) becomes smaller in magnitude; i.e. it becomes “less negative” down the group.
⇒ (B) → (I)

Case C: Atomic radius
• Across a period the nuclear charge increases while the electrons are added to the same principal shell, pulling the electronic cloud closer to the nucleus.
• Consequently the atomic radius decreases from left to right in a period.
⇒ (C) → (II)

Case D: Electropositive (metallic) character
• Down any main group the outer electrons experience more shielding and are held less tightly, so the tendency to lose electrons (electropositivity) increases.
⇒ (D) → (III)

The complete matching is:
(A) - (IV), (B) - ( I), (C) - (II), (D) - (III)

Therefore the correct option is:
Option A which is: (A)-(IV), (B)-(I), (C)-(II), (D)-(III)

We need to identify which statements are NOT true about the periodic table.

A. The properties of elements are function of atomic weights. This was Mendeleev's original idea but is incorrect. Properties are a function of atomic numbers (modern periodic law). Statement A is NOT true.

B. The properties of elements are function of atomic numbers. This is the modern periodic law. TRUE statement.

C. Elements having similar outer electronic configurations are arranged in same period. Elements with similar outer electronic configurations are in the same GROUP (not period). Elements in the same period have the same principal quantum number. Statement C is NOT true.

D. An element's location reflects the quantum numbers of the last filled orbital. This is true. The period number corresponds to the principal quantum number, and the block/group corresponds to the azimuthal and magnetic quantum numbers. TRUE statement.

E. The number of elements in a period is same as the number of atomic orbitals available in energy level that is being filled. The number of elements in a period equals the number of available orbitals times 2 (since each orbital holds 2 electrons), not just the number of orbitals. For example, Period 2 has 8 elements (4 orbitals × 2). Statement E is NOT true.

Statements that are NOT true: A, C, and E.

The correct answer is Option A: A, C and E Only.

Statement I: An element in the extreme left of the periodic table forms acidic oxides.

Statement II: Acid is formed during the reaction between water and oxide of a reactive element present in the extreme right of the periodic table.

Evaluation of Statement I:

Elements on the extreme left of the periodic table are alkali metals (Group 1) and alkaline earth metals (Group 2). These are highly electropositive metals that form basic oxides (e.g., $$Na_2O + H_2O \rightarrow 2NaOH$$), not acidic oxides. Statement I is false.

Evaluation of Statement II:

Elements on the extreme right of the periodic table (excluding noble gases) are halogens and other non-metals. Non-metals form acidic oxides. For example:

$$SO_3 + H_2O \rightarrow H_2SO_4$$ (sulphuric acid)

$$Cl_2O_7 + H_2O \rightarrow 2HClO_4$$ (perchloric acid)

So acid is indeed formed when water reacts with oxides of reactive non-metals on the right side. Statement II is true.

Statement I is false but Statement II is true, which corresponds to Option 4.

In each period, the electronic configuration of a group 13 element ends with $$ns^{2}np^{1}$$ whereas that of the adjacent group 14 element ends with $$ns^{2}np^{2}$$.

Because the group 14 atom has one more proton and one more valence-shell electron, its effective nuclear charge ($$Z_{\text{eff}}$$) is higher. A higher $$Z_{\text{eff}}$$ holds the outermost electrons more strongly, so more energy is required to remove the first electron.

Observed first-ionisation enthalpies (in $$\text{kJ mol}^{-1}$$) confirm this:
B 800 < C 1086  Al 577 < Si 786  Ga 579 < Ge 762  In 558 < Sn 709  Tl 589 < Pb 716

Thus, the first ionisation enthalpy of a group 14 element is indeed higher than that of the corresponding group 13 element. Statement I is correct.

Melting and boiling points depend mainly on the type and strength of bonding present. From group 13 to group 14 the bonding becomes more covalent and the lattice (or giant covalent network) becomes much stronger.

Typical melting points (in °C) and boiling points (in °C):
B 2300 / 2550  C 3550 / 4027  Al 660 / 2467  Si 1414 / 3265  Ga 30 / 2403  Ge 938 / 2830  In 156 / 2080  Sn 232 / 2600  Tl 304 / 1473  Pb 327 / 1749

For every pair, the group 14 element has the higher melting and boiling points. Therefore Statement II is incorrect.

Conclusion: Statement I is correct while Statement II is incorrect. Hence the correct option is Option A.

The trend of atomic (metallic) radius in a group is governed by two opposing factors.
  • Addition of a new shell on moving down the group tends to increase the radius.
  • Simultaneous entry of electrons into an inner d-sub-shell offers poor shielding; the increased effective nuclear charge $$\left(Z_{\text{eff}}\right)$$ tends to contract the radius.
For elements just after a transition series the second factor may dominate.

Group 13 illustrates this point. The electronic configurations are
  $$\text{Al : }[Ne]\,3s^{2}3p^{1}$$
  $$\text{Ga : }[Ar]\,3d^{10}4s^{2}4p^{1}$$
Because the ten $$3d$$ electrons of Ga shield the nuclear charge very poorly, $$Z_{\text{eff}}$$ rises appreciably from Al to Ga. The extra 4-shell is therefore pulled in more strongly than expected and the metallic (atomic) radius of Ga becomes slightly smaller than that of Al.
Typical metallic radii: $$r_{\text{Al}} = 143\,\text{pm}, \; r_{\text{Ga}} = 135\,\text{pm}$$.

