Nitrogen family Questions in English

(A) Among the hydrides of Group $15$ elements,$NH_3$ is considered non-combustible under ordinary conditions.
The combustion of $NH_3$ is difficult in the absence of a catalyst (like $Pt$) because the flame temperature is lower than the ignition temperature of the $NH_3$-air mixture.
Other hydrides like $PH_3$,$AsH_3$,and $SbH_3$ are highly flammable and burn readily in air.

(D) The anhydride of an acid is formed by the removal of water molecules from the acid.
For nitric acid $(HNO_3)$,the reaction is:
$2HNO_3 \rightarrow N_2O_5 + H_2O$
Thus,$N_2O_5$ is the anhydride of $HNO_3$.

(C) The industrial production of phosphorus involves heating a mixture of calcium phosphate (present in bone ash),silica $(SiO_2)$,and coke $(C)$ in an electric furnace at approximately $1770 \ K$.
The chemical reactions are as follows:
$2Ca_3(PO_4)_2 + 6SiO_2 \xrightarrow{1770 \ K} 6CaSiO_3 + P_4O_{10}$
$P_4O_{10} + 10C \xrightarrow{1770 \ K} P_4 + 10CO$
Thus,the correct mixture is bone ash,silica,and coke.

(C) Nitrogen forms $5$ oxides in different oxidation states:
$i$. $NO$ (Nitric oxide,Colourless gas)
$ii$. $N_2O$ (Nitrous oxide,Colourless gas)
$iii$. $N_2O_3$ (Dinitrogen trioxide,Blue liquid)
$iv$. $N_2O_4$ (Dinitrogen tetraoxide,Colourless liquid)
$v$. $N_2O_5$ (Dinitrogen pentaoxide,Colourless solid)

(D) Concentrated nitric acid is thermally unstable. Upon heating,it undergoes decomposition to produce nitrogen dioxide,oxygen,and water.
The balanced chemical equation is:
$4HNO_3 \rightarrow 4NO_2 + O_2 + 2H_2O$
Therefore,the correct option is $D$.

(C) The correct answer is $(C)$.
White phosphorus has a very low ignition temperature of approximately $303\,K$ $(30^o\,C)$.
Due to this,it spontaneously catches fire when exposed to air.
To prevent this,it is stored under water.

(D) The correct answer is $(d)$.
Stability of $+3$ oxidation state increases down the group due to the inert pair effect,where the $ns^2$ electrons become less available for bonding.
Reducing character of hydrides increases down the group because the bond dissociation energy of $E-H$ bonds decreases as the atomic size increases,making it easier to release hydrogen.

(A) . The Haber process is the primary industrial method for the fixation of atmospheric nitrogen into ammonia.
The chemical reaction is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ at $650-800 \ K$ and $200-350 \ atm$ pressure.

(A) . $P_4 + 5O_2 \to P_4O_{10}$.
White phosphorus is highly reactive due to angular strain in its $P_4$ molecule,which makes it easily oxidised in air to form phosphorus pentoxide.

(B) pure sample of nitrogen is prepared by the thermal decomposition of Barium azide $\left( Ba(N_3)_2 \right)$.
The reaction is as follows:
$Ba(N_3)_2 \xrightarrow{\Delta} Ba + 3N_2$.

(D) $N_2O$ is a neutral oxide.
$N_2O$ is itself non-combustible,but it supports the combustion of many substances like sulphur because it decomposes to release oxygen at high temperatures.
The reaction is: $S + 2N_2O \to SO_2 + 2N_2$.

(B) $(B)$. When $N_2O$ is inhaled in moderate quantities, it produces hysterical laughter, hence the name laughing gas.

(C) $NO_2$ is a mixed anhydride of nitrous acid $(HNO_2)$ and nitric acid $(HNO_3)$.
When $NO_2$ reacts with $H_2O$,it undergoes disproportionation:
$2NO_2 + H_2O \rightarrow HNO_2 + HNO_3$
When this mixture reacts with $NaOH$,it forms the corresponding sodium salts:
$HNO_2 + HNO_3 + 2NaOH \rightarrow NaNO_2 + NaNO_3 + 2H_2O$
Thus,it forms a mixture of nitrate and nitrite salts.

