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: $$A_2O_3$$.

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.

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): $$NH_4Cl$$.

{image}

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$$_3$$.

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 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.

Maximum allowed concentration in ppm for F− ,So4 -2, NO3-, and zn are 2, 5,50,200 respectively.

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.

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:

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).

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:

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.

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).

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.

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.

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.

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.

Let us match each reaction with its industrial catalyst:

(A) $$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$$

This is the Haber process for ammonia synthesis. The catalyst used is finely divided iron with promoters potassium oxide and aluminium oxide:

$$\text{Catalyst: } Fe_xO_y + K_2O + Al_2O_3 \rightarrow \textbf{(III)}$$

(B) $$CO(g) + 3H_2(g) \rightarrow CH_4(g) + H_2O(g)$$

This is the methanation reaction (Sabatier reaction). The catalyst used is nickel:

$$\text{Catalyst: Ni} \rightarrow \textbf{(IV)}$$

(C) $$CO(g) + H_2(g) \rightarrow HCHO(g)$$

The synthesis of formaldehyde from synthesis gas uses a copper catalyst:

$$\text{Catalyst: Cu} \rightarrow \textbf{(I)}$$

(D) $$CO(g) + 2H_2(g) \rightarrow CH_3OH(g)$$

This is methanol synthesis. The catalyst used is a mixture of copper, zinc oxide, and chromium oxide:

$$\text{Catalyst: Cu/ZnO-Cr}_2\text{O}_3 \rightarrow \textbf{(II)}$$

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

Therefore, the correct answer is Option C.

We need to identify which boron compound gives a strongly basic aqueous solution.

Analyzing each compound shows that sodium borohydride (NaBH₄) hydrolyzes slowly in water and, although a strong reducing agent, does not produce a strongly basic solution, while lithium borohydride (LiBH₄) behaves similarly without imparting strong basicity. Diborane (B₂H₆) reacts with water to form boric acid (H₃BO₃), which is weakly acidic rather than basic. Borax (Na₂B₄O₇), when dissolved in water:

$$\text{Na}_2\text{B}_4\text{O}_7 + 7\text{H}_2\text{O} \rightarrow 2\text{NaOH} + 4\text{H}_3\text{BO}_3$$

is the salt of a strong base (NaOH) and a weak acid (H₃BO₃), so its aqueous solution is strongly basic. Therefore, Na₂B₄O₇ (borax) gives a strongly basic aqueous solution. The answer is Option D: Na₂B₄O₇.

We are asked to identify the compound responsible for the blue coloured bead when borax is heated with CoO on a platinum loop.

The borax bead test is a classic qualitative analysis technique. When borax ($$Na_2B_4O_7 \cdot 10H_2O$$) is heated, it first loses water and swells up, then melts into a transparent glassy bead. The key reaction during heating is:

$$Na_2B_4O_7 \xrightarrow{\Delta} 2NaBO_2 + B_2O_3$$

The $$B_2O_3$$ (boron trioxide) produced is a Lewis acid and reacts with metal oxides to form metaborates. When CoO is introduced into the hot borax bead, the following reaction occurs:

$$CoO + B_2O_3 \rightarrow Co(BO_2)_2$$

The product $$Co(BO_2)_2$$ is cobalt(II) metaborate, which imparts the characteristic deep blue colour to the bead. This blue colour is so distinctive that the borax bead test is considered a reliable confirmatory test for cobalt.

Among the given options, $$B_2O_3$$ (Option A) is colourless and merely serves as the acidic flux. Options C and D represent other possible boron-cobalt compounds, but the standard product formed in the borax bead test is the metaborate $$Co(BO_2)_2$$.

Hence, the correct answer is Option B.

We need to identify which compound gives nitrogen gas ($$N_2$$) on thermal decomposition. Option A: $$NaNO_2$$ (Sodium nitrite) decomposes on heating but does not produce $$N_2$$ gas directly; it can decompose to give $$Na_2O$$, $$NO$$, and $$NO_2$$.

Option B: $$NaNO_3$$ (Sodium nitrate) decomposes on heating as:

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

This gives oxygen gas, not nitrogen gas.

Option C: $$Ba(N_3)_2$$ (Barium azide) decomposes on heating as:

$$Ba(N_3)_2 \xrightarrow{\Delta} Ba + 3N_2$$

This gives pure nitrogen gas ($$N_2$$).

Option D: $$Ba(NO_3)_2$$ (Barium nitrate) decomposes on heating as:

$$2Ba(NO_3)_2 \xrightarrow{\Delta} 2BaO + 4NO_2 + O_2$$

This gives $$NO_2$$ and $$O_2$$, not $$N_2$$.

Azides ($$N_3^-$$) are compounds that contain three nitrogen atoms bonded together; on thermal decomposition, they readily release nitrogen gas ($$N_2$$). Barium azide is a well-known source of pure nitrogen gas upon heating. The correct answer is Option C: $$Ba(N_3)_2$$.

We need to evaluate both statements about oxides of Group 15 elements.

Statement I: The pentavalent oxide $$E_2O_5$$ is less acidic than the trivalent oxide $$E_2O_3$$ of the same element.

In general, the higher the oxidation state of the central atom, the more acidic the oxide. This is because a higher oxidation state means a greater positive charge on the central atom, which makes it attract electron density from the O-H bond more strongly, facilitating the release of $$H^+$$ ions.

For Group 15 elements:

- In $$E_2O_5$$, the oxidation state of E is +5

- In $$E_2O_3$$, the oxidation state of E is +3

Since higher oxidation state leads to more acidic character, $$E_2O_5$$ is more acidic than $$E_2O_3$$, not less.

Therefore, Statement I is FALSE.

Statement II: The acidic character of trivalent oxide $$E_2O_3$$ decreases down the group.

As we go down Group 15 (N → P → As → Sb → Bi):

- The electronegativity of the element decreases

- The metallic character increases

- The oxides change from acidic → amphoteric → basic

For trivalent oxides specifically:

- $$N_2O_3$$ — acidic

- $$P_2O_3$$ — acidic

- $$As_2O_3$$ — amphoteric

- $$Sb_2O_3$$ — amphoteric

- $$Bi_2O_3$$ — basic

The acidic character clearly decreases down the group.

Therefore, Statement II is TRUE.

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

When an aqueous solution of ammonium chloride ($$NH_4Cl$$) is treated with sodium nitrite ($$NaNO_2$$), the following reaction takes place:

$$NH_4Cl + NaNO_2 \rightarrow NaCl + N_2 + 2H_2O$$

This is a well-known reaction used in the laboratory for the preparation of pure nitrogen gas ($$N_2$$). The ammonium ion ($$NH_4^+$$) acts as a mild reducing agent and the nitrite ion ($$NO_2^-$$) acts as a mild oxidizing agent. The nitrogen in $$NH_4^+$$ (oxidation state -3) and the nitrogen in $$NO_2^-$$ (oxidation state +3) combine to form $$N_2$$ (oxidation state 0). This is a comproportionation (disproportionation) reaction where nitrogen in two different oxidation states combines to form nitrogen in an intermediate oxidation state.

The gas produced is $$N_2$$, which matches Option B.

The answer is $$\boxed{\text{Option B: } N_2}$$.

We need to find the products when white phosphorus ($$P_4$$) reacts with thionyl chloride ($$SOCl_2$$).

Thionyl chloride ($$SOCl_2$$) is a chlorinating agent. When it reacts with white phosphorus, it converts phosphorus into a phosphorus chloride and undergoes reduction itself.

The balanced reaction is:

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

Let us verify this by analyzing the reaction. White phosphorus ($$P_4$$) has phosphorus in the 0 oxidation state. Thionyl chloride provides chlorine atoms to phosphorus. Each phosphorus atom accepts three chlorine atoms to form $$PCl_3$$ (phosphorus trichloride), where phosphorus goes to the +3 oxidation state.

Now, the sulfur in $$SOCl_2$$ is in the +4 oxidation state (since O is $$-2$$ and each Cl is $$-1$$, giving $$S + (-2) + 2(-1) = 0$$, so $$S = +4$$). In the products, sulfur appears in $$SO_2$$ (oxidation state +4) and $$S_2Cl_2$$ (oxidation state +1). The sulfur in $$S_2Cl_2$$ has been reduced from +4 to +1, while phosphorus has been oxidized from 0 to +3.

