MCQ
The species having bond order different from that in $CO$ is
- ✓$NO^-$
- B$NO^+$
- C$CN^-$
- D$N_2$
$NO ^{-}$contains $16$ electrons ( $7$ from nitrogen, $8$ from oxygen, add $1$ more for negative charge).
$NO ^{+}$has $14$ electrons ( $7$ from nitrogen, $8$ from oxygen, subtract $1$ for positive charge).
$CN ^{-1}$ has $14$ electrons ( $7$ from nitrogen, $6$ from carbon, add $1$ more for negative charge).
$N _2$ has $14$ electrons ( $7$ from nitrogen each).
The species containing the same number of electrons have the same bond order.
So $NO ^{-}$is the only one which has bond order different. So option $A$ is correct.
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take place in two steps :
$(a)$ $Br^{-} + H^{+} + H_2O_2 \xrightarrow{{slow}} HOBr + H_2O$
$(b)$ $HOBr + Br^{-} + H^{+} \xrightarrow{{fast}} H_2O + Br_2$
The order of the reaction is