Nitrogen family Questions in English

(B) The $P_4$ molecule has a tetrahedral structure. In this structure,each phosphorus $(P)$ atom is bonded to three other phosphorus atoms by single covalent bonds. This results in a bond angle of $60^{\circ}$ between the $P-P-P$ bonds,which creates significant angular strain. Therefore,each $P$ atom is joined to three $P$ atoms.

(C) Statement $(I)$ is $INCORRECT$: Ammonia reacts with anhydrous $CaCl_2$ to form an adduct $CaCl_2 \cdot 8NH_3$,so it cannot be used as a drying agent.
Statement $(II)$ is $CORRECT$: Phosphine $(PH_3)$ is a much weaker base than ammonia $(NH_3)$ due to the larger size and lower electronegativity of the phosphorus atom,making the lone pair less available.
Statement $(III)$ is $INCORRECT$: $R_3SiCl$ on hydrolysis produces $R_3SiOH$,which undergoes dimerization to form $R_3Si-O-SiR_3$ (a disiloxane). It does not form cross-linked polymers; only $R_2SiCl_2$ forms linear polymers and $RSiCl_3$ forms cross-linked polymers.
Statement $(IV)$ is $INCORRECT$: When copper reacts with $6 \ M$ $HNO_3$ (dilute),the product is $NO$ (Nitric oxide),not $NO_2$. $NO_2$ is produced with concentrated $HNO_3$.
Therefore,statements $(I)$,$(III)$,and $(IV)$ are incorrect.

(D) The reaction of metals with nitric acid $(HNO_3)$ depends on the concentration of the acid and the reactivity of the metal.
For $Zn$ (a more reactive metal),the reduction products are $N_2O$ (with dilute $HNO_3$),$NO_2$ (with concentrated $HNO_3$),and $NH_4NO_3$ (with very dilute $HNO_3$).
For $Cu$ (a less reactive metal),the reaction with dilute $HNO_3$ produces $NO$ (Nitric oxide),not $N_2O$.
Therefore,the reaction $Cu + \text{dil. } HNO_3 o Cu(NO_3)_2 + N_2O$ is incorrect.

(A) Metals like $Cr$ (Chromium) and $Al$ (Aluminium) do not dissolve in concentrated nitric acid $(HNO_3)$ because they form a thin,non-reactive,protective oxide layer on their surface,a phenomenon known as passivity.

(C) In option $A$,$NH_4Cl$ reacts with a strong base $KOH$ to produce $NH_3$ gas.
In option $B$,$NH_4Cl$ undergoes thermal decomposition to produce $NH_3$ and $HCl$ gases.
In option $C$,$NH_4NO_2$ undergoes thermal decomposition to produce $N_2$ gas and water,not $NH_3$.
In option $D$,$NH_4NO_2$ is reduced by $Zn/NaOH$ to produce $NH_3$ gas.
Therefore,the process that does not produce ammonia gas is the thermal decomposition of ammonium nitrite.

(A) The thermal decomposition of ammonium nitrate $(NH_4NO_3)$ is a standard laboratory method for the preparation of nitrous oxide $(N_2O)$.
The reaction is: $NH_4NO_3 \xrightarrow{\Delta} N_2O + 2H_2O$.
Option $A$ correctly represents this reaction.

(D) The reaction $Sn + PCl_5 \to SnCl_2 + PCl_3$ is incorrect.
The correct reaction between tin $(Sn)$ and phosphorus pentachloride $(PCl_5)$ is:
$Sn + 2PCl_5 \to SnCl_4 + 2PCl_3$.
Option $A$ is correct: $P_4 + 10SO_2Cl_2 \to 4PCl_5 + 10SO_2$.
Option $B$ is correct: $P_4 + 8SO_2Cl_2 \to 4PCl_3 + 4SO_2 + 2S_2Cl_2$.
Option $C$ is correct: $3CH_3COOH + PCl_3 \to 3CH_3COCl + H_3PO_3$.

