MCQ
The covalency of nitrogen in $HN{O_3}$ is
- A$0$
- B$3$
- C$4$
- ✓$5$
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$\mathrm{MnO}_{4}^{-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}+\mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}$
the correct coefficients of the reactants for the balanced equation are
$\mathrm{MnO}_{4}^{-} \quad \mathrm{C}_{2} \mathrm{O}_{4}^{2-}\quad \mathrm{H}^{+}$
Product $(b)$ of above reaction is
$Image$
$(II)\, O^+_2 ,NO,N^-_2$ have same bond order of $2 \frac{1}{2}$
$(III)$ Bond order can assume any value including zero upto four
$(IV)\, NO^-_3$ and $BO^-_3$ have same bond order for $X - O$ bond (where $X$ is central atom)