(B) The reaction of $Zn$ with dilute $HNO_3$ is given by the equation:
$4Zn + 10HNO_3 (\text{dil.}) \to 4Zn(NO_3)_2 + 5H_2O + N_2O$
Thus,$Zn$ produces nitrous oxide $(N_2O)$ with dilute $HNO_3$.

(A) The correct answer is $A$.
Metaphosphoric acid $(HPO_3)$ exists in a polymeric form,commonly known as polymetaphosphoric acid,represented as $(HPO_3)_n$.

(B) Superphosphate of lime is a mixture of calcium dihydrogen phosphate and gypsum,obtained by treating phosphate rock with concentrated $H_2SO_4$.
$Ca_3(PO_4)_2 + 2H_2SO_4 + 4H_2O \to Ca(H_2PO_4)_2 + 2CaSO_4.2H_2O$ (Superphosphate of lime)

(B) White phosphorus consists of discrete tetrahedral $P_4$ molecules,where each phosphorus atom is linked to three other phosphorus atoms by single covalent bonds.

(D) The catalytic oxidation of ammonia (Ostwald process) is represented by the reaction:
$4NH_3 + 5O_2 \xrightarrow{Pt, 1100 \ K} 4NO + 6H_2O$
The oxide formed is $NO$ (Nitric oxide).
This $NO$ is further oxidized to $NO_2$ and then absorbed in water to produce $HNO_3$:
$2NO + O_2 \to 2NO_2$
$4NO_2 + 2H_2O + O_2 \to 4HNO_3$
Thus,the oxide used in the preparation of $HNO_3$ is $NO$.

(C) The reaction between phosphorus and concentrated nitric acid is an oxidation-reduction reaction.
Concentrated nitric acid acts as a strong oxidizing agent and oxidizes phosphorus to phosphoric acid $(H_3PO_4)$.
The balanced chemical equation is:
$P_4 + 20HNO_3 \to 4H_3PO_4 + 20NO_2 + 4H_2O$
Therefore,the correct product of oxidation is $H_3PO_4$.

(B) The acidic or alkaline nature of oxides of Group $15$ and Group $13$ elements depends on their metallic character.
As we move down a group,the metallic character increases,which leads to an increase in the basic (alkaline) nature of the oxides.
$P_2O_3$,$As_2O_3$,and $B_2O_3$ are acidic in nature.
$Bi_2O_3$ is basic (alkaline) in nature due to the high metallic character of Bismuth $(Bi)$.

(B) The reaction of phosphorus$(III)$ oxide $(P_2O_3)$ with water results in the formation of phosphorous acid $(H_3PO_3)$.
The balanced chemical equation is:
$P_2O_3 + 3H_2O \to 2H_3PO_3$

(A) $NF_3$ is the least basic among the nitrogen trihalides.
This is due to the high electronegativity of the three $F$ atoms,which strongly pull the electron density away from the nitrogen atom.
As a result,the lone pair of electrons present on the nitrogen atom is not easily available for donation to a Lewis acid.

(D) The hydrolysis of phosphorus trichloride $(PCl_3)$ in water proceeds according to the following chemical equation:
$PCl_3 + 3H_2O \to H_3PO_3 + 3HCl$
Thus,the product formed is phosphorous acid $(H_3PO_3)$.

(C) The correct answer is $(C)$.
Safe matchsticks use red phosphorus because it is less reactive and more stable than white phosphorus,making it safer for handling and storage.

(D) $N_2O$ (nitrous oxide) is a colourless gas.
$NO$ (nitric oxide) is a colourless gas.
$N_2O_5$ (dinitrogen pentoxide) is a colourless solid.
$NO_2$ (nitrogen dioxide) is a brown coloured gas.
Therefore,the correct option is $D$.

