p-Block Elements

Option A: O₃ molecule is bent

✅ True

• The structure of ozone (O₃) is angular (bent) with a bond angle of about 117°.

• It has resonance structures and sp²-hybridized central oxygen.

Option B: Ozone is violet-black in solid state

✅ True

• In solid state, ozone is violet-black.

• In liquid state, ozone is dark blue.

• In gas state, it is pale blue.

Option C: Ozone is diamagnetic gas

✅ True

• Although O₂ is paramagnetic, ozone (O₃) has all paired electrons, hence diamagnetic.

Option D: ONCl and ONO⁻ are isoelectronic

❌ False

Let’s check the total number of electrons:

• ONCl:

  O (8) + N (7) + Cl (17) = 32 electrons

• ONO⁻:

  O (8×2 = 16) + N (7) + 1 (extra for negative charge) = 24 electrons

→ They are not isoelectronic.

✅ Correct answer: Option D is the wrong statement.

When $ \text{PCl}_5 $ is fluorinated in a polar organic solvent, it forms ionic isomers. The formation of these ionic species can be understood as follows:

1. $ \text{PCl}_5 $ in a polar organic solvent can dissociate into ionic forms, such as $ \left[\text{PCl}_4\right]^+ $ and $ \left[\text{PCl}_6\right]^- $.


2. Upon fluorination, $ \text{PCl}_5 $ can be converted into $ \left[\text{PF}_6\right]^- $ and $ \left[\text{PCl}_4\right]^+ $.


3. Further fluorination can lead to the formation of mixed ionic compounds where $ \text{PCl}_5 $ can form isomers like $ \left[\text{PCl}_4\right]^+ \left[\text{PCl}_4 \text{F}_2\right]^- $.

The two ionic isomers formed are:

1. $ \left[\text{PCl}_4\right]^+ \left[\text{PCl}_4 \text{F}_2\right]^- $, which are colorless crystals.


2. $ \left[\text{PCl}_4\right]^+ \left[\text{PF}_6\right]^- $, which are white crystals.

Hence, the ionic isomers formed when $ \text{PCl}_5 $ is fluorinated in a polar organic solvent are :

To determine the products of the disproportionation of nitrous acid ($ \mathrm{HNO}_2 $) at room temperature, let's analyze the chemical behavior of $ \mathrm{HNO}_2 $ in aqueous solution.

Disproportionation is a redox reaction in which a single substance is simultaneously oxidized and reduced, yielding two different products. In the case of nitrous acid ($ \mathrm{HNO}_2 $), the reaction can be represented as follows:

$ 3 \mathrm{HNO}_2 \rightarrow \mathrm{HNO}_3 + \mathrm{H_2O} + 2 \mathrm{NO} $

Breaking this down, we see that nitrous acid disproportionates into nitric acid ($ \mathrm{HNO}_3 $), water ($ \mathrm{H_2O} $), and nitrogen monoxide ($ \mathrm{NO} $). Additionally, in an aqueous solution, nitric acid dissociates to form $ \mathrm{H_3O}^+ $ (hydronium ion) and $ \mathrm{NO_3}^- $ (nitrate ion).

Therefore, the species formed from this disproportionation reaction are $ \mathrm{H_3O}^+ $, $ \mathrm{NO_3}^- $, and $ \mathrm{NO} $.

From the given options, the correct answer is:

Option A

$ \mathrm{H_3O}^+ $, $ \mathrm{NO_3}^- $, and $ \mathrm{NO} $

To match the reactions given in List-I with the products in List-II, let's analyze each reaction individually based on its reactants and expected products.

• Reaction (P): $\mathrm{P}_2 \mathrm{O}_3 + 3 \mathrm{H}_2 \mathrm{O} \rightarrow$ involves the hydration of phosphorus trioxide. The product of this reaction is phosphorous acid:

$ \mathrm{P}_2 \mathrm{O}_3 + 3 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}_3 \mathrm{PO}_3 $

Hence, P matches with (2) $\mathrm{H}_3 \mathrm{PO}_3$.

• Reaction (Q): $\mathrm{P}_4 + 3 \mathrm{NaOH} + 3 \mathrm{H}_2 \mathrm{O} \rightarrow$ involves the reaction of phosphorus with a base (NaOH) and water, leading to the production of phosphine and sodium hypophosphite, not exactly reflected in the products listed, but given the context and possible products, we align with phosphorus chemistry principles. Typically, reactions involving phosphorus and strong bases can generate phosphine:

$ \mathrm{P}_4 + \mathrm{NaOH} + 3 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{PH}_3 +\mathrm{NaH}_2 \mathrm{PO}_2 $

Thus, Q matches best with (3) $\mathrm{PH}_3$ given the options, acknowledging the simplicity of the given answer choices.

