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The Gibbs free energy change for the above reaction at $298\, K$ is $x \times 10^{-1} \,k\,J\, mol ^{-1}$;
The value of $x$ is ..... [nearest integer]$\left [\text { Given : } E _{ Cu ^{2} / / Cu }=0.34\, V ; E _{ Sn ^{2} / Sn }^{\ominus}=-0.14 \,V ; F=96500\, C\, mol ^{-1}\right]$
($A$) can also be prepared by reaction of $\mathrm{P}_4$ and $\mathrm{HNO}_3$
($B$) is diamagnetic
($C$) contains one $\mathrm{N}-\mathrm{N}$ bond
($D$) reacts with $\mathrm{Na}$ metal producing a brown gas