p-Block Elements

Correct order of

(i) Boiling point: $\mathrm{HF}>\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}$

As we go from $\mathrm{HCl}$ to $\mathrm{HI}$, the size and molar mass increase, so van der Waals forces increase and the boiling point increases.

But $\mathrm{HF}$ has strong hydrogen bonding because fluorine is highly electronegative, so its boiling point becomes the highest.

(ii) Melting point : $\mathrm{HI}>\mathrm{HF}>\mathrm{HBr}>\mathrm{HCl}$

In general, from $\mathrm{HCl}$ to $\mathrm{HI}$, increasing size and mass increase intermolecular forces, so melting point increases.

However, $\mathrm{HF}$ shows hydrogen bonding, so its melting point is higher than $\mathrm{HCl}$ and $\mathrm{HBr}$.

Overall, $\mathrm{HI}$ has the highest melting point among hydrogen halides due to stronger dispersion forces in the solid state.

Hydrides of group 15: $\mathrm{NH_3,\ PH_3,\ AsH_3,\ SbH_3}$

A. Stability decreases down the group — Correct

Down the group, the $\mathrm{E-H}$ bond strength decreases because atomic size increases, so overlap becomes poorer. Hence hydrides become less stable:

$\mathrm{NH_3 > PH_3 > AsH_3 > SbH_3}$

B. Basicity decreases down the group — Correct

Basicity depends on the availability of the lone pair on $\mathrm{E}$. Down the group, the lone pair becomes more diffused and less available for donation, so basicity decreases:

$\mathrm{NH_3 > PH_3 > AsH_3 > SbH_3}$

C. Reducing character increases down the group — Correct

Reducing nature increases as the $\mathrm{E-H}$ bond becomes weaker down the group, so the hydride can donate hydrogen (or get oxidised) more easily:

$\mathrm{NH_3 < PH_3 < AsH_3 < SbH_3}$

D. Boiling point increases down the group — Not correct as a general statement

Boiling points generally increase down the group due to increasing molar mass, but $\mathrm{NH_3}$ is anomalous because of hydrogen bonding, which raises its boiling point above what is expected.

So the trend is not strictly increasing from $\mathrm{NH_3}$ to $\mathrm{SbH_3}$.

Correct option: Option A (A, B & C only)

The lightest element in group 7 is manganese (Mn).

An oxoanion where Mn is in the +6 oxidation state is the manganate ion, $\text{MnO}_4^{2-}$.

Its potassium salt is potassium manganate, $\text{K}_2\text{MnO}_4$, which is green.

Correct option: D) green

Incorrect statement: Option C

• $^{13}\text{C}$ is stable (non-radioactive).

• The common radioactive isotope of carbon is $^{14}\text{C}$ (also $^{11}\text{C}$ is radioactive, but short-lived).

The other statements (A, B, D) are correct in standard chemical context.

$ \mathrm{PbO}_2+\mathrm{Pb} \rightarrow 2 \mathrm{PbO} $

In this reaction, lead changes from $+4$ (in $\mathrm{PbO}_2$) and $0$ (in $\mathrm{Pb}$) to $+2$ (in $\mathrm{PbO}$).

Since $\mathrm{Pb}^{2+}$ is more stable than $\mathrm{Pb}^{4+}$, the reaction prefers to form $\mathrm{Pb}^{2+}$.

So, this reaction is spontaneous, hence $\Delta \mathrm{G}^{\circ}(1)$ is negative.

$ \mathrm{SnO}_2+\mathrm{Sn} \rightarrow 2 \mathrm{SnO} $

In this reaction, tin changes from $+4$ (in $\mathrm{SnO}_2$) and $0$ (in $\mathrm{Sn}$) to $+2$ (in $\mathrm{SnO}$).

But $\mathrm{Sn}^{2+}$ is less stable than $\mathrm{Sn}^{4+}$, so forming $\mathrm{Sn}^{2+}$ is not preferred.

Therefore, this reaction is non-spontaneous, hence $\Delta \mathrm{G}^{\circ}(2)$ is positive.

(A) $\mathrm{B}^{+3}<\mathrm{Al}^{+3}<\mathrm{Ga}^{+3}<\mathrm{In}^{+3}<\mathrm{Tl}^{+3}$ : ionic size. This order means that the ionic radius increases as we go down the group (from $\mathrm{B}^{+3}$ to $\mathrm{Tl}^{+3}$). So the ionic radii do not decrease down the group. Therefore, statement (A) is incorrect.

(B) $\mathrm{B}>\mathrm{Tl}>\mathrm{In}>\mathrm{Ga}>\mathrm{Al}: \mathrm{EN}$. This is the actual order of electronegativity in group 13. It does not simply decrease from top to bottom; instead, there are irregular variations. So the statement that electronegativity decreases down the group is not correct. Therefore, statement (B) is incorrect.

