The species having bond order different from that in CO is

Download App

MCQ

The species having bond order different from that in $CO$ is

✓
$NO^-$
B
$NO^+$
C
$CN^-$
D
$N_2$

✓

Answer

Correct option: A.

$NO^-$

a
$CO$ contains $14$ electrons ( $6$ from carbon and $8$ from oxygen).

$NO ^{-}$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.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from 4. Chemical bonding and molecular structure→4. Chemical bonding and molecular structure — all question types→Chemistry STD 11 Science question papers→Rajasthan Board STD 11 Science question papers→Rajasthan Board question paper generator→

Similar questions

Heterolytic bond dissociation energy of alkyl halides follows the sequence

→

For a concentrated solution of a weak electrolyte ( $K _{ eq }=$ equilibrium constant) $A _2 B _3$ of concentration ' $c$ ', the degree of dissociation " $\alpha$ ' is

→

One mole of an ideal monoatomic gas undergoes two reversible processes $(A \rightarrow B$ and $B \rightarrow C)$ as shown in the given figure:

(image)

$A \rightarrow B$ is an adiabatic process. If the total heat absorbed in the entire process ( $A \rightarrow B$ and $B \rightarrow C$ ) is $R T_2 \ln 10$, the value of $2 \log V_3$ is . . . . . [Use, molar heat capacity of the gas at constant pressure, $\mathrm{C}_{\mathrm{p}, \mathrm{m}}=\frac{5}{2} \mathrm{R}$ ]