Nitrogen family Questions in English

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

(A) Nitrous oxide $(N_2O)$ is a neutral oxide. It is soluble in cold water.

(D) The industrial production of nitric acid by the Ostwald process involves the catalytic oxidation of ammonia $(NH_3)$:
$4NH_{3(g)} + 5O_{2(g)} \xrightarrow{Pt/Rh} 4NO_{(g)} + 6H_2O_{(g)}$
Then,nitric oxide $(NO)$ is oxidized to nitrogen dioxide $(NO_2)$:
$2NO_{(g)} + O_{2(g)} \to 2NO_{2(g)}$
Finally,nitrogen dioxide is absorbed in water to produce nitric acid $(HNO_3)$:
$4NO_{2(g)} + 2H_2O_{(l)} + O_{2(g)} \to 4HNO_{3(aq)}$
Thus,$NH_3$ is the starting material.

(B) Concentrated nitric acid acts as a strong oxidizing agent and oxidizes phosphorus to orthophosphoric acid $(H_3PO_4)$.
The balanced chemical equation is:
$P + 5HNO_3 \text{ (conc.)} \to H_3PO_4 + 5NO_2 + H_2O$

(A) When nitrogen dioxide $(NO_2)$ reacts with water,it undergoes disproportionation to form a mixture of nitrous acid $(HNO_2)$ and nitric acid $(HNO_3)$.
The balanced chemical equation is:
$2NO_2 + H_2O \to HNO_2 + HNO_3$

(A) White phosphorus is converted to red phosphorus by heating it in an inert atmosphere at $240 - 250 \, ^\circ C$.
Chemical reaction: $P_4 \text{ (white)} \xrightarrow{240 - 250 \, ^\circ C, \text{ inert atmosphere}} P_n \text{ (red)}$

(B) The structure of pyrophosphoric acid $(H_4P_2O_7)$ is as follows:
$HO-P(O)(OH)-O-P(O)(OH)-OH$
In the structure of pyrophosphoric acid $(H_4P_2O_7)$,there are $4$ hydroxyl $(-OH)$ groups attached to the phosphorus atoms. Therefore,it is a tetrabasic acid.

(D) In $N_2O_5$,the oxidation state of nitrogen is $+5$.
Nitrogen cannot attain an oxidation state higher than $+5$.
Therefore,it cannot act as a reducing agent.
Instead,it acts as a strong oxidizing agent.

(A) The bond angle in the hydrides of group $15$ elements decreases down the group.
The correct order is $NH_3 > PH_3 > AsH_3 > SbH_3$.
This is because as we move from $N$ to $Sb$,the electronegativity of the central atom decreases,which leads to a decrease in the bond angle.

(A) When ammonia $(NH_3)$ reacts with heated copper$(II)$ oxide $(CuO)$,it acts as a reducing agent and gets oxidized to dinitrogen gas $(N_2)$.
The chemical equation for this reaction is:
$2NH_3(g) + 3CuO(s) \xrightarrow{\Delta} N_2(g) + 3Cu(s) + 3H_2O(l)$

(C) When ammonium nitrite $(NH_4NO_2)$ is heated,it undergoes thermal decomposition to produce nitrogen gas and water vapor:
$NH_4NO_2 \xrightarrow{\Delta} N_2 + 2H_2O$
Other ammonium salts like $(NH_4)_2SO_4$ or $(NH_4)_2CO_3$ do not produce $N_2$ gas upon heating.

(C) The basicity of nitrogen trihalides depends on the availability of the lone pair on the nitrogen atom.
As the electronegativity of the halogen atom decreases from $F$ to $I$,the electron density on the nitrogen atom increases due to the decrease in the inductive effect.
Therefore,the order of basicity is $NF_3 < NCl_3 < NBr_3 < NI_3$.
Thus,$NI_3$ is the most basic among the given options.

(C) Metaphosphoric acids have the general formula $(HPO_3)_n$.
Cyclic metaphosphoric acids are formed when $n \ge 3$.
For $n = 5$,the formula is $(HPO_3)_5$ or $H_5P_5O_{15}$,which represents a cyclic structure known as cyclometaphosphoric acid.

