p-Block Elements

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.

Argon is an inert gas; hence, less reactive. It shield metal during welding. Argon is frequently blended with carbon dioxide (CO$_2$), hydrogen (H$_2$), helium (He) or oxygen (O$_2$) to enhance the arc characteristics or facilitate metal transfer in gas metal arc welding.

Hybridisation, H = $\frac{1}{2}$(V + X $-$ C + A) = $\frac{1}{2}$ (8 + 0 $-$ 0 + 0 ) = 4. So, sp$^3$ hybridised and has tetrahedral structure.

No. of lone pair = H $-$ X $-$ D = 4 $-$ 0 $-$ 3 = 1. As lone pair is present the electron pair geometry will be distorted tetrahedral and structure is pyramidal.

H = hybridisation no. (if 2 then sp, if 3 then sp$^2$ etc.)

V = valence electrons of central atom.

X = no. of monovalent atom.

C = cationic charge, A = anionic charge.

D = no. of divalent atom

6XeF$_4$ + 12H$_2$O $\to$ 4Xe + 2XeO$_3$ + 24HF + 3O$_2$

XeF$_6$ + 3H$_2$O $\to$ XeO$_3$ + 6HF

XeF$_4$ and XeF$_6$ are expected to be oxidizing because in both the reactions XeF$_4$ and XeF$_6$ are themselves reduced to xenon (Xe) and oxidises water to oxygen.

When boric acid reacts with sodium hydroxide, the products formed are sodium tetrahydroxoborate (III), sodium metaborate and water. The reaction is reversible in nature.

The reaction is difficult to proceed in forward direction due to weak monoprotic boric acid. When cis-1, 2-diol is added in the mixture of boric acid and sodium hydroxide, the weak acid gets converted to a comparatively stronger acid due to chelate complex formation. This stronger acid along with NaOH can undergo titration using phenolphthalein as an indicator. $\mathrm{B}(\mathrm{OH})_3+\mathrm{NaOH} \rightleftharpoons \mathrm{Na}\left[\mathrm{B}(\mathrm{OH})_4\right]$

Due to stable nature, it is removed from the solution during reaction. The entire reaction proceed in the same manner, disrupting the reversible chemical reaction and favouring the forward reaction completely.

(A) When $\mathrm{Bi}^{3+}$ is treated with water, it gives $[\mathrm{BiO}]^{+}$ ion and two protons.

$ \mathrm{Bi}^{3+}+\mathrm{H}_2 \mathrm{O} \rightarrow\left[\mathrm{BiO}^{+}+2 \mathrm{H}^{+}\right. $

The hydrolysis is a chemical reaction in which water is used for breaking down chemical bonds. The conversion of $\mathrm{Bi}^{+} \rightarrow[\mathrm{BiO}]^{+}$is carried out with the help of water; hence, it is a hydrolysis reaction.

Hence, option A from column I matches with option Q from column II.

(B) The conversion of $\left[\mathrm{AlO}_2\right]^{-} \rightarrow \mathrm{Al}(\mathrm{OH})_3$ is carried out by acidification. The proton is added to one of the reactant to produce aluminium hydroxide. The acid serve as a source of protons required for the acidification. The chemical reaction for the same is shown below:

$ \left[\mathrm{AlO}_2\right]^{-} \rightarrow \mathrm{Al}(\mathrm{OH})_3 $

Hence, option B from column I matches with option R from column II.

(C) Two molecules of orthosillicate $2 \mathrm{SiO}_4^{4-}$ gives pyrosilicate $\mathrm{Si}_2 \mathrm{O}_7^{6-}$ on strong heating.

$ 2 \mathrm{H}_4 \mathrm{SiO}_4 \xrightarrow{\Delta} \mathrm{H}_6 \mathrm{~S}_2 \mathrm{O}_7 $

Equation can also be written as

$ 2 \mathrm{SiO}_4^{4-} \xrightarrow{\Delta} \mathrm{Si}_2 \mathrm{O}_7^{6-} $

Hence, option C from column I matches with option P in column II.

(D) The conversion of $\left(\mathrm{B}_4 \mathrm{O}_7^{2-}\right)$ into $\left[\mathrm{B}(\mathrm{OH})_3\right]$ can be achieved either by hydrolysis or acidification.

The ( $\mathrm{B}_4 \mathrm{O}_7^{2-}$ ) is an anion in $\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7$.

When $\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7$ is treated with HCl or $\mathrm{H}_2 \mathrm{SO}_4$, it gives $\mathrm{H}_3 \mathrm{BO}_3$.

