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).

The reaction in each case is,

$ \begin{aligned} 2 \mathrm{NaBH}_4+\mathrm{I}_2 \longrightarrow & \mathrm{~B}_2 \mathrm{H}_6 +2 \mathrm{NaI}+\mathrm{H}_2 \uparrow \end{aligned} $

II. $\mathrm{B}_2 \mathrm{H}_6+3 \mathrm{O}_2 \longrightarrow \mathrm{~B}_2 \mathrm{O}_3+3 \mathrm{H}_2 \mathrm{O}$

III. $\mathrm{BF}_3+\mathrm{NaH} \longrightarrow \mathrm{B}_2 \mathrm{H}_6+6 \mathrm{NaF}$

IV. $\mathrm{B}_2 \mathrm{H}_6+6 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{~B}(\mathrm{OH})_3$$ +6 \mathrm{H}_2 \uparrow $

Hence, only in I and IV, $\mathrm{H}_2$ is released.

Statement given in option (d) is incorrect regarding the gas evolved of reaction. The correct form is,

$\mathrm{CO}_2$ is not poisonous gas.

$ \mathrm{CaCO}_3+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow $

$\mathrm{NaNO}_2$ gives more number of oxides on reacting with HCl .

Among the given statements, option (c) is incorrect. It's correct form is $\mathrm{SeO}_2$ is colourless solid at room temperature.

Step 1: Reaction with Lithium Aluminium Hydride

When boron trifluoride ($\mathrm{BF}_3$) reacts with lithium aluminium hydride ($\mathrm{LiAlH}_4$) in ether, it forms lithium fluoride ($\mathrm{LiF}$), aluminium fluoride ($\mathrm{AlF}_3$), and diborane ($\mathrm{B}_2\mathrm{H}_6$):

$ 4 \mathrm{BF}_3+3 \mathrm{LiAlH}_4 \xrightarrow{\text { Ether }} 3 \mathrm{LiF}+3 \mathrm{AlF}_3 + \mathrm{B}_2 \mathrm{H}_6 $

Step 2: Naming the Product X

So, here $X = \mathrm{B}_2\mathrm{H}_6$ (which is called diborane).

Step 3: Reaction of X with Ammonia

Diborane ($\mathrm{B}_2\mathrm{H}_6$) reacts with ammonia ($\mathrm{NH}_3$) to first form an adduct, which is a combined compound: $\left[\mathrm{B}_2\mathrm{H}_6 \cdot 2\mathrm{NH}_3\right]$.

Step 4: Heating to Get Z

When this adduct is heated, it breaks down to give borazine ($\mathrm{B}_3\mathrm{N}_3\mathrm{H}_6$), which is compound $Z$, and hydrogen gas ($\mathrm{H}_2$):

$ 2~\mathrm{B}_2 \mathrm{H}_6 + 2~\mathrm{NH}_3 \longrightarrow \left[\mathrm{B}_2 \mathrm{H}_6 \cdot 2~\mathrm{NH}_3\right] \longrightarrow 2~\mathrm{B}_3 \mathrm{N}_3 \mathrm{H}_6 (\mathrm{Z}) + \mathrm{H}_2 $

Step 5: Counting Sigma ($\sigma$) and Pi ($\pi$) Bonds in Z

Borazine ($\mathrm{B}_3\mathrm{N}_3\mathrm{H}_6$) is a ring, and its bonds are:

There are 6 sigma ($\sigma$) bonds between B and N in the ring.
There are 3 sigma ($\sigma$) bonds between B and H.
There are 3 sigma ($\sigma$) bonds between N and H.

So, the total number of $\sigma$ bonds ($x$) is: $6 + 3 + 3 = 12$.

There are 3 pi ($\pi$) bonds ($y$) in the ring.

So, $x + y = 12 + 3 = 15$.

The reactions involved are as follows.

$\mathrm{PCl}_5(B)$ in solid state exists as $\left[\mathrm{PCl}_4^{+}\right]\left[\mathrm{PCl}_6^{-}\right]$as ionic lattice. $\mathrm{PCl}_5$ on hydrolysis gives phosphoric acid.

