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(A) The total number of electrons in $O_2^+$ is $15$. The molecular orbital configuration is: $KK \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 =

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Bond order of $2.5$

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(A) The total number of electrons in $O_2^+$ is $15$. The molecular orbital configuration is: $KK \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 = \pi 2p_y^2 \pi^* 2p_x^1$. Bond order is calculated as: $	ext{Bond Order} = \frac{N_b - N_a}{2} = \frac{10 - 5}{2} = 2.5$. Since it contains one unpaired electron,it is paramagnetic. The bond order of $O_2^+$ $(2.5)$ is higher than that of $O_2$ $(2.0)$,so $O_2^+$ is more stable than $O_2$.

According to the Molecular Orbital Theory,which of the following properties does the $O_2^+$ species possess?

Similar Questions

The bond order of which of the following two species is the same?

Based on the Molecular Orbital Theory,what is the number of non-bonding electron pairs in $O_2^{2-}$?

Match the following molecules/ions in List-$I$ with their respective bond orders in List-$II$:
| List-$I$ (Molecules/ions) | List-$II$ (Bond order) |
| :--- | :--- |
| $A. N_2^+$ | $I. 1.0$ |
| $B. CO$ | $II. 1.5$ |
| $C. O_2$ | $III. 2.0$ |
| $D. O_2^-$ | $IV. 2.5$ |
| | $V. 3.0$ |

Which of the following is planar due to back bonding?

What is the bond order in $CO$ molecule?

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