Nitrogen family Questions in English

(C) $2 NaN_3 \xrightarrow{\Delta} 2 Na + 3 N_2$ (Liberates $N_2$)
$B$ $(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} Cr_2O_3 + N_2 + 4 H_2O$ (Liberates $N_2$)
$C$ $2 NH_4Cl + Ca(OH)_2 \longrightarrow CaCl_2 + 2 NH_3 + 2 H_2O$ (Liberates $NH_3$,not $N_2$)
$D$ $Ba(N_3)_2 \xrightarrow{\Delta} Ba + 3 N_2$ (Liberates $N_2$)
Thus,option $C$ is the correct answer.

(C) Ammonia $(NH_3)$ is a basic gas. To dry it,we must use a basic drying agent.
$P_4O_{10}$ and Conc $H_2SO_4$ are acidic and will react with $NH_3$ to form salts.
Anhydrous $CaCl_2$ forms a complex compound with $NH_3$ $(CaCl_2 \cdot 8NH_3)$,so it cannot be used.
Quick lime $(CaO)$ is a basic drying agent that does not react with $NH_3$ and is therefore used to remove moisture from it.

(C) The reaction of sodium nitrite with sulphuric acid is: $2NaNO_2 + H_2SO_4 \rightarrow Na_2SO_4 + 2HNO_2$. The $HNO_2$ formed is unstable and decomposes: $3HNO_2 \rightarrow HNO_3 + 2NO + H_2O$. Thus,$X$ is $NO$.
Gold dissolves in aqua regia $(HNO_3 + 3HCl)$ to form chloroauric acid and nitric oxide: $Au + HNO_3 + 4HCl \rightarrow H[AuCl_4] + NO + 2H_2O$. Thus,$Y$ is $NO$.
Therefore,both $X$ and $Y$ are $NO$.

(B) The bond energy depends on the bond length. As the size of the central atom increases down the group $(N < P < As)$,the bond length increases,and the bond energy decreases.
The order of bond energies is $N-H > P-H > As-H$.
The values are $389 \ kJ \ mol^{-1}$ for $N-H$,$318 \ kJ \ mol^{-1}$ for $P-H$,and $247 \ kJ \ mol^{-1}$ for $As-H$.
Therefore,the correct order is $389, 318, 247$.

(C) The structure of pyrophosphoric acid $(H_4P_2O_7)$ consists of two $P=O$ bonds,four $P-OH$ bonds,and one $P-O-P$ bridge.
Counting the bonds:
- Each $P=O$ bond contains $1$ $\sigma$ and $1$ $\pi$-bond (Total: $2$ $\sigma$,$2$ $\pi$).
- Each $P-OH$ bond contains $1$ $\sigma$-bond (Total: $4$ $\sigma$).
- The $P-O-P$ bridge contains $2$ $\sigma$-bonds.
- Each $O-H$ bond contains $1$ $\sigma$-bond (Total: $4$ $\sigma$).
Total $\sigma$-bonds = $2 + 4 + 2 + 4 = 12$.
Total $\pi$-bonds = $2$.
Therefore,the correct answer is $12, 2$.

(D) Phosphorus $(P)$ is a non-metal and typically forms acidic oxides such as $P_2O_3$ and $P_4O_{10}$.
$Al_2O_3$ (Aluminum oxide),$SnO$ (Tin oxide),and $Sb_2O_3$ (Antimony oxide) are known to be amphoteric,meaning they can react with both acids and bases.
Therefore,$P$ cannot form an amphoteric oxide.

(D) In a group,the ionic radius increases from top to bottom due to the addition of new shells.
Since $As$,$Sb$,and $Bi$ belong to Group $15$,the ionic radius follows the order: $As^{3+} < Sb^{3+} < Bi^{3+}$.
Therefore,the correct order is $Bi^{3+} > Sb^{3+} > As^{3+}$.

(A) The reaction of copper with dilute nitric acid is given by: $3Cu + 8HNO_3 (\text{dilute}) \rightarrow 3Cu(NO_3)_2 + 2NO + 4H_2O$.
The reaction of copper with concentrated nitric acid is: $Cu + 4HNO_3 (\text{conc.}) \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O$.
The reaction of zinc with dilute nitric acid is: $4Zn + 10HNO_3 (\text{dilute}) \rightarrow 4Zn(NO_3)_2 + 5H_2O + N_2O$.
The reaction of zinc with concentrated nitric acid is: $Zn + 4HNO_3 (\text{conc.}) \rightarrow Zn(NO_3)_2 + 2H_2O + 2NO_2$.
Therefore,the reaction that produces nitrogen $(II)$ oxide $(NO)$ is the reaction between copper and dilute nitric acid.

