MCQ
The oxidation state of nitrogen in ${N_3}H$ is
- A$ + \frac{1}{3}$
- B$+3$
- C$ - 1$
- ✓$ - \frac{1}{3}$
$\mathop {{N_3}H}\limits^{ * \,\,\,\,\,\,\,\,} $
$3x + 1 = 0$, $3x = - 1$, $x = - \frac{1}{3}$.
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Statement $I$: $\mathrm{S}_8$ solid undergoes disproportionation reaction under alkaline conditions to form $\mathrm{S}^{2-}$ and $\mathrm{S}_2 \mathrm{O}_3{ }^{2-}$
Statement $II$: $\mathrm{ClO}_4^{-}$can undergodisproportionation reaction under acidic condition. In the light of the above statements, choose the most appropriate answer from the options given below :
| Column $I$ | Column $II$ |
| $(a)\,XeO_6^{-4}$ | $(P)$ Tetrahedral |
| $(b)\,ClO_2$ | $(Q)\,V-$ Shape |
| $(c)\,NH_4^+$ | $(R)$ Trigonal Bipyramidal |
| $(d)\,XeO_3F_2$ | $(S)$ Octahedral |
| Column $-I$ | Column $-II$ |
| (Atomic number) | (Position of element in Periodic table) |
| $(A)$ $Z = 37$ | $(P)$ $p-$ block |
| $(B)$ $Z = 42$ | $(Q)$ $f-$ block |
| $(C)$ $Z = 34$ | $(R)$ $d-$ block |
| $(D)$ $Z = 92$ | $(S)$ $s-$ block |