So the products of the reaction are $$PCl_3$$, $$SO_2$$, and $$S_2Cl_2$$.

Hence, the correct answer is Option B.

We need to determine the products formed when concentrated $$HNO_3$$ reacts with iodine ($$I_2$$).

Concentrated nitric acid is a strong oxidizing agent. When it reacts with iodine, it oxidizes iodine to its highest common oxide acid form. Iodine (oxidation state 0) gets oxidized to iodic acid $$HIO_3$$ (where iodine has an oxidation state of +5).

The balanced chemical equation for this reaction is:

$$I_2 + 10HNO_3 \text{(conc.)} \rightarrow 2HIO_3 + 10NO_2 + 4H_2O$$

Let us verify the balance. On the left side: 2 iodine atoms, 10 nitrogen atoms, 10 hydrogen atoms, and 30 oxygen atoms. On the right side: 2 iodine atoms (in $$2HIO_3$$), 10 nitrogen atoms (in $$10NO_2$$), 10 hydrogen atoms (2 in $$2HIO_3$$ and 8 in $$4H_2O$$), and oxygen: 6 (in $$2HIO_3$$) + 20 (in $$10NO_2$$) + 4 (in $$4H_2O$$) = 30 oxygen atoms. The equation is balanced.

The nitrogen in $$HNO_3$$ (oxidation state +5) is reduced to $$NO_2$$ (oxidation state +4), confirming that $$HNO_3$$ acts as the oxidizing agent while $$I_2$$ is oxidized.

So the products are $$HIO_3$$, $$NO_2$$, and $$H_2O$$.

Hence, the correct answer is Option C.

We need to determine how many of the given oxides are amphoteric.

The given oxides are: $$Na_2O$$, $$As_2O_3$$, $$N_2O$$, $$NO$$, and $$Cl_2O_7$$.

1. $$Na_2O$$ (Sodium oxide): Na is a strongly electropositive alkali metal. $$Na_2O$$ is a basic oxide. It reacts with water to form NaOH and reacts with acids but not with bases.

2. $$As_2O_3$$ (Arsenic trioxide): Arsenic is a metalloid. $$As_2O_3$$ is an amphoteric oxide. It reacts with both acids and bases:

$$As_2O_3 + 6HCl \rightarrow 2AsCl_3 + 3H_2O$$ (reacts with acid)

$$As_2O_3 + 6NaOH \rightarrow 2Na_3AsO_3 + 3H_2O$$ (reacts with base)

3. $$N_2O$$ (Nitrous oxide): This is a neutral oxide. It does not react with either acids or bases.

4. $$NO$$ (Nitric oxide): This is a neutral oxide. It does not react with either acids or bases.

5. $$Cl_2O_7$$ (Dichlorine heptoxide): This is the anhydride of perchloric acid ($$HClO_4$$). It is a strongly acidic oxide.

Therefore, only $$As_2O_3$$ is amphoteric. The number of amphoteric oxides = 1.

The correct answer is Option B: 1.

Fluorine forms one oxoacid. It forms only one oxoacid, which is hypofluorous acid ($$HOF$$, also called fluoric(I) acid). All other halogens (Cl, Br, I) form multiple oxoacids (e.g., $$HOCl, HClO_2, HClO_3, HClO_4$$). The assertion is correct.

Fluorine has the smallest size amongst all halogens and is highly electronegative. Fluorine is the smallest halogen (atomic radius ~ 64 pm) and has the highest electronegativity (3.98 on the Pauling scale). This is factually correct.

Fluorine forms only one oxoacid because:

1. Due to its very small size, fluorine cannot accommodate multiple oxygen atoms around it.

2. Due to its high electronegativity (higher than oxygen), fluorine cannot exhibit positive oxidation states. In oxoacids of other halogens, the halogen shows positive oxidation states (+1, +3, +5, +7), but fluorine cannot do this because it is more electronegative than oxygen.

3. Fluorine lacks d-orbitals, so it cannot expand its octet to form higher oxoacids.

Therefore, both the small size and high electronegativity of fluorine directly explain why it forms only one oxoacid. The reason is the correct explanation of the assertion.

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

We need to determine how many of the given nitrogen oxides — N$$_2$$O, N$$_2$$O$$_3$$, N$$_2$$O$$_4$$, and N$$_2$$O$$_5$$ — contain an N-N bond. N$$_2$$O has the structure N=N=O (linear), so it contains an N-N bond.

N$$_2$$O$$_3$$ has the structure O=N-N=O with an additional oxygen on one nitrogen: ON-NO$$_2$$, so it also contains an N-N bond. N$$_2$$O$$_4$$ is the dimer of NO$$_2$$, with structure O$$_2$$N-NO$$_2$$, and it too contains an N-N bond.

N$$_2$$O$$_5$$ has the structure O$$_2$$N-O-NO$$_2$$, where the two nitrogen atoms are connected through an oxygen bridge; there is no direct N-N bond.

N$$_2$$O, N$$_2$$O$$_3$$, and N$$_2$$O$$_4$$ have N-N bonds, which makes three compounds. Hence, the correct answer is Option C.

We need to explain why dinitrogen ($$N_2$$) and dioxygen ($$O_2$$), despite being the main constituents of the atmosphere, do not react with each other under normal atmospheric conditions to form oxides of nitrogen.

The reaction between $$N_2$$ and $$O_2$$ to form nitric oxide is:

$$N_2(g) + O_2(g) \rightarrow 2NO(g)$$

This reaction is highly endothermic ($$\Delta H \approx +180 \text{ kJ/mol}$$). The reason is that the $$N \equiv N$$ triple bond is extremely strong (bond dissociation energy $$\approx 941 \text{ kJ/mol}$$), and breaking it requires a very large amount of energy. Similarly, the $$O = O$$ bond energy is about 498 kJ/mol. The energy released in forming the N-O bonds in NO is not sufficient to compensate for the energy required to break both the $$N \equiv N$$ and $$O = O$$ bonds.

Because the reaction is endothermic and has a very high activation energy, it requires extremely high temperatures (above 2000°C) to proceed at a significant rate. Such high temperatures occur in lightning strikes and in internal combustion engines, which is why small amounts of NO are produced under those conditions, but not under normal atmospheric conditions.

Let us examine the other options:

Option A states that $$N_2$$ is unreactive under atmospheric conditions — while $$N_2$$ is relatively inert, the fundamental reason is the thermodynamics of the reaction (endothermic nature), not just the inertness of nitrogen alone.

Option B states that oxides of nitrogen are unstable — this is not correct in general; $$NO_2$$ and other nitrogen oxides do exist stably.

Option C states the reaction can occur with a catalyst — while catalysts can lower activation energy, the key reason they don't react is the high temperature requirement due to the endothermic nature.

Hence, the correct answer is Option D.

When white phosphorus ($$P_4$$) is heated with concentrated NaOH solution, it undergoes disproportionation.

The reaction of white phosphorus with concentrated NaOH solution is:

$$P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$$

In this reaction:

- Phosphorus in $$P_4$$ has an oxidation state of 0.

- In $$PH_3$$ (phosphine), phosphorus has an oxidation state of $$-3$$ (reduction).

- In $$NaH_2PO_2$$ (sodium hypophosphite), phosphorus has an oxidation state of $$+1$$ (oxidation).

This is a disproportionation reaction where phosphorus is simultaneously oxidised and reduced.

The main products are phosphine ($$PH_3$$) and sodium hypophosphite ($$NaH_2PO_2$$).

Hence, the correct answer is Option D.

We need to determine the most stable trihalide of nitrogen among $$NF_3$$, $$NCl_3$$, $$NBr_3$$, and $$NI_3$$.

Key factors affecting stability of nitrogen trihalides:

1. Bond strength: The N-X bond strength decreases as the size of the halogen increases:

$$N-F > N-Cl > N-Br > N-I$$

This is because nitrogen is a small atom, and it forms the strongest bond with the smallest halogen (fluorine) due to effective orbital overlap.