Case 1 — Statement (I)
Statement (I) says “the metallic radius of Al is less than that of Ga”.
Actual data show $$r_{\text{Al}} \gt r_{\text{Ga}}$$, so Statement (I) is incorrect.

The ionic radii of the tripositive ions also increase down the group, but here the contraction produced by the 3d electrons is not strong enough to reverse the general increase. Hence
  $$r\!\left(\text{Al}^{3+}\right) \approx 53\,\text{pm},\qquad r\!\left(\text{Ga}^{3+}\right) \approx 62\,\text{pm},\qquad r\!\left(\text{Al}^{3+}\right) \lt r\!\left(\text{Ga}^{3+}\right).$$

Case 2 — Statement (II)
Statement (II) says “the ionic radius of Al$$^{3+}$$ is less than that of Ga$$^{3+}$$”. As shown above this is correct.

Combining the two cases:
Statement (I) is incorrect while Statement (II) is correct. Therefore the appropriate choice is Option B.

Statement (I) deals with the Law of Octaves proposed by Newlands. The Law of Octaves states that when elements are arranged in order of increasing atomic weight, every eighth element exhibits properties similar to the first. Therefore Newlands arranged elements by increasing atomic weight, not atomic number. Hence Statement (I) is false.

Statement (II) refers to the work of Lothar Meyer. Meyer plotted the atomic volume of elements against their atomic number and observed a periodically repeating pattern in these physical properties. This demonstrated periodicity when properties are plotted versus atomic number. Hence Statement (II) is true.

Since Statement (I) is false and Statement (II) is true, the correct choice is Option C.

We are given five successive ionization energies of an element: 800, 2427, 3658, 25024, and 32824 kJ/mol. We need to predict the group of this element.

The key to determining the group is finding where a dramatic increase occurs between consecutive ionization energies:

- IE$$_1$$ = 800 kJ/mol

- IE$$_2$$ = 2427 kJ/mol (ratio to IE$$_1$$ = 3.0)

- IE$$_3$$ = 3658 kJ/mol (ratio to IE$$_2$$ = 1.5)

- IE$$_4$$ = 25024 kJ/mol (ratio to IE$$_3$$ = 6.8) -- Huge jump!

- IE$$_5$$ = 32824 kJ/mol (ratio to IE$$_4$$ = 1.3)

There is a massive jump between IE$$_3$$ (3658) and IE$$_4$$ (25024) -- nearly a 7-fold increase. This indicates that the first three electrons are relatively easy to remove (valence electrons), while the fourth electron comes from an inner, more stable shell.

An element with 3 valence electrons belongs to Group 13 (the Boron group). The electronic configuration would be of the form $$[\text{noble gas}] \, ns^2 \, np^1$$, giving 3 valence electrons before reaching the noble gas core.

The correct answer is Option 1: Group 13.

The atomic radius decreases across a period (from left to right) and increases down a group. We need to evaluate each option to find the incorrect decreasing order.

Option A: Si > P > Cl > F
Si, P, and Cl are in period 3. Atomic radius decreases across the period: Si (143 pm) > P (110 pm) > Cl (99 pm). F is in period 2 and has a smaller radius (72 pm) than Cl. Thus, Cl > F holds. The order Si > P > Cl > F is correct.

Option B: Be > Mg > Al > Si
Be (period 2, group 2) has an atomic radius of 112 pm. Mg (period 3, group 2) has a larger radius (160 pm) due to increasing principal quantum number down the group. Thus, Mg > Be, so Be > Mg is false. For Mg > Al > Si: Mg (160 pm) > Al (143 pm) > Si (118 pm) is correct across period 3, but the initial Be > Mg is incorrect. Hence, the entire order Be > Mg > Al > Si is incorrect.

Option C: Al > B > N > F
Al (period 3, group 13) has a radius of 143 pm. B (period 2, group 13) has a radius of 85 pm. Since atomic radius increases down the group, Al > B holds. In period 2, radius decreases: B (85 pm) > N (75 pm) > F (72 pm). Thus, Al > B > N > F is correct.

Option D: Mg > Al > C > O
Mg (160 pm) > Al (143 pm) in period 3 is correct. Al (143 pm) > C (77 pm, period 2) holds as period 3 elements have larger radii than period 2. In period 2, C (77 pm) > O (73 pm) is correct. Thus, Mg > Al > C > O is correct.

Option B is the incorrect decreasing order.

Final Answer: B

Statement I: Along the period, chemical reactivity does NOT gradually increase from group 1 to group 18. Reactivity first decreases (metals less reactive), then increases for non-metals, and noble gases (group 18) are least reactive. So Statement I is FALSE.

Statement II: Group 1 elements form basic oxides (like Na$$_2$$O, K$$_2$$O) and Group 17 elements form acidic oxides (like Cl$$_2$$O$$_7$$). This is TRUE.

The answer is Option (3): Statement I is false but Statement II is true.

List I contains four statements about the periodic properties of some main-group elements, while List II supplies four specific elements. We have to pair every statement with the correct element by using the well-known periodic trends.

