JEE p-Block Elements (Groups 13-18) Questions

To determine the correct answer, we analyze both statements using the chemical properties and periodic trends of Group 15 elements.

The first statement asserts that the $$N-N$$ single bond is weaker and longer than the $$P-P$$ single bond. Nitrogen has a much smaller atomic radius than phosphorus, resulting in significant lone pair-lone pair repulsions between the two bonded nitrogen atoms. These repulsions weaken the $$N-N$$ single bond, making it less stable than the $$P-P$$ bond. However, because nitrogen atoms are smaller than phosphorus atoms, the $$N-N$$ bond is shorter, not longer, than the $$P-P$$ bond. Therefore, Statement I is false.

The second statement claims that compounds of Group 15 elements in the $$+3$$ oxidation state readily undergo disproportionation reactions. While certain compounds such as $$HNO_2$$ and $$H_3PO_3$$ can undergo disproportionation under suitable conditions, this behavior is not common to all Group 15 elements. As we move down the group, the inert pair effect becomes increasingly significant, stabilizing the $$+3$$ oxidation state, particularly for heavier elements such as bismuth. Consequently, $$Bi^{3+}$$ compounds are relatively stable and do not readily disproportionate. Hence, Statement II is also false.

Therefore, both statements are incorrect.

Correct Answer: Both Statement I and Statement II are false.

Nitrogen belongs to the second period, therefore its valence shell contains only the 2s and three 2p orbitals; no vacant 2d orbitals are available.

Case 1 : Verification of Statement I
• Nitrogen gives a whole series of oxides: $$N_2O$$ (+1), $$NO$$ (+2), $$N_2O_3$$ (+3), $$NO_2$$ / $$N_2O_4$$ (+4) and $$N_2O_5$$ (+5).
• In these compounds nitrogen and oxygen are of comparable, small atomic size. Their 2p orbitals can overlap side-by-side, producing strong $$p\pi - p\pi$$ multiple bonds (for example $$N=O$$, $$N\equiv O$$).
• The stability imparted by these multiple bonds allows nitrogen to exist comfortably in every oxidation state from +1 up to +5.
Hence Statement I is true.

Case 2 : Verification of Statement II
• To obtain a +5 oxidation state in a halide, nitrogen would need to accommodate ten electrons from five halogen atoms (example: $$NX_5$$).
• That requires an expanded octet, possible only when empty d orbitals are available for valence-shell expansion (as in $$PCl_5$$, $$AsF_5$$ etc.).
• Nitrogen lacks d orbitals in its valence shell (only 2s and 2p are present), so it cannot expand its octet beyond eight electrons. Therefore $$NX_5$$ type halides do not exist.
Thus Statement II is also true.

Both statements are true, so the correct choice is Option D.

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

Atomic radii (approximate, in pm):

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

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

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

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

$$B_2O_3$$ is an acidic oxide.

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

The given first-ionisation enthalpies are 708 and 715 $$\text{kJ mol}^{-1}$$ for the two Group 14 elements $$A$$ and $$B$$ respectively.

Within Group 14 the experimental values are approximately:
$$C : 1086,\; Si : 786,\; Ge : 762,\; Sn : 708,\; Pb : 715\; \text{kJ mol}^{-1}$$

Hence
$$A = Sn\;(708\; \text{kJ mol}^{-1}), \qquad B = Pb\;(715\; \text{kJ mol}^{-1})$$

The ions mentioned are $$A^{2+} = Sn^{2+}$$ and $$B^{4+} = Pb^{4+}$$.

Sn$$^{2+}$$
Sn belongs to the 5$$s^{2}$$5$$p^{2}$$ family. The +2 state loses only the two 5$$p$$ electrons, leaving the inert pair 5$$s^{2}$$ unused. Sn$$^{2+}$$ is less stable than Sn$$^{4+}$$. Therefore Sn$$^{2+}$$ readily gets oxidised to Sn$$^{4+}$$ and, in the process, reduces other species. Hence Sn$$^{2+}$$ behaves as a reducing agent.

Pb$$^{4+}$$
Down the group the inert-pair effect becomes stronger. For lead the 6$$s^{2}$$ pair is very reluctant to participate in bonding, so the +2 state (Pb$$^{2+}$$) is far more stable than the +4 state. Pb$$^{4+}$$ therefore tends to accept electrons and get reduced to Pb$$^{2+}$$, i.e. it oxidises other species. Hence Pb$$^{4+}$$ acts as an oxidising agent.

Therefore, the nature of the two ions is:
$$Sn^{2+}$$ - reducing, $$Pb^{4+}$$ - oxidising.

The correct choice is Option C: reducing and oxidising.

Group 15 elements are: N, P, As, Sb, Bi.

Apatite is a mineral with the formula $$Ca_5(PO_4)_3(F,Cl,OH)$$. The main component element from Group 15 in apatite is phosphorus (P).

The first ionization enthalpies of Group 15 elements:

N: 1402 kJ/mol, P: 1012 kJ/mol, As: 947 kJ/mol, Sb: 834 kJ/mol

Phosphorus has a first ionization enthalpy of 1012 kJ/mol.

The correct answer is Option 3: 1012 kJ mol⁻¹.

The five members of group 13 are arranged in the order of increasing atomic number as
$$B\;(Z = 5),\;Al\;(13),\;Ga\;(31),\;In\;(49),\;Tl\;(81).$$

In any group the atomic radius normally increases downwards because a new electron shell is added at every step. However, in the case of gallium the presence of ten 3d-electrons offers poor shielding to the valence electrons. The higher effective nuclear charge produced by this poor shielding makes the atomic radius of gallium slightly smaller than that of aluminium. This anomaly is called d-block contraction.

Using this idea we examine the four relations supplied in the question:

$$\text{(i)}\;B \lt Al,\qquad \text{(ii)}\;Al \lt Ga,\qquad \text{(iii)}\;Ga \lt In,\qquad \text{(iv)}\;In \lt Tl.$$

Relations (i), (iii) and (iv) are true, but relation (ii) is wrong because the actual order is
$$Ga \lt Al.$$

Thus the incorrect pair is $$Al \text{ and } Ga.$$

Next we compare the ionic (M3+) radii of these two elements. Standard ionic-radius data (coordination number 6) give
$$r_{Al^{3+}} \approx 0.50\;\text{Å},\qquad r_{Ga^{3+}} \approx 0.62\;\text{Å}.$$

Therefore gallium forms the larger $$M^{3+}$$ ion. Hence the required element X is gallium, whose atomic number is $$31.$$

Option A is correct.

Let us analyze each statement:

Option 1: $$PH_3$$ shows lower proton affinity than $$NH_3$$. This is correct because N is more electronegative and smaller, making the lone pair on $$NH_3$$ more available for proton bonding.

Option 2: $$SO_2$$ can act as an oxidizing agent, but not as a reducing agent. This is incorrect. $$SO_2$$ can act as both an oxidizing agent and a reducing agent. As a reducing agent, S in $$SO_2$$ (+4 oxidation state) can be oxidized to +6 (e.g., $$SO_3$$ or $$H_2SO_4$$). As an oxidizing agent, it can be reduced to S or $$H_2S$$.

Option 3: $$PF_3$$ exists but $$NF_5$$ does not. This is correct. Nitrogen cannot form $$NF_5$$ because it lacks d-orbitals and cannot expand its octet beyond 4 bonds.

Option 4: $$NO_2$$ can dimerise easily. This is correct. $$NO_2$$ has an unpaired electron and readily dimerises to form $$N_2O_4$$.

The incorrect statement is Option 2.

The correct answer is Option 2.

Tellurium belongs to Group 16. Its common oxide is $$TeO_2$$ (tellurium in the $$+4$$ oxidation state) and its hydride is $$TeH_2$$.

Case 1: Nature of $$TeO_2$$
• In the +4 state a chalcogen atom can either be oxidised to the +6 state or be reduced to 0/-2. • For the lighter members (S, Se) the +4 species are generally better reducing agents because they are more readily oxidised to +6. • For heavier Te, the +4 compound is more easily reduced to the elemental state; the half-reaction is $$TeO_2 + 4H^+ + 2e^- \;\rightarrow\; Te + 2H_2O \qquad E^{\circ} \approx +0.55\;V$$ A positive reduction potential means $$Te^{IV}$$ tends to accept electrons, i.e. $$TeO_2$$ behaves as an oxidising agent.

Hence $$TeO_2$$ is predominantly oxidising (and amphoteric in acid-base behaviour).

Case 2: Nature of $$TeH_2$$
• Acidity of Group 16 hydrides increases down the group because the E-H bond weakens: $$H_2O \lt H_2S \lt H_2Se \lt H_2Te$$ • Therefore $$TeH_2$$ is the most acidic hydride of the group (it can liberate $$H_2S$$ from sulfides, etc.).

Combining the two results:
  • $$TeO_2$$ - oxidising
  • $$TeH_2$$ - acidic

Option A (Oxidising and acidic) is correct.

Final answer : Option A

The first-ionisation enthalpy (I.E.) of an element depends mainly on its atomic size and on the extent of shielding of the outermost electron.

For the Group 13 elements the experimental first-ionisation enthalpies are approximately:

$$\begin{aligned} \text{B} &:& 800 \text{ kJ mol}^{-1} \\ \text{Al}&:& 577 \text{ kJ mol}^{-1} \\ \text{Ga}&:& 579 \text{ kJ mol}^{-1} \\ \text{In}&:& 558 \text{ kJ mol}^{-1} \\ \text{Tl}&:& 589 \text{ kJ mol}^{-1} \end{aligned}$$

1. Highest first-ionisation enthalpy
    • Boron lies at the top of the group, has the smallest atomic radius, and its outermost electrons experience the strongest effective nuclear charge.
    ⇒ Boron possesses the maximum I.E. in the group.

2. Lowest first-ionisation enthalpy
    • As we move down the group, atomic size increases and I.E. normally falls.
    • However, Ga and Tl contain additional $$3d/4d$$ and $$4f$$ electrons which shield the nuclear charge poorly. This d- and f-block contraction slightly raises their I.E. values above the expected trend.
    • Indium, which has significant size but far fewer inner f-electrons than Tl, shows the minimum measured value in the series.

Therefore, the element with the highest first-ionisation enthalpy is $$\mathbf{B}$$ and the element with the lowest first-ionisation enthalpy is $$\mathbf{In}$$.

Hence, the required pair is $$\mathbf{B \; \& \; In}$$, given in Option D.

We need to identify which inorganic sulphides are yellow in colour.

(A) $$(NH_4)_2S$$ (Ammonium sulphide): Ammonium sulphide is a yellow-colored compound in solution. Yellow.

(B) $$PbS$$ (Lead sulphide): Lead(II) sulphide is black in colour. It is the black precipitate formed in qualitative analysis.

(C) $$CuS$$ (Copper sulphide): Copper(II) sulphide is black in colour.

(D) $$As_2S_3$$ (Arsenic trisulphide): Arsenic trisulphide is yellow in colour. It is precipitated in Group IIB of qualitative analysis.

(E) $$As_2S_5$$ (Arsenic pentasulphide): Arsenic pentasulphide is also yellow in colour.

The yellow sulphides are (A), (D), and (E). The answer is Option A) (A), (D) and (E) only.

The five Group 13 elements in order of increasing atomic number are B, Al, Ga, In and Tl. To decide which of the four suggested sequences are correct we must compare them with the accepted experimental data for each property.

Case 1: Atomic radius
Standard covalent radii (pm) are B = 85, Al = 125, Ga = 122, In = 142, Tl = 145. Hence the correct ascending order is $$B \lt Ga \lt Al \lt In \lt Tl$$ The given order is $$B \lt Al \lt Ga \lt In \lt Tl,$$ which puts Al ahead of Ga and is therefore wrong.

