Match the molecules in List-I with their resp | vedclass

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(A) According to Molecular Orbital Theory $(MOT)$:Bond Order $(B.O.)$ = $\frac{1}{2} (N_b - N_a)$For $Li_2$ ($6$ electrons): $\sigma 1s^2, \sigma^* 1s

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$A-iii, B-i, C-iv, D-v$

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(A) According to Molecular Orbital Theory $(MOT)$:Bond Order $(B.O.)$ = $\frac{1}{2} (N_b - N_a)$For $Li_2$ ($6$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2$. $B.O. = \frac{1}{2} (4 - 2) = 1.0$ $(iii)$.For $N_2$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. $B.O. = \frac{1}{2} (10 - 4) = 3$ $(i)$.For $Be_2$ ($8$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2$. $B.O. = \frac{1}{2} (4 - 4) = 0$ $(iv)$.For $O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. $B.O. = \frac{1}{2} (10 - 6) = 2$ $(v)$.Therefore,the correct match is $A-iii, B-i, C-iv, D-v$.

List-$I$	List-$II$
$A. Li_2$	$i. 3$
$B. N_2$	$ii. 1.5$
$C. Be_2$	$iii. 1.0$
$D. O_2$	$iv. 0$
	$v. 2$

Match the molecules in List-$I$ with their respective bond orders in List-$II$.
List-$I$ List-$II$
$A. Li_2$ $i. 3$
$B. N_2$ $ii. 1.5$
$C. Be_2$ $iii. 1.0$
$D. O_2$ $iv. 0$
$v. 2$

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Match the molecules in List-$I$ with their respective bond orders in List-$II$.List-$I$List-$II$$A. Li_2$$i. 3$$B. N_2$$ii. 1.5$$C. Be_2$$iii. 1.0$$D. O_2$$iv. 0$$v. 2$