(A) Metaphosphoric acid has the empirical formula $HPO_3$. The cyclic trimer of metaphosphoric acid is known as cyclotrimetaphosphoric acid,which has the molecular formula $H_3P_3O_9$.

(A) Nitrogen has an electronic configuration of $1s^2 2s^2 2p^3$. It lacks $d$-orbitals in its valence shell $(n=2)$,so it cannot expand its octet to form $NCl_5$.
Phosphorus has an electronic configuration of $[Ne] 3s^2 3p^3$. It possesses vacant $3d$-orbitals,which allow it to undergo $sp^3d$ hybridization to form $PCl_5$.

(C) The basic character of the oxides of group $15$ elements increases as we move down the group from $N$ to $Bi$.

(D) The brown color of $HNO_3$ is due to the presence of dissolved $NO_2$ gas.
To remove this color,air is passed through the warm acid,which drives out the dissolved $NO_2$.

(B) The structure of $H_4P_2O_5$ (pyrophosphorous acid) contains a $P - O - P$ linkage.
The structure is as follows:
$HO-P(OH)-O-P(OH)-H$ (with one $P=O$ bond).
Specifically,$H_4P_2O_5$ has the structure $(HO)_2P-O-P(H)(OH)=O$,which features the $P-O-P$ bond.

(C) Phosphine burns in an atmosphere of chlorine to form phosphorus pentachloride and hydrogen chloride,releasing heat in the process.
$PH_3 + 4Cl_2 \to PCl_5 + 3HCl$

(D) The oxide of nitrogen $NO_2$ is a brown-coloured gas.
$N_2O_5$ is a colourless solid.
$NO$ and $N_2O_4$ are colourless gases.

(D) Thermal decomposition reactions are as follows:
$NH_4NO_2 \xrightarrow{\Delta} N_2 + 2H_2O$
$2NaN_3 \xrightarrow{\Delta} 2Na + 3N_2$
$(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} Cr_2O_3 + 4H_2O + N_2$
Since all the given compounds release $N_2$ gas upon heating,the correct option is $D$.

(D) The thermal decomposition of ammonium dichromate is given by the reaction:
$(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} N_2 + 4H_2O + Cr_2O_3$
As shown in the reaction,$N_2$ gas is evolved.

(A) The thermal decomposition of ammonium dichromate is given by: $(NH_4)_2Cr_2O_7 \to N_2 + Cr_2O_3 + 4H_2O$.
Thus,the gas released is $N_2$.
Heating ammonium nitrite also produces $N_2$ gas: $NH_4NO_2 \to N_2 + 2H_2O$.

(A) At $-30\,^oC$,$N_2O_3$ exists as a blue liquid.
It is formed by the equimolar mixture of $NO$ and $NO_2$ gases.
The reaction is as follows:
$NO(g) + NO_2(g) \xrightarrow{-30\,^oC} N_2O_3(l)$ (blue liquid).

(B) $PH_3$ (Phosphine) is a colorless,highly poisonous gas that possesses a characteristic odor of rotten fish.

(C) The chemical formula of hypophosphorous acid (phosphinic acid) is $H_3PO_2$.
In its structure,the phosphorus atom is bonded to two hydrogen atoms directly ($P-H$ bonds),one hydroxyl group $(-OH)$,and one oxygen atom via a double bond $(P=O)$.
Therefore,there are $2$ hydrogen atoms directly bonded to the phosphorus atom.

(C) $1$. In aqueous medium,the acidity of hydrogen halides depends on the bond dissociation energy. $HCl$ is a stronger acid than $HF$ because the $H-Cl$ bond is weaker than the $H-F$ bond.
$2$. $HClO_4$ (perchloric acid) is a stronger acid than $HClO_3$ (chloric acid) because the conjugate base $ClO_4^-$ is more stable due to the presence of more oxygen atoms,which disperse the negative charge.
$3$. $HNO_3$ (nitric acid) is a stronger acid than $HNO_2$ (nitrous acid) because the oxidation state of nitrogen in $HNO_3$ is $+5$,while in $HNO_2$ it is $+3$. Higher oxidation state leads to greater electron withdrawal,making the $O-H$ bond more polar and the acid stronger.
$4$. Therefore,the statement '$HNO_3$ is a stronger acid than $HNO_2$' is correct.