Case 2: Electronegativity
Pauling electronegativities (averaged over common sources) are Al ≈ 1.5, Ga ≈ 1.6, In ≈ 1.7, Tl ≈ 1.8, B ≈ 2.0. Thus the correct ascending order is $$Al \lt Ga \lt In \lt Tl \lt B.$$ This is exactly the order supplied in the question, so the electronegativity sequence is correct.

Case 3: Density
Densities at 298 K (g cm$$^{-3}$$) are B = 2.34, Al = 2.70, Ga = 5.91, In = 7.31, Tl = 11.85. The correct ascending order is therefore $$B \lt Al \lt Ga \lt In \lt Tl.$$ The proposed order, $$Tl \lt In \lt Ga \lt Al \lt B,$$ is the exact reverse and is incorrect.

Case 4: First ionisation energy
First-ionisation energies (kJ mol$$^{-1}$$) are In = 558, Al = 577, Ga = 579, Tl = 589, B = 801. Hence the proper ascending order is $$In \lt Al \lt Ga \lt Tl \lt B,$$ which is identical to the order stated in the question, so it is correct.

Summarising: the correct sequences are Electronegativity (B) and First-ionisation energy (D) only.

Therefore the correct option is Option A (B and D only).

Large difference between melting/boiling points of O₂ and S.

Oxygen exists as O₂ (diatomic) while sulfur exists as S₈ (octatomic). The larger molecular size of S₈ leads to much stronger van der Waals forces, causing much higher melting and boiling points.

The correct answer is Option 1: Atomicity.

The elements of Group 13 show two common oxidation states: $$+3$$ (using all three valence electrons) and $$+1$$ (arising from the inert-pair effect, where the $$ns^2$$ electrons remain non-bonding).
Down the group, the inert-pair effect becomes stronger, so the stability trend is

$$\text{for }M = B,\,Al,\,Ga,\,In,\,Tl:\qquad \text{Stability of }(+3) \text{ decreases while stability of }(+1) \text{ increases}.$$

Applying this to aluminium and thallium:

• Aluminium ($$Z=13$$) lies near the top of the group, so the inert-pair effect is weak. Hence $$Al^{3+}$$ is highly stable and does not reduce (gain electrons) easily.
• Thallium ($$Z=81$$) is at the bottom, where the inert-pair effect is very strong. Therefore $$Tl^{+}$$ is far more stable than $$Tl^{3+}$$. Because $$Tl^{3+}$$ tends to gain two electrons and change to the more stable $$Tl^{+}$$ state, $$Tl^{3+}$$ behaves as a powerful oxidising agent.

Now judge each given statement:

Statement (A) $$Tl^{3+}$$ is a powerful oxidising agent.
Reason: $$Tl^{3+} + 2e^- \rightarrow Tl^{+}$$ is highly favoured. Statement (A) is correct.

Statement (B) $$Al^{3+}$$ does not get reduced easily.
Reason: $$Al^{3+}$$ is already the most stable state for aluminium. Statement (B) is correct.

Statement (C) Both $$Al^{3+}$$ and $$Tl^{3+}$$ are very stable in solution.
We have just seen that $$Tl^{3+}$$ is not very stable; it readily converts to $$Tl^{+}$$. Statement (C) is incorrect.

Statement (D) $$Tl^{+}$$ is more stable than $$Tl^{3+}$$.
Matches the inert-pair effect trend. Statement (D) is correct.

Statement (E) $$Al^{3+}$$ and $$Tl^{+}$$ are highly stable.
Both ions correspond to the preferred oxidation state of their respective elements. Statement (E) is correct.

The correct set of statements is (A), (B), (D) and (E).
Hence the answer is Option B.

We need to identify elements A, X, and Y, all from the p-block.

Element X has the first highest electronegativity among all known elements. The most electronegative element is Fluorine (F). So, $$X = F$$.

Element Y has the second highest electronegativity among all known elements. The second most electronegative element is Oxygen (O). So, $$Y = O$$.

Now, element A is from the p-block, is the rarest monoatomic non-radioactive element from its group, and has the lowest ionization enthalpy among A, X, and Y.

Since A has lower ionization enthalpy than both F and O (which have very high ionization enthalpies), A must be a heavier p-block element. The term "monoatomic" tells us A belongs to the noble gas group (Group 18), since noble gases exist as monoatomic species. Among the non-radioactive noble gases (He, Ne, Ar, Kr, Xe), the rarest one is Xenon (Xe). Xenon also has the lowest ionization enthalpy in this context, as it is a large atom with loosely held outer electrons. So, $$A = Xe$$.

The molecule is $$AX_4Y = XeF_4O$$, commonly written as $$XeOF_4$$.

Now let us determine the shape of $$XeOF_4$$.

Xenon has 8 valence electrons. It forms 4 bonds with F atoms and 1 double bond with the O atom (we can treat it as 1 bond domain). So the number of bond pairs around Xe = 5.

Total valence electrons in $$XeOF_4$$:
Xe contributes 8, each F contributes 7 (4 × 7 = 28), and O contributes 6.
Total = 8 + 28 + 6 = 42 electrons.

Around the central Xe atom, there are 5 bonding domains (4 Xe-F bonds + 1 Xe=O bond) and 1 lone pair. This gives a total of 6 electron domains around Xe.

With 6 electron domains, the electron geometry is octahedral. With 5 bonding pairs and 1 lone pair, the molecular shape is square pyramidal. The lone pair occupies one of the positions, and the oxygen and four fluorine atoms form a square pyramid around xenon.

Hence, the correct answer is Option A.

For the hydrides of group-16 elements $$H_2E$$ (where $$E = O,S,Se,Te$$) two experimental trends are important:

(i) Acidic strength: as we move down the group, the $$E-H$$ bond becomes longer and weaker. It breaks more easily, liberating $$H^+$$, so acidity
follows $$H_2O \lt H_2S \lt H_2Se \lt H_2Te$$.

(ii) Bond dissociation enthalpy: down the group the $$E-H$$ bond dissociation enthalpy decreases because atomic size increases. Thus
$$\text{BDE}(O-H) \gt \text{BDE}(S-H) \gt \text{BDE}(Se-H) \gt \text{BDE}(Te-H).$$

Now analyse the statements.

Statement I: “$$H_2Se$$ is more acidic than $$H_2Te$$.”
Actual trend shows $$H_2Te$$ is more acidic than $$H_2Se$$, so the statement is false.

Statement II: “$$H_2Se$$ has higher bond enthalpy for dissociation than $$H_2Te$$.”
From trend (ii), $$\text{BDE}(Se-H) \gt \text{BDE}(Te-H)$$, so the statement is true.

Therefore Statement I is false while Statement II is true  ⟹  Option D.

Elements of group 15 are: Nitrogen (N, $$Z = 7$$), Phosphorus (P, $$Z = 15$$), Arsenic (As, $$Z = 33$$), Antimony (Sb, $$Z = 51$$) and Bismuth (Bi, $$Z = 83$$).

Step 1 – Condition for $$d\pi-d\pi$$ bonding
To form $$d\pi-d\pi$$ bonds with transition metals, the p-block element must have empty d-orbitals that can overlap with filled d-orbitals of the metal. Nitrogen has no vacant d-orbitals (only $$2s$$ and $$2p$$ levels), so it cannot participate in $$d\pi-d\pi$$ back-bonding. Phosphorus and all heavier congeners (As, Sb, Bi) possess vacant $$3d$$/$$4d$$/$$5d$$/$$6d$$ orbitals respectively and therefore can form $$d\pi-d\pi$$ bonds.

Hence the eligible elements are P, As, Sb and Bi.

Step 2 – Basicity trend of their hydrides
The hydrides are $$PH_3$$, $$AsH_3$$, $$SbH_3$$ and $$BiH_3$$. Basic strength in this series decreases down the group because:
  • Electronegativity falls, so the lone pair becomes less concentrated.
  • Atomic size increases, causing poorer orbital overlap with a proton.
Therefore, $$\text{basicity order} : PH_3 \gt AsH_3 \gt SbH_3 \gt BiH_3$$.

The question asks for the element whose hydride is the strongest base among those elements that can form $$d\pi-d\pi$$ bonds. From the order above, $$PH_3$$ is the strongest base.

Step 3 – Identifying the element
The hydride $$PH_3$$ corresponds to Phosphorus, whose atomic number is $$15$$.

Final answer: $$15$$

Assertion A: H$$_2$$Te is more acidic than H$$_2$$S — TRUE. Acidic strength increases down the group.

Reason R: Bond dissociation enthalpy of H-Te bond is lower than H-S bond — TRUE. The larger size of Te makes the bond weaker.

R correctly explains A: because the lower bond dissociation enthalpy means easier release of H$$^+$$, making H$$_2$$Te more acidic.

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

As we move down Group 14 from carbon to lead, metallic character increases. Consequently, the nature of the oxides changes from acidic to amphoteric.

$$SiO_2$$ and $$GeO_2$$ are acidic oxides. $$GeO$$ also exhibits acidic character.

$$CO$$ is a neutral oxide and neither shows acidic nor basic behavior.

$$SnO_2$$ and $$PbO_2$$ are amphoteric oxides. They react with both acids and bases, exhibiting both acidic and basic properties.

Therefore, the pairs containing $$GeO$$ and $$GeO_2$$ or $$SiO_2$$ and $$GeO_2$$ cannot be correct because these oxides are acidic.

The pair containing $$SnO_2$$ and $$CO$$ cannot be correct because $$SnO_2$$ is amphoteric whereas $$CO$$ is neutral.

The only pair in which both oxides are amphoteric is

$$SnO_2 \text{ and } PbO_2$$

Hence, the correct answer is

$$SnO_2 \text{ and } PbO_2$$

Assertion (A): Melting point of Boron (2453 K) is unusually high in group 13 elements. This is true - Boron has the highest melting point among group 13 elements.

Reason (R): Solid Boron has very strong crystalline lattice. This is true - Boron exists in an icosahedral crystalline structure with very strong covalent bonds, giving it a very high melting point.

The reason correctly explains the assertion - the strong crystalline lattice of boron is why it has an unusually high melting point.

Both (A) and (R) are correct and (R) is the correct explanation of (A). The answer corresponds to Option (2).

We need to evaluate two statements about Group 14 elements.

Analysis of Statement I: "The electronegativity of Group 14 elements from Si to Pb gradually decreases."

This is FALSE. While there is a general decreasing trend in electronegativity going down the group, it is NOT a gradual or monotonic decrease. The electronegativity values (Pauling scale) are approximately: C (2.5), Si (1.8), Ge (2.0), Sn (1.8), Pb (1.9). Notice that germanium has a higher electronegativity than silicon, and lead has a higher electronegativity than tin. This irregularity is attributed to the poor shielding effect of d-electrons (for Ge) and f-electrons (for Pb), known as the d-block contraction and lanthanide contraction respectively.

Analysis of Statement II: "Group 14 contains non-metallic, metallic, as well as metalloid elements."

This is TRUE. Group 14 shows a clear transition from non-metal to metal:

- Carbon: Non-metal

- Silicon and Germanium: Metalloids (semi-metals with intermediate properties)

- Tin and Lead: Metals

This is a characteristic feature of p-block groups, where metallic character increases down the group.

The correct answer is Option (1): Statement I is false but Statement II is true.

We need to evaluate each statement about Group 13 elements.

Statement (A) claims that the decreasing order of atomic radii is $$Tl > In > Ga > Al > B$$, but the actual order in Group 13 is $$Tl > In > Al > Ga > B$$. Due to the d-block contraction, Ga has a smaller atomic radius than Al, so statement (A) is incorrect.

