Bonding in which of the following diatomic mo | vedclass

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(A) According to $M.O.$ Theory,bond strength is directly proportional to bond order $(B.O.)$.
If an electron is removed from an antibonding molecular

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$(A), (C)$ only

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(A) According to $M.O.$ Theory,bond strength is directly proportional to bond order $(B.O.)$.
If an electron is removed from an antibonding molecular orbital $(M.O.)$,the bond order increases,making the bond stronger.
Let us analyze the molecules:
$1$. $NO$ ($15$ $e^-$): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Removing an electron from $\pi^* 2p_x$ increases $B.O.$ from $2.5$ to $3.0$.
$2$. $N_2$ ($14$ $e^-$): Removing an electron from a bonding orbital $(\sigma 2p_z)$ decreases $B.O.$ from $3.0$ to $2.5$.
$3$. $O_2$ ($16$ $e^-$): Configuration has electrons in $\pi^* 2p_x$ and $\pi^* 2p_y$. Removing an electron from $\pi^*$ increases $B.O.$ from $2.0$ to $2.5$.
$4$. $C_2$ ($12$ $e^-$): Removing an electron from a bonding orbital $(\pi 2p)$ decreases $B.O.$ from $2.0$ to $1.5$.
$5$. $B_2$ ($10$ $e^-$): Removing an electron from a bonding orbital $(\pi 2p)$ decreases $B.O.$ from $1.0$ to $0.5$.
Thus,the bond becomes stronger for $NO$ and $O_2$.

Bonding in which of the following diatomic molecule$(s)$ becomes stronger,on the basis of $M.O.$ Theory,by removal of an electron?
$(A)$ $NO$
$(B)$ $N_2$
$(C)$ $O_2$
$(D)$ $C_2$
$(E)$ $B_2$
Choose the most appropriate answer from the options given below.

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Bonding in which of the following diatomic molecule$(s)$ becomes stronger,on the basis of $M.O.$ Theory,by removal of an electron? $(A)$ $NO$ $(B)$ $N_2$ $(C)$ $O_2$ $(D)$ $C_2$ $(E)$ $B_2$ Choose the most appropriate answer from the options given below.