# Which one of the following constitutes a group of the isoelectronic species? Option 1) $N_{2},O{_{2}}^{-},NO^{+},CO$ Option 2) $C{_{2}}^{2-},O{_{2}}^{-},CO,NO$ Option 3) $NO^{+},C{_{2}}^{2-},CN^{-},N_{2}$ Option 4) $CN^{-},N_{2},O{_{2}}^{2-},C{_{2}}^{2-}$

As we learnt in

Bond strength -

The energy required to break one mole of bonds of particular type in gaseous state is called bond energy or bond strength.

- wherein

The same amount of energy is released in formation of one mol of particular bond.

Number of electrons in each species are given below

$N_{2}= 14\: \: \: \: \: CN^{-}= 14$

$O{_{2}}^{-}= 17\: \: \: \: \: C{_{2}}^{2-}= 14$

$NO^{+}= 14\: \: \: \: \: O{_{2}}^{2-}= 18$

$CO= 14\: \: \: \: \: NO= 15$

It is quite evident from the above that $NO^{+},C{_{2}}^{2-},CN^{-},N_{2}\: and \: CO$ are isoelectronic in nature.

Hence $NO^{+},C{_{2}}^{2-},CN^{-},N_{2}$ is correct

Option 1)

$N_{2},O{_{2}}^{-},NO^{+},CO$

This option is incorrrect

Option 2)

$C{_{2}}^{2-},O{_{2}}^{-},CO,NO$

This option is incorrrect

Option 3)

$NO^{+},C{_{2}}^{2-},CN^{-},N_{2}$

This option is corrrect

Option 4)

$CN^{-},N_{2},O{_{2}}^{2-},C{_{2}}^{2-}$

This option is incorrrect

N

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