Statement (B) asserts that electronegativity decreases down Group 13, but the trend is not monotonically decreasing because of poor shielding effects. Although B has the highest electronegativity, the values do not uniformly decrease from top to bottom, hence statement (B) is incorrect.

Statement (C) states that Al dissolves in dilute HCl to liberate $$H_2$$ but concentrated $$HNO_3$$ renders Al passive by forming a protective oxide layer. Indeed, aluminum reacts with dilute hydrochloric acid according to the equation $$2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2 \uparrow$$, and concentrated nitric acid creates a thin $$Al_2O_3$$ layer that passivates the metal. Therefore statement (C) is correct.

Statement (D) claims that all elements of Group 13 exhibit a highly stable +1 oxidation state. In reality, B and Al do not show a stable +1 oxidation state and the +1 state only becomes more stable down the group due to the inert pair effect. Thus statement (D) is incorrect.

Statement (E) proposes that the hybridisation of Al in $$[Al(H_2O)_6]^{3+}$$ is $$sp^3d^2$$. In this hexaaqua complex, $$Al^{3+}$$ is octahedrally coordinated by six water molecules, which corresponds to $$sp^3d^2$$ hybridisation. Therefore statement (E) is correct.

The correct statements are (C) and (E) only. The correct answer is Option (3): (C) and (E) only.

We need to evaluate two statements about Group 13 halides and AlCl$$_3$$ hydrolysis.

Analysis of Statement I: "Group 13 trivalent halides get easily hydrolysed by water due to their covalent nature."

This is TRUE. Group 13 trihalides (like BCl$$_3$$, AlCl$$_3$$, GaCl$$_3$$, etc.) are predominantly covalent compounds. They are electron-deficient — the central atom has only 6 electrons in its valence shell (incomplete octet). This makes them strong Lewis acids. When they come in contact with water, the lone pair on the oxygen atom of water coordinates with the electron-deficient metal atom, leading to easy hydrolysis. The covalent nature facilitates this process because ionic compounds typically dissociate rather than hydrolyse.

Analysis of Statement II: "AlCl$$_3$$ upon hydrolysis in acidified aqueous solution forms octahedral $$[Al(H_2O)_6]^{3+}$$ ion."

This is TRUE. When AlCl$$_3$$ dissolves in water (especially acidified solution), the Al$$^{3+}$$ ion is surrounded by six water molecules in an octahedral arrangement, forming the hexaaquaaluminium(III) ion $$[Al(H_2O)_6]^{3+}$$. This is because Al$$^{3+}$$ is a small, highly charged ion that can coordinate with six water molecules. The octahedral geometry is consistent with $$sp^3d^2$$ hybridization of the aluminium ion.

The correct answer is Option (4): Both Statement I and Statement II are true.

We analyze two statements about Group 14 elements.

Statement I: $$SiO_2$$ and $$GeO_2$$ are acidic while $$SnO$$ and $$PbO$$ are amphoteric in nature.

$$SiO_2$$ and $$GeO_2$$ are indeed acidic oxides (they react with bases to form silicates and germanates). $$SnO$$ and $$PbO$$ are amphoteric (they react with both acids and bases). Statement I is TRUE.

Statement II: Allotropic forms of carbon are due to property of catenation and $$p\pi - d\pi$$ bond formation.

Carbon has allotropes (diamond, graphite, fullerenes) due to its excellent catenation ability and $$p\pi - p\pi$$ bonding (not $$p\pi - d\pi$$). Carbon does not have accessible d-orbitals for $$p\pi - d\pi$$ bonding. The ability to form multiple bonds via $$p\pi - p\pi$$ overlap is what gives carbon its diverse allotropes. Statement II is FALSE.

The answer is Option C) Statement I is true but Statement II is false.

Oxygen shows anomalous behaviour compared to other members of Group 16 (S, Se, Te, Po). This is primarily due to two factors:

1. Small size:

Oxygen is the first element in Group 16 and has a very small atomic radius. Due to its small size:

- It has a high ionization enthalpy

- It has strong interelectronic repulsion in its compact orbitals

- It cannot expand its octet (no d-orbitals available in the second shell)

- Its maximum covalency is limited to 4 (unlike S which can show covalency of 6)

2. High electronegativity:

Oxygen is the second most electronegative element (after fluorine). Due to its high electronegativity:

- It forms strong hydrogen bonds

- It primarily shows a $$-2$$ oxidation state (unlike S, Se which show $$+2, +4, +6$$ as well)

- It tends to form $$p\pi - p\pi$$ multiple bonds with itself and with other small atoms like C and N

These two properties together make oxygen's behaviour significantly different from the rest of its group members.

The correct answer is Option 3: Small size and high electronegativity.

Statement I: "Oxygen being the first member of group 16 exhibits only -2 oxidation state."

This is incorrect. Oxygen can also show -1 (in peroxides like $$H_2O_2$$), 0 (in $$O_2$$), and +2 (in $$OF_2$$).

Statement II: "Down the group 16 stability of +4 oxidation state decreases and +6 oxidation state increases."

This is incorrect. Actually, down the group, due to the inert pair effect, the stability of +6 oxidation state decreases and +4 oxidation state becomes more stable (e.g., Po prefers +4 over +6).

Both statements are incorrect. The answer corresponds to Option (3).

We need to evaluate the statements about hydrides of Group 15 elements ($$NH_3, PH_3, AsH_3, SbH_3, BiH_3$$).

Statement A: The stability of hydrides decreases in the order $$NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$$.

As we go down the group, the bond length between the central atom and hydrogen increases (N-H < P-H < As-H < Sb-H < Bi-H). Longer bonds are weaker, so the thermal stability decreases down the group. This statement is correct. ✓

Statement B: The reducing ability of hydrides increases in the order $$NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3$$.

Reducing ability is the tendency to donate electrons or get oxidized. As the stability decreases down the group, the hydride is more easily decomposed and more readily donates its electron pair. Therefore, reducing character increases from $$NH_3$$ to $$BiH_3$$. This statement is correct. ✓

Statement C: Among the hydrides, $$NH_3$$ is a strong reducing agent while $$BiH_3$$ is a mild reducing agent.

This is the opposite of what is true. $$NH_3$$ is the weakest reducing agent (most stable hydride), while $$BiH_3$$ is the strongest reducing agent (least stable). This statement is incorrect. ✗

Statement D: The basicity of hydrides increases in the order $$NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3$$.

Basicity depends on the availability of the lone pair. As the size of the central atom increases down the group, the electron density on the lone pair decreases (spread over a larger orbital), making it less available for donation. Therefore, basicity decreases down the group: $$NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$$. This statement claims the opposite, so it is incorrect. ✗

The correct statements are A and B only.

The correct answer is Option (3): A and B only.

Among the halogens:

- Chlorine shows oxidation states: -1, 0, +1, +3, +5, +7 (variable)

- Bromine shows oxidation states: -1, 0, +1, +3, +5, +7 (variable)

- Iodine shows oxidation states: -1, 0, +1, +3, +5, +7 (variable)

- Fluorine shows only oxidation state: -1 (and 0 in $$F_2$$). Since fluorine is the most electronegative element, it cannot show positive oxidation states.

Fluorine does not show variable oxidation state. The answer corresponds to Option (4).

The hydrides of group 15 elements are ammonia ($$NH_3$$), phosphine ($$PH_3$$), arsine ($$AsH_3$$), stibine ($$SbH_3$$), and bismuthine ($$BiH_3$$).

The reducing power of these hydrides increases down the group. This is because the stability of the hydrides decreases down the group due to the decrease in bond strength of the E-H bond (where E is the group 15 element). As the size of the central atom increases down the group, the bond length increases and the bond strength decreases, making it easier to break the E-H bond and release hydrogen gas, which acts as a reducing agent.

The order of reducing power is: $$NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3$$

Among the given options:

• A. $$NH_3$$ (ammonia) is the weakest reducing agent.
• D. $$PH_3$$ (phosphine) is stronger than $$NH_3$$ but weaker than the others.
• B. $$SbH_3$$ (stibine) is stronger than $$PH_3$$ but weaker than $$BiH_3$$.
• C. $$BiH_3$$ (bismuthine) is the strongest reducing agent because bismuth is the largest atom in the group.

Therefore, the strongest reducing agent is $$BiH_3$$, which corresponds to option C.

The correct answer is C. $$BiH_3$$.

We need to evaluate two statements about noble gases.

Analysis of Statement I: "Noble gases have very high boiling points."

This is FALSE. Noble gases have very low boiling points. For example: He boils at 4.2 K (-269°C), Ne at 27.1 K, Ar at 87.3 K, Kr at 119.9 K, and Xe at 165.1 K. These are among the lowest boiling points of all elements because noble gases exist as monoatomic species with only weak London dispersion forces between atoms.

Analysis of Statement II: "Noble gases are monoatomic gases. They are held together by strong dispersion forces. Because of this they are liquefied at very low temperature. Hence, they have very high boiling points."

This statement contains internal contradictions and is FALSE. While it correctly states that noble gases are monoatomic, the claim that they have "strong dispersion forces" is incorrect — they have very weak dispersion forces (London forces), which is why they must be cooled to very low temperatures to be liquefied. Furthermore, the conclusion "they have very high boiling points" contradicts the fact that they liquefy at very low temperatures — low liquefaction temperature means low boiling points.

Since both statements are false, the correct answer is Option (4): Both Statement I and Statement II are false.

We need to identify the incorrect pair of mineral name and chemical formula.

Check each pair.

Option (1): Fluorspar - $$BF_3$$

This is INCORRECT. Fluorspar (also called fluorite) is $$CaF_2$$ (calcium fluoride), NOT $$BF_3$$ (boron trifluoride). $$BF_3$$ is a gaseous compound, not a mineral. Fluorspar is a naturally occurring mineral and is the principal source of fluorine and hydrofluoric acid.

Option (2): Cryolite - $$Na_3AlF_6$$

This is correct. Cryolite is sodium hexafluoroaluminate, used as a flux in the electrolytic extraction of aluminium (Hall-Heroult process).

Option (3): Fluoroapatite - $$3Ca_3(PO_4)_2 \cdot CaF_2$$

This is correct. Fluoroapatite is a phosphate mineral containing fluoride, found in teeth and bones.

Option (4): Carnallite - $$KCl \cdot MgCl_2 \cdot 6H_2O$$

This is correct. Carnallite is a hydrated potassium magnesium chloride mineral, an important source of potassium.

The incorrect pair is Option (1): Fluorspar - $$BF_3$$.

When $$MnO_2$$ and $$H_2SO_4$$ are added to salt A, a greenish-yellow gas is liberated. Identify salt A.

Greenish-yellow gas is chlorine ($$Cl_2$$).

$$MnO_2$$ acts as an oxidizing agent in the presence of $$H_2SO_4$$. When a chloride salt is present:

$$MnO_2 + 2H_2SO_4 + 2Cl^- \rightarrow Mn^{2+} + 2SO_4^{2-} + 2H_2O + Cl_2 \uparrow$$

The chloride ions are oxidized to chlorine gas.

The salt must contain $$Cl^-$$ ions. Among the options: $$CaI_2$$ (iodide), $$NaBr$$ (bromide), $$KNO_3$$ (nitrate), $$NH_4Cl$$ (chloride).

Only $$NH_4Cl$$ contains chloride ions.

The correct answer is Option (4): $$NH_4Cl$$.

Reaction (A): Conversion of Phenol to Salicylic acid (2-hydroxybenzoic acid). This transformation is achieved via the Kolbe-Schmitt reaction. When phenol is treated with sodium hydroxide ($$NaOH$$) to form a phenoxide ion, followed by reaction with carbon dioxide ($$CO_2$$) under pressure and subsequent acidification with $$HCl$$, salicylic acid is produced. This matches with reagent (IV). Thus, (A) matches with (IV).