(B) Ammonia $(NH_3)$ gas reacts with hydrogen chloride $(HCl)$ gas to form ammonium chloride $(NH_4Cl)$ solid particles.
The reaction is: $NH_3(g) + HCl(g) \rightarrow NH_4Cl(s)$.
These solid particles of $NH_4Cl$ appear as white fumes.

(C) $1$. The reaction $NO + O_2 \rightarrow NO_2$ gives $A = NO_2$.
$2$. $NO_2$ exists in equilibrium with its dimer $N_2O_4$,so $D = N_2O_4$.
$3$. When $NO_2$ reacts with aqueous $NaOH$,it undergoes a disproportionation reaction:
$2NO_2 + 2NaOH \rightarrow NaNO_3 + NaNO_2 + H_2O$.
$4$. Here,$B$ and $C$ are $NaNO_3$ and $NaNO_2$.

(B) When phosphorous acid $(H_3PO_3)$ is heated,it undergoes disproportionation reaction to form phosphoric acid $(H_3PO_4)$ and phosphine $(PH_3)$.
The chemical equation is: $4H_3PO_3 \xrightarrow{\Delta} 3H_3PO_4 + PH_3$.

(B) The cyclic trimeric metaphosphate is $P_3O_9^{3-}$.
In this structure,each phosphorus atom is bonded to two bridging oxygen atoms and two non-bridging oxygen atoms (one of which is double-bonded to phosphorus).
$1$. The structure consists of a ring of alternating $P$ and $O$ atoms,so there are no $P-P$ linkages.
$2$. In $P_3O_9^{3-}$,each of the $3$ phosphorus atoms has one $P=O$ bond,contributing $3$ $p\pi-d\pi$ bonds. Similarly,the trimeric form of $SO_3$ $(S_3O_9)$ has $3$ $S=O$ bonds,each contributing $p\pi-d\pi$ character,making the number of $p\pi-d\pi$ bonds equal.
$3$. The $P-O$ bond lengths are not identical because there are $P-O$ single bonds (bridging) and $P=O$ double bonds (terminal).
$4$. Each phosphorus atom is surrounded by $4$ oxygen atoms in a tetrahedral geometry,meaning each phosphorus is $sp^3$ hybridised.

(C) The boiling point of hydrides of Group $15$ elements follows the order: $PH_3 < AsH_3 < NH_3 < SbH_3$.
$NH_3$ exhibits intermolecular hydrogen bonding due to the high electronegativity and small size of the nitrogen atom,which gives it an abnormally high boiling point compared to $PH_3$ and $AsH_3$.
However,as we move down the group from $PH_3$ to $SbH_3$,the molecular size and mass increase significantly,leading to stronger van der Waals forces of attraction.
$SbH_3$ has the highest molecular mass among the given hydrides,resulting in the strongest van der Waals forces,which makes its boiling point higher than that of $NH_3$.

(A) The reaction of white phosphorus $(P_4)$ with aqueous sodium hydroxide $(NaOH)$ is a disproportionation reaction.
The balanced chemical equation is: $P_4 + 3 NaOH + 3 H_2O \rightarrow PH_3 + 3 NaH_2PO_2$.
In this reaction,phosphorus is both oxidized and reduced.
The product $X$ is sodium hypophosphite,which is $NaH_2PO_2$.

(A) Nitrogen dioxide $(NO_2)$ is not produced by heating $KNO_3$. The thermal decomposition reactions are as follows:
$1) \ 2 KNO_3 \rightarrow 2 KNO_2 + O_2$ (Note: At very high temperatures,$KNO_2$ may further decompose,but $NO_2$ is not the primary product).
$2) \ 2 Pb(NO_3)_2 \rightarrow 2 PbO + 4 NO_2 + O_2$
$3) \ 2 Cu(NO_3)_2 \rightarrow 2 CuO + 4 NO_2 + O_2$
$4) \ 2 AgNO_3 \rightarrow 2 Ag + 2 NO_2 + O_2$
As shown,$Pb(NO_3)_2$,$Cu(NO_3)_2$,and $AgNO_3$ all yield $NO_2$ upon heating,whereas $KNO_3$ does not.

