Nitrogen family Questions in English

(B) Due to the small size of $N$,it has a unique ability to form $P \pi-P \pi$ multiple bonds.
Other elements in the group have larger atomic sizes and diffuse orbitals,which makes effective sidewise overlapping difficult,thus they prefer to form single bonds.

(C) Group $15$ elements have a valence shell configuration of $ns^2 np^3$.
In the $-3$ oxidation state,the central atom possesses a lone pair of electrons,which can be donated to an electron-deficient species.
Therefore,they act as Lewis bases in the $-3$ oxidation state.

(B) In the hydrides of $15^{th}$ group elements,the bond dissociation enthalpy decreases down the group due to an increase in the size of the central atom.
As the bond dissociation enthalpy decreases,the ease of releasing hydrogen atoms increases.
Therefore,the reducing character increases down the group: $NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3$.
Thus,$BiH_3$ is the strongest reducing agent.

(D) An acidic oxide is an oxide that reacts with water to form an acid or with a base to form a salt.
$CO$,$NO$,and $N_2O$ are neutral oxides.
$N_2O_5$ is the anhydride of nitric acid $(HNO_3)$ and reacts with water as follows:
$N_2O_5 + H_2O \rightarrow 2HNO_3$.
Therefore,$N_2O_5$ is an acidic oxide.

(C) When chlorine reacts with excess ammonia,it forms ammonium chloride and dinitrogen gas.
The chemical equation for this reaction is:
$8NH_3 + 3Cl_2 \rightarrow 6NH_4Cl + N_2$

(D) When ammonia reacts with excess chlorine,the reaction is: $NH_3 + 3Cl_2 \rightarrow NCl_3 + 3HCl$.
Thus,the products are nitrogen trichloride $(NCl_3)$ and hydrochloric acid $(HCl)$.

(A) $N_2O_5$ is a non-metallic oxide of nitrogen in a high oxidation state $(+5)$,which makes it acidic in nature.
$NO$ is a neutral oxide.
$Na_2O$ is a metallic oxide,which makes it basic in nature.
$CO$ is a neutral oxide.

(A) Nitrous oxide,also known as dinitrogen oxide or dinitrogen monoxide,has the chemical formula $N_{2}O$.
It is commonly known as laughing gas.
Under room conditions,it is a colourless,non-flammable gas with a pleasant,slightly sweet odour.

(C) Nitrogen dioxide $(NO_2)$ is a paramagnetic,dark reddish-brown gas at room temperature.
$N_2O_5$ is a white solid.
$N_2O_3$ is a blue liquid at low temperatures.
$N_2O_4$ is a colourless gas or liquid.
Therefore,$NO_2$ is the correct answer.

(D) Hydrogen bonding occurs when hydrogen is covalently bonded to a highly electronegative atom like $N$,$O$,or $F$. Among the group-$15$ hydrides ($NH_3$,$PH_3$,$AsH_3$,$SbH_3$,$BiH_3$),only $NH_3$ exhibits hydrogen bonding because $N$ has a high electronegativity and small atomic size.

(C) Among the given oxides,$N_2O$ (nitrous oxide) and $NO$ (nitric oxide) are colourless gases.
$N_2O_4$ (dinitrogen tetroxide) is a colourless solid or liquid.
$NO_2$ (nitrogen dioxide) is a reddish-brown coloured gas.
Therefore,the correct option is $NO_2$.

(C) The correct answer is $Nitrogen$ $(N)$.
Nitrogen exists as a diatomic molecule $(N_2)$ with a very strong triple bond $(N \equiv N)$.
Due to the high bond dissociation energy of this triple bond and its small atomic size,nitrogen is chemically inert under these conditions and does not react with concentrated sulphuric acid $(H_2SO_4)$.

(B) The catalytic oxidation of ammonia produces nitric oxide $(NO)$.
This process is a key step in the Ostwald process for the manufacture of nitric acid.
Ammonia is oxidized by atmospheric oxygen in the presence of a platinum-rhodium catalyst at $800-900^{\circ}C$.
The chemical equation for this reaction is:
$4NH_3(g) + 5O_2(g) \xrightarrow[Pt/Rh]{800-900^{\circ}C} 4NO(g) + 6H_2O(g)$

(B) Nitrogen $(N_2)$ does not react with hot concentrated sulphuric acid $(H_2SO_4)$.
This is because the $N \equiv N$ triple bond has a very high bond dissociation energy due to the small size of the nitrogen atom,making it chemically inert under these conditions.
In contrast,other elements like $P$,$As$,and $Sb$ can be oxidized by hot concentrated $H_2SO_4$ to their respective acids or oxides.