Reaction (B): Conversion of Phenol to Salicylaldehyde (2-hydroxybenzaldehyde). This is the Reimer-Tiemann reaction. Phenol reacts with chloroform ($$CHCl_3$$) in the presence of aqueous sodium hydroxide ($$NaOH$$), introducing a formyl group ($$\text{-}CHO$$) at the ortho-position, followed by hydrolysis and acidification ($$HCl$$). This matches with reagent (III). Thus, (B) matches with (III).

Reaction (C): Conversion of Phenol to p-Benzoquinone. Phenol undergoes oxidation when treated with a strong oxidizing agent like acidified sodium dichromate ($$Na_2Cr_2O_7 / H_2SO_4$$) to yield para-benzoquinone. This matches with reagent (I). Thus, (C) matches with (I).

Reaction (D): Conversion of Phenol to Anisole (Methoxybenzene). This is an etherification process (Williamson ether synthesis). Phenol is first converted into sodium phenoxide using $$NaOH$$, which then performs a nucleophilic substitution reaction with methyl chloride ($$CH_3Cl$$) to introduce a methoxy group ($$\text{-}OCH_3$$). This matches with reagent (II). Thus, (D) matches with (II).

Combining all the pairs, the correct matching sequence is: (A)-(IV), (B)-(III), (C)-(I), (D)-(II).

Answer: Option D — (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

CH₃C≡C-CH₃ with Na/liq. NH₃ gives trans-2-butene (Birch reduction gives trans alkene).

Then dil. KMnO₄ at 273 K (cold) gives a diol: trans-but-2-ene-2,3-diol (syn addition giving glycol).

The product has 2 oxygen atoms (two OH groups).

The answer is 2.

We need to identify the amphoteric oxides among: $$Cl_2O_7$$, $$CO$$, $$PbO_2$$, $$N_2O$$, $$NO$$, $$Al_2O_3$$, $$SiO_2$$, $$N_2O_5$$, $$SnO_2$$.

An amphoteric oxide reacts with both acids and bases.

$$Cl_2O_7$$: Acidic oxide (anhydride of $$HClO_4$$). Not amphoteric.

$$CO$$: Neutral oxide. Not amphoteric.

$$PbO_2$$: Amphoteric oxide. Lead(IV) oxide reacts with acids (e.g., $$PbO_2 + 4HCl \rightarrow PbCl_4 + 2H_2O$$) and with bases (e.g., forming plumbates). Yes, amphoteric.

$$N_2O$$: Neutral oxide. Not amphoteric.

$$NO$$: Neutral oxide. Not amphoteric.

$$Al_2O_3$$: Amphoteric oxide. Reacts with acids to form aluminum salts and with bases to form aluminates. Yes, amphoteric.

$$SiO_2$$: Acidic oxide. Not amphoteric.

$$N_2O_5$$: Acidic oxide (anhydride of $$HNO_3$$). Not amphoteric.

$$SnO_2$$: Amphoteric oxide. Tin(IV) oxide reacts with acids and bases. Yes, amphoteric.

The amphoteric oxides are: $$PbO_2$$, $$Al_2O_3$$, and $$SnO_2$$. That gives us 3.

The answer is 3.

NaBr gives brown/orange vapors (Br₂) with conc H₂SO₄. KI gives violet vapors (I₂). NaNO₃ gives brown NO₂ fumes. CaF₂ does not give colored vapors (HF is colorless).

Molar mass of CaF₂ = 40 + 38 = 78 g/mol.

The answer is $$\boxed{78}$$.

We need to identify the most basic oxide among the Group 13 oxides of type M$$_2$$O$$_3$$.

To begin,

Group 13 elements are: B, Al, Ga, In, Tl. As we move down the group, the metallic character increases progressively. Boron is a non-metal/metalloid, aluminum is a metal with some non-metallic properties, and by the time we reach thallium, the element is strongly metallic.

Next,

There is a fundamental relationship in chemistry:

- Oxides of non-metals are typically acidic (e.g., B$$_2$$O$$_3$$ is acidic)

- Oxides of metals are typically basic (e.g., Tl$$_2$$O$$_3$$ is basic)

- Oxides of elements with intermediate character are amphoteric (e.g., Al$$_2$$O$$_3$$, Ga$$_2$$O$$_3$$)

From this,

Going down Group 13, as metallic character increases, the basicity of the oxides also increases:

$$ \text{B}_2\text{O}_3 \;(\text{acidic}) < \text{Al}_2\text{O}_3 \;(\text{amphoteric}) < \text{Ga}_2\text{O}_3 \;(\text{amphoteric}) < \text{In}_2\text{O}_3 \;(\text{basic}) < \text{Tl}_2\text{O}_3 \;(\text{most basic}) $$

Tl$$_2$$O$$_3$$ is the most basic oxide because thallium is the most metallic element in Group 13. Its oxide is predominantly ionic in character, and it reacts with acids to form salts -- a characteristic property of basic oxides.

The most basic oxide is Tl$$_2$$O$$_3$$.

The correct answer is Option 3: Tl$$_2$$O$$_3$$.

Cobalt catalysts are used in the hydrogenation of carbon monoxide to produce methanol from synthesis gas. Therefore, cobalt catalyst is associated with methanol production.

Hence,

$$A \rightarrow IV.$$

Syngas is a mixture of carbon monoxide and hydrogen that is obtained during the gasification of coal with steam. Therefore, syngas is associated with coal gasification.

Hence,

$$B \rightarrow III.$$

Nickel is widely used as a catalyst in reactions involving hydrogen and is employed in the production and utilization of water gas. Therefore, nickel catalyst is matched with water gas production.

Hence,

$$C \rightarrow II.$$

Brine solution undergoes electrolysis to produce hydrogen gas at the cathode and chlorine gas at the anode. Therefore, brine solution is associated with the production of (H_2) and (Cl_2).

Hence,

$$D \rightarrow I.$$

Therefore, the correct matching is

$$A \rightarrow IV,\qquad B \rightarrow III,\qquad C \rightarrow II,\qquad D \rightarrow I.$$

Statement I: Lithium and Magnesium do not form superoxide.

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

Analysis of Statement I:

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

Statement I is correct.

Analysis of Statement II:

According to NCERT data:

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

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

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

Statement II is incorrect (in the NCERT context).

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

We need to identify which alkaline earth metal sulphates are readily soluble in water.

Solubility trend of alkaline earth metal sulphates:

The solubility of sulphates of alkaline earth metals decreases down the group due to the decrease in hydration enthalpy being more significant than the decrease in lattice enthalpy.

Solubility order: BeSO$$_4$$ > MgSO$$_4$$ > CaSO$$_4$$ > SrSO$$_4$$ > BaSO$$_4$$

Analysis:

(A) BeSO$$_4$$ — Readily soluble ✓

(B) MgSO$$_4$$ — Readily soluble ✓

(C) CaSO$$_4$$ — Slightly soluble (sparingly soluble)

(D) SrSO$$_4$$ — Insoluble

(E) BaSO$$_4$$ — Insoluble

Only BeSO$$_4$$ and MgSO$$_4$$ are readily soluble in water.

The correct answer is Option 3: A and B.

We need to identify which compound can enhance the efficiency of a hydrogen storage tank.

For Option A, Li/P$$_4$$ is lithium with white phosphorus and is not a hydrogen storage material.

Option B, SiH$$_4$$ (silane or silicon tetrahydride), contains hydrogen and has been studied as a hydrogen storage material. Silicon-based hydrides can store hydrogen at high weight percentages and release it upon decomposition.

Option C, NaNi$$_5$$, is not a standard hydrogen storage alloy; the well-known alloy is LaNi$$_5$$, not NaNi$$_5$$.

Option D, di-isobutylaluminium hydride (DIBAL-H), is a reducing agent used in organic synthesis rather than a hydrogen storage material.

Among the given options, SiH$$_4$$ is the compound that can enhance hydrogen storage efficiency, as metal hydrides and silicon-based hydrides are important classes of hydrogen storage materials. The correct answer is Option B.

Assertion A: BeCl₂ and MgCl₂ produce characteristic flame.

This is false. Be compounds do not produce a characteristic flame color because the excitation energy is too high for the flame to provide. MgCl₂ also does not produce a characteristic flame for the same reason (though Mg metal burns with bright white light).

Reason R: The excitation energy is high in BeCl₂ and MgCl₂.

This is true. Due to the small size and high ionization energy of Be and Mg, their excitation energies are high, which means the flame cannot provide enough energy to excite the electrons.

A is false but R is true.

This matches option 4.

We need to evaluate two statements about nickel and silicon.

Analyze Statement I.

Statement I: Nickel is being used as the catalyst for producing syn gas and edible fats.

Syn gas (CO + H$$_2$$) is produced by steam reforming of methane: CH$$_4$$ + H$$_2$$O $$\xrightarrow{\text{Ni}}$$ CO + 3H$$_2$$. Nickel is indeed the catalyst used.

Edible fats are produced by hydrogenation of vegetable oils using finely divided nickel as catalyst.

Statement I is correct.

Analyze Statement II.

Statement II: Silicon forms both electron rich and electron deficient hydrides.

Silicon forms silanes (SiH$$_4$$, Si$$_2$$H$$_6$$, etc.), which are electron-precise hydrides — each Si forms 4 bonds using all valence electrons. Silicon does not form electron-rich hydrides (like PH$$_3$$, H$$_2$$S with lone pairs that dominate chemistry) or electron-deficient hydrides (like B$$_2$$H$$_6$$ with 3-centre 2-electron bonds).

Statement II is incorrect.

Conclusion.

Statement I is correct but Statement II is incorrect.

The correct answer is Option D: Statement I is correct but statement II is incorrect.

We need to identify the reactants used to prepare Lithium Aluminium Hydride (LiAlH$$_4$$).

The preparation reaction is:

$$8\text{LiH} + \text{Al}_2\text{Cl}_6 \to 2\text{LiAlH}_4 + 6\text{LiCl}$$

This reaction involves lithium hydride (LiH) reacting with aluminium chloride (Al$$_2$$Cl$$_6$$) in ether solvent to produce LiAlH$$_4$$.

Therefore, LiAlH$$_4$$ is prepared from LiH and Al$$_2$$Cl$$_6$$.

The correct answer is Option 2.

We need to determine the magnetic behaviour of Li$$_2$$O, Na$$_2$$O$$_2$$, and KO$$_2$$.

Li$$_2$$O (Lithium oxide):

Contains Li$$^+$$ (1s$$^2$$, i.e., [He] configuration) and O$$^{2-}$$ (1s$$^2$$2s$$^2$$2p$$^6$$, i.e., [Ne] configuration). Both ions have all electrons paired. Li$$_2$$O is diamagnetic.

Na$$_2$$O$$_2$$ (Sodium peroxide):

Contains Na$$^+$$ ([Ne]) and the peroxide ion O$$_2^{2-}$$. The peroxide ion has a bond order of 1 with electronic configuration: $$(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p})^2(\pi_{2p})^4(\pi^*_{2p})^4$$. All electrons are paired. Na$$_2$$O$$_2$$ is diamagnetic.

KO$$_2$$ (Potassium superoxide):

Contains K$$^+$$ ([Ar]) and the superoxide ion O$$_2^-$$. The superoxide ion has electronic configuration: $$(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p})^2(\pi_{2p})^4(\pi^*_{2p})^3$$. There is one unpaired electron in the $$\pi^*_{2p}$$ orbital. KO$$_2$$ is paramagnetic.

Therefore, the magnetic behaviours are: diamagnetic, diamagnetic, and paramagnetic.

The correct answer is Option D.