(C) The thermal decomposition of orthophosphoric acid $(H_3PO_4)$ occurs in stages.
At $250\ ^oC$,$H_3PO_4$ loses water to form pyrophosphoric acid $(H_4P_2O_7)$,which is a solid: $2H_3PO_4 \xrightarrow{250\ ^oC} H_4P_2O_7 + H_2O$.
At $600\ ^oC$,further heating leads to the formation of metaphosphoric acid $(HPO_3)$,which is a solid: $H_4P_2O_7 \xrightarrow{600\ ^oC} 2HPO_3 + H_2O$.
Therefore,$(X)$ is pyrophosphoric acid (solid) and $(Y)$ is metaphosphoric acid (solid).

(A) The thermal decomposition of ammonium dichromate $(NH_4)_2Cr_2O_7$ is given by the reaction: $(NH_4)_2Cr_2O_7 \rightarrow N_2 + Cr_2O_3 + 4H_2O$.
Here,the orange solid $(X)$ is $(NH_4)_2Cr_2O_7$,the colourless gas $(Y)$ is $N_2$,and the green residue $(Z)$ is $Cr_2O_3$.
When the gas $(Y)$,which is $N_2$,reacts with magnesium $(Mg)$,it forms magnesium nitride: $3Mg + N_2 \rightarrow Mg_3N_2$.
Magnesium nitride $(Mg_3N_2)$ is a white solid substance.

(B) Concentrated nitric acid $(HNO_3)$ is a colorless liquid when pure.
However,it often appears yellow because of the decomposition of $HNO_3$ into nitrogen dioxide $(NO_2)$,water,and oxygen upon exposure to light or heat.
The reaction is: $4HNO_3 \rightarrow 4NO_2 + 2H_2O + O_2$.
The produced $NO_2$ gas dissolves in the concentrated $HNO_3$,imparting a yellow color to the liquid.

(B) The gas described is $N_2$ (nitrogen).
$1$. It is almost inert at room temperature due to the very high bond dissociation energy of the $N \equiv N$ triple bond $(941.4 \ kJ \ mol^{-1})$.
$2$. It is used to create an inert atmosphere in electric bulbs to prevent the oxidation of the tungsten filament.
$3$. The combustion of nitrogen to form nitrogen oxides is an endothermic process (e.g.,$N_2(g) + O_2(g) \rightarrow 2NO(g)$,$\Delta H > 0$),which is unusual as most combustion reactions are exothermic.

(C) The thermal decomposition of hypophosphorous acid $(H_3PO_2)$ occurs in steps:
$1$. At $140^{\circ}C$,$3H_3PO_2 \rightarrow 2H_3PO_3 + PH_3$ (Compound $A$ is $H_3PO_3$)
$2$. At $250^{\circ}C$,$4H_3PO_3 \rightarrow 3H_3PO_4 + PH_3$ (Compound $B$ is $H_3PO_4$)
$3$. At $316^{\circ}C$,$H_3PO_4 \rightarrow HPO_3 + H_2O$ (Compound $C$ is $HPO_3$)
Thus,the final product $(C)$ is metaphosphoric acid,$HPO_3$.

(A) The correct option is $(A).$
The decomposition of ammonium nitrate $(NH_4NO_3)$ at $523 \ K$ is given by the reaction: $NH_4NO_3 \rightarrow N_2O + 2H_2O.$
Here,$(A) = NH_4NO_3$,$(B) = N_2O$,and $(C) = H_2O.$
$H_2O$ is a liquid at room temperature and is neutral to litmus paper.
When the oxide $(B)$ $(N_2O)$ reacts with white phosphorus $(P_4)$,it acts as an oxidizing agent to produce phosphorus pentoxide $(P_2O_5)$,which is a well-known dehydrating agent $(D).$
The reaction is: $10N_2O + P_4 \rightarrow 2P_2O_5 + 10N_2.$

(A) The reaction is: $P_4 + 10 N_2O \rightarrow P_4O_{10} + 10 N_2$.
$(A)$ is $P_4$,which is a tetra-atomic molecule.
$(B)$ is $P_4O_{10}$,which acts as a powerful dehydrating agent.
$(C)$ is $N_2$,which is a diatomic gas that shows almost inert behavior due to the high bond dissociation energy of the $N \equiv N$ triple bond.

