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.

$$Mn^{+} : [Ar]\;3d^{5}\,4s^{1}$$
$$Cr^{+} : [Ar]\;3d^{5}$$
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.

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.

$$Fe^{2+} : [Ar]\;3d^{6}$$
$$Fe^{3+} : [Ar]\;3d^{5}$$ (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.

$$Mn^{2+} : [Ar]\;3d^{5}$$ (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).

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.

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.

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.

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).

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.

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.

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.

Count the number of electrons present in the atom. This can be done by remembering the atomic number of the atom. 

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.

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.

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

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:

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.

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

Analyse the decomposition products.

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

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

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

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

Verify the formation reaction.

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

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

This is consistent because:

- BeO provides the beryllium

- HF provides the fluorine

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

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

Verify the decomposition.

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

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

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

We need to evaluate both the Assertion and the Reason.

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

The three isotopes of hydrogen are:

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

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

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

Assertion A is true.

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

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

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

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

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

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

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

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).

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.

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.

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 $$Mg(OH)_2$$.

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 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.

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.

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

The characteristic flame colours of the given cations are:

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

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

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

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

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

The correct answer is Option B: Copper.

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

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

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

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

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

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

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

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

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

Hence, the correct answer is Option 1.

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

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

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

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

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

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

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

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

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

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

“... and decrease in electron gain enthalpy”

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

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

Hence, the correct answer is Option D.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Thus the descending order of ionic radii is

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

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

Hence, the correct answer is Option A.

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

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

and

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

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

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

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

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

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

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

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

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

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

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

Expressed simply, the order is

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

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

Hence, the correct answer is Option C.

Let us examine each statement about Mendeleev.

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

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

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

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

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

We begin by recalling the usual definitions.

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

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

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

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

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

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

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

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

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

Hence, the correct answer is Option B.

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

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

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

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

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

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

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

First we compare magnesium and aluminium.

We write their electronic configurations in the ground state:

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

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

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

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

Next we compare phosphorus and sulfur. Their configurations are

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

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

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

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

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

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

Translating this inequality back into the required arrangement we obtain

Al < Mg < S < P.

This sequence exactly matches Option C in the list provided.

Hence, the correct answer is Option C.

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

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

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

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

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

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

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

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

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

Now, list the atomic numbers (nuclear charges):

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

Hence, for the isoelectronic sequence

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

the magnitude of the radii follows the opposite order:

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

We are given the actual experimental values

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

so by the above ordering we must have

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

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

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

Hence, the correct answer is Option B.

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

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

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

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

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

The correct answer is Option (4).

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

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

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

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

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

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

Now let us match this reasoning with the given options:

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

Hence, the correct answer is Option D.

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

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

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

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

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

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

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

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

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

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

Hence, the correct answer is Option B.

We match each element with its characteristic property.

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

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

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

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

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

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

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

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

For this reaction the equilibrium constant in terms of pressure is

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

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

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

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

Hence the mole fractions are

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

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

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

Substituting the expressions for the mole fractions, we get

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

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

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

and hence

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

This is

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

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

Now we briefly justify the other three statements:

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

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

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

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

Hence, the correct answer is Option B.

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

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

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

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

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

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

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

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

Collecting all the matches we have obtained:

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

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

Hence, the correct answer is Option D.

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

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

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

First, recall that slaked lime is chemically represented as $$\text{Ca(OH)}_2$$.

When carbon dioxide gas $$\text{CO}_2$$ is bubbled through slaked lime, the very first reaction that takes place is the formation of calcium carbonate. We state the reaction:

$$\text{Ca(OH)}_2 \;+\; \text{CO}_2 \;\longrightarrow\; \text{CaCO}_3 \;+\; \text{H}_2\text{O}$$

Here, the hydroxide ion $$\text{OH}^-$$ of calcium hydroxide combines with the acidic oxide $$\text{CO}_2$$ to give the neutral salt calcium carbonate $$\text{CaCO}_3$$ along with water $$\text{H}_2\text{O}$$. So the first solid product deposited in the medium is calcium carbonate.

Now we are told in the problem that an excess of carbon dioxide is passed. Once all the available $$\text{Ca(OH)}_2$$ has changed to $$\text{CaCO}_3$$, the additional $$\text{CO}_2$$ does not remain idle; it reacts further with the freshly formed calcium carbonate in the presence of water. The well-known carbonation reaction is:

$$\text{CaCO}_3 \;+\; \text{CO}_2 \;+\; \text{H}_2\text{O} \;\longrightarrow\; \text{Ca(HCO}_3)_2$$

This equation shows that calcium carbonate absorbs another molecule of carbon dioxide and one molecule of water to produce calcium hydrogen carbonate, written as $$\text{Ca(HCO}_3)_2$$. Hydrogen carbonate is soluble in water, so the white precipitate of $$\text{CaCO}_3$$ gradually disappears as it converts into the soluble bicarbonate.

