Nitrogen family Questions in English

(C) Aqua regia is a mixture of concentrated $HCl$ and concentrated $HNO_{3}$ in a $3:1$ molar ratio.
It dissolves noble metals like gold $(Au)$ and platinum $(Pt)$ by oxidizing them.
The chemical reaction for the dissolution of gold is:
$Au + 3HNO_{3} + 4HCl \rightarrow HAuCl_{4} + 3NO + 2H_{2}O$.
As seen in the reaction,the gas evolved is nitric oxide $(NO)$.

(D) Lead$(II)$ nitrate decomposes on heating to give nitrogen dioxide $(NO_2)$,which is a brown gas $(A)$.
$2Pb(NO_3)_2 \xrightarrow{\Delta} 2PbO + 4NO_2 + O_2$
On cooling,$NO_2$ dimerizes to form dinitrogen tetroxide $(N_2O_4)$,which is a colourless solid/liquid $(B)$.
$2NO_2 \xrightarrow{\text{cooling}} N_2O_4$
$N_2O_4$ reacts with nitric oxide $(NO)$ to form dinitrogen trioxide $(N_2O_3)$,which is a blue solid $(C)$.
$N_2O_4 + 2NO \rightarrow 2N_2O_3$
In $N_2O_3$,let the oxidation state of $N$ be $x$.
$2(x) + 3(-2) = 0$ $\Rightarrow 2x - 6 = 0$ $\Rightarrow x = +3$.

(D) When ammonia $(NH_3)$ reacts with an excess of chlorine $(Cl_2)$,it produces nitrogen trichloride $(NCl_3)$ and hydrogen chloride $(HCl)$.
The balanced chemical equation is:
$NH_3 + 3Cl_2 \text{ (excess)} \rightarrow NCl_3 + 3HCl$
Note: If ammonia were in excess,the products would be $NH_4Cl$ and $N_2$.

(D) The reaction between $NO$ and $N_2O_4$ at low temperatures $(250 \ K)$ results in the formation of dinitrogen trioxide $(N_2O_3)$.
The balanced chemical equation is: $2 NO + N_2O_4 \xrightarrow{250 \ K} 2 N_2O_3$.
$N_2O_3$ is a blue solid at this temperature.

(B) $1$. Liquid nitrogen is used as a refrigerant to preserve biological material,food items,and in cryosurgery.
$2$. In the iron and chemical industry,it is used as an inert diluent for reactive chemicals.
$3$. $N_2$ does not react with dioxygen at room temperature $(25^{\circ} C)$; it requires very high temperatures $(\,2000 \ K)$ to form nitric oxide $(NO)$.
$4$. $N_2$ is diamagnetic in nature because all electrons are paired.

(C) The correct increasing order of basic strength for group $15$ hydrides is: $NH_3 > PH_3 > AsH_3 > SbH_3$.
$NH_3$ is the most basic because of its small size,where the electron density of the lone pair is concentrated over a small region. As the size of the central atom increases,the electron density gets diffused over a larger surface area,decreasing the ability to donate the electron pair.
Thus,the arrangement in option $C$ is incorrect.
Option $A$ is correct as oxidising power increases down the group due to the inert pair effect.
Option $B$ is correct as acidic strength increases with the increase in bond length down the group.
Option $D$ is correct based on the electronic configuration and stability of half-filled $p$-orbitals in $N$.

(D) The structure of hypophosphoric acid $(H_4P_2O_6)$ is $(HO)_2P(=O)-P(=O)(OH)_2$,which contains a direct $P-P$ bond and no $P-O-P$ linkage.
In contrast,pyrophosphorus acid $(H_4P_2O_5)$,diphosphoric acid $(H_4P_2O_7)$,and isohypophosphoric acid ($H_4P_2O_6$ isomer) all contain a $P-O-P$ linkage.
Therefore,the correct option is $(D)$.

(A) The reaction of ammonia with chlorine depends on the relative amounts of the reactants.
When ammonia is in excess,the reaction is: $8 NH_{3} + 3 Cl_{2} \rightarrow 6 NH_{4}Cl + N_{2}$. Thus,$X = 6 NH_{4}Cl + N_{2}$.
When chlorine is in excess,the reaction is: $NH_{3} + 3 Cl_{2} \rightarrow NCl_{3} + 3 HCl$. Thus,$Y = NCl_{3} + 3 HCl$.