(D) $N_2O_5$ is the anhydride of $HNO_3$.
$2HNO_3 \to N_2O_5 + H_2O$
In $N_2O_5$,the oxidation state of nitrogen is $+5$,which is the maximum possible oxidation state for nitrogen. Therefore,it cannot be oxidized further and can only act as an oxidizing agent.

(C) The acidic character of oxides depends on the oxidation state of the central atom and its position in the periodic table.
$1$. Acidic character increases with an increase in the oxidation state of the central atom.
$2$. Acidic character decreases down the group as the metallic character increases.
Comparing the oxidation states: $P$ in $P_4O_6$ is $+3$,$P$ in $P_4O_{10}$ is $+5$,$As$ in $As_4O_6$ is $+3$,and $As$ in $As_4O_{10}$ is $+5$.
Since $As$ is less electronegative than $P$,the oxides of $As$ are less acidic than the corresponding oxides of $P$.
Among the given options,$As_4O_6$ has the lowest oxidation state $(+3)$ and belongs to a lower period than $P$,making it the least acidic.

(C) The correct order is $NH_3 > PH_3 > AsH_3 > SbH_3$.
As we move down the group,the atomic size of the central atom increases.
This leads to the dispersion of the lone pair over a larger volume,which decreases the electron density on the central atom.
Consequently,the availability of the lone pair for donation decreases,and the basic character decreases.

(A) The thermal stability of the hydrides of Group $15$ elements decreases down the group as the bond dissociation energy decreases due to an increase in the size of the central atom.
The order of stability is: $NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$.
Therefore,$BiH_3$ is the least stable hydride.

(B) $NH_3$ is the most thermally stable hydride among the group $15$ elements due to the small size of the nitrogen atom and strong $N-H$ bond.
Because of its high thermal stability,it requires the maximum temperature for electrolysis compared to the other hydrides $(PH_3, AsH_3, SbH_3)$.

(C) When water is added to a solution of $BiCl_3$ in $HCl$,hydrolysis occurs.
The reaction is as follows:
$BiCl_3 + H_2O \to BiOCl + 2HCl$
The product formed is bismuth oxychloride $(BiOCl)$,which appears as a white precipitate.

(A) The elements $N$,$P$,$As$,$Sb$,and $Bi$ are members of the nitrogen family,which is located in group $15$ of the periodic table.
In the older $IUPAC$ notation,this group is referred to as the $VA$ group.

(A) The correct answer is $A$.
Nitrogen exists as $N_2$ gas in the atmosphere,which is chemically inert and unreactive due to the very high bond dissociation enthalpy of the $N \equiv N$ triple bond $(945 \ kJ/mol)$.
Other elements like phosphorus,arsenic,and antimony are highly reactive and are found in nature only in the form of combined states (minerals/ores).

(B) The correct answer is $B$.
$Bismuth$ $(Bi)$ is a metal and does not exhibit allotropy under normal conditions.
$Nitrogen$ $(N)$ shows allotropy in solid forms ($\alpha-nitrogen$ and $\beta-nitrogen$).
$Phosphorus$ $(P)$ exhibits allotropy (White,Red,and Black phosphorus).
$Arsenic$ $(As)$ exhibits allotropy (Yellow and Grey forms).

(A) $N$ (Nitrogen) does not form complexes because of the absence of $d$-orbitals in its valence shell.
Other elements like $P$,$As$,and $Bi$ can expand their octet or utilize $d$-orbitals to form complexes.

(A) The correct answer is $(A)$.
In the group $15$ hydrides ($NH_3$ to $SbH_3$),the basic strength decreases down the group.
This is because the lone pair on the central atom is present in a larger orbital as the size of the central atom increases.
In $NH_3$,the lone pair is in a smaller $sp^3$ hybridized orbital,making it more concentrated and easily available for donation compared to the larger atoms in $PH_3$,$AsH_3$,and $SbH_3$.

(A) The stability of hydrides of group $15$ elements decreases down the group as the bond dissociation enthalpy decreases.
As we move from $N$ to $Bi$,the size of the central atom increases,which leads to a weaker $M-H$ bond.
Therefore,$NH_3$ is the most stable hydride among the given options.