$ \mathrm{PCl}_5+4 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text { (Phosphoric acid) }}{\mathrm{H}_3 \mathrm{PO}_4}+5 \mathrm{HCl} $

In $\mathrm{PCl}_5$, the all bonds are not equivalent.

Trigonal bipyramidal

Paining of electrons takes place.

Thus, option (c) is correct.

The hydrolysis of diborane $\left(\mathrm{B}_2 \mathrm{H}_6\right)$ gives boric acid $\left(\mathrm{H}_3 \mathrm{BO}_3\right)$ as the compound $X$. The reaction is

$ \mathrm{B}_2 \mathrm{H}_6+6 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}_3 \mathrm{BO}_3+6 \mathrm{H}_2 $

The properties of boric acid are as follows.

• Acidity : Boric acid is a weak monobasic Lewis acid. It does not donate protons but rather accepts a hydroxide ion from water, making it a monobasic acid. Therefore,

statement II is correct, and statement I (tribasic acid) is incorrect.

• Structure : In the solid state, boric acid has a layered structure, where planar $\mathrm{BO}_3$ units are linked by hydrogen bonds. This structure gives it a slippery feel. Therefore, statement III is correct.

• Solubility : Boric acid has low solubility in cold water, although its solubility increases with temperature. It is not considered highly soluble. Therefore, statement IV is incorrect.

Based on the analysis, the correct statements about compound $X$ are II and III.

Statement I : Graphite's structure is well-known to be layered, with sheets of hexagonal rings of carbon atoms stacked on top of one another. Thus, this statement is correct.

• Statement II : Buckminster fullerene ( $\mathrm{C}_{60}$ ) has a conjugated $\pi$-electron system spread over its spherical surface. This delocalisation of electrons gives it aromatic properties, making it an aromatic compound. Therefore, this statement is incorrect.

• Statement III The distance between the layers in graphite is approximately 335 pm . The value 141.5 pm represents the carbon-carbon bond length within a single layer, not the interlayer distance. Thus, this statement is incorrect.

• Statement IV In both graphite and buckminster fullerene, each carbon atom is bonded to three other carbon atoms, resulting in $s p^2$ hybridisation. The unhybridised $p$-orbitals form delocalised $\pi$-systems. Thus, this statement is correct.

The correct statements are I and IV.

Among the given orders, order given in options (b) i.e.

$ \mathrm{N}_2 \mathrm{O}<\mathrm{NO}<\mathrm{N}_2 \mathrm{O}_3<\mathrm{N}_2 \mathrm{O}_4<\mathrm{N}_2 \mathrm{O}_5 \text { (a } $

(Acidic Nature) is correct.

The correct order for remaining options are

(a) $\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{O}$

(Boiling point)

(c) $\mathrm{HF}<\mathrm{HCl}<\mathrm{HBr}<\mathrm{HI}$

(Acidic Nature)

(d) $\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{O}$

(Bond angle)

Helium ( $X$ ) is used as a diluent for oxygen in modern dividing apparatus and argon $(Y)$ used to provide an inert atmosphere in high temperature

metallurgical process. Thus $Y$ and $X$ respectively are $\mathrm{Ar}, \mathrm{He}$.

Orthophosphorus acid (or phosphorous acid) on heating disproportionates to orthophosphoric acid (or phosphoric acid) and phosphine. The reaction involved is as follows.

$ 4 \mathrm{H}_3 \mathrm{PO}_3 \rightarrow \underset{(A)}{3 \mathrm{H}_3} \mathrm{PO}_4+\underset{(B)}{\mathrm{PH}_3} $

Thus, $A$ and $B$ are $\mathrm{H}_3 \mathrm{PO}_4$ and $\mathrm{PH}_3$ respectively.

Among the given options, HCl and $\mathrm{H}_2 \mathrm{~S}$ are electron rich hydrides. Electron rich hydrides are those compounds where the central atom has more electron than required to form single covalent bonds with its surrounding atoms.

Among the given statements, statement given in option (c) is incorrect about compounds of group 13 elements. It's correct form is, Diborane $\left(\mathrm{B}_2 \mathrm{H}_6\right)$ is an example of electron deficient hydride.

Among the given statements, statement given in option (c) is incorrect about oxidation state of group 14 elements, it's correct form is lead in +2 state is not a good reducing agent. This is because it readily accepts two more electrons to achieve the stable +4 oxidation state.