(B) The products of the reaction of copper with $HNO_{3}$ depend upon the concentration of $HNO_{3}$ used.
Copper metal reacts with dilute $HNO_{3}$ to form nitrogen $(II)$ oxide $(NO)$:
$3Cu + 8HNO_{3} \text{ (dil.)} \rightarrow 3Cu(NO_{3})_{2} + 2NO + 4H_{2}O$
Copper metal reacts with concentrated $HNO_{3}$ to form nitrogen $(IV)$ oxide or nitrogen dioxide $(NO_{2})$:
$Cu + 4HNO_{3} \text{ (conc.)} \rightarrow Cu(NO_{3})_{2} + 2NO_{2} + 2H_{2}O$
Thus,depending on the concentration,both $NO$ and $NO_{2}$ can be obtained.

(B) The structures of the oxo-acids of phosphorus are as follows:
$1$. Orthophosphoric acid $(H_3PO_4)$: It has $0$ $P-H$ bonds and $3$ $OH$ groups. Ratio = $0/3 = 0$.
$2$. Hypophosphorous acid $(H_3PO_2)$: It has $2$ $P-H$ bonds and $1$ $OH$ group. Ratio = $2/1 = 2$.
$3$. Phosphorous acid $(H_3PO_3)$: It has $1$ $P-H$ bond and $2$ $OH$ groups. Ratio = $1/2 = 0.5$.
$4$. Pyrophosphoric acid $(H_4P_2O_7)$: It has $0$ $P-H$ bonds and $4$ $OH$ groups. Ratio = $0/4 = 0$.
Comparing the ratios,Hypophosphorous acid has the highest ratio of $2$.

(A) The reaction of zinc $(Zn)$ with nitric acid $(HNO_{3})$ depends on the concentration of the acid.
$1$. With dilute $HNO_{3}$,zinc reacts to produce nitrous oxide $(N_{2}O)$:
$4Zn 10HNO_{3} (\text{dilute}) \rightarrow 4Zn(NO_{3})_{2} 5H_{2}O N_{2}O$
$2$. With concentrated $HNO_{3}$,zinc reacts to produce nitrogen dioxide $(NO_{2})$:
$Zn 4HNO_{3} (\text{conc.}) \rightarrow Zn(NO_{3})_{2} 2H_{2}O 2NO_{2}$
Therefore,the products are $N_{2}O$ and $NO_{2}$ respectively.

(A) When ammonia $(NH_3)$ reacts with excess bleaching powder $(CaOCl_2)$,it undergoes oxidation to produce dinitrogen gas $(N_2)$.
The balanced chemical equation is:
$3 CaOCl_2 + 2 NH_3 \rightarrow 3 CaCl_2 + N_2 + 3 H_2O$

(A) Hypophosphorous acid $(H_{3}PO_{2})$: $3(+1) + x + 2(-2) = 0 \Rightarrow x = +1$.
$(b)$ Orthophosphoric acid $(H_{3}PO_{4})$: $3(+1) + x + 4(-2) = 0 \Rightarrow x = +5$.
$(c)$ Hypophosphoric acid $(H_{4}P_{2}O_{6})$: $4(+1) + 2x + 6(-2) = 0$ $\Rightarrow 2x = 8$ $\Rightarrow x = +4$.
$(d)$ Orthophosphorous acid $(H_{3}PO_{3})$: $3(+1) + x + 3(-2) = 0 \Rightarrow x = +3$.
Thus,the correct match is $(a)-(v), (b)-(i), (c)-(ii), (d)-(iii)$.

(D) In group $15$ $(N, P, As, Sb, Bi)$,the metallic character increases down the group.
$Bi$ is the most metallic element in this group.
The reducing power of hydrides $(MH_3)$ increases down the group from $NH_3$ to $BiH_3$ because the $M-H$ bond dissociation enthalpy decreases as the size of the central atom increases.
Therefore,$BiH_3$ is the strongest reducing agent among the group $15$ hydrides,and the element is $Bi$.

(A) $N_2O$ (nitrous oxide) and $NO$ (nitric oxide) are neutral oxides of nitrogen.
$N_2O_3$,$NO_2$,and $N_2O_5$ are acidic oxides of nitrogen.