(B) Nitrogen sesquioxide is an oxide containing three atoms of oxygen with two atoms of nitrogen element.
Its chemical formula is $N_2O_3$.

(B) $N_2O_3$ is known as nitrogen sesquioxide because it contains three oxygen atoms for every two nitrogen atoms.
It is obtained as a blue solid when an equimolar mixture of $NO$ and $NO_2$ is cooled below $-20^{\circ}C$.
The oxidation state of nitrogen in $N_2O_3$ is $+3$.

(D) Among the group $15$ elements,$Bi$ (Bismuth) does not exhibit allotropy.
This is primarily because $Bi$ is a metal and the inert pair effect is very prominent in $Bi$,which prevents it from showing the variable oxidation states required to form different allotropic structures.

(A) Nitrogen differs from the rest of the members of group $15$ due to its smaller size,high electronegativity,high ionization enthalpy,and the non-availability of $d$-orbitals.
Nitrogen exists as a diatomic molecule $(N_2)$ because it has the unique ability to form $p\pi - p\pi$ multiple bonds with itself.
In contrast,heavier elements like Phosphorus $(P_4)$,Arsenic $(As_4)$,and Antimony $(Sb_4)$ exist as tetra-atomic molecules because their atomic orbitals are too large and diffuse to form effective $p\pi - p\pi$ bonds,leading them to form single bonds instead.

(A) The Haber process is an industrial method used for the large-scale production of ammonia $(NH_3)$.
In this process,nitrogen gas reacts with hydrogen gas in the presence of an iron catalyst at high temperature and pressure.
The chemical equation is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

(C) Nitrous oxide,$N_{2}O$,is commonly known as laughing gas.

(B) Apatite is a group of phosphate minerals,with the most common being fluorapatite,which has the chemical formula $Ca_5(PO_4)_3F$.
Other variations include hydroxyapatite,$Ca_5(PO_4)_3(OH)$.
In all these forms,the essential elements present are $Ca$ (calcium),$P$ (phosphorus),and $O$ (oxygen),along with $F$ or $OH$.
Therefore,the set of elements present in apatite is $Ca, P$ and $O$.

(D) The oxidation state of phosphorus $(P)$ in its oxyacids ranges from $+1$ to $+5$.
In hypophosphorous acid $(H_3PO_2)$,the oxidation state of $P$ is calculated as:
$3(+1) + x + 2(-2) = 0$
$3 + x - 4 = 0$
$x - 1 = 0$
$x = +1$.
Thus,the lowest oxidation state of phosphorus in its oxyacids is $+1$.

(D) White phosphorus $(P_4)$ consists of four phosphorus atoms arranged at the corners of a regular tetrahedron.
In this structure,each phosphorus atom is bonded to three other phosphorus atoms.
Due to the geometric constraints of the tetrahedral arrangement,the $P-P-P$ bond angle is $60^{\circ}$.
This high angular strain makes white phosphorus highly reactive.

(C) $P_{4}O_{10}$ is the anhydride of phosphoric acid. When $P_{4}O_{10}$ reacts with water,it undergoes hydrolysis to form phosphoric acid $(H_{3}PO_{4})$.
The balanced chemical equation is:
$P_{4}O_{10} + 6H_{2}O \rightarrow 4H_{3}PO_{4}$

(D) Red phosphorus is non-poisonous in nature.
Unlike white phosphorus,which is highly toxic,red phosphorus is a polymeric form that is stable and non-toxic under normal conditions.
Phosgene $(COCl_2)$,gaseous chlorine $(Cl_2)$,and phosphine $(PH_3)$ are all highly toxic substances.

(A) The chemical formula of pyrophosphoric acid is $H_4P_2O_7$.
By observing its structure,we can count the bonds:
$1$. Each phosphorus atom is bonded to two $OH$ groups. Since there are two phosphorus atoms,the total number of $P-OH$ bonds is $2 \times 2 = 4$.
$2$. There is one $P-O-P$ linkage connecting the two phosphorus atoms.
Therefore,the number of $P-OH$ bonds is $4$ and the number of $P-O-P$ bonds is $1$.

(D) The highest oxidation state possessed by phosphorus in its oxyacids is $+5$.
In oxoacids,phosphorus is tetrahedrally surrounded by other atoms.
All these acids contain at least one $P=O$ bond and one $P-OH$ bond.
The oxoacids in which phosphorus has the maximum oxidation state of $+5$ are orthophosphoric acid $(H_3PO_4)$,pyrophosphoric acid $(H_4P_2O_7)$,and metaphosphoric acid $((HPO_3)_n)$.