2. Bond energy values:

N-F bond energy $$\approx$$ 272 kJ/mol

N-Cl bond energy $$\approx$$ 200 kJ/mol

N-Br bond energy $$\approx$$ 160 kJ/mol

N-I bond energy $$\approx$$ 140 kJ/mol

3. Electronegativity effect: Fluorine is the most electronegative element. The high electronegativity of F stabilizes the N-F bond by making it strongly polar. This prevents the lone pairs on F from causing significant repulsion.

4. Trend in stability:

$$NF_3 > NCl_3 > NBr_3 > NI_3$$

In fact, $$NI_3$$ is so unstable that it is explosively sensitive when dry.

The correct answer is Option A: $$NF_3$$.

We are asked to identify the oxide of nitrogen that damages plant leaves and retards photosynthesis.

At high altitudes, the high temperature from lightning or combustion causes nitrogen and oxygen to combine, forming various nitrogen oxides. Among these, nitrogen dioxide ($$NO_2$$) is a reddish-brown, pungent gas that is a major air pollutant.

$$NO_2$$ is particularly harmful to vegetation because it dissolves in the moisture on leaf surfaces to form nitric acid ($$HNO_3$$) and nitrous acid ($$HNO_2$$). These acids damage the leaf tissue, destroy chlorophyll, and thereby retard the process of photosynthesis. This leads to visible symptoms such as leaf bleaching and necrosis.

Now, NO (nitric oxide, Option A) is a colourless gas that is less directly toxic to plants compared to $$NO_2$$; it is primarily a precursor that gets oxidised to $$NO_2$$ in the atmosphere. $$NO_3^-$$ (Option B) is the nitrate ion, which is actually a plant nutrient, not a pollutant that damages leaves. $$NO_2^-$$ (Option D) is the nitrite ion, which exists in solution form and is not an atmospheric pollutant that directly damages plant leaves.

Hence, the correct answer is Option C.

We need to find what type of interhalogen compound is formed when bromine reacts with excess fluorine, and classify it by its analogous halate type.

When bromine reacts with excess fluorine, it undergoes the reaction $$Br_2 + 5F_2 \rightarrow 2BrF_5$$ resulting in the formation of $$BrF_5$$ (bromine pentafluoride), where bromine is in the +5 oxidation state.

Interhalogen compounds are classified analogous to oxoacid salts based on the oxidation state of the central halogen:

- Hypohalite: +1 oxidation state (like $$XY$$, e.g., $$ClF$$)

- Halite: +3 oxidation state (like $$XY_3$$, e.g., $$ClF_3$$)

- Halate: +5 oxidation state (like $$XY_5$$, e.g., $$BrF_5$$)

- Perhalate: +7 oxidation state (like $$XY_7$$, e.g., $$IF_7$$)

Since $$BrF_5$$ has bromine in the +5 oxidation state, it is classified as a halate.

Hence, the correct answer is Option B.

We need to find which oxoacid of sulphur has sulphur in two different oxidation states.

In H$$_2$$S$$_2$$O$$_3$$ (thiosulphuric acid), the structure is derived from H$$_2$$SO$$_4$$ by replacing one oxygen with a sulphur atom. The two sulphur atoms are not equivalent — one is in the central position bonded to oxygen atoms, and the other is the terminal sulphur replacing oxygen. For the central sulphur, it is bonded to 3 oxygen atoms and 1 sulphur atom, giving it an oxidation state of +5. For the terminal sulphur, it replaces an oxygen atom (O$$^{2-}$$), so its oxidation state is −1. Verification: (+1)$$\times$$2 + (+5) + (−1) + (−2)$$\times$$3 = 2 + 5 − 1 − 6 = 0 $$\checkmark$$. Sulphur exists in two different oxidation states (+5 and −1).

In H$$_2$$S$$_2$$O$$_6$$ (dithionic acid), the structure is HO$$_3$$S-SO$$_3$$H and both sulphur atoms are in identical environments. Using the formula: 2(+1) + 2x + 6(−2) = 0 $$\Rightarrow$$ 2 + 2x − 12 = 0 $$\Rightarrow$$ x = +5, so both sulphur atoms have oxidation state +5 and only one oxidation state.

In H$$_2$$S$$_2$$O$$_7$$ (pyrosulphuric/disulphuric acid), the structure is HO-SO$$_2$$-O-SO$$_2$$-OH and both sulphur atoms are in identical environments. Using the formula: 2(+1) + 2x + 7(−2) = 0 $$\Rightarrow$$ 2 + 2x − 14 = 0 $$\Rightarrow$$ x = +6, so both sulphur atoms have oxidation state +6 and only one oxidation state.

In H$$_2$$S$$_2$$O$$_8$$ (peroxodisulphuric acid/Marshall's acid), the structure is HO-SO$$_2$$-O-O-SO$$_2$$-OH and it contains a peroxide (O-O) linkage. Accounting for the peroxide oxygen atoms (each −1): 2(+1) + 2x + 6(−2) + 2(−1) = 0 $$\Rightarrow$$ 2 + 2x − 12 − 2 = 0 $$\Rightarrow$$ x = +6, so both sulphur atoms have oxidation state +6 and only one oxidation state.

Only H$$_2$$S$$_2$$O$$_3$$ has sulphur in two different oxidation states (+5 and −1). Hence, the correct answer is Option A.

We need to identify which elemental form is NOT present in the enamel of teeth.

Composition of tooth enamel:

Tooth enamel is primarily composed of hydroxyapatite, which has the formula:

$$Ca_5(PO_4)_3OH$$ or equivalently $$Ca_{10}(PO_4)_6(OH)_2$$

In fluoridated enamel, some hydroxyl groups are replaced by fluoride ions, forming fluorapatite:

$$Ca_5(PO_4)_3F$$

Analyzing each option:

Option A: $$Ca^{2+}$$ - Calcium is present as $$Ca^{2+}$$ ions in hydroxyapatite. This IS present in enamel.

Option B: $$P^{3+}$$ - In the phosphate group $$PO_4^{3-}$$, phosphorus is in the +5 oxidation state, not +3. $$P^{3+}$$ is NOT present in tooth enamel.

Option C: $$F^{-}$$ - Fluoride ions are present in fluorapatite, which strengthens the enamel. This IS present in enamel.

Option D: $$P^{5+}$$ - Phosphorus exists in the +5 oxidation state in the phosphate ion $$PO_4^{3-}$$. This IS present in enamel.

The correct answer is Option B: $$P^{3+}$$.

We need to find which oxoacid of phosphorus has the highest number of oxygen atoms in its chemical formula.

Writing the chemical formulae of each oxoacid yields the following:

Option A - Pyrophosphorous acid: $$H_4P_2O_5$$ — contains 5 oxygen atoms.

Option B - Hypophosphoric acid: $$H_4P_2O_6$$ — contains 6 oxygen atoms.

Option C - Phosphoric acid: $$H_3PO_4$$ — contains 4 oxygen atoms.

Option D - Pyrophosphoric acid: $$H_4P_2O_7$$ — contains 7 oxygen atoms.

Comparing the number of oxygen atoms confirms that Pyrophosphoric acid ($$H_4P_2O_7$$) has the highest number of oxygen atoms (7). It is formed by the condensation of two molecules of phosphoric acid with loss of one water molecule:

$$ 2H_3PO_4 \xrightarrow{\Delta} H_4P_2O_7 + H_2O $$

The correct answer is Option D: Pyrophosphoric acid.

We need to explain why $$PCl_5$$ exists but $$NCl_5$$ does not.

To form five bonds (as in $$PCl_5$$), the central atom needs five orbitals for bonding, which requires $$sp^3d$$ hybridization. This hybridization uses one d-orbital.

Phosphorus (P): Electronic configuration is $$[Ne]\, 3s^2\, 3p^3$$. Phosphorus is in the third period and has access to vacant 3d orbitals. It can undergo $$sp^3d$$ hybridization to form 5 bonds, giving $$PCl_5$$ a trigonal bipyramidal structure.