List I (statements)
A. Element having the most negative (highest) electron-gain enthalpy
B. Element having the greatest first ionisation enthalpy among the elements listed
C. Element having the highest electronegativity
D. Element whose hydrated cation possesses the highest (most exothermic) hydration enthalpy

List II (elements)
I. $$Li$$     II. $$Cl$$     III. $$F$$     IV. $$Ne$$

Matching the items one by one

A → II ($$Cl$$)
For the halogens, electron-gain enthalpy becomes most negative from $$F$$ to $$Cl$$ because the very small size of $$F$$ leads to strong inter-electronic repulsion in its compact 2p subshell. Hence $$Cl$$, not $$F$$, has the highest (most negative) electron-gain enthalpy in the periodic table.

B → IV ($$Ne$$)
First ionisation enthalpy increases across a period and is highest for the noble gas at the end of the period. Among the four elements listed, $$Ne$$ is the only noble gas, so it possesses the greatest first ionisation enthalpy.

C → III ($$F$$)
Electronegativity rises across a period and drops down a group; the maximum value for any element is for $$F$$ (Pauling value ≈ 4.0). Therefore the element with the highest electronegativity is $$F$$.

D → I ($$Li$$)
Hydration enthalpy of an ion is inversely related to its ionic radius; smaller cations attract water molecules more strongly. Among the alkali-metal cations the smallest (and therefore most strongly hydrated) is $$Li^+$$, giving $$Li$$ the highest hydration enthalpy.

Thus the correct correspondence is:
A → II, B → IV, C → III, D → I,

which matches Option C: A-II, B-IV, C-III, D-I.

We need elements that CANNOT form compounds with valencies matching their group valencies.

Group valencies: B(3), C(4), N(5), S(6), O(6), F(7), P(5), Al(3), Si(4)

Elements that cannot exhibit their group valency:

- N (Group 15, valency 5): Cannot form NCl₅ as it lacks d-orbitals. Cannot show valency 5 in covalent compounds. ✓

- O (Group 16, valency 6): Cannot show valency 6 due to lack of d-orbitals. ✓

- F (Group 17, valency 7): Cannot show valency 7 as it's the most electronegative element and has no d-orbitals. ✓

All other elements can exhibit their group valency. So 3 elements cannot.

The correct answer is Option 2: 3.

Electron gain enthalpy is the enthalpy change associated with the addition of an electron to an isolated gaseous atom.

Noble gases possess completely filled valence shells and strongly resist the addition of an extra electron. Therefore, electron gain enthalpy for argon is positive.

Among the given elements, argon has the highest electron gain enthalpy.

Sulfur belongs to Group 16 and releases energy upon gaining an electron. Its electron gain enthalpy is negative, but less negative than that of the halogens.

Bromine belongs to Group 17 and has a greater tendency to accept an electron than sulfur. Consequently, its electron gain enthalpy is more negative.

Fluorine is also a halogen and has an electron gain enthalpy that is slightly more negative than that of bromine. Therefore, fluorine has the lowest electron gain enthalpy value among the given elements.

Hence, the electron gain enthalpy decreases in the order

$$Ar>S>Br>F$$

Using the given labels

$$A>D>B>C$$

Hence, the correct answer is

$$A>D>B>C$$

Assertion (A): "There is a considerable increase in covalent radius from $$N$$ to $$P$$. However from $$As$$ to $$Bi$$ only a small increase in covalent radius is observed."

This is TRUE. The covalent radii are approximately: N (70 pm), P (110 pm), As (121 pm), Sb (141 pm), Bi (148 pm). There is a large jump of about 40 pm from N to P (adding a new shell from period 2 to period 3). However, from As to Bi, the increases are much smaller because the addition of d-electrons (and f-electrons for Bi) causes poor shielding, which leads to a higher effective nuclear charge. This partially offsets the increase in radius due to additional shells.

Reason (R): "Covalent and ionic radii in a particular oxidation state increases down the group."

This is TRUE as a general trend. As we go down a group, new electron shells are added, so the atomic/ionic radii generally increase.

Relationship between A and R:

While R states that radii increase down the group (which is correct), it does not explain the specific observation in A -- namely, why the increase is large from N to P but small from As to Bi. The explanation for that requires the concept of poor shielding by d and f electrons, which is not mentioned in R.

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

A: IE generally increases across a period (not decreases). A is false.

R: Increasing nuclear charge outweighs shielding across the period. R is true (this is why IE increases).

A is false but R is true. Option (3).

Statement I: Both metals and non-metals exist in p-block and d-block elements.

Analysis: p-block contains both metals (e.g., Al, Ga, Sn, Pb) and non-metals (e.g., C, N, O, F, Cl). However, d-block elements are all metals — there are no non-metals in the d-block.

Statement I is false.

Statement II: Non-metals have higher ionisation enthalpy and higher electronegativity than the metals.

Analysis: This is generally true. Non-metals tend to have higher ionization enthalpies and higher electronegativities compared to metals, as they hold onto their electrons more tightly.

Statement II is true.

The answer is Option B: Statement I is false but Statement II is true.

Statement I: The metallic radius of Na is 1.86 A and the ionic radius of $$Na^+$$ is less than 1.86 A.