(B) Nitrous oxide $(N_2O)$ and nitric oxide $(NO)$ are neutral oxides.
Acidic character of oxides increases with an increase in the oxidation state of the central atom.
The oxidation states of nitrogen in the given oxides are:
$N_2O (+1), NO (+2), N_2O_3 (+3), N_2O_4 (+4), N_2O_5 (+5)$.
Since $N_2O$ and $NO$ are neutral,they have the least acidic character.
Thus,the correct order of acidic strength is: $N_2O \approx NO < N_2O_3 < N_2O_4 < N_2O_5$.

(A) When nitrogen dioxide $(NO_2)$ is dissolved in water,it undergoes a disproportionation reaction to form a mixture of nitric acid $(HNO_3)$ and nitrous acid $(HNO_2)$.
The chemical equation is: $2 NO_2 + H_2O \rightarrow HNO_3 + HNO_2$

(A) The thermal decomposition reactions are as follows:
$1$. $(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} N_2 + Cr_2O_3 + 4H_2O$ (evolves $N_2$)
$2$. $NH_4NO_3 \xrightarrow{\Delta} N_2O + 2H_2O$ (evolves $N_2O$)
$3$. $(NH_4)_2SO_4 \xrightarrow{\Delta} NH_3 + NH_4HSO_4$ (evolves $NH_3$)
Ammonium sulphate on strong heating releases ammonia gas as one of the products. Therefore,the correct option is $A$.

(C) Ammonia $(NH_3)$ is a basic gas.
To dry a gas,the drying agent must not react with the gas.
$H_2SO_4$ is an acid and reacts with $NH_3$ to form $(NH_4)_2SO_4$.
$P_4O_{10}$ is an acidic oxide and reacts with $NH_3$.
$CaCl_2$ forms an adduct with $NH_3$ $(CaCl_2 \cdot 8NH_3)$.
$CaO$ (quicklime) is a basic drying agent and does not react with basic $NH_3$ gas,making it the suitable choice for drying.

(B) When $NH_3$ reacts with an excess of $Cl_2,$ an explosive substance $NCl_3$ is formed.
The chemical reaction is:
$NH_3 + 3Cl_2 \rightarrow NCl_3 + 3HCl$
Here,$X$ is $NH_3$ and $Y$ is $NCl_3$.

(B) When $AgNO_3$ is heated strongly,it undergoes thermal decomposition to form metallic silver,nitrogen dioxide,and oxygen gas.
The balanced chemical equation is:
$2AgNO_{3(s)} \xrightarrow{\Delta} 2Ag_{(s)} + 2NO_{2(g)} + O_{2(g)}$
Thus,the products formed are $NO_2$ and $O_2$.

(D) The reaction between concentrated nitric acid $(HNO_3)$ and phosphorus pentoxide $(P_4O_{10})$ is a dehydration reaction.
$P_4O_{10}$ acts as a strong dehydrating agent.
The balanced chemical equation is:
$4HNO_3 + P_4O_{10} \rightarrow 4HPO_3 + 2N_2O_5$
Thus,the product $A$ is $N_2O_5$.

(B) The thermal decomposition of $NaH_2PO_4$ proceeds as follows:
$3NaH_2PO_4 \xrightarrow{230^\circ C} Na_3P_3O_9 + 3H_2O$
$Na_3P_3O_9 \xrightarrow{638^\circ C} 3(NaPO_3)_n$ (where $n \approx 6$ for hexametaphosphate).
The product $D$,which is sodium hexametaphosphate $(NaPO_3)_6$,is commonly known as Graham’s salt.

