Nitrogen family Questions in English

(C) Iron $(Fe)$ becomes passive when treated with concentrated nitric acid $(conc. HNO_3)$.
This is due to the formation of a thin,protective,and non-reactive oxide layer $(Fe_2O_3)$ on the surface of the metal,which prevents further reaction with the acid.

(A) Mercury $(II)$ chloride reacts with gaseous ammonia to form a white precipitate of amino mercury $(II)$ chloride,also known as infusible white precipitate.
The chemical reaction is: $HgCl_2 + 2NH_3 \rightarrow Hg(NH_2)Cl + NH_4Cl$.

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

(A) The atmosphere consists mainly of dinitrogen $(N_2)$ and dioxygen $(O_2)$.
$N_2$ contains a very strong triple bond $(N \equiv N)$,which requires a significant amount of energy to break (high dissociation energy).
Therefore,the reaction between $N_2$ and $O_2$ is highly endothermic and only occurs at extremely high temperatures (e.g.,during lightning or in internal combustion engines).
$N_{2(g)} + O_{2(g)} \xrightarrow{3000\,^{\circ}C} 2 NO_{(g)}$

(A) White phosphorus is soluble in $CS_2$,whereas red phosphorus is insoluble in $CS_2$. Therefore,the statement that both are soluble in $CS_2$ is incorrect.

(C) The acidity of oxides decreases down the group as the metallic character increases. Therefore,oxides of $As$ are less acidic than oxides of $P$.
Furthermore,for the same element,the acidity increases with an increase in the oxidation state of the central atom.
Comparing the given oxides:
$P_4O_{10}$ ($P$ in $+5$ state) is more acidic than $P_4O_6$ ($P$ in $+3$ state).
$As_4O_{10}$ ($As$ in $+5$ state) is more acidic than $As_4O_6$ ($As$ in $+3$ state).
Since $As$ is less electronegative than $P$,$As_4O_6$ is the least acidic among the given options.

(C) Black phosphorus is the most stable and inert allotrope of phosphorus under standard conditions.

(A) $PH_3$ is less basic than $NH_3$ because the lone pair on $P$ is present in a larger orbital,making it less available for donation. Additionally,$PH_3$ acts as a stronger reducing agent compared to $NH_3$ due to the weaker $P-H$ bond.

(A) For the oxides of elements in group $15$,the acidic strength decreases as we move down the group.
Since $P$ is at the top of the group,its oxide $P_2O_3$ is the most acidic among the given options.

(B) Nitrogen forms a stable $N \equiv N$ triple bond due to effective $p\pi - p\pi$ overlapping,which is possible because of its small size.
In the case of phosphorus,the atomic size is larger,making $p\pi - p\pi$ overlapping ineffective.
Therefore,phosphorus prefers to form three $P-P$ single bonds rather than one $P \equiv P$ triple bond.
As a result,phosphorus exists as $P_4$ molecules where each phosphorus atom is linked by single bonds.

(D) Nitrogen is chemically inert due to the very high bond dissociation energy of the $N \equiv N$ triple bond.

(C) An oxoacid is an acid that contains oxygen.
$H_3PO_3$ (phosphorous acid) contains oxygen and is an oxoacid.
$Ba(OH)_2$ and $Mg(OH)_2$ are basic hydroxides.
$HCl$ is a binary acid (hydracid) and does not contain oxygen.

(A) The structure of phosphorus trioxide $(P_4O_6)$ consists of $4$ phosphorus atoms at the corners of a tetrahedron,with $6$ oxygen atoms bridging the $P-P$ edges,resulting in $6$ $P - O - P$ bridges.
Similarly,the structure of phosphorus pentoxide $(P_4O_{10})$ consists of $4$ phosphorus atoms at the corners of a tetrahedron,with $6$ oxygen atoms bridging the $P-P$ edges and $4$ additional oxygen atoms attached to each phosphorus atom via a double bond,also resulting in $6$ $P - O - P$ bridges.
Therefore,both have $6$ $P - O - P$ bridges.