Nitrogen (N): Electronic configuration is $$[He]\, 2s^2\, 2p^3$$. Nitrogen is in the second period and has only 2s and 2p orbitals in its valence shell. There are no 2d orbitals available (d-orbitals start from the 3rd shell). Therefore, nitrogen cannot expand its octet and cannot form 5 bonds.

Since nitrogen lacks vacant d-orbitals, it can form a maximum of 4 bonds and cannot accommodate 5 chlorine atoms.

The correct answer is Option A: N does not have vacant d-orbital.

We need to count the number of acidic oxides from the given list: $$NO$$, $$N_2O$$, $$B_2O_3$$, $$N_2O_5$$, $$CO$$, $$SO_3$$, $$P_4O_{10}$$.

To begin with, $$NO$$ (Nitric oxide) behaves as a neutral oxide since it does not react with water to form an acid or a base.

Likewise, $$N_2O$$ (Nitrous oxide) is also a neutral oxide and remains unreactive with water in terms of acid or base formation.

On the other hand, $$B_2O_3$$ (Boron trioxide) is an acidic oxide and reacts with water to produce boric acid according to $$B_2O_3 + 3H_2O \rightarrow 2H_3BO_3$$.

Similarly, $$N_2O_5$$ (Dinitrogen pentoxide) is acidic and yields nitric acid when combined with water: $$N_2O_5 + H_2O \rightarrow 2HNO_3$$.

In contrast, $$CO$$ (Carbon monoxide) is a neutral oxide and does not form an acid upon reaction with water.

Meanwhile, $$SO_3$$ (Sulphur trioxide) is an acidic oxide that produces sulphuric acid in water: $$SO_3 + H_2O \rightarrow H_2SO_4$$.

Finally, $$P_4O_{10}$$ (Phosphorus pentoxide) also exhibits acidic behavior by forming phosphoric acid: $$P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4$$.

From these classifications, the acidic oxides are $$B_2O_3$$, $$N_2O_5$$, $$SO_3$$, and $$P_4O_{10}$$, which gives a total of 4 acidic oxides.

Therefore, the correct answer is Option B.

We need to identify the gas evolved in each reaction.

A. $$(NH_4)_2Cr_2O_7 \xrightarrow{\Delta}$$

Ammonium dichromate decomposes on heating:

$$(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} Cr_2O_3 + N_2 + 4H_2O$$

Gas evolved: $$N_2$$

$$\Rightarrow$$ A matches with II

B. $$KMnO_4 + HCl \to$$

Potassium permanganate reacts with concentrated HCl:

$$2KMnO_4 + 16HCl \to 2KCl + 2MnCl_2 + 5Cl_2 + 8H_2O$$

Gas evolved: $$Cl_2$$

$$\Rightarrow$$ B matches with IV

C. $$Al + NaOH + H_2O \to$$

Aluminium dissolves in aqueous NaOH:

$$2Al + 2NaOH + 2H_2O \to 2NaAlO_2 + 3H_2$$

Gas evolved: $$H_2$$

$$\Rightarrow$$ C matches with I

D. $$NaNO_3 \xrightarrow{\Delta}$$

Sodium nitrate decomposes on strong heating:

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

Gas evolved: $$O_2$$

$$\Rightarrow$$ D matches with III

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

Therefore, the correct answer is Option C.

We need to find the number of non-ionisable protons in product B.

When ethanol reacts with phosphorus trichloride, $$3C_2H_5OH + PCl_3 \rightarrow 3C_2H_5Cl + H_3PO_3$$. Therefore, product A is phosphorous acid ($$H_3PO_3$$).

Since phosphorous acid exists in its tautomeric form $$HP(O)(OH)_2$$, it contains:

- One $$P=O$$ double bond

- Two $$P-OH$$ bonds (ionisable protons)

- One $$P-H$$ bond (non-ionisable proton)

Subsequently, when $$H_3PO_3$$ reacts with $$PCl_3$$, the hydroxyl groups undergo condensation with chloride atoms, releasing HCl and forming a P-O-P linkage. The product is pyrophosphorous acid ($$H_4P_2O_5$$).

Pyrophosphorous acid consists of two phosphorus atoms joined by a bridging oxygen (P-O-P). Each phosphorus atom has:

- One $$P=O$$ bond

- One $$P-OH$$ bond (ionisable proton)

- One $$P-H$$ bond (non-ionisable proton)

- Shared bridging oxygen connecting to the other P

The molecular formula $$H_4P_2O_5$$ accounts for 2 OH protons (ionisable) and 2 P-H protons (non-ionisable), giving 4 total hydrogen atoms.

Therefore, the number of non-ionisable protons in product B is $$\boxed{2}$$.

We are asked to determine how many of the given sulphur-based oxoacids — $$H_2SO_3$$, $$H_2SO_4$$, $$H_2S_2O_8$$, and $$H_2S_2O_7$$ — contain a peroxo (O-O) bond.

A peroxo bond is a direct oxygen-oxygen single bond (-O-O-). Let us examine the structure of each acid:

$$H_2SO_3$$ (Sulphurous acid): The structure has sulphur bonded to three oxygen atoms — two OH groups and one S=O double bond. There is no O-O bond present.

$$H_2SO_4$$ (Sulphuric acid): Sulphur is at the centre, bonded to two OH groups and two S=O double bonds. All oxygen atoms are bonded to sulphur only. There is no O-O bond.

$$H_2S_2O_8$$ (Peroxodisulphuric acid / Marshall's acid): This acid is formed by joining two $$HSO_4$$ units through a peroxo linkage. Its structure is $$HO_3S-O-O-SO_3H$$. The two sulphate-like halves are connected by a direct O-O bond. This is the peroxo bond. So $$H_2S_2O_8$$ contains a peroxo bond.

$$H_2S_2O_7$$ (Pyrosulphuric acid / Oleum acid): This acid is formed by the condensation of two molecules of $$H_2SO_4$$ with loss of water. Its structure is $$HO_3S-O-SO_3H$$, where the two sulphur atoms are connected through a single bridging oxygen atom (S-O-S). There is no O-O bond.

Among the four acids, only $$H_2S_2O_8$$ (peroxodisulphuric acid) contains a peroxo (O-O) bond. The count is therefore 1.

Hence, the correct answer is 1.

We need to find the number of ionisable hydrogens in the final product B.

$$PCl_5 + H_2O \to POCl_3 + 2HCl$$

Product A is phosphorus oxychloride ($$POCl_3$$).

$$POCl_3 + 3H_2O \to H_3PO_4 + 3HCl$$

Product B is phosphoric acid ($$H_3PO_4$$).

Phosphoric acid ($$H_3PO_4$$) is a triprotic acid with 3 ionisable hydrogen atoms, all bonded to oxygen atoms as O-H groups:

$$H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^-$$

$$H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-}$$

$$HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-}$$

The number of ionisable hydrogens in B is 3.

We need to identify which interhalogens from the given list have a square pyramidal structure.

A square pyramidal structure arises when a molecule has the general formula AX$$_5$$E$$_1$$ (5 bonding pairs and 1 lone pair), with sp$$^3$$d$$^2$$ hybridization.

First, consider ClF$$_3$$. Chlorine has 7 valence electrons, forming 3 bond pairs and retaining 2 lone pairs, which gives 5 electron pairs overall. Since the resulting AX$$_3$$E$$_2$$ arrangement corresponds to a T-shaped geometry, ClF$$_3$$ does not adopt a square pyramidal structure.

Next, IF$$_7$$ involves iodine having 7 valence electrons forming 7 bond pairs with no lone pairs, for a total of 7 electron pairs. The AX$$_7$$ arrangement corresponds to a pentagonal bipyramidal shape, so IF$$_7$$ is not square pyramidal.

Turning to BrF$$_5$$, bromine contributes 7 valence electrons, yielding 5 bond pairs and 1 lone pair to make 6 electron pairs. This matches the AX$$_5$$E$$_1$$ pattern and sp$$^3$$d$$^2$$ hybridization, resulting in a square pyramidal structure, so BrF$$_5$$ qualifies.

In contrast, BrF$$_3$$ has bromine with 7 valence electrons but forms only 3 bond pairs and 2 lone pairs—5 electron pairs in total—leading to an AX$$_3$$E$$_2$$ T-shaped geometry and thus not square pyramidal.