(C) $\mathrm{B}>\mathrm{Tl}>\mathrm{Ga}>\mathrm{Al}>\mathrm{In}:$ IE. From this order of first ionisation enthalpy, we can see that boron has the highest first ionisation enthalpy among the group 13 elements. So statement (C) is correct.

(D) Trichlorides and triiodides of group $13^{\text {th }}$ elements are covalent in nature. The $\mathrm{MCl_3}$ and $\mathrm{MI_3}$ (where M is a group 13 element) have high polarising power due to the small, highly charged $\mathrm{M^{3+}}$ ion, so they form covalent bonds with halide ions. Hence, statement (D) is also correct.

$ \begin{array}{ll} \mathrm{X}=\mathrm{N} & \mathrm{X}_2 \mathrm{O}_3=\mathrm{N}_2 \mathrm{O}_3(\text { Acidic }) \\ \mathrm{Y}=\mathrm{Sb} & \mathrm{Y}_2 \mathrm{O}_3=\mathrm{Sb}_2 \mathrm{O}_3(\text { Amphoteric }) \end{array} $

Statement-I is true

$ \mathrm{BCl}_3+3 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{~B}(\mathrm{OH})_3+3 \mathrm{HCl} $

Statement-II is false

$\mathrm{HF}<\mathrm{HCl}<\mathrm{HBr}<\mathrm{HI}$ (bond length order)

Bond length in $HX$ increases as we go down the halogen group because the size of the halogen atom increases. So the longest $H-X$ bond is in $\mathrm{HI}$.

$\mathrm{F}<\mathrm{Cl}<\mathrm{Br}<\mathrm{I}$ (radius order)

Covalent radius also increases down the group. So iodine has the largest covalent radius, not the smallest. Hence, the halogen that forms the longest bond with hydrogen (iodine) does not have the smallest covalent radius.

Therefore, Statement I is incorrect.

$\mathrm{PH}_3<\mathrm{AsH}_3<\mathrm{NH}_3<\mathrm{SbH}_3<\mathrm{BiH}_3$ (Boiling point order)

From this order, the hydride with the lowest boiling point is $\mathrm{PH}_3$. So in Statement II, the element $E$ is phosphorus.

Maximum possible covalency of phosphorous is 6

Phosphorus can show a maximum covalency of $6$ (for example, in $\mathrm{PCl}_5$ it shows covalency $5$, and in $\mathrm{PF}_6^-$ it shows covalency $6$). So the part “maximum covalency of $E$ is $4$” is false.

Therefore, Statement II is incorrect.

Element $E$ must be nitrogen $(\mathrm{N})$.

The binary cation $\left(\mathrm{EH}_x\right)^{+}$ formed with hydrogen should match a common stable cation. Nitrogen forms the well-known cation $\mathrm{NH}_4^{+}$ (ammonium ion), so here $E = \mathrm{N}$ and $x = 4$.

Also, $\mathrm{EH}_3$ means $\mathrm{NH}_3$ (ammonia). In alkaline medium, $\mathrm{NH}_3$ reacts with $\mathrm{K}_2\mathrm{HgI}_4$ to give a precipitate of basic mercury(II)amido-iodine, which is the characteristic test for ammonia. This again confirms $E = \mathrm{N}$.

Now we compare first ionisation enthalpy values for the first elements of groups 13, 14, 15, and 16: $\mathrm{B}$ (group 13), $\mathrm{C}$ (group 14), $\mathrm{N}$ (group 15), and $\mathrm{O}$ (group 16).

Across a period, first ionisation enthalpy generally increases due to increasing effective nuclear charge. Also, $\mathrm{N}$ has a stable half-filled configuration $2p^3$, so its first ionisation enthalpy is higher than $\mathrm{O}$.

Therefore, among $\mathrm{B}, \mathrm{C}, \mathrm{N}, \mathrm{O}$, nitrogen has the highest first ionisation enthalpy. So the correct value corresponds to $\mathrm{N}$ (approximately $1402\ \mathrm{kJ\ mol^{-1}}$).

Reaction 1: Finding A

$\mathrm{PbCl_2 + K_2CrO_4 \rightarrow A + 2KCl}$

This is a double displacement reaction (precipitation reaction).

• $\mathrm{Pb^{2+}}$ from $\mathrm{PbCl_2}$ combines with $\mathrm{CrO_4^{2-}}$ from $\mathrm{K_2CrO_4}$

• The product formed is lead chromate

$\boxed{\mathrm{A = PbCrO_4}}$

(This is a yellow precipitate)

Reaction 2: Finding B

$\mathrm{A + NaOH \rightleftharpoons B + Na_2CrO_4}$

Substituting A:

$\mathrm{PbCrO_4 + NaOH \rightleftharpoons B + Na_2CrO_4}$

Lead chromate reacts with excess NaOH. Since lead is amphoteric, it forms a plumbate complex.