(A) Pyrophosphoric acid is formed by the condensation of two molecules of orthophosphoric acid $(H_3PO_4)$ with the elimination of one water molecule $(H_2O)$.
$2H_3PO_4 \rightarrow H_4P_2O_7 + H_2O$.
As shown in the structure,there are $4$ hydrogen atoms,$2$ phosphorus atoms,and $7$ oxygen atoms.
Therefore,the molecular formula is $H_4P_2O_7$.

(C) Red phosphorus is a polymeric structure and is chemically much less reactive than white phosphorus.
$1$. Red phosphorus is insoluble in carbon disulphide $(CS_2)$,whereas white phosphorus is soluble.
$2$. Red phosphorus does not show chemiluminescence.
$3$. Red phosphorus is non-poisonous.
$4$. Red phosphorus does not react with $NaOH$ to form phosphine $(PH_3)$ under normal conditions; only white phosphorus reacts with hot $NaOH$ solution to produce phosphine gas.
Therefore,the statement that it forms phosphine when treated with hot sodium hydroxide solution is incorrect for red phosphorus.

(D) Allotropy is a property shown by elements where they exist in two or more different physical forms in the same physical state. Among the elements of Group $15$,Nitrogen,Phosphorus,Arsenic,and Antimony exhibit allotropy. Bismuth is a metal and does not exhibit allotropy.

(C) The disproportionation reaction is a reaction in which an element in an intermediate oxidation state is simultaneously oxidized and reduced.
In $H_3PO_3$ (phosphonic acid),the oxidation state of phosphorus is $+3$.
It undergoes disproportionation upon heating to form $H_3PO_4$ (oxidation state $+5$) and $PH_3$ (oxidation state $-3$).
The reaction is: $4H_3PO_3 \rightarrow 3H_3PO_4 + PH_3$.

(B) To find the oxidation state of phosphorus $(x)$ in $H_3PO_2$ (Phosphinic acid):
$3(+1) + x + 2(-2) = 0$
$3 + x - 4 = 0$
$x - 1 = 0$
$x = +1$
Therefore,phosphorus is in the $+1$ oxidation state in Phosphinic acid $(H_3PO_2)$.

(A) An oxyacid of phosphorus that contains $P-H$ bonds can act as a reducing agent.
$H_{3}PO_{3}$ (phosphorous acid) contains one $P-H$ bond,which makes it a reducing agent.
In contrast,$H_{3}PO_{4}$,$H_{2}P_{2}O_{6}$,and $H_{4}P_{2}O_{7}$ do not contain any $P-H$ bonds and therefore do not act as reducing agents.

(A) Nitrogen has two naturally occurring stable isotopes,namely,$^{14}N$ and $^{15}N$.

(B) The reaction of white phosphorus with concentrated $NaOH$ is: $P_{4} + 3NaOH + 3H_{2}O \xrightarrow{\Delta, CO_{2}} 3NaH_{2}PO_{2} + PH_{3}\uparrow$ (phosphine).
$PH_{3}$ is less basic than $NH_{3}$ because the lone pair of electrons on the phosphorus atom is less available for donation compared to nitrogen,due to the larger size of the phosphorus atom and the involvement of $s$-character in the bonding.
Therefore,the statement that $PH_{3}$ is more basic than $NH_{3}$ is incorrect.

(B) metalloid is defined as a chemical element whose physical and chemical properties lie between those of metals and non-metals.
Among the given options,only antimony $(Sb)$ is a metalloid.

(A) The brown compound of manganese is $MnO_2$ $(A)$. When $MnO_2$ reacts with $HCl$,it produces chlorine gas $(B)$:
$MnO_2 (A) + 4 HCl \rightarrow MnCl_2 + 2 H_2O + Cl_2 (B)$
When excess chlorine gas $(B)$ reacts with ammonia $(NH_3)$,it forms nitrogen trichloride $(C)$,which is an explosive compound:
$3 Cl_2 (B) + NH_3 \rightarrow NCl_3 (C) + 3 HCl$
Thus,the compounds $A$,$B$,and $C$ are $MnO_2$,$Cl_2$,and $NCl_3$ respectively.

(B) The correct statement is that phosphorous acid $(H_{3}PO_{3})$ on heating disproportionates to give orthophosphoric acid $(H_{3}PO_{4})$ and phosphine $(PH_{3})$.
$4H_{3}PO_{3} \xrightarrow{\Delta} 3H_{3}PO_{4} + PH_{3}$.
Therefore,the statement in option $B$ is incorrect because it mentions metaphosphoric acid instead of orthophosphoric acid.