Statement II: Ions are always smaller in size than the corresponding elements.

Evaluation of Statement I:

When sodium loses an electron to form $$Na^+$$, it loses the entire outermost shell (3s). The resulting ion has the electronic configuration of neon, with a much smaller radius. The ionic radius of $$Na^+$$ is approximately 0.95 A, which is indeed less than 1.86 A. Statement I is correct.

Evaluation of Statement II:

This statement claims ions are always smaller than their parent atoms. This is true for cations (positive ions lose electrons, reducing electron-electron repulsion and the outermost shell). However, for anions (negative ions), the addition of electrons increases electron-electron repulsion and the ionic radius is larger than the atomic radius. For example, $$Cl^-$$ (1.81 A) is larger than Cl (0.99 A). Therefore, Statement II is false.

Statement I is correct but Statement II is false, which corresponds to Option 4.

We need to find the correct order of first ionization enthalpy for the elements:

(A) O (Z=8), (B) N (Z=7), (C) Be (Z=4), (D) F (Z=9), (E) B (Z=5)

Electronic configurations:

(E) B: $$1s^2\,2s^2\,2p^1$$

(C) Be: $$1s^2\,2s^2$$

(A) O: $$1s^2\,2s^2\,2p^4$$

(B) N: $$1s^2\,2s^2\,2p^3$$

(D) F: $$1s^2\,2s^2\,2p^5$$

Key points about ionization enthalpy trends:

1. General trend: IE increases across a period (left to right) due to increasing nuclear charge.

2. Anomaly 1: Be ($$2s^2$$, fully filled) has higher IE than B ($$2s^2\,2p^1$$) because removing a $$2p$$ electron is easier than removing a $$2s$$ electron.

3. Anomaly 2: N ($$2p^3$$, half-filled) has higher IE than O ($$2p^4$$) because the half-filled $$2p$$ subshell has extra exchange stability.

Standard IE$$_1$$ values (kJ/mol):

B: 801 < Be: 900 < O: 1314 < N: 1402 < F: 1681

In terms of labels: E < C < A < B < D

The correct answer is Option A: E < C < A < B < D.

We need to find the correct order of first ionization enthalpy for Group 13 elements.

The first ionization enthalpies (in kJ/mol, approximate) for Group 13 elements are:

B: 801, Al: 577, Ga: 579, In: 558, Tl: 589

The general trend down a group is decreasing IE due to increasing atomic size and shielding. However, anomalies exist:

Ga vs Al: Ga (579) has slightly higher IE than Al (577) due to poor shielding by the 3d electrons in Ga, leading to higher effective nuclear charge.

Tl vs Ga: Tl (589) has higher IE than Ga (579) due to the poor shielding effect of 4f electrons (lanthanide contraction) and relativistic effects on the 6s electrons.

Among the given options, the order $$Tl > Ga > Al$$ (589 > 579 > 577) is correct.

The correct answer is Option C: $$Tl > Ga > Al$$.

We need to determine which element among Si, Al, N, and C has the highest first ionization enthalpy.

The first ionization enthalpies (approximate values) are:

Al (Z = 13): 577 kJ/mol

Si (Z = 14): 786 kJ/mol

C (Z = 6): 1086 kJ/mol

N (Z = 7): 1402 kJ/mol

General trend: Ionization enthalpy increases across a period from left to right. However, there are exceptions due to electronic configuration stability.

Why Nitrogen has the highest IE among these elements:

Nitrogen has the electronic configuration $$1s^2\, 2s^2\, 2p^3$$. The $$2p$$ subshell is exactly half-filled, with one electron in each of the three $$2p$$ orbitals. This half-filled configuration provides extra stability due to:

1. Maximum exchange energy

2. Symmetrical distribution of electrons

This extra stability makes it significantly harder to remove an electron from nitrogen compared to the other elements listed.

The correct answer is Option 3: N (Nitrogen).

• Statement I: Fluorine has most negative electron gain enthalpy in its group.
  • False. Although Fluorine is the most electronegative element, Chlorine actually has the most negative electron gain enthalpy in Group 17. This is because Fluorine's small size leads to significant inter-electronic repulsions when an extra electron is added, making the process less exothermic than in Chlorine.
• Statement II: Oxygen has least negative electron gain enthalpy in its group.
  • True. In Group 16, Oxygen has the least negative (least exothermic) electron gain enthalpy. Similar to Fluorine, its exceptionally small atomic size creates high electron density, resulting in strong repulsions that resist the addition of a new electron compared to larger atoms like Sulfur.

We need to verify two statements about ionization enthalpy and electron gain enthalpy.

• Na: 496 kJ/mol
• Li: 520 kJ/mol
• Cl: 1251 kJ/mol
• F: 1681 kJ/mol

Order: Na (496) < Li (520) < Cl (1251) < F (1681) ✓

Statement I is TRUE.

Negative electron gain enthalpy means how exothermic the process of gaining an electron is. A more negative ΔH_eg means a higher tendency to gain an electron.

• Na: −53 kJ/mol
• Li: −60 kJ/mol
• F: −328 kJ/mol
• Cl: −349 kJ/mol

Order of negative electron gain enthalpy (magnitude): Na (53) < Li (60) < F (328) < Cl (349) ✓

Statement II is TRUE.

