Nitrogen family Questions in English

(A) Nitrogen $(I)$ oxide,also known as nitrous oxide $(N_2O)$,is prepared by the thermal decomposition of ammonium nitrate $(NH_4NO_3)$.
The chemical equation is:
$NH_4NO_3 \xrightarrow{\Delta} N_2O + 2H_2O$
Therefore,the correct option is $A$.

(C) $N_2O$ (nitrous oxide) is a linear molecule with the structure $N \equiv N^+ - O^-$. Therefore,the statement that it is not a linear molecule is incorrect.

(B) The anhydride of an acid is formed by the removal of water from the acid. For nitrous acid $(HNO_2)$,the reaction is:
$2HNO_2 \to H_2O + N_2O_3$.
Thus,$N_2O_3$ is the anhydride of nitrous acid.

(A) Heating lead$(II)$ nitrate $(Pb(NO_3)_2)$ results in its thermal decomposition to produce lead$(II)$ oxide $(PbO)$,nitrogen dioxide $(NO_2)$,and oxygen $(O_2)$.
The balanced chemical equation is: $2Pb(NO_3)_2(s) \xrightarrow{\Delta} 2PbO(s) + 4NO_2(g) + O_2(g)$.
Other nitrates like $KNO_3$ and $NaNO_3$ decompose to form nitrites $(KNO_2, NaNO_2)$ and oxygen gas,but do not release $NO_2$ under standard heating conditions.

(B) The reaction of dilute $HNO_3$ with copper $(Cu)$ produces nitric oxide $(NO)$ as the primary gaseous product.
The balanced chemical equation is:
$3Cu + 8HNO_3 (\text{dilute}) \to 3Cu(NO_3)_2 + 4H_2O + 2NO$
Therefore,the correct option is $B$.

(D) During a lightning flash,the high temperature allows atmospheric nitrogen and oxygen to react to form nitric oxide $(NO)$.
${N_2(g)} + {O_2(g)} \xrightarrow{\text{lightning}} 2NO(g)$

(D) Nitrous oxide $(N_2O)$ and nitric oxide $(NO)$ are neutral oxides of nitrogen.
$N_2O_3$,$N_2O_4$,and $N_2O_5$ are acidic in nature.
The acidic character increases with an increase in the oxidation state of nitrogen.
Therefore,the correct answer is $N_2O$.

(C) Nitric oxide $(NO)$ reacts with oxygen $(O_2)$ present in the air to form nitrogen dioxide $(NO_2)$.
The chemical equation for this reaction is:
$2NO(g) + O_2(g) \to 2NO_2(g)$

(C) Nitric oxide $(NO)$ reacts with atmospheric oxygen to form nitrogen dioxide $(NO_2)$.
The chemical equation is:
$2 NO(g) + O_2(g) \rightarrow 2 NO_2(g)$
Nitrogen dioxide $(NO_2)$ is a reddish-brown coloured gas.

(D) Nitrogen dioxide $(NO_2)$ undergoes disproportionation reaction when dissolved in water.
The chemical equation for the reaction is:
$2NO_2 + H_2O \to HNO_3 + HNO_2$
Thus,it forms a mixture of nitric acid $(HNO_3)$ and nitrous acid $(HNO_2)$.

(B) The catalytic oxidation of ammonia is the first step in the Ostwald process for the manufacture of nitric acid.
In this process,a mixture of ammonia and air is passed over a $Pt$ gauze catalyst at $800\,^{\circ}C$.
The chemical reaction is:
$4NH_3 + 5O_2 \xrightarrow{Pt, \,800\,^{\circ}C} 4NO + 6H_2O$
Thus,the product formed is nitric oxide $(NO)$.

(D) Ammonia $(NH_3)$ is highly soluble in water,so it cannot be collected by the downward displacement of water.
It is less dense than air,so it is collected by the upward displacement of air.
Since it does not react with mercury,it can also be collected by the displacement of mercury.

(D) . $NH_3$ is a highly volatile compound.
When vaporized,liquid ammonia causes intense cooling.
Hence,it is used as a coolant in ice factories and cold storages.

(A) The electronic configuration of nitrogen $(N)$ is $1s^2 2s^2 2p^3$.
It lacks vacant $d$-orbitals in its valence shell,which prevents it from expanding its octet to form five covalent bonds.
In contrast,phosphorus $(P)$ has vacant $3d$-orbitals,allowing it to form $PCl_5$ by expanding its octet.

