Find out the bond order in He_2, He_2^{+}, O_ | vedclass

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Question

(A) The bond order is calculated using the formula: $	ext{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$

Vedclass · Accepted Answer

$0, 0.5, 2$ and $2.5$

Vedclass · Answer

(A) The bond order is calculated using the formula: $	ext{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of anti-bonding electrons.$1$. For $He_2$ ($4$ electrons): Configuration is $\sigma 1s^2 \sigma^* 1s^2$. $BO = \frac{2-2}{2} = 0$.$2$. For $He_2^+$ ($3$ electrons): Configuration is $\sigma 1s^2 \sigma^* 1s^1$. $BO = \frac{2-1}{2} = 0.5$.$3$. For $O_2$ ($16$ electrons): Configuration is $KK \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 = \pi 2p_y^2 \pi^* 2p_x^1 = \pi^* 2p_y^1$. $BO = \frac{10-6}{2} = 2$.$4$. For $O_2^+$ ($15$ electrons): Configuration is $KK \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 = \pi 2p_y^2 \pi^* 2p_x^1$. $BO = \frac{10-5}{2} = 2.5$.Thus,the bond orders are $0, 0.5, 2$ and $2.5$ respectively.

List-$I$ (Molecule)	List-$II$ (Number and types of bond/s between two carbon atoms)
$A.$ Ethane	$I.$ One $\sigma$-bond and two $\pi$-bonds
$B.$ Ethene	$II.$ Two $\pi$-bonds
$C.$ Carbon molecule,$C_2$	$III.$ One $\sigma$-bond
$D.$ Ethyne	$IV.$ One $\sigma$-bond and one $\pi$-bond

Find out the bond order in $He_2, He_2^{+}, O_2$ and $O_2^{+}$ respectively.

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