(A) The complete hydrolysis of $PCl_{5}$ proceeds as follows:
$PCl_{5} + 4H_{2}O \rightarrow H_{3}PO_{4} + 5HCl$
The final product is phosphoric acid,$H_{3}PO_{4}$.
In the structure of $H_{3}PO_{4}$,there are three $P-OH$ bonds.
Since hydrogen atoms attached to oxygen atoms in $P-OH$ groups are acidic and ionisable,all three hydrogen atoms in $H_{3}PO_{4}$ are ionisable.
Therefore,the number of non-ionisable hydrogen atoms is $0$.

(D) When red phosphorus is heated in a sealed tube at $803 \,\,K$, $\alpha$-black phosphorus is formed.

(D) To determine which oxide lacks a nitrogen-nitrogen $(N-N)$ bond,let us examine the structures of the given nitrogen oxides:
$1$. $N_2O$ (Nitrous oxide): The structure is $N \equiv N^+ - O^-$,which contains an $N-N$ bond.
$2$. $N_2O_4$ (Dinitrogen tetroxide): The structure is $O_2N-NO_2$,which contains an $N-N$ bond.
$3$. $N_2O_3$ (Dinitrogen trioxide): The structure is $O=N-NO_2$,which contains an $N-N$ bond.
$4$. $N_2O_5$ (Dinitrogen pentoxide): The structure is $O_2N-O-NO_2$,which contains an $N-O-N$ linkage but no $N-N$ bond.
Therefore,$N_2O_5$ is the oxide that does not contain a nitrogen-nitrogen bond.

(A) The reaction between concentrated nitric acid $(HNO_{3})$ and phosphorus pentoxide $(P_{4}O_{10})$ in a $4:1$ molar ratio is given by:
$4 HNO_{3} + P_{4}O_{10} \rightarrow 2 N_{2}O_{5} + 4 HPO_{3}$
The nitrogen oxide compound obtained is dinitrogen pentoxide $(N_{2}O_{5})$.
$N_{2}O_{5}$ is an anhydride of nitric acid and is acidic in nature.

(C) The reducing character of group-$15$ hydrides depends on the bond dissociation enthalpy of the $E-H$ bond.
As we move down the group from $N$ to $Bi$,the atomic size increases,which leads to a decrease in the bond dissociation enthalpy of the $E-H$ bond.
Consequently,the stability of the hydrides decreases and the reducing character increases.
Therefore,$BiH_3$ has the lowest bond dissociation enthalpy and is the strongest reducing agent among the group-$15$ hydrides.

(B) The structure of hypophosphorous acid $(H_3PO_2)$ contains one $P=O$ bond,two $P-H$ bonds,and one $P-OH$ bond.
The presence of two $P-H$ bonds is responsible for its strong reducing character,as these bonds can easily release hydrogen atoms.

(A) In Group $13$ elements,the stability of the $+1$ oxidation state increases down the group due to the inert pair effect.
As we move from $B$ to $Tl$,the $ns^2$ electrons become more reluctant to participate in bonding.
Therefore,the stability of the $+1$ oxidation state follows the order: $B < Al < Ga < In < Tl$.

(B) $PCl_5$ forms five bonds by using the $d-$orbitals to expand its octet.
$NCl_5$ does not exist because nitrogen belongs to the $2^{nd}$ period and lacks $d-$orbitals in its valence shell.
Therefore,nitrogen cannot expand its octet beyond a covalency of $4$.

(A) The stability of nitrogen trihalides decreases as the size of the halogen atom increases.
The bond energy of the $N-X$ bond decreases from $F$ to $I$ due to the increasing mismatch in the size of the orbitals of $N$ and $X$ atoms.
Therefore,the order of stability is: $NF_{3} > NCl_{3} > NBr_{3} > NI_{3}$.
Thus,$NF_{3}$ is the most stable trihalide.

(A) The thermal decomposition of lead nitrate is given by:
$2Pb(NO_3)_2 \xrightarrow{673 \ K} 2PbO + 4NO_2 + O_2$
Here,$A$ is $NO_2$.
On dimerisation,$NO_2$ forms $N_2O_4$ $(B)$:
$2NO_2 \rightleftharpoons N_2O_4$
The structure of $N_2O_4$ consists of two $NO_2$ units linked by a $N-N$ bond. The structure is $O_2N-NO_2$. It does not contain any bridged oxygen atoms (i.e.,no $N-O-N$ linkage).
Therefore,the number of bridged oxygen atoms in $B$ is $0$.

