Which pair among the following does not have | vedclass

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(A) The bond order is calculated using the formula: $	ext{Bond order} = \frac{	ext{Number of bonding electrons} - 	ext{Number of antibonding electr

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$CN^{-}$ and $NO^{-}$

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(A) The bond order is calculated using the formula: $	ext{Bond order} = \frac{	ext{Number of bonding electrons} - 	ext{Number of antibonding electrons}}{2}$.$A) CN^{-}$ ($14$ electrons,bond order $= 3$) and $NO^{-}$ ($16$ electrons,bond order $= 2$). These do not have the same bond order.$B) CN^{-}$ ($14$ electrons,bond order $= 3$) and $CO$ ($14$ electrons,bond order $= 3$). These have the same bond order.$C) O_2^{2-}$ ($18$ electrons,bond order $= 1$) and $B_2$ ($10$ electrons,bond order $= 1$). These have the same bond order.$D) O_2^{+}$ ($15$ electrons,bond order $= 2.5$) and $NO^{+}$ ($14$ electrons,bond order $= 3$). These do not have the same bond order.Note: In many standard competitive exams,this question is presented such that only one pair is incorrect. Given the options,both $A$ and $D$ have different bond orders. However,$O_2^{+}$ $(2.5)$ and $NO^{+}$ $(3)$ is a classic example of unequal bond orders.

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

Which pair among the following does not have the same bond order?

Similar Questions

Although $CN^-$ and $N_2$ are isoelectronic,$N_2$ is inert compared to $CN^-$,which is due to $........$.

Assertion : $C_2$ contains two $\pi$ bonds.
Reason : $C_2$ is diamagnetic in nature.

What is the number of unpaired electrons in the $NO$ molecule?

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

Provide the energy level diagram obtained by the overlapping of $2p_{z}$ orbitals and show the diagram of the resulting molecular orbitals.

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