(C) The thermal decomposition of ammonium dichromate produces nitrogen gas,chromium $(III)$ oxide,and water vapour.
$(NH_4)_2Cr_2O_7 \stackrel{\Delta}{\longrightarrow} N_2 + Cr_2O_3 + 4H_2O$

(C) Copper is a transition metal. Concentrated nitric acid $(HNO_3)$ acts as a strong oxidizing agent. When copper reacts with concentrated $HNO_3$,it gets oxidized to copper$(II)$ nitrate,and the nitric acid is reduced to nitrogen dioxide $(NO_2)$ gas.
The balanced chemical equation for the reaction is:
$Cu(s) + 4HNO_3(conc.) \rightarrow Cu(NO_3)_2(aq) + 2NO_2(g) + 2H_2O(l)$
Thus,the main products formed are copper$(II)$ nitrate $(Cu(NO_3)_2)$ and nitrogen dioxide $(NO_2)$.

(A) The reducing character of the hydrides of group $15$ elements depends upon the thermal stability of the $E-H$ bond.
As we move down the group,the bond dissociation enthalpy decreases due to an increase in the size of the central atom.
Consequently,the stability of the hydrides decreases,and the ease of releasing hydrogen atoms increases.
Therefore,the reducing character increases down the group.
The correct order is: $NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3$.

(A) The reducing property of the hydrides of $VA$ group increases from $NH_3$ to $BiH_3$ due to the decrease in bond dissociation enthalpy: $NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3$.
The tendency to donate a lone pair (basic strength) decreases from $NH_3$ to $BiH_3$ because the electron density on the central atom decreases as the size of the atom increases.
Thermal stability of $VA$ group hydrides decreases from $NH_3$ to $BiH_3$ as the $M-H$ bond strength decreases: $NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$.
Bond angle of $VA$ group hydrides decreases from $NH_3$ to $BiH_3$ due to the decrease in electronegativity of the central atom: $NH_3 (107.8^{\circ}) > PH_3 (93.6^{\circ}) > AsH_3 (91.8^{\circ}) > SbH_3 (91.3^{\circ}) > BiH_3 (90^{\circ})$.
Thus,all statements $I, II, III,$ and $IV$ are correct.

(D) The basic character of a molecule depends on the availability of the lone pair of electrons on the central atom for donation.
In $PH_3$,the lone pair is present in an orbital with high $s$-character,making it less available for donation.
In $P(CH_3)_3$,the three methyl groups are electron-donating ($+I$ effect),which increases the electron density on the phosphorus atom,making the lone pair more available for donation.
Therefore,$P(CH_3)_3$ is a stronger base than $PH_3$.

(A) The reducing character of the hydrides of group $15$ elements depends upon the thermal stability of the hydrides.
As we move down the group,the bond dissociation energy decreases due to an increase in the size of the central atom,which leads to a decrease in the stability of the hydrides.
Consequently,the ability to release hydrogen atoms increases,making the reducing character increase down the group.
Therefore,the correct order of reducing ability is: $BiH_3 > SbH_3 > AsH_3 > PH_3 > NH_3$.
Among the given options,the correct order is $BiH_3 > SbH_3 > PH_3 > NH_3$.

(D) The stability of the lower oxidation state increases down the group in Group $13$ due to the inert pair effect.
$Tl$ (Thallium) belongs to Group $13$ and exhibits a stable $+1$ oxidation state,which is lower than the group oxidation state of $+3$.
Therefore,$Tl$ forms stable compounds in the low oxidation state.

(B) Aluminium reacts with nitrogen at high temperatures $(1073 \ K - 1473 \ K)$ to form aluminium nitride $(AlN)$:
$2 \ Al + N_2 \rightarrow 2 \ AlN$
Aluminium nitride $(AlN)$ reacts with water to produce aluminium hydroxide $(Al(OH)_3)$ as a precipitate and ammonia gas $(NH_3)$:
$AlN + 3 \ H_2O \rightarrow Al(OH)_3 (ppt) + NH_3 (g)$
Therefore,the compound $C$ is ammonia $(NH_3)$.

(C) In group $13$ elements,the stability of the $+1$ oxidation state increases down the group due to the inert pair effect.
For $Tl$ (Thallium),the $+1$ oxidation state is more stable than the $+3$ oxidation state because the $6s^2$ electrons are reluctant to participate in bonding.