(B) To determine which oxide contains an odd electron,we calculate the total number of valence electrons for each molecule:
$1$. $N_{2}O$: $(2 \times 5) + 6 = 16$ valence electrons (even).
$2$. $NO_{2}$: $5 + (2 \times 6) = 17$ valence electrons (odd).
$3$. $N_{2}O_{3}$: $(2 \times 5) + (3 \times 6) = 28$ valence electrons (even).
$4$. $N_{2}O_{5}$: $(2 \times 5) + (5 \times 6) = 40$ valence electrons (even).
Since $NO_{2}$ has an odd number of valence electrons $(17)$,it is a paramagnetic molecule with an odd electron on the nitrogen atom.

(A) In group $13$,the stability of the $+1$ oxidation state increases as we move down the group due to the inert pair effect.
The inert pair effect is the reluctance of the $ns^2$ electrons to participate in bonding.
Therefore,the correct order of stability for the $+1$ oxidation state is $Al < Ga < In < Tl$.

(A) The reaction of white phosphorus $(P_{4})$ with an alkali (like $NaOH$) is a disproportionation reaction.
The reaction is: $P_{4} + 3NaOH + 3H_{2}O \rightarrow PH_{3} + 3NaH_{2}PO_{2}$.
However,the question specifies that $B$ is an oxoacid of phosphorus with no $P-H$ bond.
When white phosphorus $(P_{4})$ reacts with oxygen,it forms $P_{4}O_{10}$.
$P_{4}O_{10} + 6H_{2}O \rightarrow 4H_{3}PO_{4}$.
$H_{3}PO_{4}$ (phosphoric acid) is an oxoacid of phosphorus that contains no $P-H$ bonds.
Thus,$A$ is $P_{4}$ (white phosphorus) which undergoes oxidation to form $P_{4}O_{10}$,followed by hydrolysis to form $H_{3}PO_{4}$.

(B) When an aqueous solution of ammonium chloride $(NH_4Cl)$ is treated with sodium nitrite $(NaNO_2)$,ammonium nitrite $(NH_4NO_2)$ is formed as an intermediate.
$NH_4Cl(aq) + NaNO_2(aq) \rightarrow NH_4NO_2(aq) + NaCl(aq)$
Ammonium nitrite is unstable and decomposes upon heating to produce nitrogen gas $(N_2)$ and water $(H_2O)$.
$NH_4NO_2(aq) \rightarrow N_2(g) + 2H_2O(l)$
Thus,the gas produced is $N_2$.

(D) When white phosphorus $(P_4)$ is heated with concentrated $NaOH$ solution,it undergoes a disproportionation reaction.
The chemical equation for this reaction is:
$P_4 + 3 NaOH + 3 H_2O \rightarrow 3 NaH_2PO_2 + PH_3$
Here,$PH_3$ (phosphine) and $NaH_2PO_2$ (sodium hypophosphite) are the main products formed.

(B) To determine the number of compounds having an $N-N$ bond,we examine the structures of the given nitrogen oxides:
$1$. $N_2O$: The structure is $N \equiv N^+ - O^-$. There is no $N-N$ bond (it has an $N=N$ bond).
$2$. $N_2O_3$: The structure is $O_2N-NO$. It contains an $N-N$ bond.
$3$. $N_2O_4$: The structure is $O_2N-NO_2$. It contains an $N-N$ bond.
$4$. $N_2O_5$: The structure is $O_2N-O-NO_2$. It contains an $N-O-N$ linkage,not an $N-N$ bond.
Thus,the compounds $N_2O_3$ and $N_2O_4$ have an $N-N$ bond.
The total number of such compounds is $2$.

(B) Pure nitrogen gas is prepared by the thermal decomposition of sodium or barium azide. The reaction for barium azide is:
$Ba(N_3)_2 \xrightarrow{\Delta} Ba + 3N_2(g)$

(D) The acidic character of an oxide depends on the oxidation state of the central atom. As the oxidation state increases,the electronegativity of the central atom increases,which increases the acidic character of the oxide. Therefore,$E_2O_5$ is more acidic than $E_2O_3$.
Statement $-I$ is false.
Down the group,the metallic character increases and the non-metallic character decreases. Since acidic character is associated with non-metallic oxides,the acidic character of $E_2O_3$ decreases down the group.
Statement $-II$ is true.