(C) The stability of phosphorus allotropes follows the order: $Black > Red > White$.
$White$ phosphorus is highly reactive due to high angular strain in its $P_4$ tetrahedral structure.
$Red$ phosphorus is polymeric and more stable than $white$ phosphorus.
$Black$ phosphorus is the most thermodynamically stable allotrope due to its layered structure and lack of angular strain.

(A) Red phosphorus $(P_n)$ is obtained from white phosphorus by heating it to $300$ $^{\circ}C$ in an inert atmosphere (absence of air) in the presence of a catalyst.

(C) Bones contain calcium phosphate. When exposed to air,the white phosphorus present in the bones undergoes slow oxidation (combustion) in contact with atmospheric oxygen. This process releases energy in the form of light,causing the bones to glow in the dark,a phenomenon known as phosphorescence.

(D) White phosphorus exhibits phosphorescence in air,whereas red phosphorus does not.
White phosphorus reacts with hot aqueous $NaOH$ to produce phosphine $(PH_3)$,whereas red phosphorus does not react under these conditions.
White phosphorus is soluble in carbon disulphide $(CS_2)$,whereas red phosphorus is insoluble.
Both white phosphorus and red phosphorus burn when heated in air to form phosphorus pentoxide $(P_4O_{10})$.

(A) In the modern process,phosphorus is manufactured by heating a mixture of phosphorite mineral $(Ca_3(PO_4)_2)$ with sand $(SiO_2)$ and coke $(C)$ in an electric furnace at $1773 \text{ K}$.
The chemical reaction is: $2Ca_3(PO_4)_2 + 6SiO_2 + 10C \rightarrow 6CaSiO_3 + 10CO + P_4$.

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

(B) The reaction of white phosphorus $(P_4)$ with $NaOH$ solution produces phosphine $(PH_3)$ via a disproportionation reaction.
$Ca_3P_2$ reacts with water to yield $PH_3$ and $Ca(OH)_2$.
$P_4O_6$ reacts with water to produce $H_3PO_3$ (phosphorous acid),not $PH_3$.
Red phosphorus is chemically inert compared to white phosphorus and does not react with $NaOH$ solution under normal conditions.
Therefore,the correct option is $(B)$.

(A) The reaction between phosphonium iodide $(PH_4I)$ and sodium hydroxide $(NaOH)$ is a base-acid reaction that produces phosphine gas $(PH_3)$,water $(H_2O)$,and sodium iodide $(NaI)$.
The balanced chemical equation is:
$PH_4I + NaOH \rightarrow PH_3 + H_2O + NaI$

(C) The correct answer is $(C)$.
Phosphine $(PH_3)$ is prepared by the hydrolysis of calcium phosphide $(Ca_3P_2)$.
The chemical reaction is as follows:
$Ca_3P_2 + 6H_2O \to 3Ca(OH)_2 + 2PH_3$

(B) Aluminium phosphide $(AlP)$ reacts with dilute sulphuric acid $(H_2SO_4)$ to produce aluminium sulphate and phosphine gas $(PH_3)$.
The balanced chemical equation is:
$2AlP + 3H_2SO_4 \rightarrow Al_2(SO_4)_3 + 2PH_3$

(B) The basicity of hydrides of Group $15$ elements depends on the availability of the lone pair on the central atom for donation.
In $NH_3$,the lone pair is present in an $sp^3$ hybrid orbital which is small and concentrated,making it readily available for donation.
In $PH_3$,the lone pair is present in a large $3s$ orbital with more $s$-character,making it less available for donation.
Therefore,$PH_3$ is less basic than $NH_3$.

(D) The reaction of phosphorus $(III)$ oxide $(P_2O_3)$ with water is given by:
$P_2O_3 + 3H_2O \to 2H_3PO_3$
In this reaction,phosphorus is in the $+3$ oxidation state.
$H_3PO_3$ is phosphorous acid.
$HPO_3$,$H_4P_2O_7$,and $H_3PO_4$ are derived from phosphorus $(V)$ oxide $(P_2O_5)$.

(D) $P_2O_5$ reacts with water to form orthophosphoric acid.
The chemical equation is: $P_2O_5 + 3H_2O \to 2H_3PO_4$.
Therefore,the correct option is $D$.