Balancing the equation:

$\mathrm{PbCrO_4 + 4NaOH \rightarrow Na_2[Pb(OH)_4] + Na_2CrO_4}$

Here, $\mathrm{Pb^{2+}}$ is oxidised to $\mathrm{Pb^{4+}}$ in the plumbate ion, or more accurately, lead forms the tetrahydroxoplumbate(II) complex which gets oxidised to tetrahydroxoplumbate(IV).

$\boxed{\mathrm{B = Na_2[Pb(OH)_4]}}$

Reaction 3: Finding X

$\mathrm{PbSO_4 + 4CH_3COONH_4 \rightarrow (NH_4)_2SO_4 + X}$

Here, ammonium acetate acts as a complexing agent.

• 4 acetate ions ($\mathrm{CH_3COO^-}$) coordinate with $\mathrm{Pb^{2+}}$

• This forms a tetraacetatoplumbate(II) complex

$\mathrm{PbSO_4 + 4CH_3COONH_4 \rightarrow (NH_4)_2SO_4 + (NH_4)_2[Pb(CH_3COO)_4]}$

$\boxed{\mathrm{X = (NH_4)_2[Pb(CH_3COO)_4]}}$

$\Rightarrow \mathrm{SiO}_2, \mathrm{CO}_2, \mathrm{GeO}, \mathrm{GeO}_2$ are acidic oxides.

$\Rightarrow \mathrm{SnO}, \mathrm{SnO}_2, \mathrm{PbO}, \mathrm{PbO}_2$ are amphoteric oxides; they can react with both acids and bases.

Counting the required pairs

1. $[\mathrm{SiO}_2,\mathrm{CO}_2]$ – both acidic

2. $[\mathrm{SnO},\mathrm{SnO}_2]$ – both amphoteric

3. $[\mathrm{PbO},\mathrm{PbO}_2]$ – both amphoteric

4. $[\mathrm{GeO},\mathrm{GeO}_2]$ – both acidic

Only two pairs (2 and 3) contain oxides that are both amphoteric, so Statement I is correct.

$\Rightarrow \mathrm{BF}_3$ is electron-deficient (only six electrons around B) and therefore acts as a Lewis acid.

The boron atom in $\mathrm{BF}_3$ is $\mathrm{sp}^2$ hybridised, giving the molecule a trigonal planar geometry.

$\mathrm{BF}_3$ accepts the lone pair from $\mathrm{NH}_3$ to form an adduct, confirming its Lewis-acid behaviour. Hence Statement II is also correct.

(i) True: Tl¹⁺ is more stable than Tl³⁺ due to the inert pair effect. Therefore, Tl³⁺ acts as a powerful oxidizing agent.

(ii) True: The standard reduction potential for Al^3+/Al is -1.66 V, indicating that Al³⁺ is difficult to reduce and thus highly stable.

(iii) False: Tl³⁺ is not stable in solution.

(iv) True: Tl¹⁺ is more stable than Tl³⁺.

(v) True: Both Al³⁺ and Tl¹⁺ are highly stable in their respective oxidation states.

Statement 1 is correct since left to right 1 E increases in general in periodic table.

Statement 2 is incorrect since M.P. of group 14 elements is more than group 13 elements.

In oxide of nitrogen it can achieve +5 oxidation state because it can form $\mathrm{p} \pi-\mathrm{p} \pi$ bond with oxygen e.g. $\mathrm{N}_2 \mathrm{O}_5$

Nitrogen cannot form halide in +5 oxidation state because it does not contain d-orbital.

e.g. $\quad \mathrm{NX}_5$ does not exist

X = halide

Size order

$\begin{aligned} & \mathrm{Al}>\mathrm{Ga} \\ & \mathrm{Al}^{3+}<\mathrm{Ga}^{3+} \end{aligned}$

Atomic number of Ga is 31

single bond weaker than

due to more $\ell \mathrm{p}-\ell \mathrm{p}$ repulsion.

Bond length $\Rightarrow d_{p-p}>d_{\mathrm{N}-\mathrm{N}}(\operatorname{size} \uparrow$, B.L. $\uparrow)$

In group 15 elements only N \& P show disproportionation in +3 oxidation state, As, $\mathbf{S b} \&$ Bi have almost inert for disproportionation in +3 oxidation state.

So both statements are false.

Tellurium dioxide ($\mathrm{TeO}_2$) acts as an oxidizing agent. This is because it can accept electrons and be reduced from its +4 oxidation state to a lower oxidation state.

Tellurium hydride ($\mathrm{TeH}_2$) is considered acidic. This is due to its relatively low bond dissociation energy, which makes it prone to breaking, releasing protons (H$^+$).