(A) The reaction of zinc with concentrated nitric acid is given by: $Zn + 4HNO_3 (\text{conc.}) \rightarrow Zn(NO_3)_2 + 2NO_2 + 2H_2O$. Here,the oxide of nitrogen $(A)$ is $NO_2$. The oxidation number of $N$ in $NO_2$ is $x + 2(-2) = 0$,so $x = +4$.
The reaction of zinc with dilute nitric acid is given by: $4Zn + 10HNO_3 (\text{dil.}) \rightarrow 4Zn(NO_3)_2 + N_2O + 5H_2O$. Here,the oxide of nitrogen $(B)$ is $N_2O$. The oxidation number of $N$ in $N_2O$ is $2x + (-2) = 0$,so $2x = +2$,$x = +1$.
Therefore,the oxidation numbers of nitrogen in $(A)$ and $(B)$ are $+4$ and $+1$ respectively.

(C) Phosphorus has vacant $d$-orbitals in its valence shell,which allows it to expand its octet and form $PCl_5$ (phosphorus$(V)$ chloride) in addition to $PCl_3$ (phosphorus$(III)$ chloride).
Nitrogen,being in the second period,lacks vacant $d$-orbitals and cannot expand its octet beyond a covalency of $4$. Therefore,it cannot form $NCl_5$.
The assertion $(A)$ is correct.
The reason $(R)$ states that the electronegativity of nitrogen is higher than that of phosphorus,which is a true statement.
However,the inability of nitrogen to form $NCl_5$ is due to the absence of vacant $d$-orbitals,not due to its electronegativity.
Thus,$(R)$ is a true statement but not the correct explanation for $(A)$.

(D) The thermal decomposition of ammonium dichromate $(NH_4)_2Cr_2O_7$ is a classic laboratory reaction often used to demonstrate a volcanic eruption effect.
The balanced chemical equation for this reaction is:
$(NH_4)_2Cr_2O_7(s) \xrightarrow{\Delta} N_2(g) + Cr_2O_3(s) + 4H_2O(g)$
Thus,the products formed are nitrogen gas $(N_2)$,chromium$(III)$ oxide $(Cr_2O_3)$,and water vapor $(H_2O)$.

(B) The thermal decomposition reactions are as follows:
$(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} Cr_2O_3 + N_2 \uparrow + 4H_2O$
$Ba(N_3)_2 \xrightarrow{\Delta} Ba + 3N_2$
$2NaNO_3 \xrightarrow{\Delta} 2NaNO_2 + O_2$
From the above reactions,it is clear that ammonium dichromate $(ii)$ and barium azide $(iii)$ liberate dinitrogen gas upon heating,whereas sodium nitrate $(i)$ liberates dioxygen gas.
Therefore,the correct option is $ii$ and $iii$ only.

(C) The boiling point of the hydrides of the nitrogen family $(Group \ 15)$ follows the order: $PH_3 < AsH_3 < NH_3 < SbH_3 < BiH_3$.
$NH_3$ has an anomalously high boiling point due to intermolecular hydrogen bonding,but the boiling point increases down the group due to an increase in molecular mass and van der Waals forces.
Comparing the given hydrides: $PH_3$ has the lowest boiling point $(Y = PH_3)$.
$BiH_3$ has the highest boiling point $(X = BiH_3)$.
Therefore,$X = BiH_3$ and $Y = PH_3$.

(B) Nitrogen cannot form pentahalides because it lacks vacant $d$-orbitals in its valence shell to expand its octet. Thus,Assertion $(A)$ is true.
Nitrogen can exhibit a $+5$ oxidation state,for example,in $N_2O_5$ or $HNO_3$. Thus,Reason $(R)$ is true.
However,the inability to form pentahalides is due to the absence of $d$-orbitals,not because it cannot exhibit a $+5$ oxidation state. Therefore,$(R)$ is not the correct explanation of $(A)$.

(D) The order in '$A$' is correct because the bond angle decreases as the electronegativity of the central atom decreases down the group,leading to less repulsion between bond pairs.
The order in '$B$' is correct because the electron density on the central atom decreases as the size of the atom increases,making the lone pair less available for donation.
The order in '$C$' is incorrect because thermal stability decreases down the group due to the increase in bond length and decrease in bond dissociation energy. The correct order is $NH_3 > PH_3 > AsH_3 > SbH_3$.
The order in '$D$' is correct. $NH_3$ has a high boiling point due to intermolecular $H$-bonding,while for the others $(PH_3, AsH_3, SbH_3)$,the boiling point increases down the group due to increasing van der Waals forces (molecular weight).
Thus,the correct orders are $A, B$ and $D$.