(C) The chemical formula of hypophosphorous acid is $H_3PO_2$.
In its structure,there is only one $-OH$ group attached to the phosphorus atom.
Since only the hydrogen atoms bonded to oxygen atoms are ionizable,it acts as a monobasic acid.
Therefore,the correct option is $C$.

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

(B) The structure of $H_3PO_3$ (phosphorous acid) contains two $P-OH$ bonds and one $P-H$ bond.
Only the hydrogen atoms attached to oxygen atoms are ionizable.
Since there are two $OH$ groups attached to the phosphorus atom,$H_3PO_3$ is a dibasic acid.
Therefore,the correct option is $(B)$.

(B) The reaction of phosphorus with hot concentrated $H_2SO_4$ is an oxidation-reduction reaction.
$P_4 + 10H_2SO_4 \to 4H_3PO_4 + 10SO_2 + 4H_2O$
In this reaction,phosphorus $(P_4)$ is oxidized to orthophosphoric acid $(H_3PO_4)$ and sulfuric acid is reduced to sulfur dioxide $(SO_2)$.

(C) The cyanamide process involves the hydrolysis of calcium cyanamide $(CaCN_2)$ to produce ammonia $(NH_3)$.
The chemical reaction is as follows:
$CaCN_2 + 3H_2O \to CaCO_3 + 2NH_3$

(B) The chemical formula of pyrophosphoric acid is $H_4P_2O_7$.
The structure consists of two $P=O$ groups and an $O-P-O-P-O$ backbone,where each phosphorus atom is bonded to two hydroxyl $(-OH)$ groups.
The structure is $(HO)_2P(O)-O-P(O)(OH)_2$.
Therefore,there are $4$ hydroxyl groups present in the molecule.

(C) The acid strength of oxoacids of phosphorus depends on the number of $P=O$ and $P-OH$ groups. However,the overall difference in their acid strength is relatively small because phosphorus is not a highly electronegative element. This results in the $P-OH$ bond being less polarized compared to other non-metal oxoacids,making the release of $H^+$ ions less sensitive to the oxidation state of phosphorus.

(B) When $BiCl_3$ undergoes hydrolysis,it reacts with water to form a white precipitate of bismuth oxychloride $(BiOCl)$ and hydrochloric acid $(HCl)$.
The balanced chemical equation is:
$BiCl_3 + H_2O \to BiOCl + 2HCl$

(A) $NO$ (Nitric oxide) is a neutral oxide.
Anhydride of nitrous acid $(HNO_2)$ is $N_2O_3$.
$NO$ has a dipole moment of $0.15 \ D$ (not $0.22 \ D$),but the most chemically incorrect statement among the options is that it is the anhydride of nitrous acid.
$NO$ is paramagnetic due to the presence of an unpaired electron.
$NO$ does not form a stable dimer in the gas phase; it remains as a monomer.

(B) The reaction of phosphorus with sodium hydroxide is a disproportionation reaction.
White phosphorus $(P_4)$ reacts with concentrated sodium hydroxide $(NaOH)$ solution in an inert atmosphere to produce phosphine $(PH_3)$ and sodium hypophosphite $(NaH_2PO_2)$.
The balanced chemical equation is: $P_4 + 3NaOH + 3H_2O \to PH_3 + 3NaH_2PO_2$.

(A) Among the given hydrides of group $15$ elements,$NH_3$ is the only one capable of forming intermolecular hydrogen bonds with water molecules.
This strong interaction leads to its high solubility in water.
Other hydrides like $PH_3$,$AsH_3$,and $SbH_3$ do not form hydrogen bonds and are only sparingly soluble in water.

(D) The boiling point of hydrides of Group $15$ elements generally increases down the group due to an increase in molecular mass and the resulting increase in van der Waals forces of attraction. However,$NH_3$ has an anomalously high boiling point due to intermolecular hydrogen bonding. Comparing the values,$SbH_3$ has the highest boiling point among the given options.