The significant difference in the melting and boiling points of oxygen and sulfur can be explained by considering atomicity.

Oxygen exists as $\mathrm{O}_2$ (Atomicity = 2).

Sulfur exists as $\mathrm{S}_8$ (Atomicity = 8).

Due to sulfur's higher atomicity (forming S8 molecules), its melting and boiling points are considerably higher than those of oxygen.

$\mathrm{PH_3} \text{ shows lower proton affinity than } \mathrm{NH_3}.$

Ammonia (NH₃) has a more available lone pair on nitrogen than PH₃ on phosphorus due to smaller size and higher electronegativity. Thus, NH₃ is a stronger base and has a higher proton affinity than PH₃.

This statement is correct.

$\mathrm{SO_2} \text{ can act as an oxidizing agent, but not as a reducing agent.}$

Sulfur dioxide (SO₂) is quite versatile in redox chemistry. In many reactions, it actually behaves as a reducing agent by being oxidized (e.g., to sulfate, where sulfur goes from +4 to +6 oxidation state). While under some conditions it may act as an oxidizing agent, it is well known and widely used for its reducing properties.

Hence, the claim that it “cannot act as a reducing agent” is incorrect.

$\mathrm{NO_2} \text{ can dimerise easily.}$

Nitrogen dioxide (NO₂) is known to dimerise to form dinitrogen tetroxide (N₂O₄), especially at lower temperatures.

This statement is correct.

$\mathrm{PF_3} \text{ exists but } \mathrm{NF_5} \text{ does not.}$

Phosphorus trifluoride (PF₃) is a known stable compound. In contrast, a compound like nitrogen pentafluoride (NF₅) is not observed, largely due to the limitations of nitrogen’s size and bonding capabilities in forming such a structure.

This statement is correct.

Based on the analysis, the incorrect statement is:

$\textbf{Option B.}$

Among the group 15 elements (N, P, As, Sb, Bi), the lightest two (N and P) are commonly considered nonmetals. Of these, nitrogen (N) is known to have the weakest single bond to itself (N–N).

The N–N single‐bond enthalpy ($\approx 160$\,kJ/mol) is lower (weaker) than the P–P single‐bond enthalpy ($\approx 200$\,kJ/mol).

Therefore, the “nonmetallic group 15 element with the weakest E–E bond” is nitrogen.

Maximum Covalency of Nitrogen

Although nitrogen typically forms three covalent bonds in neutral compounds (e.g., $\mathrm{NH_3}$), it can expand to four bonds in certain cationic species such as ammonium $\mathrm{NH_4}^+$ or $\mathrm{NF_4}^+$. In such species, nitrogen has a formal positive charge but is still forming four covalent bonds.

Hence, the maximum covalency of nitrogen is 4.

Answer: 4 (Option C).

Oxygen exists as $\mathrm{O}_2$

Sulphur exists as $\mathrm{S}_8$

Because of small size of oxygen, oxygen can form $2 \mathrm{p}_\pi-2 \mathrm{p}_\pi$ bond.

$\Rightarrow$ Assertion (A) is correct.

$\Rightarrow$ Reason (R) is correct.

Reason (R) is correct explanation of Assertion (A).

When lead sulphide (PbS) reacts with dilute nitric acid, the reaction is a redox process where lead sulphide is oxidized and nitric acid is reduced. The overall reaction involves the transformation of PbS and HNO₃ to form lead nitrate (Pb(NO₃)₂), sulphur, and nitrogen oxides, along with water. The specific nitrogen oxide formed depends on the reaction conditions, including the concentration of nitric acid and the temperature. Typically, with dilute nitric acid, nitric oxide (NO) and nitrogen dioxide (NO₂) are more common products than nitrous oxide (N₂O).

The general reaction can be represented as:

Or, under some conditions, sulphur (S) may also be one of the products instead of or in addition to SO₂:

From the reaction above, you can see:

• Sulphur (Option A) can be formed depending on the specific conditions of the reaction, making it a potentially correct product.
• Nitric oxide (Option B) is typically produced when dilute nitric acid reacts with PbS, making it a correct product.
• Lead nitrate (Option C) is definitely formed in the reaction, making it a correct product.
• However, Nitrous oxide (Option D) is not a typical product of the reaction of lead sulphide with dilute nitric acid. It could be formed under special circumstances in reactions involving nitrogen compounds, but it's not a standard outcome for the conditions described.

Therefore, the correct answer to the question is Option D: Nitrous oxide. It is not a typical product of the reaction of lead sulphide with dilute nitric acid under normal conditions.

Let's analyze each statement provided about p-block elements and determine their correctness:

(A) Non metals have higher electronegativity than metals.

