The bond order and magnetic property of acety | vedclass

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(A) The acetylide ion is $C_2^{2-}$.Total number of electrons in $C_2^{2-} = (6 	imes 2) + 2 = 14 \ e^-$.According to Molecular Orbital Theory,the el

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$NO^{+}$

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(A) The acetylide ion is $C_2^{2-}$.Total number of electrons in $C_2^{2-} = (6 	imes 2) + 2 = 14 \ e^-$.According to Molecular Orbital Theory,the electronic configuration of $C_2^{2-}$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$.Bond order $= \frac{1}{2} (N_b - N_a) = \frac{1}{2} (10 - 4) = 3$.Since all electrons are paired,it is diamagnetic.Now,check $NO^{+}$: Total electrons $= 7 + 8 - 1 = 14 \ e^-$.$NO^{+}$ also has a bond order of $3$ and is diamagnetic.Therefore,the properties match $NO^{+}$.

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

The bond order and magnetic property of acetylide ion are same as that of $........$.

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For a stable molecule,the value of bond order must be

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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