Note: Chlorine has a more negative electron gain enthalpy than fluorine because fluorine's very small size causes high electron-electron repulsion in its compact 2p orbitals.

The answer is Option D: Both Statement I and Statement II are true.

We need to match each element in List-I with its correct ground-state electronic configuration in List-II.

Matching Analysis:

• (A) Nitrogen ($$\text{N}$$): Atomic number 7. Its valence configuration is $$2s^2 2p^3$$.
  $$\implies \text{(A)} \rightarrow \text{(III)}$$
• (B) Sulfur ($$\text{S}$$): Atomic number 16. Its valence configuration is $$3s^2 3p^4$$.
  $$\implies \text{(B)} \rightarrow \text{(II)}$$
• (C) Bromine ($$\text{Br}$$): Atomic number 35. Its valence configuration is $$3d^{10} 4s^2 4p^5$$.
  $$\implies \text{(C)} \rightarrow \text{(I)}$$
• (D) Krypton ($$\text{Kr}$$): Atomic number 36. Its valence configuration is $$3d^{10} 4s^2 4p^6$$.
  $$\implies \text{(D)} \rightarrow \text{(IV)}$$

1. Octane: The parent chain has 8 carbons.
2. Trimethyl: There are three methyl branches attached to the main chain.
3. 2,5,6: This numbering provides the lowest set of numbers for the branches compared to the alternative (3,4,7). IUPAC rules prioritize the lowest number at the first point of difference.

In normal scenario with direct addition of water in acidic medium markovnikov's rule is followed and hence more stable carbocation is preferred hence, 2-propanol is the major product. In presence of peroxide in basic medium with B2H6 a concerted syn-additon takes place hence, anti-markovnikov's rule is followed leading to the formation of propanol-1.

Statement I: The 4f (lanthanides) and 5f (actinides) series are placed separately at the bottom of the Periodic table to preserve the principle of classification and to maintain the structure of the table. This is true.

Statement II: s-block elements (alkali and alkaline earth metals) are highly reactive and are not found in pure (free) form in nature. They always occur in combined forms (as compounds). This is false.

Statement I is true but Statement II is false. The answer corresponds to Option (3).

The IUPAC systematic naming convention for elements uses the following roots for digits:

0 = nil, 1 = un, 2 = bi, 3 = tri, 4 = quad, 5 = pent, 6 = hex, 7 = sept, 8 = oct, 9 = enn

The name "Unununnium" breaks down as:

$$\text{Un-un-un-ium} = 1\text{-}1\text{-}1 = \text{Element } 111$$

Element 111 is Roentgenium (Rg). To determine its group, we need to find its position in the periodic table.

Element 111 lies in the 7th period. The electron configuration follows the pattern:

$$[Rn]\, 5f^{14}\, 6d^{9}\, 7s^{2}$$

Wait — for element 111, we count from element 87 (Fr, Group 1):

87 (s-block, 2 elements) → 88, then 89-103 (f-block, 15 elements), then 104-111 (d-block).

In the d-block of period 7: element 104 is in Group 4, 105 in Group 5, ..., 111 is in Group $$4 + (111 - 104) = 4 + 7 = 11$$.

So element 111 belongs to Group 11 of the periodic table.

The answer is $$n = 11$$.

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.

We need to evaluate two statements about ionization enthalpy and electron configuration.

Statement I: The decrease in first ionization enthalpy from B to Al is much larger than that from Al to Ga.

The first ionization enthalpies (approximate values in kJ/mol) are:

B $$\approx 801$$, Al $$\approx 577$$, Ga $$\approx 579$$

Decrease from B to Al: $$801 - 577 = 224$$ kJ/mol

Decrease from Al to Ga: $$577 - 579 = -2$$ kJ/mol (actually a slight increase!)

The decrease from B to Al is indeed much larger than from Al to Ga. Statement I is TRUE.

The reason: From Al to Ga, the intervening $$3d$$ electrons provide poor shielding, so the effective nuclear charge increases, counteracting the effect of the additional shell.

Statement II: The d orbitals in Ga are completely filled.

Ga has the electron configuration: $$[\text{Ar}] \, 3d^{10} \, 4s^2 \, 4p^1$$

The $$3d$$ subshell has all 10 electrons, so it is completely filled. Statement II is TRUE.

Conclusion: Both Statement I and Statement II are correct.

The correct answer is Option B: Both the statements I and II are correct.

We need to match atomic numbers with their blocks in the periodic table.

(A) Z = 37: This is Rubidium (Rb), which is in Group 1 → s-block (IV)

(B) Z = 78: This is Platinum (Pt), which is a transition metal → d-block (II)

(C) Z = 52: This is Tellurium (Te), which is in Group 16 → p-block (I)

(D) Z = 65: This is Terbium (Tb), which is a lanthanide → f-block (III)

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

The correct answer is Option 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

$$\text{Option B}$$

Let us match the common names with their chemical formulas:

(A) Slaked lime - Slaked lime is the common name for calcium hydroxide, $$Ca(OH)_2$$ (II). It is produced by adding water to quicklime (CaO).

(B) Dead burnt plaster - Dead burnt plaster is anhydrous calcium sulphate, $$CaSO_4$$ (IV). It is obtained by heating plaster of Paris above 393 K.