• Reaction (R): $\mathrm{PCl}_5 + \mathrm{CH}_3 \mathrm{COOH} \rightarrow$ involves phosphorus pentachloride and acetic acid. The key reaction here is that $\mathrm{PCl}_5$ reacts with $\mathrm{CH}_3\mathrm{COOH}$ to replace $\mathrm{OH}$ groups in acetic acid with $\mathrm{Cl}$, creating acetyl chloride $\mathrm{CH}_3\mathrm{COCl}$ and $\mathrm{POCl}_3$:

$ \mathrm{PCl}_5 + \mathrm{CH}_3 \mathrm{COOH} \rightarrow \mathrm{CH}_3\mathrm{COCl} + \mathrm{POCl}_3 $

Therefore, R matches with (4) $\mathrm{POCl}_3$.

• Reaction (S): $\mathrm{H}_3 \mathrm{PO}_2 + 2 \mathrm{H}_2 \mathrm{O} + 4 \mathrm{AgNO}_3 \rightarrow$ involves hypophosphorous acid reacting in an oxidative environment given by silver nitrate, leading to the production of orthophosphoric acid among other products:

$ \mathrm{H}_3 \mathrm{PO}_2 + 2 \mathrm{H}_2 \mathrm{O} + 4 \mathrm{AgNO}_3 \rightarrow \mathrm{H}_3 \mathrm{PO}_4 +\mathrm{Ag}+\mathrm{HNO}_3 $

S, therefore, matches with (5) $\mathrm{H}_3 \mathrm{PO}_4$.

Given these reactions and the best fits for the products listed,

• P matches with 2,


• Q matches with 3,


• R matches with 4,


• S matches with 5.

Thus, the correct option is Option D: $\mathrm{P} \rightarrow 2 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 4 ; \mathrm{S} \rightarrow 5$.

The increasing order of atomic radii is as follows:

Ga < Al < In < Tl

The reaction involved is

P₄ + 8SOCl₂ $\to$ 4PCl₃ + 4SO₂ + 2S₂Cl₂

The reaction involved is

The reactions involved are

Cl₂ + 2NaOH (cold, dil.) $\to$ $\mathop {NaOCl}\limits_{(P)} $ + NaCl + H₂O

3Cl₂ + 6NaOH (hot, conc.) $\to$ $\mathop {NaCl{O_3}}\limits_{(Q)} $ + 5NaCl + 3H₂O

where NaOCl and NaClO₃ are salts of hypochlorous acid (HOCl) and chloric acid (HClO₃), respectively.

The reactions involved are

$S{O_2} + C{l_2}\buildrel {Charcoal} \over \longrightarrow \mathop {S{O_2}C{l_2}}\limits_{(R)} $

$\mathop {10S{O_2}C{l_2}}\limits_{(R)} + {P_4} \to \mathop {4PC{l_5}}\limits_{(S)} + 10S{O_2}$

$\mathop {PC{l_5}}\limits_{(S)} + 4{H_2}O \to \mathop {{H_3}P{O_4}}\limits_{(T)} + 5HCl$

The reactions involved are as follows:

(P) $Pb{O_2} + {H_2}S{O_4}\buildrel {Warm} \over \longrightarrow PbS{O_4} + {H_2}O + {1 \over 2}{O_2}$

(Q) $2N{a_2}{S_2}{O_3} + 2{H_2}O + C{l_2} \to 2NaCl + 2NaHS{O_4} + 2S$

(R) ${N_2}{H_4} + 2{I_2} \to {N_2} + 4HI$

(S) $Xe{F_2} + 2NO \to Xe + 2NOF$

The reaction involved is 4HNO₃ $\to$ 4NO₂ + 2H₂O + O₂. The yellow-brown color appears due to the formation of NO₂.

According to the reaction

P₄ + NaOH + 3H₂O $\to$ PH₃ + 3NaH₂PO₂

oxidation state of phosphorus in P₄ is zero while in PH₃ it is $-$3 and in NaH₂PO₂, it is +1. This shows that this is a type of disproportionation reaction because there is an increase as well as decrease in the oxidation state of phosphorus.

Bleaching powder is CaOCl₂ which contains HOCl as an oxoacid and the anhydride of which is Cl₂O.

