Question

Consider the following reactions (unbalanced)
$Zn+\text{hot conc.}{H}_{2}S{O}_4\to \text{G+R+X}$
$Zn+\text{conc.}NaOH\to \text{T+Q}$
$G+H_{2}S+N{H}_{4}OH\to Z\left(aprecipitate\right)+X+Y$
Choose the correct option(s).

• A. The oxidation state of $Zn$ in T is + 1
• B. R is a V-shaped molecule
• C. Z is dirty white in colour
• D. Bond order of Q is 1 in its ground state

Answer 1

Answer: (B), (C), (D)

Bond order of Q is 1 in its ground state

$Zn+\underset{\text{hot and conc.}}{H_{2}S{O}_4}\to \underset{\left(G\right)}{ZnS{O}_4}+\underset{\left(R\right)}{S{O}_2}+\underset{\left(X\right)}{H_{2}O}$

$Zn+\text{conc. NaOH}\to \underset{\left(T\right)}{N{a}_{2}Zn{O}_2}+\underset{\left(Q\right)}{H_2}$

$ZnS{O}_4+H_{2}S+2N{H}_{4}OH(aq.)\to \underset{\left(Z\right)}{ZnS}\downarrow +\underset{\left(Y\right)}{(N{H}_4{)}_{2}S{O}_4}+\underset{\left(X\right)}{2H_{2}O}$
a) The oxidation state of $Zn$ in T i.e. in $N{a}_{2}Zn{O}_2$ is +2.
b) R i.e. $S{O}_2$ is a V-shaped molecule.
c) $ZnS$ is white in colour.
d) Bond order of q i.e. $H_2$ in its ground state is 1.