\text{Substance}	$B.pt$ \text{ in } $(K)$
$NH_3$	$238.5$
$PH_3$	$185.5$
$AsH_3$	$210.6$
$SbH_3$	$254.6$

(A) The oxidation states of phosphorus in the given oxyacids are as follows:
$1$. Hypophosphorous acid $(H_3PO_2)$: $3(+1) + x + 2(-2) = 0 \implies 3 + x - 4 = 0 \implies x = +1$
$2$. Orthophosphoric acid $(H_3PO_4)$: $3(+1) + x + 4(-2) = 0 \implies 3 + x - 8 = 0 \implies x = +5$
$3$. Pyrophosphoric acid $(H_4P_2O_7)$: $4(+1) + 2x + 7(-2) = 0 \implies 4 + 2x - 14 = 0 \implies 2x = 10 \implies x = +5$
$4$. Metaphosphoric acid $(HPO_3)$: $1(+1) + x + 3(-2) = 0 \implies 1 + x - 6 = 0 \implies x = +5$
Comparing these values,the lowest oxidation state of phosphorus $(+1)$ is found in Hypophosphorous acid.

(A) In a $P_4$ molecule of white phosphorus,the four phosphorus atoms are linked to each other by single covalent bonds.
These atoms are arranged at the four corners of a regular tetrahedron,as shown in the structure.
Each phosphorus atom is bonded to three other phosphorus atoms,and the bond angle is $60^\circ$.

(D) . The most common minerals of phosphorus are hydroxy apatite,$Ca_9(PO_4)_6 \cdot Ca(OH)_2$,and fluorapatite,$Ca_9(PO_4)_6 \cdot CaF_2$.

(A) The three important oxidation states of phosphorus are $-3, +3$,and $+5$.
$1$. $-3$: Phosphorus has $5$ electrons in its valence shell $(3s^2 3p^3)$. To attain a stable octet configuration,it gains $3$ electrons,resulting in a $-3$ oxidation state.
$2$. $+3$: Phosphorus uses its $3$ unpaired $p$-electrons for bonding,resulting in a $+3$ oxidation state.
$3$. $+5$: Phosphorus uses all $5$ valence electrons ($3s^2$ and $3p^3$) for bonding,resulting in a $+5$ oxidation state.

(A) Nitrogen $(N)$ has the electronic configuration $1s^2 2s^2 2p^3$. It lacks vacant $d$-orbitals in its valence shell,so it cannot expand its octet to form $NCl_5$.
Phosphorus $(P)$ has the electronic configuration $[Ne] 3s^2 3p^3 3d^0$. Due to the presence of vacant $3d$-orbitals,phosphorus can expand its octet and form $PCl_5$ by promoting an electron from the $3s$ orbital to the $3d$ orbital.
Therefore,the correct reason is the availability of vacant $d$-orbitals in $P$ but not in $N$.

(D) When metallic tin $(Sn)$ reacts with concentrated nitric acid $(HNO_3)$,it gets oxidized to metastannic acid $(H_2SnO_3)$.
The balanced chemical equation is:
$Sn + 4HNO_3 (\text{conc.}) \rightarrow H_2SnO_3 + 4NO_2 + H_2O$
Thus,the correct product is meta stannic acid.

(C) The reaction of $Copper$ with cold and dilute $Nitric \ acid$ $(HNO_3)$ is the standard laboratory method for the preparation of $Nitric \ oxide$ $(NO)$.
The balanced chemical equation is:
$3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 4H_2O + 2NO$.

(C) The pentavalency in phosphorus is more stable than that of nitrogen due to the larger size of the phosphorus atom.
Nitrogen has a small atomic size and lacks $d$-orbitals,which prevents it from expanding its octet to form pentavalent compounds.
Phosphorus,being larger,can accommodate five bonds.

(D) $PH_3$ (phosphine) is a colorless gas with a rotten fish smell. It is slightly soluble in water. However,compared to $NH_3$,it is much less soluble because it does not form hydrogen bonds with water molecules. Among the given options,the statement that $PH_3$ is insoluble in water is often considered the intended answer in many textbooks to highlight the lack of hydrogen bonding compared to $NH_3$.

(A) The reaction of $NaNO_2$ with $NH_4Cl$ produces ammonium nitrite $(NH_4NO_2)$ as an intermediate,which is thermally unstable and decomposes to give nitrogen gas.
Step $1$: $NH_4Cl + NaNO_2 \longrightarrow NH_4NO_2 + NaCl$
Step $2$: $NH_4NO_2 \overset{\Delta}{\longrightarrow} N_2 + 2H_2O$
Therefore,the correct option is $A$.

(A) Nitrogen reacts with metals to form nitrides.
For example,lithium reacts with nitrogen to form lithium nitride:
$6Li + N_2 \xrightarrow{450^\circ C} 2Li_3N$.