(B) The reaction of white phosphorus $(P_{4})$ with alkali $(NaOH)$ is given by:
$P_{4} + 3 NaOH + 3 H_{2}O \rightarrow PH_{3} + 3 NaH_{2}PO_{2}$
The product $NaH_{2}PO_{2}$ is the salt of hypophosphorous acid,also known as phosphinic acid $(H_{3}PO_{2})$.
In the structure of phosphinic acid $(H_{3}PO_{2})$,there are two $P-H$ bonds,one $P=O$ bond,and one $P-OH$ bond.
Therefore,the correct answer is phosphinic acid.

(D) The chemical formulas for the given oxoacids of phosphorus are as follows:
$1$. Pyrophosphorous acid: $H_4P_2O_5$ ($5$ oxygen atoms)
$2$. Hypophosphoric acid: $H_4P_2O_6$ ($6$ oxygen atoms)
$3$. Phosphoric acid: $H_3PO_4$ ($4$ oxygen atoms)
$4$. Pyrophosphoric acid: $H_4P_2O_7$ ($7$ oxygen atoms)
Comparing the number of oxygen atoms,Pyrophosphoric acid $(H_4P_2O_7)$ contains the highest number of oxygen atoms $(7)$.

(C) The reactions are as follows:
$A$. $(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} N_2 + Cr_2O_3 + 4H_2O$. Thus,$A-II$.
$B$. $2KMnO_4 + 16HCl \rightarrow 2KCl + 2MnCl_2 + 5Cl_2 + 8H_2O$. Thus,$B-IV$.
$C$. $2Al + 2NaOH + 6H_2O \rightarrow 2Na[Al(OH)_4] + 3H_2$. Thus,$C-I$.
$D$. $2NaNO_3 \xrightarrow{\Delta} 2NaNO_2 + O_2$. Thus,$D-III$.
Therefore,the correct matching is $A-II, B-IV, C-I, D-III$.

(B) The reaction of white phosphorus $(P_{4})$ with thionyl chloride $(SOCl_{2})$ is a standard chemical reaction used to produce phosphorus trichloride $(PCl_{3})$,sulfur dioxide $(SO_{2})$,and disulfur dichloride $(S_{2}Cl_{2})$.
The balanced chemical equation is:
$P_{4} + 8 SOCl_{2} \rightarrow 4 PCl_{3} + 4 SO_{2} + 2 S_{2}Cl_{2}$

(D) The reaction between $N_{2}$ and $O_{2}$ is highly endothermic,requiring a very high temperature (approximately $1483-2000 \ K$) to proceed:
$N_{2}(g) + O_{2}(g) \xrightarrow{1483-2000 \ K} 2NO(g)$
Since such high temperatures are not available in the atmosphere,they do not react under normal conditions.

(B) The reducing property of phosphorus oxoacids is primarily determined by the number of $P-H$ bonds present in the molecule.
$H_3PO_2$ (hypophosphorous acid) contains two $P-H$ bonds.
$H_3PO_3$ (phosphorous acid) contains one $P-H$ bond.
$H_3PO_4$ (phosphoric acid) contains zero $P-H$ bonds.
$H_4P_2O_7$ (pyrophosphoric acid) contains zero $P-H$ bonds.
Since $H_3PO_2$ has the maximum number of $P-H$ bonds,it acts as the strongest reducing agent among the given options.

(B) The correct option is $B$.
Phosphorus reacts with chlorine gas to form phosphorus trichloride,$PCl_3$,which is a colourless liquid.
$P_4 + 6Cl_2 \longrightarrow 4PCl_3$ (colourless liquid)
When $PCl_3$ is exposed to moist air,it undergoes hydrolysis to produce $HCl$ fumes and phosphorous acid,$H_3PO_3$.
$PCl_3 + 3H_2O \longrightarrow H_3PO_3 + 3HCl$

(D) The thermal decomposition reactions of the given compounds are as follows:
$A) (NH_4)_2Cr_2O_7 \xrightarrow{\Delta} Cr_2O_3 + N_2 \uparrow + 4H_2O$
$B) 2NaN_3 \xrightarrow{\Delta} 2Na + 3N_2 \uparrow$
$C) NH_4NO_2 \xrightarrow{\Delta} N_2 \uparrow + 2H_2O$
$D) (NH_4)_2(C_2O_4) \xrightarrow{\Delta} 2NH_3 + H_2C_2O_4$
As shown above,$(NH_4)_2(C_2O_4)$ produces ammonia $(NH_3)$ and oxalic acid $(H_2C_2O_4)$ upon heating,not nitrogen gas $(N_2)$. Therefore,the correct option is $D$.