Considering the dimer I$$_2$$Cl$$_6$$, each iodine is bonded to four bridging chlorine atoms and retains one lone pair, producing a square planar arrangement around each iodine and giving rise to an edge-shared bioctahedral framework. Consequently, I$$_2$$Cl$$_6$$ is not square pyramidal.

For IF$$_5$$, iodine’s 7 valence electrons form 5 bond pairs plus 1 lone pair, totaling 6 electron pairs. The resulting AX$$_5$$E$$_1$$ configuration with sp$$^3$$d$$^2$$ hybridization yields a square pyramidal structure, so IF$$_5$$ is included.

As for ClF, the simple diatomic molecule has a linear geometry and obviously does not have a square pyramidal shape.

Finally, ClF$$_5$$ involves chlorine with 7 valence electrons forming 5 bond pairs and 1 lone pair, for 6 electron pairs in an AX$$_5$$E$$_1$$ arrangement. Therefore, ClF$$_5$$ also adopts a square pyramidal structure.

From the above analysis, the interhalogens with square pyramidal structures are BrF$$_5$$, IF$$_5$$, and ClF$$_5$$. Therefore, the number of interhalogens with square pyramidal structure = 3.

We are given an inorganic compound 'X' that satisfies several chemical tests. First, it produces brown fumes with concentrated H$$_2$$SO$$_4$$, which indicates the presence of nitrate ions (NO$$_3^-$$) — concentrated H$$_2$$SO$$_4$$ reacts with nitrates to release brown NO$$_2$$ gas. Second, the dark brown ring test with FeSO$$_4$$ and concentrated H$$_2$$SO$$_4$$ is the classic brown ring test confirming the presence of nitrate ions.

These two tests confirm compound X is a nitrate salt. The options remaining are Cu(NO$$_3$$)$$_2$$, Co(NO$$_3$$)$$_2$$, and Pb(NO$$_3$$)$$_2$$.

Third, when a solution of X in dilute HCl is treated with H$$_2$$S gas, a precipitate Y forms. In the presence of dilute HCl (acidic conditions), only group II ions precipitate as sulfides with H$$_2$$S. Cu$$^{2+}$$ forms black CuS, Pb$$^{2+}$$ forms black PbS, while Co$$^{2+}$$ precipitates only in ammoniacal medium (group IV). This rules out Co(NO$$_3$$)$$_2$$.

Fourth, precipitate Y treated with concentrated HNO$$_3$$ followed by excess NH$$_4$$OH gives a deep blue coloured solution. The deep blue colour with excess ammonia is the characteristic reaction of the tetraamminecopper(II) complex, $$[\text{Cu(NH}_3)_4]^{2+}$$. Lead sulfide treated with concentrated HNO$$_3$$ gives Pb$$^{2+}$$, which does not form a deep blue complex with ammonia.

Therefore, the precipitate Y is CuS, and compound X is $$\text{Cu(NO}_3)_2$$, which is option 3.

We need to identify the set in which the compounds have different chemical natures from each other.

Option 1: B(OH)$$_3$$ (boric acid) is a Lewis acid (acidic), and H$$_3$$PO$$_3$$ (phosphorous acid) is also acidic. Both are acids, so they have the same nature.

Option 2: B(OH)$$_3$$ is acidic (it acts as a Lewis acid, accepting electron pairs from water). Al(OH)$$_3$$ is amphoteric — it can act as both an acid and a base. Since B(OH)$$_3$$ is purely acidic and Al(OH)$$_3$$ is amphoteric, they have different natures.

Option 3: NaOH is a strong base and Ca(OH)$$_2$$ is also a base (though slightly less soluble). Both are basic in nature, so they have the same nature.

Option 4: Be(OH)$$_2$$ is amphoteric and Al(OH)$$_3$$ is also amphoteric. Both have the same nature.

The set where the compounds have different natures is option 2: B(OH)$$_3$$ (acidic) and Al(OH)$$_3$$ (amphoteric).

In Group 15, the elements are N, P, As, Sb, and Bi. Among these, the metallic character increases as we go down the group. Bismuth (Bi) is the most metallic element in Group 15.

The hydrides of Group 15 elements are $$NH_3$$, $$PH_3$$, $$AsH_3$$, $$SbH_3$$, and $$BiH_3$$. The reducing power of these hydrides increases down the group because the bond strength between the central atom and hydrogen decreases as the size of the central atom increases. Since $$BiH_3$$ (bismuthine) has the weakest Bi-H bond, it is the most easily decomposed and hence the strongest reducing agent among all Group 15 hydrides.

Therefore, the element that is a metal and forms the hydride with the strongest reducing power is Bi (Bismuth). The correct answer is Option (4): Bi.

We need to evaluate both the Assertion and the Reason.

Assertion A states that in $$TlI_3$$, which is isomorphous to $$CsI_3$$, the metal is present in the +1 oxidation state. In $$CsI_3$$, cesium is in the +1 state as $$Cs^+$$ with $$I_3^-$$ (triiodide ion). Since $$TlI_3$$ is isomorphous to $$CsI_3$$, it has the same crystal structure, meaning thallium is also in the +1 state as $$Tl^+$$ with $$I_3^-$$. This is consistent with the inert pair effect, where the 6s electrons of Tl are reluctant to participate in bonding. So Assertion A is correct.

Reason R states that Tl has fourteen f electrons in its electronic configuration. The electronic configuration of Tl (Z = 81) is $$[Xe] \, 4f^{14} \, 5d^{10} \, 6s^2 \, 6p^1$$. Indeed, Tl has 14 f electrons. So Reason R is also correct.

However, the fact that Tl has 14 f electrons is not the direct reason for Tl existing in the +1 state in $$TlI_3$$. The +1 state is due to the inert pair effect. While the poor shielding by f electrons does contribute to the enhanced inert pair effect, the Reason as stated merely notes the presence of f electrons without explaining the mechanism. Therefore, R is not the correct explanation of A.

The correct answer is Option (4): Both A and R are correct but R is NOT the correct explanation of A.

We begin by recalling the simple laboratory reaction that actually produces hydrogen peroxide from an oxide and water. The well-known reaction is with sodium peroxide. The balanced chemical equation is

$$\mathrm{Na_2O_2 + 2\,H_2O \;\longrightarrow\; 2\,NaOH + H_2O_2}$$

Here we can see dissociation and recombination step by step. First, water hydrates the peroxide ion, and because the peroxide ion $$\mathrm{O_2^{2-}}$$ is present in $$\mathrm{Na_2O_2}$$, one of its oxygen atoms combines with two hydrogen atoms drawn from two water molecules to form the $$\mathrm{OOH^-}$$ (hydroperoxide) intermediate; two such intermediates then couple to give one complete $$\mathrm{H_2O_2}$$ molecule, while the sodium cations are simultaneously balanced by hydroxide ions. Writing every ionic detail, we have

$$\mathrm{Na_2O_2 \; \xrightarrow{\;H_2O\;} \; 2\,Na^+ + O_2^{2-}}$$

$$\mathrm{O_2^{2-} + 2\,H_2O \;\longrightarrow\; 2\,OH^- + H_2O_2}$$

Now summing the ionic expressions, we arrive again at

$$\mathrm{Na_2O_2 + 2\,H_2O \;\longrightarrow\; 2\,NaOH + H_2O_2}$$

This reaction proceeds smoothly at room temperature, illustrating that $$\mathrm{Na_2O_2}$$ “gives” $$\mathrm{H_2O_2}$$ most readily simply by contact with water.

Let us briefly compare the other oxides mentioned:

$$\mathrm{BaO_2\cdot 8H_2O}$$ liberates $$\mathrm{H_2O_2}$$ only when treated with dilute acids, not merely with water, because the peroxide anion needs to be protonated while the barium ion must be converted to a soluble salt to shift the equilibrium.

$$\mathrm{SnO_2}$$ and $$\mathrm{PbO_2}$$ are higher oxides of tin and lead respectively. They do not contain the peroxide linkage $$\mathrm{-O-O-}$$, so when they encounter water they undergo hydrolysis to other oxides or hydroxides rather than forming $$\mathrm{H_2O_2}$$.