Thus, in chronological order, the sequence of products obtained is

$$\boxed{\text{First: } \text{CaCO}_3 \quad\longrightarrow\quad \text{Then: } \text{Ca(HCO}_3)_2}$$

Comparing this sequence with the options provided, we find that Option C lists exactly this order: “CaCO$$_3$$, Ca(HCO$$_3$$)$$_2$$”.

Hence, the correct answer is Option C.

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

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

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

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

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

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

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

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

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

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

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

Hence, the correct answer is Option B.

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

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

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

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

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

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

In the scheme of qualitative inorganic analysis, metal cations are separated group-wise by taking advantage of differences in the solubility of their salts. The sequence of groups and the reagents employed are fixed, so each radical always appears in the same group. Before matching, let us recall the standard group distribution:

We have

Group I: insoluble chlorides with dilute $$\mathrm{HCl},$$ which contains $$\mathrm{Ag^+,\; Pb^{2+},\; Hg_2^{2+}}.$$

Next,

Group II: sulphides precipitated by passing } \mathrm{H_2S} \text{ in acidic medium. This group is split into two sub-groups:

Group IIA (Cu group): $$\mathrm{Cu^{2+},\; Cd^{2+},\; Pb^{2+},\; Bi^{3+},\; Hg^{2+}},$$

Group IIB (As group): $$\mathrm{As^{3+/5+},\; Sb^{3+/5+},\; Sn^{2+/4+}}.$$

Then, on adding $$\mathrm{NH_4Cl}$$ and excess $$\mathrm{NH_4OH},$$ the cations whose hydroxides are insoluble are brought down as follows:

Group III: $$\mathrm{Fe^{3+},\; Al^{3+},\; Cr^{3+}}.$$

If any cations still remain in solution, bubbling $$\mathrm{H_2S}$$ through the alkaline filtrate (containing $$\mathrm{NH_4OH}$$ and $$\mathrm{NH_4Cl}$$) precipitates sulphides of

Group IV: $$\mathrm{Zn^{2+},\; Mn^{2+},\; Co^{2+},\; Ni^{2+}}.$$

Using this standard information, let us place each ion from List - I into its correct qualitative group from List - II.

(a) $$\mathrm{Mn^{2+}}$$
From the above summary, $$\mathrm{Mn^{2+}}$$ is precipitated as $$\mathrm{MnS}$$ in the alkaline medium of Group IV. So, $$\mathrm{Mn^{2+}} \longrightarrow \text{Group IV} \; (iii).$$

(b) $$\mathrm{As^{3+}}$$
Arsenic belongs to the As sub-group; its sulphide $$\mathrm{As_2S_3}$$ is precipitated by $$\mathrm{H_2S}$$ in acid medium and is soluble in yellow ammonium sulphide, a characteristic of Group IIB. Thus, $$\mathrm{As^{3+}} \longrightarrow \text{Group IIB} \; (iv).$$

(c) $$\mathrm{Cu^{2+}}$$
Copper falls in the Cu sub-group; its black sulphide $$\mathrm{CuS}$$ precipitates directly with $$\mathrm{H_2S}$$ in acid medium and does not dissolve in yellow ammonium sulphide, placing it firmly in Group IIA. Hence, $$\mathrm{Cu^{2+}} \longrightarrow \text{Group IIA} \; (ii).$$

(d) $$\mathrm{Al^{3+}}$$
Aluminium hydroxide $$\mathrm{Al(OH)_3}$$ is gelatinous and precipitates on adding $$\mathrm{NH_4OH}$$ in the presence of $$\mathrm{NH_4Cl},$$ i.e. during Group III testing. Therefore, $$\mathrm{Al^{3+}} \longrightarrow \text{Group III} \; (i).$$

Collecting all the pairings, we obtain

$$(a)-(iii)$$, $$\; (b)-(iv)$$, $$\; (c)-(ii)$$, $$\; (d)-(i).$$

This sequence corresponds exactly to Option A in the given choices.

Hence, the correct answer is Option A.

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

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

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

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

We begin by recalling how ozone $$(\mathrm{O_3})$$ is naturally produced in the Earth’s stratosphere. Molecular oxygen $$(\mathrm{O_2})$$ absorbs electromagnetic radiation of a suitable energy and undergoes a process called photodissociation, breaking into two oxygen atoms:

$$\mathrm{O_2}\;+\;h\nu\;\longrightarrow\;2\,\mathrm{O}$$

Here $$h\nu$$ denotes a photon whose energy is $$E = h\nu$$, where $$h$$ is Planck’s constant and $$\nu$$ is the frequency of the radiation.