(C) Caustic soda - Caustic soda is the common name for sodium hydroxide, $$NaOH$$ (I).

(D) Washing soda - Washing soda is hydrated sodium carbonate, $$Na_2CO_3 \cdot 10H_2O$$ (III).

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

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 find which element remains liquid inside pure boiling water (at 100°C).

For an element to be liquid at 100°C, its melting point must be below 100°C and its boiling point must be above 100°C.

Ga (Gallium): Melting point = 29.8°C, Boiling point = 2204°C. At 100°C, Ga is liquid. ✓

Br (Bromine): Melting point = -7.2°C, Boiling point = 58.8°C. At 100°C, Br₂ would be a gas. ✗

Li (Lithium): Melting point = 180.5°C. At 100°C, Li is still a solid. ✗

Cs (Caesium): Melting point = 28.5°C, Boiling point = 671°C. At 100°C, Cs is liquid. However, Cs reacts violently with water, so it would not "remain" as a liquid inside boiling water.

The correct answer is Ga (Gallium).

We need to identify the correct statements about Beryllium.

Statement A says Beryllium oxide is purely acidic in nature. This is false — Beryllium oxide (BeO) is actually amphoteric. It reacts with both acids and bases:

$$BeO + 2HCl \rightarrow BeCl_2 + H_2O$$

$$BeO + 2NaOH \rightarrow Na_2BeO_2 + H_2O$$

Statement B says Beryllium carbonate is kept in the atmosphere of CO$$_2$$. This is true — BeCO$$_3$$ is unstable and decomposes readily, so it is kept in an atmosphere of CO$$_2$$ to prevent decomposition.

Statement C says Beryllium sulphate is readily soluble in water. This is true — unlike other alkaline earth metal sulphates (like BaSO$$_4$$), BeSO$$_4$$ is readily soluble in water due to the high hydration energy of the small Be$$^{2+}$$ ion.

Statement D says Beryllium shows anomalous behavior. This is true — Beryllium shows anomalous behavior compared to other members of Group 2 due to its very small size, high ionization energy, and high electronegativity, and it shows a diagonal relationship with aluminium.

So, the correct statements are B, C, and D only.

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 $$\boxed{\text{Option C}}$$.

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): $$(NH_4)_2BeF_4$$.

The reducing strength of an alkali metal in aqueous solution depends on its standard reduction potential ((E^\circ)). A more negative reduction potential corresponds to a stronger reducing agent, while a less negative value corresponds to a weaker reducing agent.

The standard reduction potential is influenced by sublimation energy, ionization energy, and hydration energy. Although lithium has the highest ionization energy among the alkali metals, the exceptionally large hydration energy of the small (Li^+) ion more than compensates for it. As a result, lithium has the most negative standard reduction potential ((E^\circ = -3.04\ \text{V})) and is therefore the strongest reducing agent in aqueous solution.

Among the given options, potassium, sodium, and rubidium have less negative standard reduction potentials than lithium. Hence, lithium cannot be the weakest reducing agent.

Therefore, the answer key indicating (Li) as the weakest reducing agent is incorrect. In aqueous solution, (Li) is actually the strongest reducing agent due to its very high hydration enthalpy.

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 $$\boxed{\text{Option A}}$$.

Lime (CaO, quicklime) reacts exothermally with water:

$$CaO + H_2O \to Ca(OH)_2$$ (slaked lime)

$$Ca(OH)_2$$ has low solubility in water. Its aqueous solution is lime water.

Test for CO₂: $$Ca(OH)_2 + CO_2 \to CaCO_3 \downarrow + H_2O$$

CaCO₃ (B) is the white insoluble precipitate (milky appearance).

Further reaction with CO₂: $$CaCO_3 + CO_2 + H_2O \to Ca(HCO_3)_2$$ (soluble)

Therefore, 'A' is slaked lime [Ca(OH)₂].

The correct answer is Option 2.

We need to find the number of water molecules in washing soda and soda ash respectively.

Washing soda: Washing soda is the common name for hydrated sodium carbonate. Its chemical formula is $$Na_2CO_3 \cdot 10H_2O$$ (sodium carbonate decahydrate). The "deca" prefix indicates 10 water molecules. Washing soda is obtained by recrystallizing soda ash from aqueous solution, and the crystals that form contain 10 molecules of water of crystallization per formula unit. Therefore, washing soda has 10 water molecules.

Soda ash: Soda ash is the common name for anhydrous sodium carbonate, $$Na_2CO_3$$. The term "ash" comes from its historical production by burning plant material and extracting the residue. Being anhydrous, soda ash contains 0 (zero) water molecules. It is essentially the same compound as washing soda but without the water of crystallization.

Key distinction: Both washing soda and soda ash are forms of sodium carbonate ($$Na_2CO_3$$). The difference is that washing soda is the hydrated form ($$Na_2CO_3 \cdot 10H_2O$$) while soda ash is the anhydrous form ($$Na_2CO_3$$). When washing soda is heated, it loses its water of crystallization and becomes soda ash.

The correct answer is Option 3: 10 and 0.

This describes the Solvay process for making sodium hydrogen carbonate (NaHCO₃).