This statement is correct. Non-metals, especially those in the upper right of the periodic table like fluorine, oxygen, and nitrogen, have higher electronegativity values compared to metals.

(B) Non metals have lower ionisation enthalpy than metals.

This statement is incorrect. Non-metals generally have higher ionization enthalpy compared to metals because they hold onto their electrons more tightly due to higher nuclear charge and smaller atomic radii.

(C) Compounds formed between highly reactive nonmetals and highly reactive metals are generally ionic.

This statement is correct. Highly reactive nonmetals (like halogens) tend to gain electrons while highly reactive metals (like alkali metals and alkaline earth metals) tend to lose electrons, resulting in the formation of ionic compounds.

(D) The non-metal oxides are generally basic in nature.

This statement is incorrect. Non-metal oxides are generally acidic in nature. They react with bases to form salts and water.

(E) The metal oxides are generally acidic or neutral in nature.

This statement is incorrect. Metal oxides are usually basic in nature, although some metal oxides can be amphoteric (both acidic and basic). Most metal oxides react with acids to form salts and water.

After evaluating all the statements, the correct statements are (A) and (C).

Therefore, the correct answer is:

Option D: (A) and (C) only

Let's analyze each statement step by step to identify the incorrect ones:

(A) Dinitrogen is a diatomic gas which acts like an inert gas at room temperature.

This statement is generally correct. Dinitrogen ($N_2$) is indeed a diatomic gas and is very inert at room temperature due to the strong triple bond between the nitrogen atoms.

(B) The common oxidation states of these elements are $-3,+3$ and +5.

This statement is also correct. The group 15 elements commonly exhibit oxidation states of $-3$ (in nitrides), $+3$ (as in $\mathrm{NH}_3$), and +5 (as in $\mathrm{HNO_3}$).

(C) Nitrogen has unique ability to form $\mathrm{p}\pi-\mathrm{p}\pi$ multiple bonds.

This statement is correct. Nitrogen has a unique ability to form $p\pi-p\pi$ multiple bonds, which is why it can form a triple bond ($N_2$).

(D) The stability of +5 oxidation states increases down the group.

This statement is incorrect. The stability of +5 oxidation states actually decreases down the group due to the effect of inert pair effect. For example, bismuth in +5 state is less stable than nitrogen in +5 state.

(E) Nitrogen shows a maximum covalency of 6.

This statement is incorrect. Nitrogen can exhibit a maximum covalency of 4 (as in $\mathrm{NH_4^+}$). It cannot expand its octet to show a covalency of 6.

Based on the above analysis, the incorrect statements are (D) and (E). Therefore, the correct answer is:

Option B: (D) and (E) only

For test of phosphorus:

$\mathrm{{H_3}P{O_4} + 12{(N{H_4})_2}Mo{O_4} + 21HN{O_3} \to \mathop {{{(N{H_4})}_3}\,.\,12Mo{O_3}}\limits_{(Yellow\,ppt)} + 21N{H_4}N{O_3} + 12{H_2}O}$

To evaluate the statements provided, let's first understand the concepts involved:

Assertion A: The stability order of the +1 oxidation state of $\mathrm{Ga}$, In, and $\mathrm{Tl}$ is Ga < In < Tl.

Reason R: The inert pair effect stabilizes the lower oxidation state down the group.

The elements $\mathrm{Ga}$ (Gallium), $\mathrm{In}$ (Indium), and $\mathrm{Tl}$ (Thallium) belong to Group 13 of the periodic table. As we move down this group, the +3 oxidation state becomes less stable while the +1 oxidation state becomes more stable. This trend is attributed to the inert pair effect.

The inert pair effect refers to the reluctance of the s-electrons to participate in bonding as we move down the group in the periodic table. For Group 13 elements, this effect causes the lower +1 oxidation state to be more stable compared to the +3 oxidation state down the group.

Given this, the Reason R correctly explains why the +1 oxidation state is more stable for $\mathrm{Tl}$ compared to $\mathrm{Ga}$ and $\mathrm{In}$. Therefore, the stability order of the +1 oxidation state should indeed be Ga < In < Tl.

Conclusion: Both Assertion A and Reason R are true, and Reason R is the correct explanation for Assertion A.

Therefore, the correct answer is:

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

To determine which halogens among $\mathrm{F}_2, \mathrm{Cl}_2, \mathrm{Br}_2$, and $\mathrm{I}_2$ can undergo disproportionation reactions, we need to understand what a disproportionation reaction is.

In a disproportionation reaction, a single substance is simultaneously oxidized and reduced, forming two different products. The halogen X₂ disproportionates in water as follows:

$ \mathrm{X}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HX} + \mathrm{HXO} $

Here, X represents a halogen.

