Question

what is the bond order of O2^2+ ?​

Answer 1

Answer:

Explanation:

In normal O2, there are 6 bonding electrons and 2 antibonding electrons, making the bond order 2. By removing the 2 highest electrons, which reside in antibonding orbitals, to make O2^(2+), the calculation becomes (6–0)/2 = 3.

Answer 2

Answer:

The bond order of $O_{2} ^{2+}$ is  $3.$

Explanation:

Bord order is defined as half the difference between the number of electrons in bonding molecular orbital ($N_{b}$) and the number of electrons in the antibonding molecular orbitals ($N_{a}$)·

⇒    Bord order  $=\ \ \frac{1}{2} (N_{b}\ -\ N_{a})$

The electronic configuration of $O_{2} ^{2+}$ is $:$

  $O_{2} ^{2+}$  $=$  σ$1s^{2}$  σ$*1s^{2}$  σ$2s^{2}$  σ$*2s^{2}$  σ$2p_{z} ^{2}$   $\pi 2p_{x}^{2}$  $=\ \pi 2p_{y}^{2}$    

The above configuration indicate that $O_{2} ^{2+}$ has $10$ electrons in bonding molecular orbital ($N_{b}$) and $4$ electrons in antibonding molecular orbitals ($N_{a}$)

On substituting the values,

⇒ Bond order   $=\ \ \frac{1}{2}\ (\ 10\ -\ 4\ )$

⇒ Bond order   $=\ \ 3$

Hence the bond order of $O_{2} ^{2+}$ is  $3.$