(C) $NO_2$ (Nitrogen dioxide) gas is brown in colour and acidic in nature.

(C) Among the given compounds,only ammonium nitrate $(NH_4NO_3)$ does not produce $N_2$ gas on decomposition; instead,it produces nitrous oxide $(N_2O)$.
$Ba(N_3)_2 \xrightarrow{\Delta} Ba + 3N_2$
$NH_4NO_2 \xrightarrow{\Delta} N_2 + 2H_2O$
$NH_4NO_3 \xrightarrow{\Delta} N_2O + 2H_2O$
$(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} Cr_2O_3 + N_2 + 4H_2O$

(A) The thermal decomposition of ammonium dichromate $(NH_4)_2Cr_2O_7$ produces nitrogen gas $(N_2)$:
$(NH_4)_2Cr_2O_7 \stackrel{\Delta}{\longrightarrow} N_2 \uparrow + Cr_2O_3 + 4H_2O$
Similarly,ammonium nitrite $(NH_4NO_2)$ on heating also liberates nitrogen gas $(N_2)$:
$NH_4NO_2 \stackrel{\Delta}{\longrightarrow} N_2 \uparrow + 2H_2O$
Thus,the same gas is obtained by heating $NH_4NO_2$.

(B) The thermal stability of the hydrides of group $15$ elements decreases down the group as $NH_{3} > PH_{3} > AsH_{3} > SbH_{3} > BiH_{3}$.
As we move down the group,the atomic size of the central atom increases (from $N$ to $Bi$).
This leads to an increase in the bond length and a decrease in the bond dissociation enthalpy (bond strength).
Consequently,the thermal stability decreases.

(C) The reactions are as follows:
$1$. $2NO(g) + O_2(g) \rightarrow 2NO_2(g)$ (Reddish brown gas $Q$)
$2$. $2NO_2(g) \rightleftharpoons N_2O_4(g)$ (Colourless gas $R$ on cooling)
$3$. $NO(g) + NO_2(g) \rightarrow N_2O_3(s)$ (Blue solid $S$)
Thus,$P = NO$,$Q = NO_2$,$R = N_2O_4$,and $S = N_2O_3$.

(D) Very pure nitrogen gas is obtained by the thermal decomposition of sodium azide $(NaN_3)$.
The chemical reaction is:
$2NaN_3 \rightarrow 2Na + 3N_2$
This method is commonly used in automobile airbags.

(D) The electronic configuration of $N$ $(Z=7)$ is $1s^{2} 2s^{2} 2p^{3}$.
Since the principal quantum number $n=2$,it does not have $d$-orbitals available.
Due to the stable half-filled $2p$ subshell,its ionization enthalpy is very high.
Nitrogen is the third most electronegative element in the periodic table.
Due to its small size,nitrogen has a unique ability to form $p\pi-p\pi$ multiple bonds with itself (forming $N \equiv N$).

(D) The basicity of group-$15$ hydrides depends on the availability of the lone pair of electrons on the central atom for donation.
As we move down the group from $N$ to $Bi$,the size of the central atom increases.
This leads to the dispersion of the lone pair over a larger volume,making it less available for donation.
Therefore,the basicity decreases down the group.
The correct decreasing order is $NH_3 > PH_3 > AsH_3 > SbH_3$.

(A) Hypophosphorus acid $(H_3PO_2)$ acts as a strong reducing agent because it contains two $P-H$ bonds.
It can effectively reduce $Ag^{+}$ ions to metallic $Ag$ and itself gets oxidized to phosphorus acid $(H_3PO_3)$.
The chemical reaction is as follows:
$3AgNO_3 + H_3PO_2 + 2H_2O \longrightarrow 3Ag + H_3PO_3 + 3HNO_3$

(B) The strong reducing property of hypophosphorus acid $(H_3PO_2)$ is due to the presence of two $P-H$ bonds. The hydrogen atoms directly bonded to the phosphorus atom are responsible for its reducing character. The structure of $H_3PO_2$ is as follows:
$O=P(H)(H)(OH)$

(A) Phosphorus pentachloride $(PCl_{5})$ on hydrolysis gives phosphoric acid $(H_{3}PO_{4})$,which is a tribasic acid.
$PCl_{5} + 4H_{2}O \longrightarrow H_{3}PO_{4} + 5HCl$
In $H_{3}PO_{4}$,there are three $P-OH$ groups,making it tribasic.