Since only $$\mathrm{Na_2O_2}$$ forms $$\mathrm{H_2O_2}$$ directly, it is the oxide that “gives $$\mathrm{H_2O_2}$$ most readily on treatment with $$\mathrm{H_2O}$$.”

Hence, the correct answer is Option A.

Disproportionation is a redox reaction in which the same element is simultaneously oxidised and reduced, giving two different products with different oxidation states.

For a species to undergo disproportionation, the element must be in an intermediate oxidation state that can go both to a higher and a lower oxidation state.

BrO$$^-$$ has Br in $$+1$$ oxidation state: it can disproportionate as $$3\text{BrO}^- \rightarrow \text{BrO}_3^- + 2\text{Br}^-$$, where Br goes to $$+5$$ and $$-1$$.

BrO$$_2^-$$ has Br in $$+3$$ oxidation state: it can disproportionate as $$2\text{BrO}_2^- \rightarrow \text{BrO}_3^- + \text{BrO}^-$$, with Br going to $$+5$$ and $$+1$$.

BrO$$_3^-$$ has Br in $$+5$$ oxidation state: it can disproportionate as $$4\text{BrO}_3^- \rightarrow 3\text{BrO}_4^- + \text{Br}^-$$, with some Br going up to $$+7$$ and one Br going down to $$-1$$.

BrO$$_4^-$$ has Br in $$+7$$ oxidation state, which is the maximum possible oxidation state for bromine. Since there is no higher oxidation state available for Br, BrO$$_4^-$$ cannot be oxidised further and therefore cannot undergo disproportionation.

The species that does NOT show disproportionation is BrO$$_4^-$$, which is option 1.

We are told that $$Al_2O_3$$ is leached with alkali (NaOH). This is the Bayer's process for extraction of aluminium.

When $$Al_2O_3$$ is treated with NaOH solution, it dissolves to form sodium aluminate: $$Al_2O_3 + 2NaOH + 3H_2O \to 2NaAl(OH)_4$$. So X is $$NaAl(OH)_4$$ (sodium aluminate).

Now, when $$CO_2$$ gas is passed through the sodium aluminate solution, the weakly acidic $$CO_2$$ neutralises the solution and precipitates aluminium hydroxide: $$NaAl(OH)_4 + CO_2 \to Al(OH)_3 + NaHCO_3$$. So Y is $$CO_2$$.

The precipitate $$Al(OH)_3$$ is the hydrated form of alumina, written as $$Al_2O_3 \cdot xH_2O$$. So Z is $$Al_2O_3 \cdot xH_2O$$.

Hence, X = $$NaAl(OH)_4$$, Y = $$CO_2$$, Z = $$Al_2O_3 \cdot xH_2O$$.

Hence, the correct answer is Option 4.

First we recall that the ease with which an ionic solid dissolves in water is governed by the thermodynamic relation

$$\Delta H_{\text{solution}} \;=\; \Delta H_{\text{lattice}} \;+\; \Delta H_{\text{hydration}},$$

where

$$\Delta H_{\text{lattice}} > 0 \quad (\text{energy required to break the crystal}),$$

and

$$\Delta H_{\text{hydration}} < 0 \quad (\text{energy released when ions are hydrated}).$$

A salt dissolves readily when the magnitude of the exothermic hydration enthalpy outweighs the endothermic lattice enthalpy, giving a small or negative $$\Delta H_{\text{solution}}$$. Conversely, very large positive lattice enthalpy or very small (less negative) hydration enthalpy will make the solid sparingly soluble.

For an alkali metal halide $$\text{M}^+\text{X}^-$$ both ions may vary in size. The lattice enthalpy is inversely proportional to the distance between the ion centres, stated qualitatively by

$$\Delta H_{\text{lattice}} \;\propto\; \dfrac{z_+\,z_-}{r_+ + r_-},$$

where $$z_+$$ and $$z_-$$ are the ionic charges (both unity in the present case) and $$r_+$$, $$r_-$$ are the ionic radii. Hence, smaller ions give larger (more positive) lattice enthalpies.

Within the alkali metal ions the order of size is

$$\text{Li}^+ \lt \text{Na}^+ \lt \text{K}^+ \lt \text{Rb}^+ \lt \text{Cs}^+,$$

while within the halide ions

$$\text{F}^- \lt \text{Cl}^- \lt \text{Br}^- \lt \text{I}^-.$$

Therefore for the fluoride series $$\text{LiF},\,\text{NaF},\,\text{KF},\,\dots$$ the compound $$\text{LiF}$$ contains the smallest cation and the smallest anion, so $$r_+ + r_-$$ is minimum and

$$\Delta H_{\text{lattice}}(\text{LiF})$$

is the largest (most positive) among all alkali metal fluorides. A very large positive lattice term requires a large negative hydration term before the salt can dissolve. Although $$\text{Li}^+$$ is indeed strongly hydrated, the extreme lattice enthalpy still dominates, giving a very small (and actually positive) value of $$\Delta H_{\text{solution}}$$. Hence $$\text{LiF}$$ is sparingly soluble in water.

Now we examine each assertion:

Option A claims that $$\text{LiF}$$ has the least negative standard enthalpy of formation among the fluorides. Because of its very high lattice enthalpy, the enthalpy of formation of $$\text{LiF}$$ is, in fact, the most negative; therefore the statement is false.

Option B says that the enthalpy of formation of alkali metal bromides becomes less negative down the group. In reality the trend is irregular; lattice enthalpy decreases, but other terms in the Born-Haber cycle change as well, so the statement cannot be generalised as written. Thus the option is incorrect.

Option C attributes the low solubility of $$\text{CsI}$$ to a high lattice enthalpy. Here both $$\text{Cs}^+$$ and $$\text{I}^-$$ are large; consequently $$r_+ + r_-$$ is large and $$\Delta H_{\text{lattice}}$$ is comparatively small. The real reason for poor solubility is the very low (less negative) hydration enthalpy of these large ions. Hence this statement is also false.

Option D states that, among the alkali metal halides, $$\text{LiF}$$ is least soluble in water. We have already reasoned that its extraordinarily high lattice enthalpy overrides the hydration energy advantage, giving the smallest solubility. This statement is true.

Therefore the only correct choice is the fourth one.

Hence, the correct answer is Option 4.

We need to evaluate each statement about diborane ($$\text{B}_2\text{H}_6$$).

Statement (a): Diborane is prepared by the oxidation of $$\text{NaBH}_4$$ with $$\text{I}_2$$. The reaction is $$2\text{NaBH}_4 + \text{I}_2 \to \text{B}_2\text{H}_6 + 2\text{NaI} + \text{H}_2$$. Here iodine acts as the oxidising agent and $$\text{NaBH}_4$$ is oxidised to produce diborane. This statement is correct.

Statement (b): Each boron atom in diborane is $$sp^3$$ hybridised (not $$sp^2$$), since each boron forms four bonds — two terminal B-H bonds and two bridging 3-centre-2-electron bonds. This statement is incorrect.

Statement (c): Diborane has two banana-shaped 3-centre-2-electron (3c-2e) bridging bonds, not one. This statement is incorrect.

Statement (d): Diborane is not planar. The four terminal hydrogen atoms and the two boron atoms lie in one plane, but the two bridging hydrogen atoms lie above and below this plane, giving the molecule a non-planar structure. This statement is incorrect.

Since only statement (a) is correct, the answer is option (2): (a) only.

We have to judge the truth of both statements and also decide whether the Reason really explains the Assertion. Let us examine them one by one, carrying every algebraic and conceptual step along.