(C) The reaction between $KNO_3$ and concentrated $H_2SO_4$ produces $HNO_3$:
$KNO_3 + H_2SO_4 \xrightarrow{\Delta} KHSO_4 + HNO_3$
Upon further heating,$HNO_3$ undergoes thermal decomposition to produce brown fumes of nitrogen dioxide $(NO_2)$:
$4HNO_3 \xrightarrow{\Delta} 4NO_2 + 2H_2O + O_2$
Therefore,the brown fumes are due to $NO_2$.

(B) White phosphorus reacts with $NaOH$ to produce $PH_3$.
$Ca_3P_2$ undergoes hydrolysis to produce $PH_3$.
$PH_4I$ reacts with $KOH$ to produce $PH_3$.
Red phosphorus is chemically inert towards alkalis and does not react with $NaOH$ to produce $PH_3$.

(C) When white phosphorus $(P_4)$ is boiled with a strong solution of caustic soda $(NaOH)$,it undergoes disproportionation to form phosphine $(PH_3)$ and sodium hypophosphite $(NaH_2PO_2)$.
The chemical equation is: $P_4 + 3NaOH + 3H_2O \to PH_3 + 3NaH_2PO_2$.
Thus,the product formed is phosphine $(PH_3)$.

(D) Due to the inert pair effect,the stability of the $+5$ oxidation state decreases down the group $15$.
$Bi$ (Bismuth) is the heaviest element in the group,and its $+3$ oxidation state is much more stable than its $+5$ oxidation state.
Therefore,$Bi$ does not form a stable pentahalide like $BiF_5$ under normal conditions; instead,it prefers to form trihalides.

(A) Among the group $15$ hydrides $(NH_3, PH_3, AsH_3, SbH_3, BiH_3)$,the basic strength decreases down the group.
This is because the lone pair on the central atom is more concentrated in a smaller orbital for $N$ $(NH_3)$,making it the most available for donation.
Therefore,$NH_3$ (ammonia) is the most basic compound.

(B) As we move down the group,the electronegativity of the central atom decreases.
Consequently,the bond pair of electrons moves away from the central atom.
This results in a decrease in the $H-M-H$ bond angle,which approaches $90^o$.
This indicates that the $M-H$ bonds involve almost pure $p$-orbitals,as there is very little hybridization.

(A) The oxidation states of phosphorus in the given oxoacids are as follows:
$1$. Hypophosphorous acid $(H_3PO_2)$: $3(+1) + x + 2(-2) = 0 \implies x = +1$.
$2$. Orthophosphoric acid $(H_3PO_4)$: $3(+1) + x + 4(-2) = 0 \implies x = +5$.
$3$. Pyrophosphoric acid $(H_4P_2O_7)$: $4(+1) + 2x + 7(-2) = 0 \implies 2x = +10 \implies x = +5$.
$4$. Metaphosphoric acid $(HPO_3)$: $1(+1) + x + 3(-2) = 0 \implies x = +5$.
Thus,hypophosphorous acid has the lowest oxidation state of $+1$.

(D) Nitrogen exhibits various oxidation states ranging from $-3$ to $+5$.

(C) $NCl_3$ undergoes hydrolysis to give $NH_3$ and $HOCl$ $(NCl_3 + 3H_2O \rightarrow NH_3 + 3HOCl)$.
$PH_3$ is less stable than $NH_3$ due to the smaller size of the $N$ atom compared to the $P$ atom.
$NH_3$ is a weaker reducing agent than $PH_3$ because the $N-H$ bond is stronger than the $P-H$ bond.
Nitric oxide $(NO)$ is paramagnetic in the gaseous state due to the presence of an unpaired electron,but it forms a dimer $(N_2O_2)$ in the solid state,which is diamagnetic.
Therefore,all statements are technically true in their respective contexts,but statement $C$ is often phrased as $NH_3$ being a weaker reducing agent,which is correct. Upon reviewing the options,if the question implies a false statement,there might be a typo in the source. However,based on standard chemistry,all these statements are factually correct.