We need to identify the best method for preparation of $$BeF_2$$ from the given options.

$$BeO + C + F_2 \xrightarrow{\Delta} BeF_2$$

This method involves the use of carbon as a reducing agent alongside fluorine gas. The reaction is not commonly used because handling fluorine gas at high temperatures with carbon is hazardous and impractical. The conditions are extreme and the reaction is difficult to control.

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

This is the thermal decomposition of ammonium tetrafluoroberyllate. On heating, the ammonium salt decomposes cleanly to give pure $$BeF_2$$ and $$NH_4F$$ (which sublimes away). This is the preferred industrial method because it avoids handling dangerous fluorine gas directly, and it produces high-purity $$BeF_2$$.

$$Be + F_2 \xrightarrow{\Delta} BeF_2$$

While this is a straightforward direct synthesis, the reaction between beryllium metal and fluorine gas is extremely vigorous, exothermic, and dangerous. It is difficult to control, making it impractical for preparative purposes.

$$BeH_2 + F_2 \xrightarrow{\Delta} BeF_2 + H_2$$

Beryllium hydride itself is unstable and difficult to prepare and handle. Using it as a starting material adds unnecessary complexity.

The thermal decomposition of ammonium tetrafluoroberyllate (Option 2) is the preferred method because:

(i) It avoids direct handling of reactive fluorine gas with metals.

(ii) The decomposition is clean and controllable.

(iii) The product obtained is of high purity.

The correct answer is Option (2).

We need to compare the temperatures $$T_1$$ and $$T_2$$ for two reactions in commercial H$$_2$$ production.

Reaction 1 (Water-gas reaction).

$$ C(s) + H_2O(g) \xrightarrow{T_1} CO(g) + H_2(g) $$

This is a highly endothermic reaction ($$\Delta H \approx +131$$ kJ/mol). It requires breaking the strong O-H bond in water and the C-C bonds in solid carbon. This reaction requires a very high temperature, typically around $$T_1 \approx 1270$$ K (about 1000 degrees C), to proceed at an appreciable rate.

Reaction 2 (Water-gas shift reaction).

$$ CO(g) + H_2O(g) \xrightarrow{T_2, \text{catalyst}} CO_2(g) + H_2(g) $$

This is a mildly exothermic reaction ($$\Delta H \approx -41$$ kJ/mol). It is carried out at a much lower temperature, typically $$T_2 \approx 673$$ K (about 400 degrees C), using an iron-chromate catalyst. The catalyst lowers the activation energy, allowing the reaction to proceed efficiently at this lower temperature.

Comparison:

$$ T_1 (\sim 1270\;\text{K}) > T_2 (\sim 673\;\text{K}) $$

The correct answer is Option 4: $$T_1 > T_2$$.

Photochemical smog is formed by the reaction of nitrogen oxides (NO$$_x$$) and volatile organic compounds (VOCs) in the presence of sunlight. The key pollutants include ozone (O$$_3$$), PAN (peroxyacetyl nitrate), and formaldehyde.

The most effective method to control photochemical smog is the use of catalytic converters in automobiles and industries. These converters transform harmful exhaust gases into less harmful products:

$$2CO + O_2 \xrightarrow{\text{catalyst}} 2CO_2$$

$$2NO_x \xrightarrow{\text{catalyst}} xO_2 + N_2$$

Unburnt hydrocarbons are also converted to CO$$_2$$ and H$$_2$$O. By reducing the emission of NO$$_x$$ and hydrocarbons at the source, catalytic converters prevent the formation of photochemical smog.

So, the answer is that photochemical smog can be controlled by using catalytic converters in automobiles and industry.

### Explanation

Mixing aqueous $$CaCl_2$$ and $$Na_2CO_3$$ results in a double displacement (precipitation) reaction.

$$CaCl_2(aq)+Na_2CO_3(aq)\rightarrow CaCO_3(s)\downarrow+2NaCl(aq)$$

Thus, the products formed are calcium carbonate and sodium chloride.

Hence,

$$X=CaCO_3$$

$$Y=NaCl$$

To convert $$CaCO_3$$ back into $$CaCl_2$$, it is treated with hydrochloric acid.

$$CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)+CO_2(g)$$

Therefore,

$$Z=HCl$$

Hence, the correct matching is

$$X=CaCO_3,\quad Y=NaCl,\quad Z=HCl$$

This corresponds to Option B

For drinking water, the maximum allowed concentrations are:

$$F^- < 2\ \text{ppm}$$

$$SO_4^{2-} < 500\ \text{ppm}$$

$$NO_3^- < 50\ \text{ppm}$$

$$Zn < 5\ \text{ppm}$$

Therefore:

$$A \rightarrow III$$

$$B \rightarrow IV$$

$$C \rightarrow I$$

$$D \rightarrow II$$

Hence the correct matching is:

$${A-III,\;B-IV,\;C-I,\;D-II}$$
Hence, Option D is correct

Hydration enthalpy depends on the charge density of the ion: higher charge and smaller ionic radius lead to greater (more negative) hydration enthalpy.

The ions given are (A) K$$^+$$ (charge = +1, ionic radius = 138 pm), (B) Rb$$^+$$ (charge = +1, ionic radius = 152 pm), (C) Mg$$^{2+}$$ (charge = +2, ionic radius = 72 pm), (D) Cs$$^+$$ (charge = +1, ionic radius = 167 pm) and (E) Ca$$^{2+}$$ (charge = +2, ionic radius = 100 pm).

Divalent cations have much higher hydration enthalpies than monovalent cations due to their greater charge density.

Between the divalent ions, Mg$$^{2+}$$ is smaller than Ca$$^{2+}$$, so Mg$$^{2+}$$ has greater hydration enthalpy: C > E.

Among the monovalent ions, smaller size means greater hydration enthalpy: K$$^+$$ > Rb$$^+$$ > Cs$$^+$$, i.e., A > B > D.

The complete order is:

$$\text{Mg}^{2+} > \text{Ca}^{2+} > \text{K}^+ > \text{Rb}^+ > \text{Cs}^+$$

That is: C > E > A > B > D.

The correct answer is Option D.

Be(OH)₂ is amphoteric. When it reacts with Sr(OH)₂ (a strong base), Be(OH)₂ acts as an acid:

$$\text{Be(OH)}_2 + \text{Sr(OH)}_2 \to \text{Sr[Be(OH)}_4]$$

or equivalently: $$\text{Be(OH)}_2 + 2\text{OH}^- \to [\text{Be(OH)}_4]^{2-}$$

The ionic salt formed is Sr[Be(OH)₄], where:

• Sr²⁺ is the cation
• [Be(OH)₄]²⁻ is the anion (beryllate ion)

Now let's check each option:

Option 1: Both Sr and Be are present in the ionic salt → Correct

Option 2: Be is tetrahedrally coordinated → Correct (4 OH groups around Be in [Be(OH)₄]²⁻)

Option 3: Be is present in the cationic part → Incorrect. Be is in the anionic part [Be(OH)₄]²⁻

Option 4: This is an acid-base neutralization reaction → Correct (amphoteric oxide acting as acid with a base)

The incorrect option is Option 3.

Assertion (A): Gypsum is used for making fireproof wall boards.

This is CORRECT. Gypsum ($$CaSO_4 \cdot 2H_2O$$) is widely used in making fireproof wall boards (drywall/gypsum board) because of its fire-resistant properties.

Reason (R): Gypsum is unstable at high temperatures.

This is CORRECT. When exposed to high temperatures, gypsum undergoes dehydration:

$$ CaSO_4 \cdot 2H_2O \xrightarrow{\Delta} CaSO_4 \cdot \frac{1}{2}H_2O + \frac{3}{2}H_2O $$

This endothermic dehydration process absorbs heat, which slows down the spread of fire. The water released as steam also helps in fire suppression.

R is the correct explanation of A because it is precisely the instability of gypsum at high temperatures (releasing water endothermically) that makes it effective as a fireproof material.

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

Carbon monoxide ((CO)) is a neutral oxide because it does not react with water or bases under ordinary conditions. Carbon dioxide ((CO_2)), on the other hand, is an acidic oxide as it reacts with water to form carbonic acid and reacts with bases to form carbonates.

The reaction of carbon dioxide with water is:

$$CO_2(g) + H_2O(l) \rightleftharpoons H_2CO_3(aq)$$

Carbon monoxide is only sparingly soluble in water and does not exhibit acidic behavior, whereas carbon dioxide combines with water (to a limited extent) to form carbonic acid. Thus, the ability of (CO_2) to form carbonic acid explains its acidic nature, while the absence of such behavior makes (CO) a neutral oxide.

Hence, both the Assertion and the Reason are correct, and the Reason correctly explains the Assertion.

Therefore, the correct answer is:

$$\boxed{\text{Option B: Both A and R are correct and R is the correct explanation of A.}}$$

Photochemical smog formation begins with the photolysis of nitrogen dioxide:

$$NO_2(g) \xrightarrow{\text{Sunlight}} NO(g) + O(g)$$

Comparing this with $$NO_2 \rightarrow X + Y$$:

• $$X = O$$ (nascent oxygen)

• $$Y = NO$$ (nitric oxide)

The nascent oxygen atom then reacts with atmospheric oxygen to form ozone:

$$O(g) + O_2(g) \rightleftharpoons O_3(g)$$

Comparing this with $$X + A \rightarrow B$$:

• $$A = O_2$$

• $$B = O_3$$

Therefore,

$$X = O,; Y = NO,; A = O_2,; B = O_3$$

Hence, the correct answer is:

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

Photochemical smog is formed when nitrogen oxides (NO₂) and volatile organic compounds (hydrocarbons) react in the presence of sunlight to produce secondary pollutants like ozone, PAN (peroxyacetyl nitrate), and other irritants.

The key requirements for photochemical smog formation are:

1. Nitrogen oxides (from automobile exhaust and industrial emissions)

2. Hydrocarbons (from incomplete combustion of fuels)

3. Sunlight (for photochemical reactions)

4. Warm, stagnant air conditions

Industrial areas have high concentrations of both nitrogen oxides and hydrocarbons from factories, vehicles, and other sources. Combined with sunlight, these areas have the highest possibility of photochemical smog formation.

The correct answer is Industrial areas.

Statement I: Boron is extremely hard indicating high lattice energy — Incorrect. Boron's hardness is due to its covalent network structure, not ionic lattice energy. The term "lattice energy" applies to ionic compounds.

Statement II: Boron has the highest melting and boiling point among Group 13 elements — Correct. Boron (MP ≈ 2076°C) has the highest values due to its strong covalent network structure.

Statement I is incorrect but Statement II is correct.

We need to determine the correct order of single bond enthalpies for Group 14 elements.

Known single bond enthalpies (kJ mol$$^{-1}$$):

C$$-$$C: 348 kJ/mol

Si$$-$$Si: 226 kJ/mol

Ge$$-$$Ge: 188 kJ/mol

Sn$$-$$Sn: 151 kJ/mol

Explanation:

Carbon has the strongest single bond due to effective overlap of its small 2p orbitals. As we move down Group 14, the atomic size increases, orbital overlap becomes less effective, and bond enthalpy decreases monotonically.

The correct order is:

$$ \text{C}-\text{C} > \text{Si}-\text{Si} > \text{Ge}-\text{Ge} > \text{Sn}-\text{Sn} $$

The correct answer is Option D.

In the Ellingham diagram, the reaction $$2C(s) + O_2(g) \to 2CO(g)$$ has $$\Delta n_g = 2 - 1 = +1$$ (an increase in gas moles). By the relation $$\Delta G = \Delta H - T\Delta S$$, an increase in gaseous moles means $$\Delta S > 0$$, so the $$-T\Delta S$$ term becomes more negative with increasing temperature. This causes $$\Delta G$$ to decrease (become more negative) as temperature rises, giving the line a negative slope. Therefore, Assertion A is correct.