First, consider the Assertion (A): “Lithium salts are hydrated.”
The lithium ion is $$\text{Li}^+$$. Its ionic radius $$r_{\text{Li}^+}$$ is the smallest in the entire alkali-metal group. Hydration depends upon the interaction between an ion and water dipoles, and the energy released is called hydration enthalpy $$\Delta H_{\text{hyd}}$$. A standard qualitative rule is

$$\Delta H_{\text{hyd}}\propto \dfrac{z^{+}}{r^{2}}$$

where $$z^{+}$$ is the charge on the cation and $$r$$ is its radius. Because $$r_{\text{Li}^+}$$ is minimum while $$z^{+}=1$$ for all alkali-metal ions, the term $$\dfrac{z^{+}}{r^{2}}$$ is maximum for $$\text{Li}^+$$. Hence $$\Delta H_{\text{hyd}}(\text{Li}^+)$$ is largest, water molecules are strongly attracted, and solid lithium salts often crystallise with molecules of water, i.e. they are hydrated. So the Assertion is correct.

Now consider the Reason (R): “Lithium has higher polarising power than other alkali metal group members.”
Polarising power is the ability of a cation to distort the electron cloud of an anion. The classical expression is

$$\text{Polarising power}\propto\dfrac{z^{+}}{r^{2}}$$

The numerator $$z^{+}$$ again equals 1 for all these monovalent cations, while the denominator $$r^{2}$$ is smallest for $$\text{Li}^+$$. Therefore $$\dfrac{1}{r^{2}}$$ is largest for lithium, making its polarising power the greatest among the alkali metals. Hence the Reason is also correct.

We must now decide whether the Reason truly explains the Assertion. Hydration stems from strong ion-dipole attraction, quantified by hydration enthalpy, which again involves the same $$\dfrac{z^{+}}{r^{2}}$$ factor. Polarising power, however, describes distortion of an anion’s electron cloud and leads chiefly to increased covalent character of the salt (e.g. LiI is more covalent than KI). While both high hydration enthalpy and high polarising power arise from the small size of $$\text{Li}^+$$, polarising power itself is not the direct cause of water of crystallisation. In other words, the Reason is a correct statement in its own right but it does not logically account for why lithium salts are hydrated.

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

Hence, the correct answer is Option C.

We need to evaluate both statements about Group 2 (alkaline earth metal) compounds.

Consider Statement I: Both $$CaCl_2 \cdot 6H_2O$$ and $$MgCl_2 \cdot 8H_2O$$ undergo dehydration on heating. First, the standard hydrate of magnesium chloride is $$MgCl_2 \cdot 6H_2O$$, not $$MgCl_2 \cdot 8H_2O$$, so the formula given in the statement itself is incorrect. Moreover, when $$CaCl_2 \cdot 6H_2O$$ is heated, it does not undergo simple dehydration to give anhydrous $$CaCl_2$$. Instead, upon heating it first partially dehydrates but at higher temperatures it undergoes hydrolysis, releasing HCl gas and forming calcium oxychloride or $$CaO$$. Similarly, $$MgCl_2 \cdot 6H_2O$$ on heating undergoes hydrolysis: $$MgCl_2 \cdot 6H_2O \xrightarrow{\Delta} MgO + 2HCl + 5H_2O$$, rather than simple dehydration to anhydrous $$MgCl_2$$. To obtain anhydrous $$MgCl_2$$, the hydrate must be heated in a stream of dry HCl gas to prevent hydrolysis. Therefore, Statement I is false.

Consider Statement II: BeO is amphoteric, whereas the oxides of other elements in the same group are acidic. It is true that BeO is amphoteric — it reacts with both acids and bases. However, the oxides of other Group 2 elements (MgO, CaO, SrO, BaO) are basic in nature, not acidic. In fact, their basic character increases going down the group, with BaO being strongly basic. Therefore, Statement II is also false because it incorrectly states that other Group 2 oxides are acidic.

Since both statements are false, the correct answer is Option (2): Both statement I and statement II are false.

First, consider $$H_3PO_3$$ (phosphorous acid). This is a diprotic acid with the structure $$HP(O)(OH)_2$$. It has two ionizable $$O-H$$ groups and one non-ionizable $$P-H$$ bond. So 1 mole of $$H_3PO_3$$ reacts with 2 moles of NaOH.

For 50 mL of 1 M $$H_3PO_3$$: moles of $$H_3PO_3 = 0.05$$ mol. Moles of NaOH required $$= 2 \times 0.05 = 0.1$$ mol. Volume of 1 M NaOH $$= \frac{0.1}{1} = 0.1$$ L $$= 100$$ mL.

Next, consider $$H_3PO_2$$ (hypophosphorous acid). This is a monoprotic acid with the structure $$H_2P(O)(OH)$$. It has only one ionizable $$O-H$$ group and two non-ionizable $$P-H$$ bonds. So 1 mole of $$H_3PO_2$$ reacts with 1 mole of NaOH.

For 100 mL of 2 M $$H_3PO_2$$: moles of $$H_3PO_2 = 0.2$$ mol. Moles of NaOH required $$= 1 \times 0.2 = 0.2$$ mol. Volume of 1 M NaOH $$= \frac{0.2}{1} = 0.2$$ L $$= 200$$ mL.

The volumes required are 100 mL and 200 mL, which corresponds to option (3).

First, we recall the chemical names and formulae of the three calcium sulphate materials mentioned:

We have $$\text{gypsum} = \mathrm{CaSO_4\cdot 2H_2O}.$$

Now, the coefficient of $$\mathrm{H_2O}$$ in the formula tells us exactly how many molecules of water are present in one formula unit. Here the coefficient is $$2$$, so gypsum contains $$2$$ molecules of water of crystallisation.

When gypsum is heated strongly enough to drive out all its water, the product is called dead-burnt plaster. The formula becomes simply $$\mathrm{CaSO_4}$$ with no attached water. So the number of water molecules in dead-burnt plaster is $$0$$.

If, however, gypsum is only partially dehydrated—specifically, if one and a half molecules of water are removed from every two molecules originally present—we obtain plaster of Paris. The well-known formula for plaster of Paris is written as

$$\mathrm{CaSO_4\cdot\frac{1}{2}H_2O}.$$

Here the fraction $$\dfrac{1}{2}$$ in front of $$\mathrm{H_2O}$$ means that in the average crystal lattice there is half a molecule of water per formula unit. Thus plaster of Paris contains $$0.5$$ molecule of water.

Collecting these results together, we have:

$$ \begin{aligned} \text{Gypsum} &:& 2 \text{ water molecules},\\ \text{Dead-burnt plaster} &:& 0 \text{ water molecules},\\ \text{Plaster of Paris} &:& 0.5 \text{ water molecule}. \end{aligned} $$

Looking at the options, the sequence $$2,\;0,\;0.5$$ appears in Option A.

Hence, the correct answer is Option A.

First, we recall that the Group-14 elements are $$\text{C, Si, Ge, Sn, Pb}$$. In these elements the common higher oxidation state is $$+4$$, and many of their complex anions can be viewed as having the general formula $$[\text{E}X_6]^{2-}$$ where every ligand $$X$$ carries a charge $$-1$$.

To see whether a particular hexacoordinate ion is possible, we examine two points:

(i) the oxidation state balance We write the usual charge-balance equation $$6(-1) + (\text{oxidation state of E}) = -2.$$ Solving, we have $$\text{oxidation state of E} = +4.$$ So for every choice of central atom $$\text{E}$$ the required oxidation state is indeed the normal $$+4$$, which is accessible to $$\text{Si, Ge}$$ and $$\text{Sn}$$. Hence charge balance causes no difficulty for any of the listed species.

(ii) the steric (size) requirement for an octahedral arrangement In an octahedral $$[\text{E}X_6]^{2-}$$ ion all six ligands must crowd around the central atom. The larger the ligands, the more space they demand; the larger the central atom, the more easily it can accommodate them. Fluoride $$F^-$$ is the smallest normal monodentate ligand, chloride $$Cl^-$$ is appreciably larger, and hydroxide $$OH^-$$ lies in between. Among the central atoms, the size trend is $$r(\text{Si}) \lt r(\text{Ge}) \lt r(\text{Sn}).$$

Now we check each option in turn.

Option A : $$[\text{GeCl}_6]^{2-}$$. Germanium is larger than silicon, so six chloride ions can fit around it. Indeed, hexachlorogermanate(IV) salts such as $$\text{K}_2[\text{GeCl}_6]$$ are known. Therefore this ion exists.