Consider the titration reaction

CaOCl₂ + 2KI $\to$ I₂ + Ca(OH)₂ + KCl

According to this reaction, 25 mL of CaOCl₂ reacts with 30 mL of 0.50 M KI

I₂ + 2Na₂S₂O₃ $\to$ Na₂S₄O₆ + 2NaI

Given that 48 mL of 0.25 N Na₂S₂O₃ was used to reach the end point. So, the number of moles of I₂ produced = 48 $\times$ 0.25/2 = 6.

According to the reaction,

Number of millimoles of bleaching powder = Number of moles of I₂ = (1/2) $\times$ Number of moles of Na₂S₂O₃ = 6

So, the molarity of CaOCl₂ is

${{Number\,of\,moles\,of\,bleaching\,powder} \over {Volume\,of\,solution}} = {{6\,mmol} \over {25\,mL}} = 0.24M$

In HNO₃, the oxidation state of N is +5 while in NO, it is +2. In N₂, it is zero and in NH₄Cl, the oxidation state of nitrogen is $-$3. So, the decreasing order of oxidation state of nitrogen is

HNO₃ > NO > N₂ > NH₄Cl

Extra pure N₂ can be obtained by heating Ba(N₃)₂.

Explanation

Thermal Decomposition of Barium Azide

Thermal decomposition of barium azide produces very pure nitrogen gas (N₂).

$\text{Ba(N}_3)_2 \xrightarrow{\Delta} \text{Ba}(s) + 3\text{N}_2(g)$

This reaction yields pure nitrogen gas.

Decomposition of Ammonium Dichromate

Decomposition of ammonium dichromate ((NH₄)₂Cr₂O₇) produces chromium (III) oxide as impurities.

$(\text{NH}_4)_2\text{Cr}_2\text{O}_7 \xrightarrow{\Delta} \text{N}_2(g) + 4\text{H}_2\text{O} + \text{Cr}_2\text{O}_3(s)$

While nitrogen gas is produced, chromium oxide impurities also form.

Heating Ammonium Nitrate

Heating ammonium nitrate (NH₄NO₃) results in the formation of nitrous oxide and water, not pure nitrogen gas.

$\text{NH}_4\text{NO}_3 \xrightarrow{\Delta} \text{N}_2\text{O}(g) + \text{H}_2\text{O}$

No nitrogen gas is produced in this reaction.

Reaction of Ammonia with Copper Oxide

The reaction between ammonia (NH₃) and copper oxide (CuO) gives nitrogen gas along with copper as an impurity.

$2\text{NH}_3(g) + 3\text{CuO}(s) \rightarrow \text{N}_2(g) + 3\text{Cu}(s) + 3\text{H}_2\text{O}(l)$

While nitrogen gas is produced, copper impurities remain.

(A) Dimethyl silyl chloride undergoes hydration followed by polymerisation owing to the loss of water molecule.

(A) → P, S

$ \begin{aligned} & \text { (B) } 6 \mathrm{XeF}_4+12 \mathrm{H}_2 \mathrm{O} \rightarrow 4 \mathrm{Xe}+2 \mathrm{XeO}_3 +24 \mathrm{HF} \text { (Hydrogen fluoride) }+3 \mathrm{O}_2 \text { (Oxygen) } \\ & \end{aligned} $

Here, xenon undergoes disproportionation reaction (which is a kind of redox reaction) with xenon (in +4 oxidation state) gets oxidized to xenon trioxide $\left(\mathrm{XeO}_3\right)$ and reduced to xenon (Xe) at the same time.

HF reacts with glass as shown :

$ \mathrm{Na}_2 \mathrm{SiO}_3+6 \mathrm{HF} \rightarrow \mathrm{Na}_2 \mathrm{SiF}_6+3 \mathrm{H}_2 \mathrm{O} $

(B) → P, Q, R, T

(C)

$2 \mathrm{Cl}_2+2 \mathrm{H}_2 \mathrm{O} \longrightarrow {4 \mathrm{HCl} \text { (Hydrogen } \text { halide )}+\mathrm{O}_2}$

(C) $\rightarrow$ P, Q

(D) $\mathrm{VCl}_5+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{VOCl}_3+2 \mathrm{HCl}$ (Hydrogen chloride)

(D) $\rightarrow P$

The reactions are as follows:

(A) 3Cu + 8HNO$_3$ (dil.) $\to$ 3Cu(NO$_3$)$_2$ + 4H$_2$O + 2NO

(B) Cu + 4HNO$_3$ (conc.) $\to$ Cu(NO$_3$)$_2$ + 4H$_2$O + 2NO$_2$

(C) 4Zn + 10HNO$_3$ (dil.) $\to$ 4Zn(NO$_3$)$_2$ + 5H$_2$O + N$_2$O

(D) Zn + HNO$_3$ (conc.) $\to$ Zn(NO$_3$)$_2$ + 2H$_2$O + 2NO$_2$

Phosphorus reacts with limited supply of oxygen to form the given product.