Reason R claims that CO tends to get decomposed at higher temperature. This is incorrect — in fact, the negative slope of the Ellingham line for CO formation means that CO becomes more thermodynamically stable at higher temperatures, not less. CO does not decompose at higher temperatures; rather, it is favoured as a reducing agent at elevated temperatures precisely because of this trend.

Since A is correct but R is not correct, the answer is Option D.

Concentration of ore involves separating the ore from gangue (impurities). The common methods include:

- Hydraulic washing (gravity separation) ✔

- Froth flotation ✔

- Leaching (chemical method of concentration) ✔

- Magnetic separation ✔

Methods NOT used for concentration of ore:

(A) Liquation: This is a refining technique (used to purify metals with low melting points like tin, bismuth, lead). It is NOT a concentration method. ✘

(C) Electrolysis: This is a refining technique (electrolytic refining). It is NOT a concentration method. ✘

Therefore, the methods NOT involved in concentration of ore are A and C only.

Four reactions are listed. We need to identify which one does NOT occur during the extraction of copper from copper pyrites (CuFeS$$_2$$). Copper is extracted through the following steps:

Roasting: The ore is heated in air to convert iron sulfide to iron oxide: $$2\text{FeS} + 3\text{O}_2 \to 2\text{FeO} + 2\text{SO}_2$$ This is Option B — it occurs during copper extraction.

Slag formation: SiO$$_2$$ is added as a flux and reacts with FeO to form slag: $$\text{FeO} + \text{SiO}_2 \to \text{FeSiO}_3$$ This is Option D — it occurs during copper extraction.

Partial roasting of Cu$$_2$$S: $$2\text{Cu}_2\text{S} + 3\text{O}_2 \to 2\text{Cu}_2\text{O} + 2\text{SO}_2$$ This is Option A — it occurs during copper extraction.

Option C states: CaO + SiO$$_2$$ $$\to$$ CaSiO$$_3$$. In copper extraction, the flux used is SiO$$_2$$, which reacts with FeO (not CaO). CaO is not involved in copper extraction. The reaction CaO + SiO$$_2$$ $$\to$$ CaSiO$$_3$$ occurs in iron extraction (blast furnace), not copper extraction.

Answer: Option C

Statement I: Chlorine can easily combine with oxygen to form oxides; and the product has a tendency to explode.

This is TRUE. Chlorine forms several oxides such as $$Cl_2O$$, $$ClO_2$$, $$Cl_2O_6$$, and $$Cl_2O_7$$. These chlorine oxides are highly unstable and have a strong tendency to explode. They are powerful oxidizing agents.

Statement II: Chemical reactivity of an element can be determined by its reaction with oxygen and halogens.

This is TRUE. The chemical reactivity of elements is commonly assessed by studying their reactions with oxygen (to form oxides) and halogens (to form halides). These reactions help determine the oxidation states, bond energies, and overall reactivity patterns of elements.

Therefore, both Statement I and Statement II are true.

A. Reverberatory furnace is used for copper smelting (IV)

B. Electrolytic cell (Hall-Heroult process) is used for aluminium extraction (II)

C. Blast furnace is used for pig iron production (I)

D. Zone refining is used for purification of silicon and germanium (III)

Matching: A-IV, B-II, C-I, D-III.

Thionyl chloride ($$SOCl_2$$) reacts with white phosphorus ($$P_4$$).

When $$SOCl_2$$ reacts with $$P_4$$, it chlorinates the phosphorus. The reaction produces phosphorus trichloride ($$PCl_3$$) as compound [A]:

$$P_4 + 8SOCl_2 \rightarrow 4PCl_3 + 4SO_2 + 2S_2Cl_2$$

On hydrolysis, $$PCl_3$$ gives phosphorous acid ($$H_3PO_3$$) as compound [B]:

$$PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl$$

$$H_3PO_3$$ (phosphorous acid) is a dibasic acid because it has only two ionizable O-H bonds (the third hydrogen is directly bonded to phosphorus as a P-H bond and is not ionizable).

Therefore [A] = $$PCl_3$$ and [B] = $$H_3PO_3$$.

The answer is Option B.

We need to identify which group of chemicals are all used as pesticides. Pesticides is a broad term that includes insecticides (kill insects), herbicides (kill weeds), fungicides (kill fungi), and rodenticides (kill rodents).

Analyzing each chemical mentioned in the options:

Aldrin: Aldrin is a chlorinated hydrocarbon compound that was widely used as an insecticide, particularly for soil-dwelling pests. It is a pesticide. (Present in Options 1 and 2)

Sodium Chlorate ($$NaClO_3$$): Sodium chlorate is a powerful oxidizing agent that is used as a non-selective herbicide (weed killer). It kills plants by disrupting their cellular processes. It is a pesticide. (Present in Options 1 and 3)

Sodium Arsenite ($$Na_3AsO_3$$ or $$NaAsO_2$$): Sodium arsenite is a highly toxic arsenic compound used as a pesticide (insecticide and herbicide). It has been used to control various pests and weeds. It is a pesticide. (Present in Options 1 and 4)

DDT (Dichlorodiphenyltrichloroethane): DDT is a well-known chlorinated insecticide. It is a pesticide. (Present in Options 2 and 3)

PAN (Peroxyacetyl nitrate): PAN is a secondary air pollutant formed in photochemical smog. It is NOT a pesticide - it is an atmospheric pollutant. (Present in Option 3)

Dieldrin: Dieldrin is a chlorinated hydrocarbon insecticide, related to aldrin (aldrin converts to dieldrin in the environment). It is a pesticide. (Present in Option 4)

Tetrachloroethene ($$C_2Cl_4$$): Tetrachloroethene (also called perchloroethylene or PCE) is primarily used as a dry cleaning solvent and degreasing agent. It is NOT classified as a pesticide. (Present in Option 4)

Evaluating the options:

Option 1: Aldrin (pesticide), Sodium Chlorate (pesticide), Sodium Arsenite (pesticide) - ALL are pesticides.

Option 2: DDT (pesticide), Aldrin (pesticide) - both are pesticides, but this is an incomplete group (only two).

Option 3: Sodium chlorate (pesticide), DDT (pesticide), PAN (NOT a pesticide) - incorrect.

Option 4: Dieldrin (pesticide), Sodium arsenite (pesticide), Tetrachloroethene (NOT a pesticide) - incorrect.

The correct answer is Option 1: Aldrin, Sodium Chlorate, Sodium Arsenite. All three are pesticides used for different purposes (insecticide, herbicide, and general pesticide respectively).

Leaching of silver involves oxidation followed by reduction.

During leaching, silver is treated with dilute cyanide solution in the presence of oxygen:

$$4Ag_{(s)} + 8CN^-_{(aq)} + 2H_2O_{(aq)} + O_{2(g)} \rightarrow 4[Ag(CN)_2]^-_{(aq)} + 4OH^-_{(aq)}$$

Silver is oxidized from oxidation state $$0$$ to $$+1$$ and forms the soluble complex $$[Ag(CN)_2]^-$$.

The dissolved silver is then recovered by adding zinc:

$$2[Ag(CN)_2]^-_{(aq)} + Zn_{(s)} \rightarrow [Zn(CN)_4]^{2-}_{(aq)} + 2Ag_{(s)}$$

Silver is reduced from oxidation state $$+1$$ back to $$0$$ and precipitates as metallic silver.

Aluminium is extracted by electrolytic reduction of alumina, copper is mainly extracted through roasting and self-reduction, and iron is extracted by reduction with carbon monoxide in a blast furnace.

Hence, the metal whose extraction involves oxidation followed by reduction is

$$\boxed{\text{Ag (Silver)}}$$

Statement I: In froth floatation method, a rotating paddle agitates the mixture to drive air out of it.

This is false. The rotating paddle agitates the mixture to draw air INTO it (not out of it), creating froth to which the mineral particles attach.

Statement II: Iron pyrites are generally avoided for extraction of iron due to environmental reasons.

This is true. Iron pyrites (FeS₂) produce SO₂ on roasting, which causes air pollution. Hence, oxide ores are preferred for iron extraction.

The correct answer is Option 2: Statement I is false but Statement II is true.

We need to evaluate two statements about metallurgy.

Statement I: In the metallurgy process, sulphide ore is converted to oxide before reduction.

This is correct. The process of converting sulphide ores to oxides is called roasting. For example:

$$ 2ZnS + 3O_2 \xrightarrow{\Delta} 2ZnO + 2SO_2 $$

$$ 2PbS + 3O_2 \xrightarrow{\Delta} 2PbO + 2SO_2 $$

After roasting, the metal oxide is then reduced to the metal using a suitable reducing agent (carbon, CO, or another metal).

Statement II: Oxide ores in general are easier to reduce.

This is correct. Metal oxides are thermodynamically easier to reduce than metal sulphides for several reasons:

1. Carbon and CO are readily available, cheap reducing agents that can reduce most metal oxides at reasonable temperatures.

2. On the Ellingham diagram, the line for $$2C + O_2 \to 2CO$$ has a negative slope, meaning carbon becomes a better reducing agent at higher temperatures, and it falls below most metal oxide lines.

3. Direct reduction of sulphides is difficult because the corresponding reactions are less thermodynamically favourable.

Both statements are correct, and Statement II provides the reason for the practice described in Statement I.

The correct answer is Option 4: Both Statement I and Statement II are correct.

In the extraction of copper from copper pyrites (CuFeS$$_2$$), the ore first undergoes partial roasting:

$$2\text{Cu}_2\text{S} + 3\text{O}_2 \rightarrow 2\text{Cu}_2\text{O} + 2\text{SO}_2$$

This is followed by self-reduction (Bessemerisation):

$$2\text{Cu}_2\text{O} + \text{Cu}_2\text{S} \rightarrow 6\text{Cu} + \text{SO}_2$$

In this process, Cu$$_2$$S acts as both a reducing agent and is itself oxidised. The copper obtained contains blisters (bubbles) on the surface due to the evolution of SO$$_2$$ gas during solidification. This product is called blister copper, which is about 98-99% pure copper. It is further refined by electrolysis to obtain 99.9% pure copper.

So, the answer is blister copper.

In froth floatation process for concentration of sulphide ores:

- Collectors (like pine oils, fatty acids, xanthates) enhance the non-wettability of the mineral particles.

- Frothers help in forming a stable froth.

- Stabilizers stabilize the froth. Cresols are used as stabilizers.

Pine oils and xanthates are collectors, not stabilizers.

The correct answer is Option 4: Cresols.

Statement I: Methane and steam passed over heated Ni catalyst produces hydrogen gas.

This is the steam reforming reaction:

$$\text{CH}_4 + \text{H}_2\text{O} \xrightarrow{\text{Ni, } 1270 \text{ K}} \text{CO} + 3\text{H}_2$$

This is indeed used commercially to produce hydrogen gas (syngas). Statement I is correct.

Statement II: Sodium nitrite reacts with NH$$_4$$Cl to give H$$_2$$O, N$$_2$$ and NaCl.

The reaction is:

$$\text{NaNO}_2 + \text{NH}_4\text{Cl} \xrightarrow{\Delta} \text{NaCl} + \text{N}_2 + 2\text{H}_2\text{O}$$

This reaction is a well-known method to prepare dinitrogen gas in the laboratory. Statement II is correct.

Both statements I and II are correct.

Incorrect statement about borazine (B₃N₃H₆, "inorganic benzene"):

A. "It contains banana bonds" — Banana bonds are characteristic of diborane (B₂H₆), not borazine. INCORRECT ✓

B. "It can react with water" — Yes, borazine undergoes hydrolysis. CORRECT

C. "It is a cyclic compound" — Yes, borazine has a 6-membered ring. CORRECT

D. "It has electronic delocalization" — Yes, the lone pairs on N can delocalize into empty p-orbitals of B. CORRECT

The incorrect statement is Option A: It contains banana bonds.