(A) The reaction between sodium nitrite $(NaNO_2)$ and hydrochloric acid $(HCl)$ is as follows:
$2 NaNO_2 + 2 HCl \rightarrow 2 NaCl + NO + NO_2 + H_2O$
In this reaction,the two nitrogen oxides produced are nitric oxide $(NO)$ and nitrogen dioxide $(NO_2)$.

(C) $(A)\ NO_2^-(aq) + NH_4^+(s) \xrightarrow{\Delta} N_{2(g)} + 2H_2O$. This reaction liberates $N_2$ gas.
$(B)\ CO(NH_2)_{2(s)} + 2HNO_{2(l)} \longrightarrow 2N_{2(g)} + CO_2 + 3H_2O$. This reaction liberates $N_2$ gas.
$(C)\ 2NH_{3(g)} + 3NaOCl_{(aq)} \xrightarrow{\text{gelatine}} N_2H_4 + 3NaCl + H_2O$. This reaction produces hydrazine $(N_2H_4)$ and does not liberate $N_2$ gas.
$(D)\ (NH_4)_2Cr_2O_{7(s)} \xrightarrow{\Delta} Cr_2O_3 + N_{2(g)} + 4H_2O$. This reaction liberates $N_2$ gas.

(D) Reaction $(A): NH_4 NO_3 \xrightarrow{\Delta} N_2 O + 2 H_2 O$. $N_2 O$ is diamagnetic.
Reaction $(B): 2 NaNO_2 + H_2 SO_4 \rightarrow Na_2 SO_4 + H_2 O + NO_2 + NO$. (Note: The reaction produces $NO$ and $NO_2$,both are paramagnetic).
Reaction $(C): 2 Pb(NO_3)_2 \xrightarrow{673 \ K} 2 PbO + 4 NO_2 + O_2$. $NO_2$ and $O_2$ are paramagnetic.
The paramagnetic gaseous products are $NO$ (from $B$),$NO_2$ (from $B$ and $C$),and $O_2$ (from $C$).
Total number of unique paramagnetic gaseous products is $3$.

(B) The reaction described is the Brown Ring Test,which is used to detect the presence of nitrate ions $(NO_3^-)$ in a solution.
When dilute $FeSO_4$ is added to a solution containing nitrate ions,followed by the careful addition of concentrated $H_2SO_4$,the following reaction occurs:
$2NO_3^- + 3Fe^{2+} + 4H^+ \rightarrow 2NO + 3Fe^{3+} + 2H_2O$
The nitrogen oxide formed "in situ" is nitric oxide $(NO)$.
This $NO$ then reacts with the hydrated ferrous sulfate complex to form the brown-colored complex $[Fe(H_2O)_5(NO)]SO_4$.

(A) The reaction between $N_2$ and $O_2$ is highly endothermic and requires a very high temperature to proceed.
This reaction is part of the Birkeland-Eyde process,which uses an electric arc furnace to provide temperatures greater than $3000 \ K$ to facilitate the formation of $NO$ from $N_2$ and $O_2$.

(A) $(A)\ NO + NO_2 \xrightarrow{< 250 K} N_2O_3$. $N_2O_3$ is blue in liquid or solid state $(P = \text{Blue})$.
$(B)\ 2 NO_2 \longleftrightarrow N_2O_4$. $N_2O_4$ is a colourless gas/liquid $(Q = \text{Colourless})$.
$(C)\ 2 Pb(NO_3)_2 \xrightarrow{673 K} 4 NO_2 + 2 PbO + O_2$. $NO_2$ is a reddish-brown gas $(R = \text{Brown})$.
$(D)\ 4 HNO_3 + P_4O_{10} \longrightarrow 2 N_2O_5 + 4 HPO_3$. $N_2O_5$ is a colourless crystalline solid $(S = \text{Colourless})$.
Thus,the correct sequence is $P$: Blue,$Q$: Colourless,$R$: Brown,$S$: Colourless.

(D) $NO_2$ (Nitrogen dioxide) is not a colourless compound.
It is a brown gas,highly reactive and paramagnetic.
The oxidation state of $N$ in $NO_2$ is $+4$.