Statment given in I, II and IV are correct, while III is incorrect. Its correct form is given in statement II.

The reaction involved are as follows.

Option A: Chlorine oxidises ferrous salts to ferric salts in acidic medium.

• Ferrous (Fe²⁺) has an oxidation state of +2. Ferric (Fe³⁺) has an oxidation state of +3.

• Chlorine (Cl₂) is a strong oxidizing agent.

• In acidic medium, chlorine readily oxidizes Fe²⁺ to Fe³⁺:

  2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻

• This statement is correct.

Option B: Chlorine oxidises iodine to periodic acid in water.

• Iodine (I₂) has an oxidation state of 0.

• Periodic acid (HIO₄) has iodine in a +7 oxidation state.

• When chlorine reacts with iodine in the presence of water, it typically oxidizes iodine to iodic acid (HIO₃), where iodine is in a +5 oxidation state:

  I₂ + 5Cl₂ + 6H₂O → 2HIO₃ + 10HCl

• To obtain periodic acid (HIO₄), a much stronger oxidizing agent than chlorine (like hot concentrated nitric acid, or by electrolysis, or certain peroxo compounds) is generally required, or specific conditions like strongly alkaline medium with hypochlorite. Simple "chlorine in water" does not usually produce periodic acid from iodine.

• Therefore, this statement is incorrect.

Option C: Chlorine acts as a bleaching agent due to oxidation.

• When chlorine dissolves in water, it forms hypochlorous acid (HOCl):

  Cl₂ + H₂O ⇌ HCl + HOCl

• Hypochlorous acid is a powerful oxidizing agent. It decomposes to release nascent oxygen ([O]), which oxidizes colored substances into colorless ones, thereby achieving bleaching.

  HOCl → HCl + [O]

  Colored substance + [O] → Colorless substance

• Thus, the bleaching action of chlorine is indeed due to oxidation.

• This statement is correct.

Option D: Chlorine is manufactured by Deacon's process.

• Deacon's process is an industrial method for the production of chlorine. It involves the catalytic oxidation of hydrogen chloride gas with atmospheric oxygen, usually using copper(II) chloride as a catalyst at high temperatures:

  4HCl(g) + O₂(g) ⇌ 2Cl₂(g) + 2H₂O(g)

• While electrolysis of brine is the predominant modern method, Deacon's process is a known and historically significant method for chlorine manufacturing.

• This statement is correct.

Based on the analysis, Option B is the incorrect statement.

The final answer is $\boxed{B}$

Both the Assertion and the Reason are correct statements, but the Reason does not explain the Assertion properly.

Phosphorus can make both phosphorus(III) chloride and phosphorus(V) chloride because phosphorus has empty $d$-orbitals in its outer shell. This lets phosphorus use these orbitals to form more than four bonds, so it can form compounds like $\mathrm{PCl}_5$ (phosphorus(V) chloride).

Nitrogen cannot form nitrogen(V) chloride ($\mathrm{NCl}_5$) because nitrogen does not have $d$-orbitals in its valence shell. It can only make up to four bonds, so it cannot expand its octet to form five bonds like phosphorus can.

Therefore, the main reason why phosphorus can form $\mathrm{PCl}_5$ but nitrogen cannot form $\mathrm{NCl}_5$ is the availability of $d$-orbitals in phosphorus, not the difference in electronegativity.

Statement given in I and II are correct regarding boron, while statement III and IV are incorrect.

Their correct forms are,

III. in diborane oxidation state of hydrogen is -1 .

IV. boric acid is not a tribasic acid, it is monobasic acid.

The complete reaction is as follows,

$ \underset{\substack{(X) \\ \text { Hypophosphorus acid }}}{\mathrm{H}_3 \mathrm{PO}_2(a q)}+2 \mathrm{AgNO}_3(a q) \xrightarrow{\mathrm{H}_2 \mathrm{O}} \underset{(Y)}{\mathrm{H}_3 \mathrm{PO}_4}+2 \mathrm{Ag}+2 \mathrm{HNO}_3 $

