MCQ
The oxidation state of nitrogen is highest in
- ✓${N_3}H$
- B$N{H_2}OH$
- C${N_2}{H_4}$
- D$N{H_3}$
✓
Answer
Correct option: A.
${N_3}H$
(a) $3 \times x + 1\,(1) = 0$
$3x + 1 = 0$
$3x = - 1,\, \Rightarrow x = - \frac{1}{3}\,\,{\rm{in}}\,\,{N_3}H$
$x + 2\,( + 1) + 1\,( - 2) + 1(1) = 0$
$x = - 1\,\,{\rm{in}}\,\,N{H_2}OH$
$x \times 2 + 4\,(1) = 0$$x = - \frac{4}{2} = - 2\,{\rm{in}}\,\,{N_2}{H_4}$
$x + 3\,(1) = 0$$x = - 3\,\,in\,\,N{H_3}$
Hence, highest in ${N_3}H$.
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