(A) $1$. The stability of group-$15$ hydrides decreases from $NH_3$ to $BiH_3$ because the bond dissociation energy decreases as the size of the central atom increases. Thus,the statement in option $A$ is incorrect.
$2$. Nitrogen can form $p\pi - p\pi$ multiple bonds but cannot form $d\pi - p\pi$ bonds due to the absence of $d$-orbitals. However,the statement in option $B$ is technically ambiguous as it implies nitrogen cannot form $d\pi - p\pi$ bonds,which is true,but option $A$ is factually incorrect regarding stability trends.
$3$. Due to high inter-electronic repulsion of the non-bonding electrons,the $N-N$ single bond is weaker than the $P-P$ single bond.
$4$. $N_2O_4$ exhibits resonance,and the statement is generally accepted in the context of its structure.

(B) Anhydrous $AlCl_3$ can be prepared from $Al_2O_3$ by heating a mixture of $Al_2O_3$ and carbon in a stream of dry chlorine gas at high temperature $(1000 \ ^\circ C)$.
The chemical reaction is: $Al_2O_3 + 3C + 3Cl_2 \xrightarrow{1000 \ ^\circ C} 2AlCl_3 + 3CO$.
The resulting $AlCl_3$ vapor is then cooled to obtain solid anhydrous $AlCl_3$.

(B) The chemical formula of phosphonic acid (also known as phosphorous acid) is $H_3PO_3$.
In its structure,there are two $P-OH$ bonds and one $P-H$ bond.
The hydrogen atoms attached to oxygen atoms are ionizable,whereas the hydrogen atom attached directly to the phosphorus atom is not ionizable.
Since there are two $P-OH$ groups,it can release two $H^+$ ions in an aqueous solution.
Therefore,phosphonic acid is dibasic.

(A) The reaction between ammonium perchlorate and dilute nitric acid is a double displacement reaction:
$NH_4ClO_4 + HNO_3 (\text{dil}) \to HClO_4 + NH_4NO_3$
Thus,$(X)$ is $NH_4NO_3$.
Upon heating,ammonium nitrate undergoes thermal decomposition:
$NH_4NO_3 \xrightarrow{\Delta} N_2O_{(g)} + 2H_2O_{(g)}$
Thus,$(Y)$ is $N_2O$.
Therefore,$(X)$ and $(Y)$ are $NH_4NO_3$ and $N_2O$ respectively.
Hence,option $A$ is correct.

(C) $NH_4NO_2$ does not produce ammonia $(NH_3)$ upon heating. Instead,it undergoes thermal decomposition to produce nitrogen gas $(N_2)$ and water vapor $(H_2O)$.
The reaction is: $NH_4NO_{2(s)} \rightarrow N_{2(g)} + 2H_2O_{(g)}$
In contrast,$(NH_4)_2SO_4$,$(NH_4)_2CO_3$,and $NH_4Cl$ all release $NH_3$ gas upon heating.
Therefore,option $C$ is correct.

(D) Very pure nitrogen gas is obtained by the thermal decomposition of sodium azide $(NaN_3)$ or barium azide $(Ba(N_3)_2)$.
The reaction for sodium azide is:
$2 NaN_3 \xrightarrow{\Delta} 2 Na + 3 N_2$
While other reactions like the decomposition of ammonium nitrite also produce $N_2$,the thermal decomposition of azides is the standard laboratory method for preparing ultra-pure nitrogen.

(B) The reaction of $Zn$ with concentrated $HNO_3$ is:
$Zn + 4HNO_3 \text{ (conc.)} \to Zn(NO_3)_2 + 2NO_2 + 2H_2O$
Thus,$X = NO_2$.
The reaction of $Zn$ with dilute $HNO_3$ is:
$4Zn + 10HNO_3 \text{ (dil.)} \to 4Zn(NO_3)_2 + N_2O + 5H_2O$
Thus,$Y = N_2O$.
Therefore,the compounds $X$ and $Y$ are $NO_2$ and $N_2O$ respectively.

