- A$ClO_3^-< ClO_4^- < ClO_2^-< ClO^-$
- B$ClO^-< ClO_4^-< ClO_3^-< ClO_2^-$
- ✓$ClO^-< ClO_2^-< ClO_3^-< ClO_4^-$
- D$ClO_4^-< ClO_3^-< ClO_2^-< ClO^-$
$ClO_2^-$ $B.O.=1.5$
$ClO_3^-$ $B.O.=1.66$
$ClO_4^-$ $B.O=1.75$
$OR$
$Cl{O^ - } < ClO_2^ - < ClO_3^ - < ClO_4^ - $
$B.O.$ $1.0$ $1.5$ $1.66$ $1.75$
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$' A ^{\prime} \stackrel{ HNO _3}{\longrightarrow} X ^{\prime} \stackrel{ NaBH _4}{\longrightarrow} \underset{\text { Chiral compound }}{\text { } B ^{}}$ $X$ is $.......$.
$O ^{2-}, N ^{3-}, F ^{-}, Mg ^{2+}, Na ^{+}$ and $Al ^{3+}$ is
$N_2 + 3H_2 \rightleftharpoons 2NH_3 \,;$ $K_1$
$N_2 + O_2 \rightleftharpoons 2NO\,;$ $K_2$
$H_2 + 2 O_2 \rightleftharpoons H_2O\,;$ $K_3$
The equilibrium constant $(K)$ of the reaction :
$2NH_3 + \frac{5}{2} \overset K \leftrightarrows 2NO + 3H_2O$