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Compound A on hydrolysis gives gaseous compound B and alkaline solution. When B reacts with Nassler's reagent it gives brown ppt. When 'B' reacts with $H_{2} [{PtCl}_{6}]$ it gives yellow ppt of C. Hybridised state of central atom of the ionisation sphere of compound C is :

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$s p^{3} d^{2}$

$s p^{3}$

$d^{2} s p^{3}$

$d s p^{2}$

answer is A.

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Detailed Solution

Here,

$\begin{aligned} [{NH}_{4}^{+}] + H_{2} O \to ({NH}_{3}) + Akali \end{aligned} {[{HgI}_{4}]}^{2 -} K_{2} {HI}_{4}$

${NH}_{3} + H_{2} [PtCl 6] ⟶ {({NH}_{4})}_{2} [{PtCl}_{6}] ↓$

${({NH}_{4})}_{2} [{PtCl}_{6}]$

Here N is $s p^{3}$ hyb.

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