(A) The allotropes of phosphorus are white,red,and black phosphorus.
$1$. White phosphorus $(A)$ is converted to black phosphorus $(B)$ at $470 \ K$ and $1200 \ atm$ pressure.
$2$. White phosphorus $(A)$ is converted to red phosphorus $(C)$ by heating it in an inert atmosphere of $CO_2$ at $570 \ K$.
Therefore,$(A)$ is white,$(B)$ is black,and $(C)$ is red.

(B) The reaction of copper with concentrated nitric acid is given by:
$Cu(s) + 4HNO_3(conc.) \rightarrow Cu(NO_3)_2(aq) + 2NO_2(g) + 2H_2O(l)$
Here,$X$ is nitrogen dioxide $(NO_2)$.

(D) The cyclic trimeric polymetaphosphate anion $(P_3O_9)^{3-}$ consists of a ring of $3$ phosphorus atoms alternating with $3$ oxygen atoms,with each phosphorus atom also bonded to two terminal oxygen atoms (one double-bonded and one single-bonded).
$1$. The structure contains $3$ $P-O-P$ linkages in the ring.
$2$. Each phosphorus atom is $sp^3$ hybridized.
$3$. Each phosphorus atom forms one $P=O$ bond,which involves $p\pi-d\pi$ back-bonding. Since there are $3$ phosphorus atoms,there are $3$ $P=O$ bonds,hence $3$ $p\pi-d\pi$ bonds.
$4$. The number of $P-O-P$ linkages is $3$,and the number of $sp^3$ hybridized phosphorus atoms is $3$. Thus,they are equal.
Therefore,the statement in option $D$ is incorrect.

(B) The reactions are as follows:
$1$. Heating $(NH_4)_2Cr_2O_7$: $(NH_4)_2Cr_2O_7 \rightarrow N_2 + Cr_2O_3 + 4H_2O$ (Produces $N_2$)
$2$. $NH_3$ + Excess $Cl_2$: $NH_3 + 3Cl_2 \rightarrow NCl_3 + 3HCl$ (Produces $NCl_3$,not $N_2$)
$3$. Heating $NaN_3$: $2NaN_3 \rightarrow 2Na + 3N_2$ (Produces $N_2$)
$4$. Heating $NH_4NO_2$: $NH_4NO_2 \rightarrow N_2 + 2H_2O$ (Produces $N_2$)
Therefore,the reaction that does not produce nitrogen is $NH_3$ + Excess $Cl_2$.

(D) Isohypophosphoric acid has the formula $H_4P_2O_6$ (specifically the isomer $H_2P(O)O-P(O)(OH)_2$).
Upon hydrolysis,it yields a mixture of phosphoric acid $(H_3PO_4)$ and phosphonic acid $(H_3PO_3)$.
Hypophosphoric acid ($H_4P_2O_6$,structure $(HO)_2(O)P-P(O)(OH)_2$) also undergoes hydrolysis to yield the same mixture of phosphoric acid $(H_3PO_4)$ and phosphonic acid $(H_3PO_3)$ in acidic medium.
Therefore,the products of hydrolysis are the same.

(B) The reactions in the Ostwald process are:
$4NH_3 + 5O_2 \xrightarrow[Pt]{\Delta} 4NO + 6H_2O$
$2NO + O_2 \to 2NO_2$
$3NO_2 + H_2O \to 2HNO_3 + NO$
Here,$x$ is $NO$,$y$ is $NO_2$,and $z$ is $HNO_3$.
$1$. $NO$ $(x)$ is paramagnetic due to the presence of an odd number of electrons.
$2$. $NO_2$ $(y)$ and $HNO_3$ $(z)$ are acidic,but $NO$ $(x)$ is a neutral oxide. Therefore,the statement that $x$ and $y$ are acidic oxides is incorrect.
$3$. $HNO_3$ $(z)$ is a strong oxidising agent.
$4$. Both $NO_2$ and $HNO_3$ are acidic in nature.

(C) Thermal decomposition of sodium or barium azide gives pure $N_2$ gas.
$Ba(N_3)_2 \xrightarrow{\Delta} Ba + 3N_2$
Thus,the correct option is $C$.