Compound A: Ca(OH)₂ (slaked lime). It reacts with NH₄Cl:

$$Ca(OH)_2 + 2NH_4Cl \to CaCl_2 + 2NH_3 + 2H_2O$$

Compound B: NH₃ (ammonia).

NH₃ reacts with H₂O and excess CO₂:

$$NH_3 + H_2O + CO_2 \to NH_4HCO_3$$

Compound C: NH₄HCO₃ (ammonium hydrogen carbonate).

NH₄HCO₃ reacts with saturated NaCl solution:

$$NH_4HCO_3 + NaCl \to NaHCO_3 + NH_4Cl$$

NaHCO₃ precipitates out (being less soluble).

The correct answer is Option 3: Ca(OH)₂, NH₃, NH₄HCO₃.

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 $$\mathbf{5}$$.

Reaction 1: $$Na_2O + H_2O \to 2NaOH$$

So X = NaOH (1 oxygen atom per molecule)

Reaction 2: $$Cl_2O_7 + H_2O \to 2HClO_4$$

So Y = HClO₄ (4 oxygen atoms per molecule)

Total oxygen atoms in X and Y = 1 + 4 = 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 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.

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.

We need to determine the products when $$H_2SO_4$$ is added to $$BaO_2$$ (barium peroxide). $$BaO_2$$ is barium peroxide, which contains the peroxide ion $$O_2^{2-}$$.

When barium peroxide reacts with dilute sulfuric acid, it undergoes a double displacement reaction:

$$BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2$$

The $$Ba^{2+}$$ ion combines with $$SO_4^{2-}$$ to form the insoluble salt $$BaSO_4$$. The peroxide ion $$O_2^{2-}$$ combines with $$2H^+$$ to form $$H_2O_2$$ (hydrogen peroxide). This is actually one of the historical methods for producing hydrogen peroxide.

The correct answer is Option D.

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.

Lithium aluminum hydride ($$LiAlH_4$$) acts as a strong reducing agent and hydride donor. When it reacts with beryllium chloride ($$BeCl_2$$), a metathesis (double displacement) reaction takes place, yielding beryllium hydride, lithium chloride, and aluminum chloride.

The balanced chemical equation for the reaction is:

$$2BeCl_2 + LiAlH_4 \longrightarrow 2BeH_2 + LiCl + AlCl_3$$

• $$BeH_2$$ is formed as a polymeric hydride.
• $$LiCl$$ and $$AlCl_3$$ are formed as byproducts.

The correct option is C: $$BeH_2 + LiCl + AlCl_3$$

We need to determine the products formed when temporary hard water is boiled.

Temporary hardness is caused by the presence of calcium hydrogen carbonate $$Ca(HCO_3)_2$$ and magnesium hydrogen carbonate $$Mg(HCO_3)_2$$.

For calcium hydrogen carbonate:

$$Ca(HCO_3)_2 \xrightarrow{\Delta} CaCO_3 \downarrow + H_2O + CO_2 \uparrow$$

Calcium hydrogen carbonate decomposes to give insoluble calcium carbonate.

For magnesium hydrogen carbonate:

$$Mg(HCO_3)_2 \xrightarrow{\Delta} Mg(OH)_2 \downarrow + 2CO_2 \uparrow$$

Magnesium hydrogen carbonate, on prolonged boiling, gives magnesium hydroxide (not magnesium carbonate), because $$MgCO_3$$ is more soluble and further decomposes.

The products are $$CaCO_3$$ and $$Mg(OH)_2$$.

The correct answer is Option A: $$CaCO_3$$ and $$Mg(OH)_2$$.

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.

We need to identify which compound is used as a chemical in certain types of fire extinguishers. Let us examine each option:

Option A: Washing soda ($$Na_2CO_3 \cdot 10H_2O$$) - Used in cleaning and water softening, not in fire extinguishers.

Option B: Baking soda ($$NaHCO_3$$) - Sodium bicarbonate is used in soda-acid type fire extinguishers. When heated or mixed with acid, it decomposes as: $$2NaHCO_3 \to Na_2CO_3 + H_2O + CO_2$$. The $$CO_2$$ gas released helps extinguish the fire by cutting off the oxygen supply.

Option C: Caustic soda ($$NaOH$$) - Used in soap making and industrial processes, not in fire extinguishers.

Option D: Soda ash ($$Na_2CO_3$$) - Anhydrous washing soda, used in glass manufacturing, not in fire extinguishers.

The correct answer is Option B: Baking soda.

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 identify which among baking soda, caustic soda, and washing soda contains the carbonate anion (CO₃²⁻). Writing the chemical formulae, baking soda is NaHCO₃ (sodium bicarbonate) which contains the bicarbonate ion HCO₃⁻, not the carbonate ion. Caustic soda is NaOH (sodium hydroxide) containing the hydroxide ion OH⁻, and washing soda is Na₂CO₃·10H₂O (sodium carbonate decahydrate) which contains the carbonate ion CO₃²⁻.

Only washing soda contains the carbonate anion (CO₃²⁻). Note: baking soda contains bicarbonate (HCO₃⁻), which is different from carbonate (CO₃²⁻).

The answer is Option A: Washing Soda only.

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.