Let's analyze each halogen to see if disproportionation is possible:

Fluorine ($\mathrm{F}_2$): Fluorine is the most electronegative element and has a very high oxidation potential. It does not undergo disproportionation because it prefers to remain in the $-1$ oxidation state and does not form higher oxidation states easily. Therefore, $\mathrm{F}_2$ cannot undergo a disproportionation reaction.

Chlorine ($\mathrm{Cl}_2$): Chlorine can undergo disproportionation. In water, chlorine disproportionates as follows:

$ \mathrm{Cl}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HCl} + \mathrm{HOCl} $

Here, chlorine is both reduced to $\mathrm{HCl}$ (chloride, -1 oxidation state) and oxidized to $\mathrm{HOCl}$ (hypochlorite, +1 oxidation state).

Bromine ($\mathrm{Br}_2$): Bromine can also undergo disproportionation. In water, bromine disproportionates as follows:

$ \mathrm{Br}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HBr} + \mathrm{HOBr} $

Here, bromine is both reduced to $\mathrm{HBr}$ (bromide, -1 oxidation state) and oxidized to $\mathrm{HOBr}$ (hypobromite, +1 oxidation state).

Iodine ($\mathrm{I}_2$): Iodine can also undergo disproportionation. In alkaline solutions, iodine disproportionates as follows:

$ \mathrm{I}_2 + \mathrm{OH}^- \rightarrow \mathrm{I}^- + \mathrm{IO}^- $

Here, iodine is both reduced to $\mathrm{I}^-$ (iodide, -1 oxidation state) and oxidized to $\mathrm{IO}^-$ (hypoiodite, +1 oxidation state).

Based on the above analysis, all halogens except $\mathrm{F}_2$ can undergo disproportionation. Therefore, the correct answer is:

Option B

$\mathrm{Cl}_2, \mathrm{Br}_2$ and $\mathrm{I}_2$

(A) Covalent radius of group 14 elements increases from $\mathrm{C}$ to $\mathrm{Pb}$.

(B) Electronegativity decreases from $\mathrm{C}$ to $\mathrm{Si}$ and thereafter it remains more or less constant.

(C) Maximum covalency of $\mathrm{C}$ is 4 as the outermost shell of $\mathrm{C}$ is $2^{\text {nd }}$ shell and it cannot expand its octet. But other elements can expand their covalency due to the availability of vacant d-orbitals.

(D) Heavier elements do not form $\mathrm{p} \pi-\mathrm{p} \pi$ bonds due to their larger atomic size.

(E) Carbon can show negative oxidation states when it is covalently bonded to less electronegative elements like $\mathrm{CH}_4$.

S-1 is correct as gallium is used in manufacturing of thermometers.

S-2 is incorrect

To determine which of the following materials is not a semiconductor, we must understand what a semiconductor is and then look at the properties of each material listed.

Semiconductors are materials that have a conductivity level between conductors (such as metals) and insulators (such as most ceramic materials). They can conduct electricity under certain conditions or when doped with impurities. The most common semiconductors are silicon and germanium, but some compound materials like gallium arsenide and silicone carbide also behave as semiconductors.

Now, analyzing each option given:

Option A: Copper oxide - Copper oxide can behave as a semiconductor, particularly in some specific structures or when used in thin-film solar cells. Thus, it can exhibit semiconductor properties.

Option B: Graphite - Graphite is a form of carbon where the carbon atoms are bonded together in a sheet-like form. It is known for its excellent electrical conductivity due to the delocalization of pi electrons across the carbon layers. Its behavior is more characteristic of a conductor rather than a semiconductor.

Option C: Silicon - Silicon is the most well-known and widely used semiconductor material in electronic devices, including transistors, diodes, and integrated circuits. Its semiconducting properties are due to its band structure, which can be altered by doping with other elements.

Option D: Germanium - Germanium is another classic semiconductor material used in the early development of electronic devices. Though not as commonly used as Silicon in modern electronics, due to its similar crystal structure and semiconducting properties, Germanium remains an important semiconductor material.

Based on the provided information, Option B: Graphite is not a semiconductor. Its properties align more with those of a conductor rather than a semiconductor, distinguishing it from the other options listed which all have semiconducting capabilities under certain conditions.

(A) Group-13 size : $\mathrm{B}<\mathrm{Ga}<\mathrm{Al}<\mathrm{In} \simeq$ $\mathrm{TI}$

(B) In Group-13, Al has exceptional lower EN

$\mathrm{EN}: \mathrm{B}>\mathrm{TI}>\mathrm{In}>\mathrm{Ga}>\mathrm{Al}$

$\begin{aligned} \text{(C)} \quad & 2 \mathrm{Al}+6 \mathrm{HCl} \rightarrow 2 \mathrm{AlCl}_3+3 \mathrm{H}_2 \\ & \mathrm{Al}+\mathrm{HNO}_3 \rightarrow \text { No reaction (Due to protective oxide layer)} \end{aligned}$

(D) +1 oxidation state is observed mainly for heavier Group-13 element like $\mathrm{TI}$.