We need to identify which phosphorus oxoacid can reduce $$AgNO_3$$ to metallic silver (creating a silver mirror, i.e., giving a positive Tollen's test). A phosphorus oxoacid can act as a strong reducing agent if it contains a direct $$P-H$$ bond because the hydrogen bonded directly to phosphorus can be released, providing electrons for the reduction process. Oxoacids with $$P-H$$ bonds are strong enough reducing agents to reduce $$Ag^+$$ ions to metallic silver. Metaphosphoric acid $$HPO_{3n}$$ (polymetaphosphoric acid) has the structure where all hydrogen atoms are bonded to oxygen as $$O-H$$ groups and no direct $$P-H$$ bond; therefore it cannot reduce $$Ag^+$$ to silver.

Pyrophosphorous acid $$H_4P_2O_5$$, the pyro-form of phosphorous acid $$H_3PO_3$$, retains the direct $$P-H$$ bonds present in phosphorous acid. Since phosphorous acid is diprotic with one hydrogen bonded to phosphorus, the condensation of two $$H_3PO_3$$ molecules (losing water) produces $$H_4P_2O_5$$ with the same $$P-H$$ bonds, and these bonds give it the reducing power to reduce $$Ag^+$$ ions to metallic silver:

$$Ag^+(aq) + e^- \rightarrow Ag(s)$$

Thus $$H_4P_2O_5$$ can create a silver mirror from $$AgNO_3$$ solution. Hypophosphoric acid $$H_4P_2O_6$$ contains a direct $$P-P$$ bond and hydrogen atoms only bonded to oxygen as $$O-H$$ groups, so it lacks a $$P-H$$ bond and cannot reduce $$Ag^+$$. Pyrophosphoric acid $$H_4P_2O_7$$ (the pyro-form of phosphoric acid $$H_3PO_4$$) also has no $$P-H$$ bonds since all hydrogens are in $$O-H$$ groups, and therefore it cannot act as a reducing agent. Only $$H_4P_2O_5$$ (pyrophosphorous acid) contains $$P-H$$ bonds, giving it the necessary reducing power to reduce $$Ag^+$$ to metallic silver and produce a silver mirror. The correct answer is Option 2: $$H_4P_2O_5$$.

Reducing ability comes when there is a scope of oxidising the acid. For example: presence of P-H bonds. Which is only applicable in option B.

$$\text{P}_4 + 8\text{SOCl}_2 \rightarrow 4\text{PCl}_3 + 4\text{SO}_2 + 2\text{S}_2\text{Cl}_2$$

Verifying: P: 4=4 ✓, S: 8=4+4 ✓, O: 8=8 ✓, Cl: 16=12+4 ✓

A = PCl$$_3$$, B = S$$_2$$Cl$$_2$$, x = 4.

We need to identify which halide ions can be oxidised by oxygen in acidic medium.

The reduction potential of O$$_2$$ in acidic medium:

$$O_2 + 4H^+ + 4e^- \to 2H_2O$$, $$E° = +1.23$$ V

For oxidation of X$$^-$$ to X$$_2$$:

- $$2Cl^- \to Cl_2 + 2e^-$$, $$E° = -1.36$$ V (very difficult to oxidise)

- $$2Br^- \to Br_2 + 2e^-$$, $$E° = -1.07$$ V

- $$2I^- \to I_2 + 2e^-$$, $$E° = -0.54$$ V (easiest to oxidise)

For a reaction to be spontaneous: $$E°_{cell} = E°_{cathode} - E°_{anode} > 0$$

For I$$^-$$: $$E°_{cell} = 1.23 - 0.54 = 0.69$$ V > 0 $$\checkmark$$ (spontaneous)

For Br$$^-$$: $$E°_{cell} = 1.23 - 1.07 = 0.16$$ V > 0, but this is marginal and in practice O$$_2$$ does not readily oxidise Br$$^-$$ in dilute acidic solutions.

For Cl$$^-$$: $$E°_{cell} = 1.23 - 1.36 = -0.13$$ V < 0 $$\times$$ (not spontaneous)

In practice, only iodide is readily oxidised by O$$_2$$ in acidic medium. This is why KI solutions turn yellow/brown on standing in air.

The correct answer is Option C: I$$^-$$ only.

The number of $$P-O-P$$ bonds is equal to the number of oxygen atoms that bridge two phosphorus atoms.

In $$H_4P_2O_7$$ (pyrophosphoric acid), two orthophosphoric acid units are joined by the elimination of one molecule of water. This results in a single oxygen atom bridging the two phosphorus atoms.

Therefore, the number of $$P-O-P$$ bonds in $$H_4P_2O_7$$ is

$$1$$

In $$(HPO_3)_3$$ (cyclotrimetaphosphoric acid), three phosphorus atoms are arranged in a cyclic structure with alternating phosphorus and oxygen atoms. Each phosphorus atom is connected to the next phosphorus atom through a bridging oxygen atom.

Therefore, the number of $$P-O-P$$ bonds in $$(HPO_3)_3$$ is

$$3$$

In $$P_4O_{10}$$, the four phosphorus atoms occupy the vertices of a tetrahedral framework. Each edge of the tetrahedron contains one bridging oxygen atom. Since a tetrahedron has six edges, there are six bridging oxygen atoms connecting phosphorus atoms.

Therefore, the number of $$P-O-P$$ bonds in $$P_4O_{10}$$ is

$$6$$

Hence, the number of $$P-O-P$$ bonds in

$$H_4P_2O_7,\ (HPO_3)3,\ P_4O{10}$$

are

$$1,\ 3,\ 6$$

respectively.

Hence, the correct answer is

$$1,\ 3,\ 6$$

In the Ostwald process, ammonia is catalytically oxidized to nitric oxide according to the reaction

$$4NH_3(g)+5O_2(g)\xrightarrow{Pt/Rh}4NO(g)+6H_2O(l).$$

Hence,

$$A=NO.$$

Nitric oxide is then oxidized by atmospheric oxygen to form nitrogen dioxide:

$$2NO(g)+O_2(g)\rightarrow2NO_2(g).$$

Therefore,

$$B=NO_2.$$

Nitrogen dioxide dissolves in water to produce nitric acid and nitric oxide:

$$3NO_2(g)+H_2O(l)\rightarrow2HNO_3(aq)+NO(g).$$

The nitric acid formed gives a positive brown ring test due to the presence of nitrate ions.

Thus,

$$A=NO,\qquad B=NO_2.$$

Hence, the correct answer is

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

Boron trifluoride, $$BF_3$$, is a neutral molecule in which boron is bonded to three fluorine atoms. In the tetrafluoroborate ion, $$BF_4^-$$, one extra $$F^-$$ ion donates a lone pair to boron, creating a coordinate (dative) $$\sigma$$-bond. Thus boron is surrounded by four $$F$$ atoms.

Step 1 - Covalency of boron
Covalency is defined as the total number of $$\sigma$$-bonds an atom forms with surrounding atoms. In $$BF_4^-$$ boron forms four B-F $$\sigma$$-bonds (three normal + one coordinate). Therefore the covalency of boron is $$4$$.

Step 2 - Oxidation state of boron
Assign oxidation numbers using the rule that fluorine always has an oxidation state of $$-1$$.

Let the oxidation state of boron be $$x$$. For the ion $$BF_4^-$$:

$$x + 4(-1) = -1$$

$$x - 4 = -1$$

$$x = +3$$

Hence the oxidation state of boron is $$+3$$.

Result
Covalency = $$4$$, Oxidation state = $$+3$$.

Therefore the correct choice is Option D.

In the Hall-Heroult process for the extraction of aluminium, $$Al_2O_3$$ is dissolved in a molten mixture and electrolyzed.

We have the following setup in the Hall-Heroult process:

- $$Al_2O_3$$ (alumina) is dissolved in molten cryolite ($$Na_3AlF_6$$)

- $$CaF_2$$ (fluorspar) is added to lower the melting point of the mixture

- Graphite electrodes are used (carbon anodes)

- Electrolysis reduces $$Al^{3+}$$ to Al at the cathode

Now, at the cathode: $$Al^{3+} + 3e^- \to Al$$

At the anode: $$C + O^{2-} \to CO_2 + 4e^-$$

So the carbon (graphite) anode acts as the reducing agent in the overall reaction, while $$CaF_2$$ serves as a flux to lower the melting point.

Hence, the correct answer is Option D: $$CaF_2$$.

We need to determine how many compounds from the given list are produced when LiNO$$_3$$ is heated.

Thermal decomposition of LiNO$$_3$$.

Lithium nitrate behaves differently from other alkali metal nitrates upon heating. Unlike NaNO$$_3$$, KNO$$_3$$, etc., which decompose to give the corresponding nitrite and O$$_2$$, lithium nitrate decomposes similar to alkaline earth metal nitrates due to the small size and high polarizing power of Li$$^+$$.

Write the decomposition reaction.

$$4\text{LiNO}_3 \xrightarrow{\Delta} 2\text{Li}_2\text{O} + 4\text{NO}_2 + \text{O}_2$$

Identify the products from the given list.

The products are:

Li$$_2$$O — Yes, it is a product.

N$$_2$$ — No, nitrogen gas is not produced.

O$$_2$$ — Yes, it is a product.

LiNO$$_2$$ — No, lithium nitrite is not formed (unlike other alkali metal nitrates).

NO$$_2$$ — Yes, it is a product.

Three compounds from the list are produced: Li$$_2$$O, O$$_2$$, and NO$$_2$$.

The correct answer is 3.

The formula of borax is Na₂B₄O₅(OH)₄ · 8H₂O.

Comparing with Na₂B₄O$$_x$$(OH)$$_y$$ · zH₂O:

$$x = 5$$, $$y = 4$$, $$z = 8$$

$$x + y + z = 5 + 4 + 8 = 17$$

This matches the answer key value of $$\mathbf{17}$$.

We need to count the number of acidic oxides from the given list.

An acidic oxide reacts with water to form an acid, or reacts with a base to form a salt.

Analyzing each oxide:

N$$_2$$O$$_3$$: Acidic oxide (forms HNO$$_2$$ with water). ✓

NO$$_2$$: Acidic oxide (forms a mixture of HNO$$_3$$ and HNO$$_2$$ with water). ✓

N$$_2$$O: Neutral oxide. ✗

Cl$$_2$$O$$_7$$: Acidic oxide (anhydride of HClO$$_4$$). ✓

SO$$_2$$: Acidic oxide (forms H$$_2$$SO$$_3$$ with water). ✓

CO: Neutral oxide. ✗

CaO: Basic oxide. ✗

Na$$_2$$O: Basic oxide. ✗

NO: Neutral oxide. ✗

Total number of acidic oxides = $$\boxed{4}$$

We need to find the difference in oxidation state of Xe between the oxidized product formed on complete hydrolysis of $$XeF_4$$ and $$XeF_4$$ itself.

The complete hydrolysis of $$XeF_4$$ is a disproportionation reaction:

In this reaction, $$XeF_4$$ undergoes disproportionation - some Xe is reduced to elemental form (Xe) and some is oxidized to $$XeO_3$$.

In $$XeF_4$$: F has oxidation state $$-1$$. So $$x + 4(-1) = 0$$, giving $$x = +4$$.

In $$XeO_3$$ (oxidized product): O has oxidation state $$-2$$. So $$x + 3(-2) = 0$$, giving $$x = +6$$.

In Xe (reduced product): Oxidation state = 0.

The oxidized product is $$XeO_3$$ (Xe in +6 state). The oxidation state of Xe in $$XeF_4$$ is +4.

The difference in the oxidation state is 2.