$ \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \xrightarrow{\mathrm{HCl} \text { or } \mathrm{H}_2 \mathrm{SO}_4} \mathrm{H}_3 \mathrm{BO}_3 \text { or } \mathrm{B}(\mathrm{OH})_3 $

Similarly, hydrolysis of $\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7$ yields $\mathrm{H}_3 \mathrm{BO}_3$.

$ \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \xrightarrow{\mathrm{H}_2 \mathrm{O}} \mathrm{H}_3 \mathrm{BO}_3 $

The $\mathrm{H}_3 \mathrm{BO}_3$ is a chemical formula of boric acid and is written as $\left[\mathrm{B}(\mathrm{OH})_3\right]$.

Hence, option D from column I matches with options Q and R in column II.

The correct answer is Option A: N: tetrahedral, sp³; B: tetrahedral, sp³

• BF₃ (Boron trifluoride): In its isolated state, BF₃ has a trigonal planar geometry around the boron atom with sp² hybridization.


• NH₃ (Ammonia): In its isolated state, NH₃ has a trigonal pyramidal geometry around the nitrogen atom with sp³ hybridization.

However, when BF₃ and NH₃ form a 1:1 complex, the lone pair of electrons on nitrogen donates to the empty p-orbital of boron. This leads to the formation of a coordinate bond between nitrogen and boron.

• Nitrogen (N): Now has four bonding pairs of electrons (three with hydrogen and one with boron), resulting in a tetrahedral geometry and sp³ hybridization.


• Boron (B): Also has four bonding pairs of electrons (three with fluorine and one with nitrogen), leading to a tetrahedral geometry and sp³ hybridization.

The oxidation number of an atom in a molecule or an ion is essentially a theoretical charge that the atom would have if all the bonds to atoms of different elements were completely ionic. To determine the oxidation numbers of sulfur in the given compounds: S₈, S₂F₂, and H₂S, let's analyze them one by one.

1. In S₈: This is an elemental form of sulfur, comprising of sulfur atoms only. In elemental forms, the atoms are not bonded to atoms of a different element, but only to themselves. In such cases, the oxidation number is always 0. Hence, the oxidation number of sulfur in S₈ is 0.

2. In S₂F₂: This is a molecule consisting of sulfur and fluorine. Fluorine is more electronegative than sulfur and virtually always has an oxidation number of -1 in its compounds. Let's denote the oxidation number of sulfur as $x$. Since there are two fluorine atoms, each contributing -1, and the overall molecule is neutral, we get:

$2x + (2 \times -1) = 0$

$2x - 2 = 0$

$2x = 2$

$x = +1$

Therefore, the oxidation number of sulfur in S₂F₂ is +1.

3. In H₂S: In this compound, hydrogen has an oxidation number of +1 (except when bonded to metals where it can be -1). Since there are two hydrogens, their total contribution is +2. Let's denote the oxidation number of sulfur in this compound as $x$. The compound is neutral overall, so:

$2(+1) + x = 0$

$2 + x = 0$

$x = -2$

Therefore, the oxidation number of sulfur in H₂S is -2.

Comparing with the given options, the correct answers are 0, +1, and -2 for S₈, S₂F₂, and H₂S, respectively. Therefore, the correct option is Option A: 0, +1, and -2.

Option (A) : Correct. The basicity of oxides usually increases on descending a group. Therefore, Bi₂O5 is more basic than N₂O₅.

Option (B) : Correct. Covalent nature of a molecule depends on the electronegativity difference between bonded atoms.

Option (C) : Correct. Boiling point of NH₃ is more than that of PH₃ due to hydrogen bonding.

Option (D) : Incorrect. P$-$P single bond is stronger than N$-$N single bond. This is due to the fact that N is small in size, due to smaller size of atoms lone pair of repulsion will be more.

(A) $N{H_4}N{O_3}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{below\,300^\circ C}} {N_2}O + 2{H_2}O$

${N_2}O\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{above\,600^\circ C}} {N_2} + {O_2}$

(B) ${(N{H_4})_2}C{r_2}{O_7}\buildrel \Delta \over \longrightarrow {N_2} + C{r_2}{O_3} + 4{H_2}O$

(C) $Ba{({N_3})_2}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{180^\circ C}^\Delta } Ba + 3{N_2}$

(D) $M{g_3}{N_2}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{above\,700^\circ C}} 3Mg + {N_2}$

Hence, only ${(N{H_4})_2}C{r_2}{O_7}$ and $Ba{({N_3})_2}$ can provide N₂ gas on heating below $300^\circ C$