$\mathrm{B}$ and $\mathrm{Al}$ do not show +1 oxidation state

(E) $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} \rightarrow$ Hybridisation : $s p^3 d^2$

The most abundant isotope of Boron is ${ }_5 \mathrm{~B}^{11}$.

No. of neutrons in it $=x=6$

$2 \mathrm{B}(\mathrm{s})+\mathrm{N}_2(\mathrm{g}) \xrightarrow{\Delta} 2 \mathrm{BN}(\mathrm{s})$

Oxidation state of boron in $\mathrm{B N=y=+3}$

So, $x+y=6+3$

$=9$

Both statements need to be evaluated to determine which (if any) are true:

Statement (I) : $\mathrm{SiO}_2$ and $\mathrm{GeO}_2$ are acidic while $\mathrm{SnO}$ and $\mathrm{PbO}$ are amphoteric in nature.

This statement is generally true. The oxides of silicon ($\mathrm{SiO}_2$) and germanium ($\mathrm{GeO}_2$) are indeed acidic. These elements belong to group 14 of the periodic table and as you move down the group, the acidic character of the oxides decreases while the basic character increases. Tin oxide ($\mathrm{SnO}$) and lead oxide ($\mathrm{PbO}$) are lower in the group and they are amphoteric, which means they show both acidic and basic properties depending on the reacting substance.

Statement (II) : Allotropic forms of carbon are due to the property of catenation and $\mathrm{p} \pi-\mathrm{d} \pi$ bond formation.

This statement is false. Catenation is indeed the property that carbon has to form long chains and rings with other carbon atoms, which is responsible for the vast number of organic compounds. However, in the context of pure carbon allotropes—such as diamond, graphite, and fullerene—the different physical structures and properties are mainly due to the $\mathrm{sp}^3$, $\mathrm{sp}^2$, and $\mathrm{sp}$ hybridizations of carbon, not to $\mathrm{p} \pi-\mathrm{d} \pi$ bonding. Carbon does not use d-orbitals in forming the allotropes. The $\mathrm{p} \pi-\mathrm{d} \pi$ bonding is typically discussed in the context of transition metal complexes, not allotropic forms of carbon.

Thus, the correct answer is Option D : Statement I is true but Statement II is false.

The reducing strength of a substance is generally associated with its ability to donate electrons to other substances. In the context of hydrides of Group 15 elements, such as $\mathrm{SbH}_3$, $\mathrm{NH}_3$, $\mathrm{BiH}_3$, and $\mathrm{PH}_3$, the reducing strength can be considered based on their chemical reactivity, stability, and tendency to donate electrons.

In general, the stability of hydrides decreases as we move down the group in the periodic table, which means that heavier hydrides are less stable and hence act as stronger reducing agents. This trend is primarily due to the increase in the size of the central atom as we move down the group, leading to weaker bonds between the central atom and hydrogen in the hydride. As the bond strength decreases, the hydrides can more readily release hydrogen in the form of protons (H⁺) or hydride ions (H⁻), making them stronger reducing agents.

The order of reducing strength for Group 15 hydrides is typically as follows:

$\mathrm{BiH}_3 > \mathrm{SbH}_3 > \mathrm{PH}_3 > \mathrm{NH}_3$

Among the options given, $\mathrm{BiH}_3$ (bismuthine) is the strongest reducing agent. It is the least stable of the listed hydrides, due to the large size and low electronegativity of bismuth compared to antimony, phosphorus, and nitrogen. This makes $\mathrm{BiH}_3$ more prone to releasing electrons and acting as a reducing agent.

In contrast, $\mathrm{NH}_3$ (ammonia) is the most stable among the given hydrides and, therefore, the weakest reducing agent. Its bond with hydrogen is comparatively stronger, and nitrogen's higher electronegativity makes the molecule more resistant to donating electrons.

Thus, the correct answer is Option C, $\mathrm{BiH}_3$.

In trivalent state most of the compounds being covalent are hydrolysed in water. Trichlorides on hydrolysis in water form tetrahedral $\left[\mathrm{M}(\mathrm{OH})_4\right]^{-}$ species, the hybridisation state of element $\mathrm{M}$ is $\mathrm{sp}^3$.

In case of aluminium, acidified aqueous solution forms octahedral $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ ion.

$\mathrm{S}_1: \mathrm{S}_8+12 \mathrm{OH}^{\ominus} \rightarrow 4 \mathrm{~S}^{2-}+2 \mathrm{~S}_2 \mathrm{O}_3^{2-}+6 \mathrm{H}_2 \mathrm{O}$

$\mathrm{S}_2: \mathrm{ClO}_4^{\ominus}$ cannot undergo disproportionation reaction as chlorine is present in it's highest oxidation state.