We need to determine the product when dihydrogen ($$H_2$$) reacts with copper(II) oxide ($$CuO$$). Hydrogen gas is a reducing agent; when it reacts with metal oxides, it reduces the metal oxide, so this redox reaction involves $$H_2$$ acting as a reducing agent and $$CuO$$ getting reduced.

$$CuO(s) + H_2(g) \xrightarrow{\Delta} Cu(s) + H_2O(l)$$

Copper in $$CuO$$ has an oxidation state of +2, hydrogen reduces $$Cu^{2+}$$ to $$Cu^0$$ (metallic copper) and is itself oxidized from 0 to +1 in $$H_2O$$. This reaction is used as a standard test for the reducing property of hydrogen gas. When $$H_2$$ is passed over heated $$CuO$$ (black), the black powder turns reddish-brown as metallic copper ($$Cu$$) is formed.

Other possible products such as $$Cu(OH)_2$$ would require an aqueous base environment rather than a direct reaction of $$H_2$$ with $$CuO$$; $$Cu_2O$$ would indicate partial reduction, but $$H_2$$ completely reduces $$CuO$$ to metallic $$Cu$$; and copper hydride ($$CuH_2$$) is not formed under these conditions.

Therefore, the correct answer is Option B: $$Cu(s)$$.

We need to identify the industrial process that produces molecular hydrogen ($$H_2$$) as a by-product.

Option A: $$Na_2CO_3$$ - Sodium carbonate is manufactured by the Solvay process, which involves $$NaCl$$, $$NH_3$$, and $$CO_2$$. No $$H_2$$ is produced as a by-product.

Option B: $$NaOH$$ - Sodium hydroxide is manufactured by the Chlor-alkali process, which involves the electrolysis of concentrated brine (NaCl solution): $$2NaCl(aq) + 2H_2O(l) \xrightarrow{\text{electrolysis}} 2NaOH(aq) + Cl_2(g) + H_2(g)$$ In this process, $$H_2$$ gas is liberated at the cathode as a by-product, along with $$Cl_2$$ at the anode. The main product is $$NaOH$$.

Option C: Na metal - Sodium metal is produced by the Downs process (electrolysis of molten NaCl). No water is involved, so no $$H_2$$ is produced.

Option D: $$NaHCO_3$$ - Sodium bicarbonate is produced in the Solvay process. No $$H_2$$ is produced.

The correct answer is Option B: $$NaOH$$.

In the metallurgical extraction of copper, the reaction is: $$\text{FeO} + \text{SiO}_2 \rightarrow \text{FeSiO}_3$$.

In metallurgy, gangue refers to the earthy impurities (unwanted material) associated with the ore, flux is a substance added to remove gangue by combining with it, and slag is the fusible product formed when flux reacts with gangue.

In copper extraction, the ore is copper pyrite (CuFeS$$_2$$). During smelting, FeS is converted to FeO, which is the unwanted impurity (gangue) that needs to be removed. SiO$$_2$$ (silica) is added as flux to combine with the gangue (FeO), producing FeSiO$$_3$$ (iron silicate) as the slag, which is easily separable.

Answer: FeO is the gangue and FeSiO$$_3$$ is the slag; hence, the correct answer is Option D.

Silver is commonly extracted using the MacArthur-Forrest cyanide process. In this method, metallic silver reacts with sodium cyanide in the presence of water and oxygen to form the soluble complex sodium dicyanoargentate. This reaction can be represented as follows: $$ 4Ag + 8NaCN + 2H_2O + O_2 \rightarrow 4Na[Ag(CN)_2] + 4NaOH $$ Once the silver is in solution, zinc dust is added to precipitate pure silver metal, which demonstrates that silver extraction does involve cyanide.

Similarly, gold extraction employs the cyanide process, in which gold metal reacts with sodium cyanide, water, and oxygen to yield sodium dicyanoaurate as the soluble complex: $$ 4Au + 8NaCN + 2H_2O + O_2 \rightarrow 4Na[Au(CN)_2] + 4NaOH $$ The precipitated gold metal is then recovered by adding zinc dust, confirming that gold extraction also involves cyanide.

Although zinc is typically obtained by roasting zinc sulfide ore (ZnS) to zinc oxide followed by carbon reduction, certain hydrometallurgical methods utilize cyanide. In these processes, zinc forms the soluble complex $$[Zn(CN)_4]^{2-}$$, showing that cyanide can participate in zinc extraction under specific conditions.

In contrast, copper extraction does not rely on cyanide. Copper is obtained from ores such as copper pyrite ($$CuFeS_2$$) through a sequence of roasting, smelting with silica flux in a blast furnace, Bessemerisation to produce blister copper, and electrolytic refining to yield pure copper. Because copper(I) cyanide is insoluble, cyanide leaching is unsuitable for copper.

The correct answer is Option D: copper.

We need to match each ore with its correct composition.

Recall the compositions of each ore. Siderite is an iron ore with the formula FeCO₃, which corresponds to (I). Malachite is a copper ore with the formula CuCO₃·Cu(OH)₂ (basic copper carbonate), matching (II). Sphalerite, also called zinc blende, is a zinc ore with the formula ZnS and corresponds to (III). Calamine is a zinc ore with the formula ZnCO₃, matching (IV).

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