(A) When copper metal is treated with cold and dilute nitric acid,it yields copper nitrate,water,and nitric oxide.
$3Cu + 8HNO_3 \text{ (dilute)} \longrightarrow 3Cu(NO_3)_2 + 4H_2O + 2NO$

(D) $Li$,$Na$ and $Mg$ belong to the $s$-block family and do not form $p \pi-p \pi$ multiple bonds.
$N$ (Nitrogen) belongs to the $p$-block family (Group $15$).
Due to its small atomic size and high electronegativity,$N$ has a unique ability to form $p \pi-p \pi$ multiple bonds with itself (as in $N_2$) and with other small $p$-block elements like $C$ and $O$.

(D) Statement $(A)$: $NO$ has $11$ valence electrons (odd number),making it paramagnetic. $NO_2$ has $17$ valence electrons (odd number),making it paramagnetic. Thus,statement $(A)$ is correct.
Statement $(B)$: In the gaseous state,$NO$ exists as a monomer with an unpaired electron,making it paramagnetic. In the liquid and solid states,$NO$ dimerizes to form $N_2O_2$,where all electrons are paired,making it diamagnetic. Thus,statement $(B)$ is correct.
Therefore,both statements are correct,and option $(D)$ is the correct answer.

(C) Aqua regia is a mixture of concentrated $HNO_3$ and concentrated $HCl$ in a $1:3$ molar ratio.
It dissolves noble metals like gold and platinum by oxidizing them.
The chemical reactions are:
$Au + 4H^+ + NO_3^- + 4Cl^- \longrightarrow AuCl_4^- + NO + 2H_2O$
$3Pt + 16H^+ + 4NO_3^- + 18Cl^- \longrightarrow 3PtCl_6^{2-} + 4NO + 8H_2O$
In both reactions,$NO$ (Nitric oxide) is produced as the byproduct.
Therefore,$X = NO$ and $Y = NO$.

(C) $(a) \ (NH_4)_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2NH_3 + 2H_2O$ (No $N_2$ gas)
$(b) \ 8NH_3 + 3Cl_2 \rightarrow N_2 + 6NH_4Cl$ ($N_2$ is liberated when $NH_3$ is in excess)
$(c) \ (NH_4)_2Cr_2O_7 \xrightarrow{\Delta} N_2 + 4H_2O + Cr_2O_3$ ($N_2$ is liberated)
$(d) \ NH_4NO_3 \xrightarrow{\Delta} N_2O + 2H_2O$ (No $N_2$ gas)
$(e) \ NH_4Cl_{(aq)} + NaNO_{2_{(aq)}} \xrightarrow{\Delta} NaCl + N_2 + 2H_2O$ ($N_2$ is liberated)
Thus,the reactions that liberate $N_2$ are $b, c, e$.

(C) The acidity of non-metal oxides increases with an increase in the oxidation state of the central atom.
Calculating the oxidation states of nitrogen in the given oxides:
$NO_2$: $x + 2(-2) = 0 \implies x = +4$
$N_2O_4$: $2x + 4(-2) = 0 \implies x = +4$
$N_2O_5$: $2x + 5(-2) = 0 \implies x = +5$
$N_2O_3$: $2x + 3(-2) = 0 \implies x = +3$
Since $N_2O_5$ has the highest oxidation state of $+5$,it is the most acidic oxide among the given compounds.

(D) $Zn$ reacts with very dilute nitric acid to form 'laughing gas' $(N_2O)$.
$4 Zn + 10 HNO_3 (\text{very dilute}) \longrightarrow 4 Zn(NO_3)_2 + N_2O + 5 H_2O$

(B) The thermal decomposition of ammonium nitrate $(NH_4NO_3)$ at $250^{\circ}C$ yields nitrous oxide $(N_2O)$ and water vapor.
$NH_4NO_3 \xrightarrow{\Delta, 250^{\circ}C} N_2O + 2H_2O$
Thus,the correct oxide formed is $N_2O$.

(B) Let's analyze each option:
$1$. Ammonia $(NH_3)$ is widely used as a refrigerant due to its high latent heat of vaporization. This statement is correct.
$2$. Nitrolim is a mixture of calcium cyanamide $(CaCN_2)$ and carbon $(C)$. The option states $Ca(CN)_2$ (calcium cyanide),which is incorrect. Thus,this statement is incorrect.
$3$. Superphosphate of lime is indeed a mixture of $Ca(H_2PO_4)_2$ and $CaSO_4 \cdot 2H_2O$. This statement is correct.
$4$. The hydrolysis of nitrogen trichloride $(NCl_3)$ proceeds as follows: $NCl_3 + 3H_2O \rightarrow NH_3 + 3HOCl$. This statement is correct.
Therefore,the incorrect statement is option $B$.