(B) Phosphine $(PH_3)$ is prepared by heating white phosphorus with concentrated sodium hydroxide solution $(NaOH)$ in an inert atmosphere of $CO_2$.
The chemical reaction is:
$P_4 + 3NaOH + 3H_2O \to 3NaH_2PO_2 + PH_3$
Thus,the correct option is $B$.

(C) The structure of $P_4O_6$ consists of four phosphorus atoms at the corners of a tetrahedron. Each phosphorus atom is bonded to three oxygen atoms,which bridge the phosphorus atoms. Thus,each phosphorus atom is attached to $3$ oxygen atoms.

(B) The hydrolysis of phosphorus trichloride $(PCl_3)$ with water proceeds as follows:
$PCl_3 + 3H_2O \to H_3PO_3 + 3HCl$
Note: $PCl_3$ reacts with water to form phosphorous acid $(H_3PO_3)$ and hydrogen chloride $(HCl)$.

(B) The formula of metaphosphoric acid is $HPO_3$. It is formed by the dehydration of orthophosphoric acid $(H_3PO_4)$.

(B) The chemical formula of orthophosphoric acid is $H_3PO_4$.
In its structure,there are three $P-OH$ bonds.
The basicity of an oxoacid is determined by the number of $P-OH$ groups present in its structure,as these hydrogen atoms are ionizable.
Since $H_3PO_4$ has three $P-OH$ groups,it is tribasic,meaning its basicity is $3$.

(A) The reaction of phosphorus pentoxide with water is given by:
$P_4O_{10} + 6H_2O \to 4H_3PO_4$
Here,$H_3PO_4$ is orthophosphoric acid.
Thus,$P_4O_{10}$ is the anhydride of orthophosphoric acid.

(C) The atomic number of nitrogen $(N)$ is $7$.
The nitrogen family belongs to group $15$. The atomic numbers of the members are determined by adding the magic numbers $8, 18, 18, 32$.
First member $(N)$: $Z = 7$
Second member $(P)$: $Z = 7 + 8 = 15$
Third member $(As)$: $Z = 15 + 18 = 33$
Therefore,the atomic number of the third member is $33$.

(C) The molecular formula $H_3PO_2$ corresponds to hypophosphorous acid.
In its structure,there is only one $P-OH$ bond,which means only one hydrogen atom is ionizable.
Therefore,its basicity is $1$ (monobasic).

(A) Hypophosphorous acid,also known as phosphinic acid,has the chemical formula $H_3PO_2$.
In its structure,the phosphorus atom is bonded to one oxygen atom via a double bond $(P=O)$,two hydrogen atoms directly $(P-H)$,and one hydroxyl group $(P-OH)$.
This corresponds to the structure shown in option $A$.

(D) When orthophosphoric acid $(H_3PO_4)$ is heated to a high temperature (approx. $600 \ ^oC$),it undergoes dehydration to form metaphosphoric acid $(HPO_3)$.
The chemical reaction is:
$H_3PO_4 \xrightarrow{600 \ ^oC} HPO_3 + H_2O$

(C) $N_2O_5$ is a colorless,deliquescent ionic solid.

(B) Pure $HNO_3$ is a colorless liquid. On exposure to light,it undergoes decomposition to form nitrogen dioxide $(NO_2)$,which dissolves in the remaining $HNO_3$ and imparts a light brown color to it.
The decomposition reaction is:
$4HNO_3 \xrightarrow{hv} 4NO_2 + O_2 + 2H_2O$

(A) The $HNO_3$ is produced on a large scale by the Ostwald process,which provides a good yield.

(C) $NF_3$ is the only stable trihalide of nitrogen.
It does not undergo hydrolysis by water because the $N-F$ bond is very strong and the nitrogen atom is sterically hindered by three fluorine atoms.
Additionally,due to the high electronegativity of fluorine,the lone pair on nitrogen is not easily available for donation,making it a very poor donor.

