Question

How many electrons should $X_2{H}_4$ liberate so that in the new compound X shows oxidation number of -1/2? $(E,N.X>H)$

• A. 10
• B. 4
• C. 3
• D. 2

Answer 1

Answer: (C)

3

$X_2{H}_4$ En of $X>H$

$\therefore H$ is in +1

and X in $-\frac{1}{2}$

$-\frac{1}{2}\times 2+(+1)\times 4=-1+4=3$

$\therefore 3e^{-}$ must be liberated.