Properties and Reaction Products of Zinc Chemical Processes

To determine the correct statement(s), we first identify all the chemical species in the given reaction sequences.

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Step 1: Identifying the Chemical Species

1. Reaction with hot concentrated $\text{H}_2\text{SO}_4$: $\text{Zn} + 2\text{H}_2\text{SO}_4 \,(\text{hot conc.}) \longrightarrow \text{ZnSO}_4 + \text{SO}_2 + 2\text{H}_2\text{O}$

   • $\text{G} = \text{ZnSO}_4$
   • $\text{R} = \text{SO}_2$
   • $\text{X} = \text{H}_2\text{O}$

2. Reaction with concentrated $\text{NaOH}$: $\text{Zn} + 2\text{NaOH} \,(\text{conc.}) \longrightarrow \text{Na}_2\text{ZnO}_2 + \text{H}_2$

   • $\text{T} = \text{Na}_2\text{ZnO}_2 \quad (\text{Sodium Zincate})$
   • $\text{Q} = \text{H}_2$

3. Precipitation Reaction: $\text{ZnSO}_4 + \text{H}_2\text{S} + 2\text{NH}_4\text{OH} \longrightarrow \text{ZnS}\downarrow + (\text{NH}_4)_2\text{SO}_4 + 2\text{H}_2\text{O}$

   • $\text{Z} = \text{ZnS} \quad (\text{Zinc Sulfide precipitate})$
   • $\text{Y} = (\text{NH}_4)_2\text{SO}_4$

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Step 2: Evaluation of Options

• Option (A): "Z has a dirty white colour"
  $\text{Z}$ is $\text{ZnS}$ (Zinc Sulfide). It precipitates as a characteristic dirty white solid in alkaline media.
  $\mathbf{A\text{ is CORRECT.}}$

• Option (B): "The oxidation state of Zn in T is +1"
  $\text{T}$ is sodium zincate, $\text{Na}_2\text{ZnO}_2$. Calculating the oxidation state of $\text{Zn}$ ($x$): $2(+1) + x + 2(-2) = 0 \implies x = +2$ Thus, the oxidation state of $\text{Zn}$ in $\text{T}$ is $+2$.
  $\mathbf{B\text{ is INCORRECT.}}$

• Option (C): "R is a V-shaped molecule"
  $\text{R}$ is sulfur dioxide ($\text{SO}_2$). The central sulfur atom undergoes $sp^2$ hybridization, containing two bonding pairs and one lone pair of electrons. This steric arrangement results in a V-shaped (bent/angular) molecular geometry with a bond angle of approximately $119^\circ$.
  $\mathbf{C\text{ is CORRECT.}}$

• Option (D): "Bond order of Q is 1 in its ground state"
  $\text{Q}$ is dihydrogen gas ($\text{H}_2$). According to Molecular Orbital Theory, its ground-state electronic configuration is $(\sigma_{1s})^2$.
  $\text{Bond Order} = \frac{N_b - N_a}{2} = \frac{2 - 0}{2} = 1$
  $\mathbf{D\text{ is CORRECT.}}$

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Conclusion

The correct options are A, C, and D.