(B) $Al$ $(X)$ is an amphoteric metal that dissolves in both dilute $HCl$ and dilute $NaOH$ to liberate $H_2$ gas.
$2Al + 6HCl \longrightarrow 2AlCl_3 + 3H_2$
$2Al + 2NaOH + 2H_2O \longrightarrow 2NaAlO_2 + 3H_2$
When $NH_4Cl$ and excess $NH_4OH$ are added to the $AlCl_3$ solution,$Al(OH)_3$ $(Y)$ is precipitated.
$AlCl_3 + 3NH_4OH \longrightarrow Al(OH)_3 \downarrow + 3NH_4Cl$
Similarly,adding $NH_4Cl$ to the $NaAlO_2$ solution (formed by $Al$ in $NaOH$) also precipitates $Al(OH)_3$ $(Y)$ due to the common ion effect and hydrolysis of the aluminate ion.
$NaAlO_2 + NH_4Cl + H_2O \longrightarrow Al(OH)_3 \downarrow + NaCl + NH_3$
Thus,$(X)$ is $Al$ and $(Y)$ is $Al(OH)_3$.

(C) $N_2O$ is a neutral oxide,which is neither acidic nor basic.
$As_2O_3$ and $Sb_4O_{10}$ are amphoteric oxides.
$Na_2O$ is a basic oxide.

(C) $PbO_2$ is prepared by the action of nitric acid on red lead $(Pb_3O_4)$.
$(a)$ $PbO + 2HCl \longrightarrow PbCl_2 + H_2O$ (This reaction does not produce $PbO_2$)
$(b)$ $2Pb(NO_3)_2 \xrightarrow{200^{\circ}C} 2PbO + 4NO_2 + O_2$ (This reaction produces $PbO$)
$(c)$ $Pb_3O_4 + 4HNO_3 \longrightarrow 2Pb(NO_3)_2 + PbO_2 + 2H_2O$ (This reaction produces $PbO_2$)
$(d)$ $Pb +$ air (contains $O_2, H_2O$ and $CO_2$) forms a protective layer of $PbCO_3$ on the surface.

(A) .
White phosphorus is highly reactive and catches fire spontaneously when exposed to air to produce dense white fumes of phosphorus pentoxide,$P_4O_{10}$.
The chemical reaction is: $P_4 + 5O_2 \rightarrow P_4O_{10}$

(A)
$P_2O_5$ and $As_2O_3$ are acidic oxides,$Sb_2O_3$ is amphoteric,while $Bi_2O_3$ is a basic oxide.
$P_2O_5$ is the most acidic among them all.
This is because down the group,metallic character increases and non-metallic character decreases,which leads to an increase in the basicity of oxides and a decrease in acidity.

(C) .
The products formed in each reaction given in the options are as follows:
$(i) \ NH_4Cl + KOH \longrightarrow KCl + NH_3 + H_2O$
$(ii) \ AlN + 3H_2O \longrightarrow Al(OH)_3 + NH_3$
$(iii) \ NH_4Cl + NaNO_2 \longrightarrow NaCl + N_2 + 2H_2O$
$(iv) \ 2NH_4Cl + Ca(OH)_2 \longrightarrow CaCl_2 + 2NH_3 + 2H_2O$
Thus,ammonia is not produced in the reaction given in option $(C)$.

(D) .
The thermal decomposition of $(NH_4)_2Cr_2O_7$ yields chromium$(III)$ oxide $(Cr_2O_3)$,nitrogen gas $(N_2)$,and water vapor $(H_2O)$.
The balanced chemical equation is:
$(NH_4)_2Cr_2O_7 \stackrel{\Delta}{\longrightarrow} Cr_2O_3 + N_2 + 4H_2O$

(B) Phosphorus oxoacids containing at least one $P-H$ bond act as strong reducing agents and can reduce $AgNO_3$ solution to metallic silver $(Ag)$,forming a silver mirror.
Among the given options,$H_4P_2O_5$ (pyrophosphorous acid) contains two $P-H$ bonds.
Therefore,it can reduce $AgNO_3$ to $Ag$.