Consider the following reactions (unbalanced)
Zn + hot conc. $H_{2} {SO}_{4}$ ® G + R + X
Zn + conc. NaOH ® T + Q
G + $H_{2} S + {NH}_{4} OH$ ® Z (a precipitate) + X + Y
Choose the correct option(s)

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The oxidation state of Zn in T is + 1

Bond order of Q is 1 in its ground state

R is a V-shaped molecule

Z is dirty white in colour

answer is B, C, D.

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Detailed Solution

$Z n + \underset{H o t & C o n c .}{H_{2} S O_{4}} \to G + R + X$

$Z n + C o n c . N a O H \to \underset{(T)}{N a_{2} Z n O_{2}} + \underset{(Q)}{H_{2}}$

$\underset{(G)}{Z n S O_{4}} + H_{2} S + N H_{4} O H \to \underset{(Z)}{Z n S} + \underset{(Y)}{{(N H_{4})}_{2}} S O_{4} + \underset{(X)}{H_{2} O}$

1) Oxidation state of Zn in Na2ZnO2 is +2

2) Bond order of Q is one for $H_{2} (Q)$

3) ZnS is White in color.

4) ${SO}_{2}$ is angular in shape.

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