The pair of species that has the same bond or | vedclass

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

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$CO, NO^{+}$

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(A) The bond order is calculated using the formula: $Bond \ order = \frac{N_b - N_a}{2}$,where $N_b$ is the number of electrons in bonding molecular orbitals and $N_a$ is the number of electrons in antibonding molecular orbitals.For $CO$ ($6+8=14$ electrons): Bond order $= \frac{10-4}{2} = 3$.For $NO^{+}$ ($7+8-1=14$ electrons): Bond order $= \frac{10-4}{2} = 3$.Since both $CO$ and $NO^{+}$ have $14$ electrons,they have the same bond order of $3$.For other options:$NO^{-}$ ($16$ electrons,$BO=2$),$CN^{-}$ ($14$ electrons,$BO=3$).$O_2$ ($16$ electrons,$BO=2$),$N_2$ ($14$ electrons,$BO=3$).$O_2$ ($16$ electrons,$BO=2$),$B_2$ ($10$ electrons,$BO=1$).

Pair of species	Identical property
$A$. $B_2 \& O_2$	$P$. Bond order $= 2.5$
$B$. $Be_2 \& H_2^{2-}$	$Q$. Paramagnetic nature
$C$. $N_2^{+} \& N_2^{-}$	$R$. Diamagnetic nature
$D$. $O_2^{+} \& O_2^{-}$	$S$. Doesn't exist

The pair of species that has the same bond order in the following is

Similar Questions

According to the Molecular Orbital Theory,which of the following does not exist?

Match the following:

Pair of species Identical property

$A$. $B_2 \& O_2$ $P$. Bond order $= 2.5$

$B$. $Be_2 \& H_2^{2-}$ $Q$. Paramagnetic nature

$C$. $N_2^{+} \& N_2^{-}$ $R$. Diamagnetic nature

$D$. $O_2^{+} \& O_2^{-}$ $S$. Doesn't exist

The bond order is not $3$ for

Which of the following species,$O_2^-$ or $O_2^{2-}$,has a higher bond order?

What is bond order? Explain with examples.

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