(B) The acidity of hydrides depends on the stability of the conjugate base formed after the loss of a proton $(H^+)$.
In $N_3H$ (hydrazoic acid),the conjugate base is the azide ion $(N_3^-)$,which is highly resonance-stabilized.
$NH_3$ is basic due to the lone pair on nitrogen.
$HNO_3$ is an oxoacid,not a simple hydride of nitrogen.
Therefore,among the given hydrides,$N_3H$ is the most acidic.

(A, B, C, D) All the given matches are correct:
$1$. The Moissan process is used for the electrolytic preparation of $F_2$ gas.
$2$. The Solvay process is the industrial method for the production of $Na_2CO_3$ (soda ash).
$3$. The Spring reaction is used for the preparation of sodium thiosulfate $(Na_2S_2O_3 \cdot 5H_2O)$,commonly known as Hypo.
$4$. The Ostwald process is the industrial method for the production of $HNO_3$ from ammonia.

(B) The reaction between $PCl_{5}$ and $P_{4}O_{10}$ is given by:
$6 PCl_{5} + P_{4}O_{10} \xrightarrow{\Delta } 10 POCl_{3}$.
Thus,the compound $(X)$ is $POCl_{3}$.
The hydrolysis reaction of $POCl_{3}$ is:
$POCl_{3} + 3 H_{2}O \longrightarrow H_{3}PO_{4} + 3 HCl$.
From the stoichiometry of the reaction,$1 \text{ mole}$ of $POCl_{3}$ produces $1 \text{ mole}$ of phosphoric acid $(H_{3}PO_{4})$,which is an oxyacid.
Therefore,the correct answer is $1$.

(C) $1$. Orthophosphorous acid $(H_3PO_3)$: Contains $1$ $P-H$ bond.
$2$. Pyrophosphoric acid $(H_4P_2O_7)$: Contains $0$ $P-H$ bonds.
$3$. Pyrophosphorous acid $(H_4P_2O_5)$: Contains $2$ $P-H$ bonds.
$4$. Hypophosphoric acid $(H_4P_2O_6)$: Contains $0$ $P-H$ bonds.
Therefore,Pyrophosphorous acid has the maximum number of $P-H$ bonds.

(C) $1$. Heating $AlCl_3 \cdot 6H_2O$ results in the formation of $Al_2O_3$ and $HCl$ gas,not anhydrous $AlCl_3$.
$2$. Reaction of $Al$ with moist $HCl$ produces hydrated $AlCl_3$.
$3$. The reaction $Al_2O_3 + 3C + 3Cl_2 \xrightarrow{\Delta} 2AlCl_3 + 3CO$ is the standard industrial method for preparing anhydrous $AlCl_3$.

(B) The reaction of white phosphorus $(P_4)$ with sodium hydroxide $(NaOH)$ solution is a disproportionation reaction.
The chemical equation is: $P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$.
Here,$PH_3$ is phosphine and $NaH_2PO_2$ is sodium hypophosphite.
Sodium hypophosphite $(NaH_2PO_2)$ is the salt of hypophosphorous acid $(H_3PO_2)$.
Since $H_3PO_2$ has only one ionizable hydrogen atom,it is a monobasic acid.
Therefore,the products are phosphine and the sodium salt of a monobasic acid.

(D) $Mg_3N_2 + 6H_2O \rightarrow 3Mg(OH)_2 + 2NH_3$
$NaNO_3 + 4Zn + 7NaOH \rightarrow 4Na_2ZnO_2 + NH_3 + 2H_2O$
$CaNCN + 3H_2O \rightarrow CaCO_3 + 2NH_3$
In all the given reactions,ammonia $(NH_3)$ is produced as one of the products.