Now, each oxygen atom that has just been formed is highly reactive. It quickly combines with another molecule of oxygen to yield ozone:

$$\mathrm{O}\;+\;\mathrm{O_2}\;+\;M\;\longrightarrow\;\mathrm{O_3}\;+\;M$$

In the above step $$M$$ represents a third body (often $$\mathrm{N_2}$$ or $$\mathrm{O_2}$$ itself) which carries away the excess energy and stabilises the newly formed ozone molecule.

The crucial point is the wavelength or, equivalently, the energy of the radiation that can split $$\mathrm{O_2}$$. Only photons whose energy exceeds the bond dissociation energy of $$\mathrm{O_2}$$ can accomplish this. The bond dissociation energy of $$\mathrm{O_2}$$ is approximately $$498\;\text{kJ mol}^{-1}$$. Converting this energy to wavelength using the relation $$E = h c/\lambda$$, we get

$$\lambda \;=\;\frac{h\,c}{E}$$

Substituting $$h = 6.626\times10^{-34}\,\text{J s}$$, $$c = 3.00\times10^{8}\,\text{m s}^{-1}$$, and $$E = 498\times10^{3}\,\text{J mol}^{-1}\Big/\;N_\text{A}$$ where $$N_\text{A} = 6.022\times10^{23}\,\text{mol}^{-1}$$, the calculated wavelength comes out to be roughly $$\lambda \approx 240\;\text{nm}$$.

A wavelength of about $$240\;\text{nm}$$ lies squarely in the ultraviolet (UV) region of the electromagnetic spectrum, which spans roughly $$10\;\text{nm} \lt \lambda \lt 400\;\text{nm}$$. Hence, ultraviolet radiation has exactly the right amount of energy to break the $$\mathrm{O_2}$$ bond and initiate ozone formation.

Neither $$\gamma$$-rays nor cosmic rays are required for this routine atmospheric chemistry; they are far more energetic and occur much less frequently in the stratosphere. Visible light, on the other hand, does not possess sufficient energy to cleave the $$\mathrm{O_2}$$ bond.

So, the dominant agent that assists ozone formation in the stratosphere is ultraviolet radiation.

Hence, the correct answer is Option D.

We begin by recalling that transition metals, especially those belonging to the earlier part of the d-block, can absorb hydrogen atoms into the voids (interstices) of their metallic lattices, thereby giving what are called interstitial hydrides. In such hydrides, the small $$H$$ atoms occupy the tetrahedral or octahedral holes of the metal lattice without markedly disturbing the close-packed metal framework.

The general representation of an interstitial hydride is written as

$$M + \dfrac{x}{2}\,H_2 \;\longrightarrow\; MH_x,$$

where $$M$$ stands for the metal and $$x$$ is a non-stoichiometric, often fractional, value because the exact number of hydrogen atoms accommodated depends on temperature, pressure and the radius of the interstices.

For the formation of such hydrides two main factors are crucial:

1. The metal lattice must possess sufficiently large holes so that the radius ratio $$\dfrac{r_H}{r_M}$$ is small enough (about 0.4 or less) to let the hydrogen atoms slip in.
2. The metal should have a relatively low electronegativity so that the bonding is predominantly metallic with a partial $$M \, \delta^{+}$$ and $$H \, \delta^{-}$$ character.

Metals situated toward the left of the first transition series (Sc, Ti, V, Cr, etc.) meet these requirements better than the later members (Fe, Co, Ni, Cu, Zn) because their atomic radii are larger and their lattices provide more spacious interstitial sites.

Among the four given choices—$$\text{Cr}$$, $$\text{Fe}$$, $$\text{Mn}$$ and $$\text{Co}$$—chromium is the earliest element in the series, lying in Group 6 with a comparatively large atomic radius (128 pm) and a bcc lattice that can readily take up hydrogen. Text-book data show that chromium forms non-stoichiometric hydrides approximately of composition $$\text{CrH}_{0.8}$$ under moderate conditions, whereas manganese, iron and cobalt require much more severe pressures and still absorb less hydrogen.

So, when ease of hydride formation is compared, chromium stands out:

$$\text{Cr} + x\,H_2 \;\longrightarrow\; \text{CrH}_x \quad\text{(occurs most readily)}.$$

Hence, the correct answer is Option A.

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

First, recall two fundamental periodic facts.

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

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

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

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

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

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

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

Summarising the matches we obtained:

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

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

Hence, the correct answer is Option A.