The pair of species with the same bond order | vedclass

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(A) To find the bond order,we use the formula: $	ext{Bond Order} = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$

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$O_2^{2-}, B_2$

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(A) To find the bond order,we use the formula: $	ext{Bond Order} = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.$1$. For $O_2^+$: Total electrons = $15$. Configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Bond order = $\frac{1}{2}(10 - 5) = 2.5$.$2$. For $NO^+$: Total electrons = $14$. Configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. Bond order = $\frac{1}{2}(10 - 4) = 3.0$.$3$. For $O_2^{2-}$: Total electrons = $18$. Bond order = $1.0$.$4$. For $B_2$: Total electrons = $10$. Bond order = $1.0$.Since both $O_2^{2-}$ and $B_2$ have a bond order of $1.0$,they form the correct pair.

The pair of species with the same bond order is

Similar Questions

Which of the following species,$O_2^-$ or $O_2^{2-}$,has a higher bond order?

What is bond order? Explain with examples.

The bond order in $NO$ is $2.5$ while that in $NO^{+}$ is $3$. Which of the following statements is true for these two species?

What is the number of electrons present in antibonding orbitals of $N_2$ molecule according to molecular orbital theory?

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