(B) The boiling point of hydrides of Group $15$ elements depends on the molecular mass and the presence of hydrogen bonding.
$NH_3$ has the highest boiling point due to intermolecular hydrogen bonding.
For the remaining hydrides $(PH_3, AsH_3, SbH_3, BiH_3)$,the boiling point increases with an increase in molecular mass due to stronger van der Waals forces.

\text{Hydride}	\text{Boiling point } $(K)$
$NH_3$	$238.5$
$PH_3$	$185.5$
$AsH_3$	$210.6$
$SbH_3$	$254.6$
$BiH_3$	$290.0$

Thus,$PH_3$ has the lowest boiling point.

(A) $NCl_3$ (Nitrogen trichloride) is highly reactive and thermally unstable due to the weak $N-Cl$ bond and the small size of the nitrogen atom,which leads to significant steric repulsion between the chlorine atoms.
Consequently,it decomposes explosively into nitrogen gas and chlorine gas.
Therefore,the correct option is $A$.

(B) The acidic character of oxides of Group $15$ elements decreases down the group.
$P_2O_3$ is the most acidic oxide among the given options because $P$ is at the top of the group.
$N_2O_3$ and $P_2O_3$ are acidic,$As_2O_3$ and $Sb_2O_3$ are amphoteric,and $Bi_2O_3$ is basic.
Therefore,the correct option is $B$.

(C) $SbCl_2$ does not exist because $V^{th}$ group elements (Group $15$) typically exhibit $+3$ and $+5$ oxidation states.
$Sb$ in $SbCl_2$ would be in the $+2$ oxidation state,which is not a stable or common oxidation state for elements of this group.

(B) The thermal decomposition of ammonium nitrite $(NH_4NO_2)$ yields pure $N_2$ gas.
$NH_4Cl_{(aq)} + NaNO_{2(aq)} \to NH_4NO_{2(aq)} + NaCl_{(aq)}$
$NH_4NO_{2(aq)} \xrightarrow{\Delta} N_{2(g)} + 2H_2O_{(l)}$
Note: While $(NH_4)_2Cr_2O_7$ also produces $N_2$,the reaction with $NH_4Cl$ and $NaNO_2$ is the standard laboratory method for preparing pure $N_2$.

(C) Pure nitrogen is prepared by the thermal decomposition of ammonium nitrite $(NH_4NO_2)$.
The reaction is: $NH_4NO_2 \xrightarrow{\Delta} N_2 + 2H_2O$.
Alternatively,very pure nitrogen can be obtained by the thermal decomposition of sodium azide $(NaN_3)$ or barium azide $(Ba(N_3)_2)$.

(D) Nitrogen is a highly electronegative element that reacts with metals to form ionic compounds known as nitrides,where the nitrogen atom exists as the nitride ion $(N^{3-})$.
Examples of such reactions include:
$6Li + N_2 \to 2Li_3N$ (Lithium nitride)
$3Mg + N_2 \to Mg_3N_2$ (Magnesium nitride)

(D) . The $N \equiv N$ bond dissociation energy is very high,approximately $945 \ kJ \ mol^{-1}$,which makes the $N_2$ molecule chemically inert under normal conditions.

(D) The electronic configuration of nitrogen ($N$,$Z=7$) is $1s^2, 2s^2, 2p^3$.
Nitrogen belongs to the second period and lacks $d$-orbitals in its valence shell.
Therefore,the statement that $d$-orbitals are available for bonding is incorrect.

(C) $Molybdenum$ is essential for nitrogen fixation in plants.
Nitrate metabolism depends upon the enzyme nitrate reductase.
$Molybdenum$ acts as an activator for the enzyme nitrate reductase,which is crucial for the process of nitrogen fixation.

(D) The correct option is $D$.
Laughing gas,which is nitrous oxide $(N_2O)$,is prepared by the thermal decomposition of ammonium nitrate $(NH_4NO_3)$.
The chemical reaction is:
$NH_4NO_3 \xrightarrow{\Delta} N_2O + 2H_2O$.