"Isoelectronic molecules and ions have identical bond orders." Explain by examples.

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Bond order in Isoelectronic molecules/ions :

If two or more diatomic molecule or ion are isoelectronic so its bond order is same. Example-$1$: $\mathrm{F}_{2}$ and $\mathrm{O}_{2}^{-2}$ both have same $18$ electrons and bond order is also same $1$ . $\mathrm{F}_{2}: \mathrm{F}-\mathrm{F}$ and $\mathrm{O}_{2}^{-2}(\mathrm{O}-\mathrm{O})^{2-}$

Example-$2$: $\mathrm{N}_{2}, \mathrm{CO}$ and $\mathrm{NO}^{+}$are isoelectronic and containing $14$ electron and triple bond in it.

Molecule/ion $\mathrm{N}_{2}$ $\mathrm{CO}$ $\mathrm{NO}^{+}$
Bond structure $\mathrm{N} \equiv \mathrm{N}$ $\mathrm{C} \equiv \mathrm{O}$ $\mathrm{N} \equiv \mathrm{O}^{+}$

 

Similar Questions

The linear combination of atomic orbitals to form molecular orbitals takes place only when the combining atomic orbitals

$A$. have the same energy

$B$. have the minimum overlap

$C$. have same symmetry about the molecular axis

$D$. have different symmetry about the molecular axis

Choose the most appropriate from the options given below:

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Which of the following has unpaired electron $(s)$ ?

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The total number of electrons in all bonding molecular orbitals of $O_{2}^{2-}$ is $....\,.$

(Round off to the Nearest Integer).

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In $O_2^-$ , $O_2$ and $O_2^{2-}$ , the total number of antibonding electrons respectively are

After understanding the assertion and reason, choose the correct option.
Assertion : In the bonding molecular orbital $(MO)$ of $H_2,$ electron density is increased between the nuclei.
Reason : The bonding $MO$ is ${\psi _A}\, + \,\,{\psi _{B,}}$ which shows destructive interference of the combining electron waves.

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