Consider the following reactions (unbalanced).

$Zn+Hotconc.H_{2}S{O}_4\to G+R+X$

$Zn+conc.NaOH\to T+Q$

$G+H_{2}S+N{H}_{4}OH\to Z\left(aprecipitate\right)+X+Y$

Choose the correct option(s).

When Zn react with hot conc. H₂SO₄ then SO₂ is released and ZnSO₄ is obtained.

$\begin{array}{l}Zn+2H_{2}S{O}_4\to ZnS{O}_4+S{O}_2\uparrow +2H_{2}O\\ \left(Hot+Conc.\right)\left(G\right)\left(R\right)\left(X\right)\end{array}$

R(SO₂) molecule is V-Shaped

Thus, option (b) is correct.

When Zn is react with conc. NaOH then H₂ gas is evolved and Na₂ZnO₂ is obtained.

$\mathrm{Zn}+2\mathrm{NaOH}\left(\text{conc.}\right)\to \underset{\left(T\right)}{{\mathrm{Na}}_2{\mathrm{ZnO}}_2}+\underset{\left(Q\right)}{{\mathrm{H}}_2}\uparrow$

In ground state, H-H(Q) (bond order=1) Thus, option (c)  is correct.

The oxidation state of Zn in T (Na₂ZnO₂) is +2

Thus, option (a) is incorrect.

$\underset{\left(G\right)}{{\mathrm{ZnSO}}_4}+{\mathrm{H}}_2\mathrm{S}+{\mathrm{NH}}_4\mathrm{OH}$$\stackrel{ }{\to }\underset{\left(Z\right)}{\mathrm{ZnS}}\downarrow +\underset{\left(X\right)}{2{\mathrm{H}}_2\mathrm{O}}+\underset{\left(Y\right)}{{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4}$

ZnS (Z) compound is dirty white coloured. Thus, option (d) is correct.