The reaction is,

$ \mathrm{Zn}+\underset{\text { (conc.) }}{4 \mathrm{HNO}_3} \longrightarrow \mathrm{Zn}\left(\mathrm{NO}_3\right)_2+\underset{(A)}{2 \mathrm{NO}_2}+2 \mathrm{H}_2 \mathrm{O} $

$ 4 \mathrm{Zn}+\underset{\text { (dil.) }}{10 \mathrm{HNO}_3} \longrightarrow 4 \mathrm{Zn}\left(\mathrm{NO}_3\right)_2+\underset{(B)}{\mathrm{N}_2 \mathrm{O}}+5 \mathrm{H}_2 \mathrm{O} $

Oxidation state of nitrogen,

(A) $\mathrm{NO}_2=+4$

(B) $\mathrm{N}_2 \mathrm{O}=+1$

Deacon's process is a chemical method use to produce chlorine gas from hydrochloric acid.

The reaction is, $4 \mathrm{HCl}+\mathrm{O}_2 \xrightarrow[723 \mathrm{~K}]{\mathrm{CuCl}_2} 2 \mathrm{Cl}_2+2 \mathrm{H}_2 \mathrm{O}$

The complete reaction is as follows,

Complete hydrolysis, Thus, $X=\mathrm{PCl}_5, Y=\mathrm{H}_3 \mathrm{PO}_4$.

$ \text {The reaction involved is, } $

$ \mathrm{XeF}_6+\mathrm{H}_2 \mathrm{O} \longrightarrow \underset{(\text { Product } X)}{\mathrm{XeOF}_4}+2 \mathrm{HF} $

Thus the shape of $(X) \mathrm{XeF}_4$ is square pyramidal.

$\mathrm{H}_3 \mathrm{PO}_3$ and $\mathrm{H}_2 \mathrm{SO}_4$ both have basicity of $2 . \mathrm{H}_3 \mathrm{PO}_3$ has two $\mathrm{P}-\mathrm{OH}$ bond meaning it can donate two $\left[\mathrm{OH}^{-}\right]$ions when in solution.

$\mathrm{H}_2 \mathrm{SO}_4$ also have two ionisation H -atom attached to oxygen, giving it a basicity of 2 .

The conjugate base of phosphorus acid is $\mathrm{H}_2 \mathrm{PO}_3^{-}$conjugate base of oleum is, $\mathrm{HS}_2 \mathrm{O}_7^{-}$

$ \begin{aligned} & \mathrm{H}_3 \mathrm{PO}_3(\mathrm{aq}) \rightleftharpoons \mathrm{H}_2 \mathrm{PO}_3^{-}+\mathrm{H}^{+} \\ & \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7 \rightleftharpoons \mathrm{HS}_2 \mathrm{O}_7^{-}+\mathrm{H}^{+} \end{aligned} $

Statement I:

CCl₄ does not undergo hydrolysis but SiCl₄ undergoes hydrolysis.

Explanation:

• In SiCl₄, the central atom Si has vacant 3d orbitals.

  → Hence, water (a nucleophile) can attack the Si atom forming hydrolyzed products like silicic acid (via intermediate Si(OH)₄).

• In CCl₄, the central atom C does not have vacant d orbitals (carbon is limited to 2s and 2p orbitals only).

  → Therefore, nucleophilic attack by water is not possible, so CCl₄ does not undergo hydrolysis.

✅ Statement I is correct.

Statement II:

Thermal and chemical stability of GeX₄ is more than GeX₂.

Explanation:

• Germanium belongs to Group 14.

  → As we go down the group, +2 oxidation state becomes more stable than +4 (due to the inert pair effect).

• However, for Ge, the +4 oxidation state is still more stable than the +2 state (the inert pair effect is not very pronounced yet).

So:

• GeX₄ (Ge⁴⁺) is more stable thermally and chemically than GeX₂ (Ge²⁺).

✅ Statement II is also correct.

✅ Final Answer:

Option A: Both Statement-I and Statement-II are correct.

$ \text { The disproportion reaction is, } $

$ 4 \mathrm{H}_3 \mathrm{PO}_3 \longrightarrow \underset{\text { (Phosphoric acid) }}{3 \mathrm{H}_3 \mathrm{PO}_4}+\mathrm{PH}_3 $