(C) Among the allotropes of phosphorus,$Black \ phosphorus$ is the most stable and least reactive form.
It possesses a layered structure similar to $Graphite$,which allows it to conduct electricity.

(B) To determine the number of $P-H$ bonds,we examine the structures of the given oxoacids of phosphorus:
$1$. $H_4P_2O_7$ (Pyrophosphoric acid): It has zero $P-H$ bonds.
$2$. $H_3PO_2$ (Hypophosphorous acid): It has two $P-H$ bonds.
$3$. $H_3PO_3$ (Phosphorous acid): It has one $P-H$ bond.
$4$. $H_3PO_4$ (Orthophosphoric acid): It has zero $P-H$ bonds.
Comparing these,$H_3PO_2$ contains the highest number of $P-H$ bonds,which is $2$.

(A) Active nitrogen is produced by passing $N_2$ gas through an electric spark at very low pressure (at $2 \ mm$ of $Hg$).

(C) The oxide $N_2O_4$ (dinitrogen tetroxide) acts as a mixed anhydride of nitrous acid $(HNO_2)$ and nitric acid $(HNO_3)$.
The chemical reaction is:
$N_2O_4 + H_2O \to HNO_2 + HNO_3$

(A) $N_2O_3$ is an acidic oxide and is the anhydride of nitrous acid $(HNO_2)$.
The reaction is as follows:
$2HNO_2 \to N_2O_3 + H_2O$

(A) The reaction between 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:
$4HNO_3 + P_4O_{10} \to 2N_2O_5 + 4HPO_3$
Thus,the product formed is dinitrogen pentoxide $(N_2O_5)$.

(D) Heating $NH_4NO_2$ results in the evolution of nitrogen gas,not ammonia.
The reaction is: $NH_4NO_2 \to N_2 + 2H_2O$.
In contrast,alkaline $NH_4Cl$ and $(NH_4)_2SO_4$ produce $NH_3$ when heated with a base or under specific conditions.

(C) $FeSO_4$ reacts with $NO$ to form a hydrated nitrosyl complex $[Fe(H_2O)_5NO]SO_4$,which is responsible for the brown ring observed in the test.

(B) Pure $N_2$ gas is prepared by the thermal decomposition of sodium azide or barium azide. However,in the laboratory,it is prepared by heating an aqueous solution of ammonium chloride and sodium nitrite:
$NH_4Cl(aq) + NaNO_2(aq) \to N_2(g) + 2H_2O(l) + NaCl(aq)$
This reaction produces pure $N_2$ gas.

(B) Nitrogen does not have $d-$ orbitals available for bonding,which is why it cannot expand its octet.

(C) $N_2O$ (nitrous oxide) is a linear molecule with the structure $N \equiv N^+ - O^- \leftrightarrow N^- = N^+ = O$.
Since the statement in option $C$ claims it is not a linear molecule,it is the incorrect statement.

(D) $NH_3$ (ammonia) is used as a refrigerant in cold storage and as a starting material for the production of nitric acid $(HNO_3)$ via the Ostwald process.
It is $NOT$ used as an anesthetic in medicines.

(C) Statement $C$ is incorrect because the formula for superphosphate of lime is $Ca(H_2PO_4)_2$ and $CaSO_4 \cdot 2H_2O$,not $Ca(H_2SO_4)_2$. Statement $D$ is also incorrect as the hydrolysis of $NCl_3$ yields $NH_3$ and $HOCl$ is a common misconception; it actually produces $NH_3$ and $HOCl$ only in specific conditions,but the standard reaction is $NCl_3 + 3H_2O \rightarrow NH_3 + 3HOCl$. However,in many textbooks,the formula provided in option $C$ is chemically invalid ($Ca(H_2SO_4)_2$ does not exist as a standard compound),making $C$ the most clearly incorrect statement.