(A) All group 16 elements form oxides of the $\mathrm{EO}_2$ and $\mathrm{EO}_3$ type where $\mathrm{E}=\mathrm{S}, \mathrm{Se}, \mathrm{Te}$ or $\mathrm{Po}$.

(B) $\mathrm{SO}_2$ is reducing while $\mathrm{TeO}_2$ is an oxidising agent.

(C) The reducing property increases from $\mathrm{H}_2 \mathrm{S}$ to $\mathrm{H}_2$ Te down the group.

(D)

SnO$_2$ and PbO$_2$ are amphoteric.

Option A is the correct answer. Both Statement I and Statement II are false.

Noble gases actually have low boiling points, not high. This is because noble gases are indeed monoatomic gases that exist as single atoms. They have a completely filled valence shell and do not easily form chemical bonds with other atoms. As a result, the intermolecular forces present in noble gases are only temporary dipole-induced dipole attractions, also known as dispersion forces (or London forces). However, these dispersion forces are very weak compared to other types of intermolecular forces like hydrogen bonding or dipole-dipole interactions that occur in other substances.

The dispersion forces in noble gases are weak because these gases are non-polar and since they are monoatomic, the electron cloud around the atom is symmetrical, meaning that there are no permanent dipoles within the molecules. As the atomic size of noble gases increases down the group, the dispersion forces do increase in strength due to the larger electron cloud that can be more easily distorted, resulting in higher boiling points for the heavier noble gases. Still, compared to other substances, the overall strength of the dispersion forces in noble gases is low, which means they can be liquefied at low temperatures and hence their boiling points are low, not high.

Therefore, the correct statement would be: "Noble gases are monoatomic gases. They are held together by weak dispersion forces. Because of this they are liquefied at very low temperatures. Hence, they have very low boiling points."

$\mathrm{Cr{O_2} + 4NaOH \to \mathop {N{a_2}Cr{O_4}}\limits_{(A)} + 2NaCl + 2{H_2}O}$

$\mathrm{N{a_2}Cr{O_4} + 2{H_2}{O_2} + 2HCl \to \mathop {Cr{O_5}}\limits_{(B)} + \mathop {2NaCl}\limits_{(Mis\sin g\,from\,balanced\,equation)} + 3{H_2}O}$

On moving down the group, bond strength of $\mathrm{M}-\mathrm{H}$ bond decreases, which reduces the thermal stability but increases reducing nature of hydrides, hence A and B are correct statements.

If nitrogen or sulphur is also present in the compound, the sodium fusion extract is first boiled with concentrated nitric acid to decompose cyanide or sulphide of sodium during Lassaigne's test

Nessler's Reagent Reaction :

$\underset{\text { (Nessler's Reagent) }}{2 \mathrm{~K}_2 \mathrm{HgI}_4}+\mathrm{NH}_3+3 \mathrm{KOH} \rightarrow \underset{\substack{\text { (lodine of Millon's base } \\ \text { (Brown precipitate }}}{\mathrm{HgO} . \mathrm{Hg}\left(\mathrm{NH}_2\right) \mathrm{I}}+7 \mathrm{KI}+2 \mathrm{H}_2 \mathrm{O}$

(1) Fluorspar is $\mathrm{CaF}_2$

Statement-I: Oxygen can have oxidation state from -2 to +2 , so statement I is incorrect

Statement- II: On moving down the group stability of +4 oxidation state increases whereas stability of +6 oxidation state decreases down the group, according to inert pair effect.

So both statements are wrong.

Fluorine is the element among the options that does not exhibit a variable oxidation state, unlike its fellow halogens. This is because fluorine is the most electronegative element, and it always forms compounds in the oxidation state of -1 as it has a strong tendency to gain one electron to achieve a stable electronic configuration.

Solid Boron has very strong crystalline lattice so its melting point unusually high in group 13 elements

During the water-gas shift reaction, carbon monoxide (CO) + H₂ reacts with steam (H₂O) to produce carbon dioxide (CO₂) and hydrogen gas (H₂). The chemical equation for the reaction is :

Therefore, option B is correct, which states that carbon monoxide is oxidized to carbon dioxide. This reaction is typically catalyzed by metal catalysts such as iron, nickel, or copper, which promote the reaction rate by providing a surface for the reactants to adsorb and react.

Option A is not correct as water is not evaporated during the reaction, but rather reacts with carbon monoxide to produce hydrogen gas and carbon dioxide.

Option C is not correct as carbon is not directly involved in the reaction.

Option D is also not correct as carbon dioxide is not reduced to carbon monoxide during the water-gas shift reaction.