P$_4$ + 3O$_2$ $\to$ P$_4$O$_6$

N$_2$ is used to retard the further oxidation.

In p-block elements, inert pair effect is observed, due to which lower oxidation state becomes more stable on going down the group. In 14^th group (C, Si, Ge, Sn, Pb) Pb is placed at lower position than Sn, that's why lower oxidation state is more stable for lead (Pb$^{2+}$) and Sn$^{+4}$ is more stable than Pb$^{+4}$. So, Pb$^{+4}$ compounds are stronger oxidising agents than Sn$^{+4}$ compounds. Lower oxidation for the group 14^th elements are more stable for the heavier members.

Due to greater solubility and nature to be prone to microbial action, nitrates are less abundant in earth's crust. Nitrates are soluble. In nitrates, nitrogen has maximum oxidation state and cannot be further oxidised.

Both NH$_3$ and PH$_3$ have one lone pair of electron on it. The lone pair in case of ammonia occupy sp$^3$ orbital is directional.

Hence, the lone pair in case of ammonia is easily available and can be easily donated. But for PH$_3$ the lone pair of electrons occupy spherical s-orbital which is less directional and thus its lone pair is not easily available neither it can be donated easily.

Also the lone pair of electrons in sp$^3$ orbital has lesser s-character and less attracted by nucleus and available for easy donation than in PH$_3$ with pure s orbital, the lone pair in which lone pair are more strongly attracted by nucleus. As we go down a group, the p-character in bond pair increases and the s-character in lone pair increases.

Thus ammonia is a better base than PH$_3$.

White phosphorus on reaction with NaOH gives PH$_3$ as one of the products.

P$_4$ + 3NaOH + 3H$_2$O $\to$ 3NaH$_2$PO$_2$ + PH$_3$

The oxidation state of P in P$_4$, NaH$_2$PO$_2$ and PH$_3$ is 0, +1 and $-$3 respectively. It is an example of disproportionation reaction.

Statement‑1:

“Band gap in germanium is small.”

✅ True.

Germanium is a semiconductor with an energy band gap of about 0.66 eV at 300 K, which is much smaller than that of silicon (~1.1 eV).

Statement‑2:

“The energy gap of each germanium atomic energy level is infinitesimally small.”

❌ False.

In a single germanium atom, the energy levels are discrete with finite energy differences between them — not infinitesimally small.

In a solid, when many atoms come together, each discrete atomic level splits into closely spaced energy states due to interactions between atoms, forming continuous bands; the gap between valence and conduction bands (the band gap) is small but finite.

So, the “infinitesimally small” part is incorrect.

✅ Correct Option: C

Statement‑1 is True, Statement‑2 is False.

A molecule is said to be chiral if it cannot be super imposed on its mirror image by any combination of rotations or translation. A chiral molecule exists as two stereoisomers that are mirror images of each other called enantiomers. Chirality is based on molecular symmetry. A chiral molecule cannot have an improper axis of rotation which includes planes of symmetry and inversion center.

Statement- 1 is true as molecules that are not superimposable on their mirror images are a symmetric and hence chiral. Such molecules are known as enantiomers.

Statement- 2: All chiral molecules don’t have to have a stereogenic center. A molecule can be chiral even if it does not have a stereocenter. A molecule’s chirality depends entirely on whether it is a symmetrical. Example-Biphenyls do not have a chiral carbon but they cannot rotate freely due to steric crowding. Hence, they are a symmetrical. Therefore, we can conclude that Statement-2 is false.

P = 5 Value electrons

Steric number = 3 + 1 = 4

Geometry - Tetrahedral hybridisation $-$ sp$^3$

P character $=\frac{3}{4}\times100=75\%$

H$_3$BO$_3$ (orthoboric acid) is a weak Lewis acid.

H$_3$BO$_3$ + H$_2$O $\rightleftharpoons$ B(OH)$^{-}_{4}$ + H$^ \oplus $

It does not donate proton rather it accepts OH$^-$ form water.

So, Statement 1 is True, Statement 2 is False.