Option B : $$[\text{Sn(OH)}_6]^{2-}$$. Tin(IV) is the largest central atom among the ones listed, and hydroxide is smaller than chloride. The well-known stannate ion $$[\text{Sn(OH)}_6]^{2-}$$ (or its oxide form $$\text{SnO}_3^{2-}$$ after dehydration) is easily formed in alkali. Thus this species also exists.

Option C : $$[\text{SiCl}_6]^{2-}$$. Here the problem is spatial. Silicon is the smallest of the three possible centres, while chloride is relatively bulky. Trying to pack six $$Cl^-$$ ions around the small $$\text{Si}^{4+}$$ produces prohibitive ligand-ligand repulsion, so the octahedral complex cannot be stabilised. No salts of the “hexachlorosilicate(IV)” ion are known.

Option D : $$[\text{SiF}_6]^{2-}$$. Although silicon is small, fluoride is even smaller, allowing the required octahedral arrangement. The ion $$[\text{SiF}_6]^{2-}$$ is perfectly stable; commercial compounds such as $$\text{Na}_2[\text{SiF}_6]$$ (sodium hexafluorosilicate) are common.

Therefore, among the four species, the only one that cannot exist is $$[\text{SiCl}_6]^{2-}$$.

Hence, the correct answer is Option C.

Calgon is sodium hexametaphosphate with the formula $$Na_6P_6O_{18}$$ (also written as $$(NaPO_3)_6$$). Let us examine each statement.

Option (1): Calgon is a polymeric compound and is water soluble. This is true — sodium hexametaphosphate is a polymeric phosphate that dissolves readily in water.

Option (2): Calgon contains the 2nd most abundant element by weight in the Earth's crust. The most abundant element in Earth's crust by weight is oxygen, and the second most abundant is silicon. Calgon contains sodium (Na), phosphorus (P), and oxygen (O), but does not contain silicon. Therefore, this statement is NOT true.

Option (3): It is also known as Graham's salt. This is true — sodium hexametaphosphate is indeed known as Graham's salt.

Option (4): It does not remove $$Ca^{2+}$$ ion by precipitation. This is true — Calgon sequesters $$Ca^{2+}$$ ions by forming a soluble complex $$[Ca_2(PO_3)_6]^{2-}$$, not by precipitation.

The statement that is NOT true is Option (2).

We have four boron trihalides: $$\text{BF}_3,\; \text{BCl}_3,\; \text{BBr}_3,\; \text{BI}_3$$. In every one of these molecules the central boron atom possesses only six valence electrons, so it is electron-deficient and can accept electron density. Each halogen atom, on the other hand, has three lone pairs and can donate one of those lone pairs into the empty $$2p$$ orbital of boron. This donation of a filled orbital on the halogen into an empty orbital on boron is called back bonding (or $$p\;(\pi)\!-\!p\;(\pi)$$ back donation when both orbitals are $$p$$ orbitals).

The strength of such back donation depends primarily on the efficiency of orbital overlap. The guiding principle is:

$$\text{Better size and energy match } \Longrightarrow \text{stronger overlap } \Longrightarrow \text{stronger back bonding}.$$

For boron, the relevant empty orbital is the $$2p$$ orbital. Therefore the ideal donor would also have a $$2p$$ orbital, because orbitals of the same principal quantum number $$n$$ (and therefore similar size and energy) overlap most effectively. Let us compare each halogen:

• In $$\text{F}$$ the lone pair resides in a $$2p$$ orbital (same principal quantum number as boron’s $$2p$$).
• In $$\text{Cl}$$ the lone pair is in a $$3p$$ orbital (larger and higher in energy than $$2p$$).
• In $$\text{Br}$$ the lone pair is in a $$4p$$ orbital (even larger and higher in energy).
• In $$\text{I}$$ the lone pair is in a $$5p$$ orbital (largest mismatch of all).

Because the radial size and energy separation between $$2p$$ and $$2p$$ orbitals are minimal, the overlap in $$\text{BF}_3$$ is far superior to that in the other trihalides. As the halogen becomes heavier (from F to I) the difference in size and energy between its lone-pair orbital and boron’s empty $$2p$$ orbital keeps increasing, so the overlap - and hence the back donation - keeps weakening.

Therefore the order of back-bond strength is

$$\text{BF}_3 \; > \; \text{BCl}_3 \; > \; \text{BBr}_3 \; > \; \text{BI}_3.$$

So the molecule with the strongest halide &rightarrow; boron back donation is $$\text{BF}_3$$, which corresponds to option 4.

Hence, the correct answer is Option 4.

In the Solvay process, ammonia is recovered by treating the ammonium chloride ($$\text{NH}_4\text{Cl}$$) by-product with calcium hydroxide ($$\text{Ca(OH)}_2$$, obtained from slaking quicklime). The reaction is: $$2\text{NH}_4\text{Cl} + \text{Ca(OH)}_2 \to 2\text{NH}_3 + \text{CaCl}_2 + 2\text{H}_2\text{O}$$.

From this reaction, when ammonia is recovered, calcium chloride ($$\text{CaCl}_2$$) is formed as a by-product. Therefore, one of the by-products formed during the recovery of $$\text{NH}_3$$ from the Solvay process is $$\text{CaCl}_2$$.

$$C_{60}$$ (Buckminsterfullerene) has the structure of a truncated icosahedron, resembling a soccer ball. Let us evaluate each statement:

(1) The six-membered rings are fused to both six and five-membered rings. This is correct. In $$C_{60}$$, each hexagonal ring is surrounded by alternating pentagons and hexagons.

(2) Each carbon atom forms three sigma bonds. This is correct. Every carbon atom in $$C_{60}$$ is $$sp^2$$ hybridised and bonded to three other carbon atoms through sigma bonds.

(3) The five-membered rings are fused only to six-membered rings. This is correct. By the isolated pentagon rule, no two pentagons share an edge in $$C_{60}$$. Each pentagonal ring is completely surrounded by hexagonal rings.

(4) It contains 12 six-membered rings and 24 five-membered rings. This is incorrect. The actual composition of $$C_{60}$$ is 12 pentagonal (five-membered) rings and 20 hexagonal (six-membered) rings. This can be verified using Euler's formula for polyhedra. The statement reverses and incorrectly states the numbers.

The incorrect statement is option (4).

Dihydrogen ($$\text{H}_2$$) has several industrial applications including the synthesis of ammonia, production of metal hydrides, as rocket fuel, and in the synthesis of various chemicals. However, the single largest industrial application needs to be identified.

The Haber-Bosch process for the synthesis of ammonia consumes the largest quantity of dihydrogen industrially. In this process, nitrogen and hydrogen react at high temperature and pressure in the presence of an iron catalyst: $$\text{N}_2 + 3\text{H}_2 \xrightarrow{\text{Fe catalyst}} 2\text{NH}_3$$.

Ammonia production is one of the most important industrial chemical processes in the world, as ammonia is the starting material for fertilisers, which feed a large fraction of the global population. The scale of ammonia production far exceeds any other use of dihydrogen, making it the single largest industrial application.

The correct answer is option (3): in the synthesis of ammonia.

We need to match each substance with the correct industrial process.

(a) Sodium Carbonate is manufactured by the Solvay process (also called the ammonia soda process), which involves the reaction of sodium chloride, ammonia, and carbon dioxide. This matches (iv).

(b) Titanium of high purity is obtained by the van-Arkel method (also known as the crystal bar process), where crude titanium is converted to volatile $$TiI_4$$ which is then decomposed on a hot tungsten filament. This matches (iii).

(c) Chlorine is manufactured by the Deacon process, which involves the catalytic oxidation of HCl by atmospheric oxygen in the presence of $$CuCl_2$$ catalyst: $$4HCl + O_2 \xrightarrow{CuCl_2} 2Cl_2 + 2H_2O$$. This matches (i).

(d) Sodium hydroxide is produced by the Castner-Kellner process, which is an electrolytic process using a mercury cathode cell for the electrolysis of brine. This matches (ii).

The correct matching is (a)$$\to$$(iv), (b)$$\to$$(iii), (c)$$\to$$(i), (d)$$\to$$(ii), which is Option (4).