(A) The reaction of ammonia $(NH_3)$ with chlorine $(Cl_2)$ depends on the relative amounts of the reactants:
$1$. When $NH_3$ is in excess,it reacts with $Cl_2$ to form nitrogen gas $(N_2)$ and ammonium chloride $(NH_4Cl)$:
$8NH_3 + 3Cl_2 \rightarrow 6NH_4Cl + N_2$
$2$. When $Cl_2$ is in excess,it reacts with $NH_3$ to form nitrogen trichloride $(NCl_3)$ and hydrogen chloride $(HCl)$:
$NH_3 + 3Cl_2 \rightarrow NCl_3 + 3HCl$
Therefore,statement $(a)$ is correct ($NCl_3$ is formed when $Cl_2$ is in excess) and statement $(c)$ is correct ($N_2$ is formed when $NH_3$ is in excess).

(B) $1$. Basic strength: Metallic character increases down the group and across the period from right to left. $Cs_2O$ is the most basic,followed by $K_2O$,$NiO$,and $MgO$. Thus,$MgO < NiO < K_2O < Cs_2O$ is the correct order.
$2$. Thermal stability: For hydrides of group $15$ elements,thermal stability decreases down the group as the bond dissociation energy decreases due to an increase in bond length $(NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3)$. The given order $NH_3 < PH_3 < AsH_3$ is incorrect.
$3$. Solubility: Solubility of alkali metal carbonates increases down the group due to the decrease in lattice energy being more significant than the decrease in hydration energy. Thus,$Li_2CO_3 < Na_2CO_3 < K_2CO_3$ is the correct order.
$4$. Since $(B)$ is the only incorrect order,the correct answer is $(B)$.

(C) Nitric oxide $(NO)$ is prepared in the laboratory by the action of dilute nitric acid on copper turnings.
The chemical reaction is as follows:
$3Cu + 8HNO_3 (\text{dilute}) \rightarrow 3Cu(NO_3)_2 + 4H_2O + 2NO$

(B) Red phosphorus is a polymeric structure and is non-poisonous in nature.
It is much less reactive than white phosphorus because it does not contain the high angular strain present in the $P_4$ tetrahedral units of white phosphorus.
White phosphorus exhibits chemiluminescence,meaning it glows in the dark due to the slow oxidation of phosphorus vapor.
Red phosphorus does not glow in the dark.
Therefore,the statement that red phosphorus glows in the dark is false.

(C) The melting points of group $15$ hydrides generally increase down the group due to an increase in molecular mass and van der Waals forces.
However,$NH_3$ has an anomalously high melting point due to the presence of intermolecular hydrogen bonding.
The correct order of melting points for group $15$ hydrides is $PH_3 < AsH_3 < SbH_3 < NH_3$.
Comparing the given options,$NH_3$ has the highest melting point,and $PH_3$ has a lower melting point than $AsH_3$.
Thus,the order $PH_3 < AsH_3 < NH_3$ is represented by the relationship $NH_3 > AsH_3 > PH_3$,which implies $NH_3 > PH_3$ and $AsH_3 > PH_3$.
Looking at the options,option $C$ represents the correct relative trend $NH_3 > PH_3$ and $AsH_3 > PH_3$.

(A) When ammonium nitrate $(NH_4NO_3)$ is heated,it undergoes thermal decomposition to produce nitrous oxide $(N_2O)$ and water vapor.
The chemical equation for the reaction is:
$NH_4NO_3(s) \xrightarrow{\Delta} N_2O(g) + 2H_2O(g)$
Therefore,the correct option is $A$.

(B) $NO_2$ is an odd electron molecule containing $17$ valence electrons,which makes it paramagnetic due to the presence of an unpaired electron.
Upon dimerisation,two $NO_2$ molecules combine to form $N_2O_4$ $(2NO_2 \rightleftharpoons N_2O_4)$.
In $N_2O_4$,all electrons are paired,making the molecule diamagnetic.
Therefore,the process of dimerisation results in a decrease in paramagnetism.

(A) Phosphinic acid is $H_3PO_2$.
In its structure,there is one $P-OH$ group and two $P-H$ bonds.
Only the hydrogen atoms attached to oxygen atoms are ionizable.
Therefore,it is a monobasic acid,and its basicity is $1$.