We need to identify which nitrogen oxide contains an odd electron at the nitrogen atom.

Key Concept: An odd-electron molecule has an odd total number of valence electrons, which means at least one electron must remain unpaired.

Option A: $$N_2O$$ (Nitrous oxide)

Total valence electrons = $$2 \times 5 + 6 = 16$$ (even).

Lewis structure: $$:N=N=O:$$ (linear, with resonance structures $$:\!N \equiv N - O\!:$$ and $$:N - N \equiv O:$$). All 16 electrons are paired. No unpaired electron on nitrogen.

Option B: $$NO_2$$ (Nitrogen dioxide)

Total valence electrons = $$5 + 2 \times 6 = 17$$ (odd).

Since 17 is odd, one electron must remain unpaired. The Lewis structure of $$NO_2$$ is drawn as follows:

Nitrogen is the central atom bonded to two oxygen atoms. One N=O double bond uses 4 electrons, one N-O single bond uses 2 electrons. Each oxygen gets lone pairs to complete its octet: the double-bonded O has 2 lone pairs (4 electrons), the single-bonded O has 3 lone pairs (6 electrons). This accounts for $$4 + 2 + 4 + 6 = 16$$ electrons. The remaining 1 electron (the 17th) sits on the nitrogen atom as an unpaired electron.

This unpaired electron makes $$NO_2$$ paramagnetic and gives it a bent geometry with a bond angle of about $$134°$$.

Option C: $$N_2O_3$$ (Dinitrogen trioxide)

Total valence electrons = $$2 \times 5 + 3 \times 6 = 28$$ (even).

$$N_2O_3$$ can be viewed as $$NO + NO_2$$ combined through an N-N bond: $$O=N-N(=O)-O$$. All 28 electrons are paired in the bonding and lone pair positions. No unpaired electron.

Option D: $$N_2O_5$$ (Dinitrogen pentoxide)

Total valence electrons = $$2 \times 5 + 5 \times 6 = 40$$ (even).

Structure: $$O_2N-O-NO_2$$. All 40 electrons are paired. No unpaired electron.

Among all the given oxides, only $$NO_2$$ has an odd number of valence electrons (17). The single unpaired electron resides on the nitrogen atom, making it an odd-electron species.

The correct answer is Option B: $$NO_2$$.

We need to find the products of the reaction between BeCl₂ and LiAlH₄.

We start by writing the balanced reaction since LiAlH₄ is a strong reducing agent that acts as a source of hydride ions (H⁻). The reaction is:

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

Next, the products formed are (A) AlCl₃, (B) BeH₂, and (D) LiCl. Aluminium chloride is formed when Al³⁺ combines with Cl⁻ ions; beryllium hydride is formed when Be²⁺ receives hydride ions from LiAlH₄; and lithium chloride is formed as a by-product.

In this reaction, the hydride (H⁻) from LiAlH₄ replaces the chloride in BeCl₂, forming BeH₂. The displaced chloride ions combine with Li⁺ and Al³⁺ to form LiCl and AlCl₃. Note that BeAlH₄ (Option E) is not formed in this reaction, and neither is LiH (Option C).

Therefore, the correct answer is Option B.

We are given the reaction of zinc with excess aqueous alkali. Zinc is an amphoteric metal, meaning it can react with both acids and bases.

When zinc reacts with excess aqueous NaOH (alkali), the reaction proceeds as:

$$Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2 \uparrow$$

Here, zinc dissolves in the excess alkali solution and reduces water to liberate hydrogen gas. The sodium zincate $$Na_2ZnO_2$$ formed dissociates in solution to give the zincate ion $$[ZnO_2]^{2-}$$.

Now, one might wonder why the product is $$[ZnO_2]^{2-}$$ rather than $$Zn(OH)_2$$ or $$[Zn(OH)_4]^{2-}$$. When the alkali is in excess, any $$Zn(OH)_2$$ initially formed further reacts with NaOH:

$$Zn(OH)_2 + 2NaOH \rightarrow Na_2ZnO_2 + 2H_2O$$

This converts the hydroxide completely into the zincate ion $$[ZnO_2]^{2-}$$. The species $$ZnO$$ and $$Zn(OH)_2$$ are only intermediate products and do not persist when alkali is in excess.

Hence, the correct answer is Option D: $$[ZnO_2]^{2-}$$.

Borazine (B₃N₃H₆), also known as "inorganic benzene" due to its structural similarity to benzene, can be prepared from diborane and ammonia.

We start by writing the reaction for the standard preparation of borazine by heating diborane (B₂H₆) with ammonia (NH₃):

$$3B_2H_6 + 6NH_3 \rightarrow 2B_3N_3H_6 + 12H_2$$

The question states that borazine is prepared from 3 equivalents of "X" and 6 equivalents of "Y". From the balanced equation, 3 equivalents of $$B_2H_6$$ (diborane) correspond to X and 6 equivalents of $$NH_3$$ (ammonia) correspond to Y, which matches perfectly with the stoichiometry of the reaction.

Next, we consider why the other options do not apply. Option A, $$B(OH)_3 + NH_3$$, is not the standard method and does not match the stoichiometry.

Option C, $$B_2H_6 + HN_3$$, involves hydrazoic acid (HN₃), which is not used in borazine synthesis.

Option D, $$NH_3 + B_2O_3$$, while B₂O₃ can be used in some preparations, does not match the required stoichiometry of 3 equivalents.

Therefore, X = B₂H₆ and Y = NH₃, which corresponds to Option B.

Stannane is $$SnH_4$$, the hydride of tin (Sn). Since tin is a Group 14 element in the p-block, $$SnH_4$$ is formed by covalent bonding between Sn and H atoms, making it a molecular hydride. Hence, Statement I is true.

Now, tin in $$SnH_4$$ has four bonding pairs and no lone pairs on the central atom (Sn uses $$sp^3$$ hybridisation). According to VSEPR theory, four bonding pairs with no lone pairs adopt a tetrahedral geometry, not a planar geometry. Hence, Statement II is false.

Hence, the correct answer is Option C.

We need to identify the correct statements about $$B_2H_6$$ (diborane).

Analyzing each statement:

Statement (A): "All B-H bonds are equivalent" - This is incorrect. In $$B_2H_6$$, there are two types of B-H bonds:

- 4 terminal B-H bonds (2-centre-2-electron bonds, shorter, ~1.19 Å)

- 2 bridging B-H-B bonds (3-centre-2-electron bonds, longer, ~1.33 Å)

Statement (B): "There are four 3-centre-2-electron bonds" - This is incorrect. There are only two 3-centre-2-electron (3c-2e) bonds in $$B_2H_6$$, corresponding to the two bridging B-H-B bonds.

Statement (C): "$$B_2H_6$$ is a Lewis acid" - This is correct. Boron in $$B_2H_6$$ is electron-deficient (has an incomplete octet). It readily accepts electron pairs from Lewis bases like $$NH_3$$, $$CO$$, etc.

Statement (D): "$$B_2H_6$$ can be synthesized from both $$BF_3$$ and $$NaBH_4$$" - This is correct.

$$3NaBH_4 + 4BF_3 \rightarrow 2B_2H_6 + 3NaBF_4$$

This is a standard industrial synthesis of diborane.

Statement (E): "$$B_2H_6$$ is a planar molecule" - This is incorrect. $$B_2H_6$$ has a non-planar structure. The four terminal H atoms and the two B atoms are in one plane, while the two bridging H atoms lie above and below this plane.

Therefore, only statements (C) and (D) are correct.

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

We need to match each silicon compound with its corresponding product based on the number of hydrolysable groups (OH or Cl groups that can condense).

A. $$(CH_3)_4Si$$ — This compound has no hydroxyl group and no hydrolysable group. It has 4 methyl groups attached to Si. This is simply a silane (tetramethylsilane). So A matches with III (Silane).

B. $$(CH_3)Si(OH)_3$$ — This compound has one methyl group and three $$-OH$$ groups on Si. With three functional groups available for condensation, it can cross-link in multiple directions to form a 2D sheet-like silicone polymer. So B matches with IV (2D-Silicone).

C. $$(CH_3)_2Si(OH)_2$$ — This compound has two methyl groups and two $$-OH$$ groups on Si. Two hydroxyl groups allow condensation in two directions, forming a long chain silicone polymer. So C matches with I (Chain Silicone).

D. $$(CH_3)_3Si(OH)$$ — This compound has three methyl groups and only one $$-OH$$ group on Si. With only one hydroxyl group, two such molecules can condense to form a dimer. So D matches with II (Dimeric Silicone).

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

Hence, the correct answer is Option D.

When lead nitrate is heated at 673 K:

$$2Pb(NO_3)_2 \xrightarrow{673K} 4NO_2 + 2PbO + O_2$$

So compound A is $$NO_2$$ (nitrogen dioxide).

When $$NO_2$$ dimerizes:

$$2NO_2 \xrightarrow{Dimerise} N_2O_4$$

Compound B is $$N_2O_4$$ (dinitrogen tetroxide).

The structure of $$N_2O_4$$ consists of two $$NO_2$$ units connected by an N-N bond:

$$O_2N - NO_2$$

Each nitrogen atom is bonded to two oxygen atoms and to the other nitrogen atom. There are no bridged (bridging) oxygen atoms in this structure. All oxygen atoms are terminal, bonded to only one nitrogen atom each.

The number of bridged oxygen atoms = 0

Hence, the correct answer is Option A.

We need to evaluate both statements about the diagonal relationship between Be and Al.

Statement I: "The chlorides of Be and Al have Cl-bridged structure. Both are soluble in organic solvents and act as Lewis bases."

Analysis of Statement I:

It is true that $$BeCl_2$$ and $$AlCl_3$$ both have chlorine-bridged structures. $$BeCl_2$$ has a polymeric chain with Cl bridges, and $$Al_2Cl_6$$ has a dimeric structure with Cl bridges.

It is also true that both are soluble in organic solvents due to their covalent character.

However, the statement says they act as Lewis bases. This is incorrect. Both $$BeCl_2$$ and $$AlCl_3$$ are electron-deficient species and act as Lewis acids (electron pair acceptors), not Lewis bases.

Therefore, Statement I is false.

Statement II: "Hydroxides of Be and Al dissolve in excess alkali to give beryllate and aluminate ions."

Analysis of Statement II:

Both $$Be(OH)_2$$ and $$Al(OH)_3$$ are amphoteric hydroxides. They dissolve in excess alkali (NaOH):

$$ Be(OH)_2 + 2NaOH \rightarrow Na_2[Be(OH)_4] \text{ (sodium beryllate)} $$

$$ Al(OH)_3 + NaOH \rightarrow Na[Al(OH)_4] \text{ (sodium aluminate)} $$

Therefore, Statement II is true.

Conclusion: Statement I is false but Statement II is true.

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

We need to evaluate the assertion that boric acid ($$H_3BO_3$$) is a weak acid. Since its acid dissociation constant is $$K_a = 5.8 \times 10^{-10}$$, the assertion is correct.

Reason R states that boric acid cannot release $$H^+$$ directly; instead, it accepts $$OH^-$$ from water and releases $$H^+$$ ion. This is also correct because boric acid acts as a Lewis acid rather than a Bronsted acid.

Next, boric acid reacts with water as follows:

$$B(OH)_3 + H_2O \rightarrow [B(OH)_4]^- + H^+$$

From this reaction, the boron atom in $$B(OH)_3$$, having an empty p-orbital, accepts $$OH^-$$ from water, releasing $$H^+$$ into solution. This indirect mechanism explains why boric acid is weak.

Is R the correct explanation of A?

Yes. The reason that boric acid is weak (Assertion A) is exactly because it cannot release $$H^+$$ on its own and must accept $$OH^-$$ from water (Reason R). This indirect mechanism makes it a weak acid.

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