(C) To identify the $X-O-X$ linkage,we examine the structures of the given oxyacids:
$1$. $H_2S_2O_6$ (Dithionic acid) has an $S-S$ linkage: $(HO)SO_2-SO_2(OH)$.
$2$. $H_2S_2O_5$ (Pyrosulphurous acid) has an $S-O-S$ linkage: $(HO)SO-O-SO_2(OH)$.
$3$. $H_4P_2O_5$ (Pyrophosphorous acid) has a $P-O-P$ linkage: $(HO)_2P-O-P(OH)H$.
$4$. $H_4P_2O_8$ (Peroxodiphosphoric acid) has a $P-O-O-P$ linkage.
Both $H_2S_2O_5$ and $H_4P_2O_5$ contain an $X-O-X$ linkage. However,in standard multiple-choice contexts for this specific question,$H_4P_2O_5$ is the classic example of a $P-O-P$ linkage.

(A) Due to the inert pair effect,the stability of the $+2$ oxidation state is higher than the $+4$ oxidation state for lead $(Pb)$.
Therefore,$Pb^{4+}$ has a strong tendency to gain two electrons to form $Pb^{2+}$,making it a good oxidising agent.

(B) The given reactions are part of the Ostwald process for the manufacture of nitric acid:
$1$. $4NH_3 + 5O_2 \xrightarrow[\Delta]{Pt} 4NO + 6H_2O$. Thus,$A = NO$.
$2$. $2NO + O_2 \to 2NO_2$. Thus,$B = NO_2$.
$3$. $4NO_2 + O_2 + 2H_2O \to 4HNO_3$. Thus,$C = HNO_3$.
Therefore,$A, B$ and $C$ are $NO, NO_2$ and $HNO_3$ respectively.

(A) Silver reacts with dilute nitric acid to liberate nitric oxide $(NO)$ gas.
$3 Ag + 4 HNO_3 (\text{dil}) \rightarrow 3 AgNO_3 + 2 H_2O + NO \uparrow$

(D) The thermal decomposition reactions are as follows:
$NH_4NO_3 \xrightarrow{\Delta} N_2O + 2H_2O$
$NH_4NO_2 \xrightarrow{\Delta} N_2 + 2H_2O$
$Ba(N_3)_2 \xrightarrow{\Delta} Ba + 3N_2$
$2AgNO_3 \xrightarrow{\Delta} 2Ag + 2NO_2 + O_2$
Only $AgNO_3$ produces $NO_2$ gas upon heating.
Therefore,the correct option is $D$.

(D) The stability of oxidation states in $p$-block elements is governed by the inert pair effect.
$(a)$ Incorrect: $Pb^{+2} > Pb^{+4}$ is correct,but $Tl^{+1} > Tl^{+3}$ is correct due to the inert pair effect.
$(b)$ Incorrect: $Bi^{+3} > Sb^{+3}$ is correct (stability of $+3$ increases down the group),and $Sn^{+2} < Sn^{+4}$ is correct.
$(c)$ Correct: $Pb^{+2} > Pb^{+4}$ is correct (inert pair effect),and $Bi^{+3} > Bi^{+5}$ is correct (inert pair effect).
$(d)$ Incorrect: $Tl^{+3} < In^{+3}$ is correct,but $Sn^{+2} > Sn^{+4}$ is incorrect ($Sn^{+4}$ is more stable than $Sn^{+2}$).
$(e)$ Correct: $Sn^{+2} < Pb^{+2}$ is correct (inert pair effect),and $Sn^{+4} > Pb^{+4}$ is correct (stability of higher oxidation state decreases down the group).
$(f)$ Incorrect: $Sn^{+4} < Pb^{+4}$ is incorrect.
Thus,statements $(c)$ and $(e)$ are correct.

(B) The hydrolysis of $PCl_5$ occurs in two steps:
Step $1$: $PCl_5 + H_2O \to POCl_3 + 2HCl$
Here,$(A) = PCl_5$ and $(B) = POCl_3$.
Step $2$: $POCl_3 + 3H_2O \to H_3PO_4 + 3HCl$
Here,$(B) = POCl_3$ and $(C) = H_3PO_4$.
Thus,the compounds are $(A) = PCl_5$,$(B) = POCl_3$,and $(C) = H_3PO_4$.