(A) When $PCl_5$ reacts with excess water,it undergoes hydrolysis to form phosphoric acid $(H_3PO_4)$ and hydrogen chloride $(HCl)$.
The balanced chemical equation is:
$PCl_5 + 4H_2O \longrightarrow H_3PO_4 + 5HCl$

(A) Phosphine $(PH_3)$ is produced by the hydrolysis of calcium phosphide $(Ca_3P_2)$.
The chemical reaction is as follows:
$Ca_3P_2 + 6H_2O \rightarrow 3Ca(OH)_2 + 2PH_3$

(D) The nitrogen atom has the electronic configuration $1s^2 2s^2 2p^3$.
It does not have vacant $d-$orbitals in its valence shell.
Therefore,it cannot expand its octet to form $NF_5$,whereas $P$,$As$,and $S$ can expand their octets due to the presence of vacant $d-$orbitals.

(D) The structures of the given nitrogen oxides are as follows:
$1$. $N_2O$ (Nitrous oxide): The structure is $N \equiv N^+ - O^-$,which contains an $N-N$ bond.
$2$. $N_2O_4$ (Dinitrogen tetroxide): The structure is $O_2N-NO_2$,which contains an $N-N$ bond.
$3$. $N_2O_3$ (Dinitrogen trioxide): The structure is $O=N-NO_2$,which contains an $N-N$ bond.
$4$. $N_2O_5$ (Dinitrogen pentoxide): In the gaseous state,the structure is $O_2N-O-NO_2$,which contains an $N-O-N$ linkage and does not contain an $N-N$ bond.

(C) The stability of phosphorus allotropes depends on the bond angle strain and the structure of the molecules.
$White$ phosphorus consists of discrete $P_4$ tetrahedral units with a bond angle of $60^{\circ}$,which causes high angular strain,making it the least stable and most reactive.
$Red$ phosphorus is a polymeric chain structure formed by breaking one $P-P$ bond in the $P_4$ tetrahedron,making it more stable than $White$ phosphorus.
$Black$ phosphorus has a layered,puckered structure and is the most thermodynamically stable allotrope of phosphorus.
Therefore,the correct order of stability is $Black > Red > White$.

(A) The chemical formula of phosphorous acid is $H_3PO_3$.
In its structure,there is one $P=O$ bond,two $P-OH$ bonds,and one $P-H$ bond.
Wait,let us re-examine the structure: The structure of $H_3PO_3$ consists of one $P=O$ bond,two $P-OH$ groups,and one $P-H$ bond.
Therefore,the number of hydrogen atoms directly attached to the phosphorus atom is $1$.

(D) The central atom in $NF_3$ is nitrogen. Nitrogen has a configuration of $1s^2 2s^2 2p^3$ and lacks vacant $d$-orbitals in its valence shell. Therefore,it cannot expand its octet to accommodate incoming water molecules,making $NF_3$ resistant to hydrolysis. In contrast,$AsCl_3$,$PF_3$,and $SbCl_3$ have vacant $d$-orbitals and undergo hydrolysis.

(C) The given reactions are:
$1$. $4NH_3 + 5O_2 \xrightarrow[500 \ K]{Pt} 4NO_{(W)} + 6H_2O$
$2$. $2NO_{(W)} + O_2 \rightleftharpoons 2NO_{2(X)}$
$3$. $3NO_2 + H_2O \rightleftharpoons 2HNO_{3(Y)} + NO_{(W)}$
Analysis of options:
- $W$ is $NO$ (Nitric oxide),not an anaesthetic ($N_2O$ is).
- Conc. $HNO_3 + Cu \to Cu(NO_3)_2 + 2NO_2 + 2H_2O$ (Produces $NO_2$,not $NO$).
- Dil. $HNO_3 + Zn \to Zn(NO_3)_2 + N_2O + H_2O$ (Produces $N_2O$).
- Conc. $HNO_3 + I_2 \to HIO_3 + NO_2 + H_2O$ (Produces $HIO_3$,not $HIO_4$).
Thus,the correct statement is $C$.