$N H_{3} + A + C O_{2} \to {(N H_{4})}_{2} C O_{3}$

${(N H_{4})}_{2} C O_{3} + H_{2} O + B \to N H_{4} H C O_{3}$

$N H_{4} H C O_{3} + N a C l \to N H_{4} C l + C$

Q. Find A, B and C in the following reactions

N H_{3} + A + C O_{2} \to (N H_{4}) C O_{3}

(N H_{4})_{2} C O_{3} + B \to C a C O_{3} ↓ + 2 N H_{4} C l

N H_{4} H C O_{3} + N a C l \to N H_{4} C l + C

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Bond length can be calculated by merely adding covalent bond radii which are H = 0.28 ˚A, N = 0.70 ˚A, O = 0.66 ˚A, Cl = 0.99 ˚A, (C=)=0.67˚A, (C≡)=0.61 ˚A, (N≡)= 0.55 ˚A and (C−)=0.77Ao. Calculate bond lengths in HCN.

The correct option is B H−C1.05˚A;C≡N1.16˚ABond length can be calculated by merely adding covalent bond radii.The C−H bond length is 0.77+0.28=1.05A0The C≡N bond length is 0.55+0.61=1.16A0

Bond length can be calculated by merely adding covalent bond radii which are H = 0.28 $˚ A$ , N = 0.70 $˚ A$ , O = 0.66 $˚ A$ , Cl = 0.99 $˚ A$ , (C=)=0.67 $˚ A$ , (C $\equiv$ )=0.61 $˚ A$ , (N $\equiv$ )= 0.55 $˚ A$ and $(C -) = 0.77 A^{o}$ .

Calculate bond lengths in $H C N$ .

The correct option is B $H - C 1.05 ˚ A; C \equiv N 1.16 ˚ A$
Bond length can be calculated by merely adding covalent bond radii.
The $C - H$ bond length is $0.77 + 0.28 = 1.05 A^{0}$
The $C \equiv N$ bond length is $0.55 + 0.61 = 1.16 A^{0}$