Number of molecules having bond order 2 from | vedclass

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(B) The bond order is calculated using the formula: $	ext{B.O.} = \frac{N_b - N_a}{2}$.$C_2$ $(12 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2,

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The bond order is calculated using the formula: $	ext{B.O.} = \frac{N_b - N_a}{2}$.$C_2$ $(12 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2$. $	ext{B.O.} = \frac{8-4}{2} = 2$.$O_2$ $(16 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. $	ext{B.O.} = \frac{10-6}{2} = 2$.$Be_2$ $(8 \ e^-)$: $	ext{B.O.} = 0$.$Li_2$ $(6 \ e^-)$: $	ext{B.O.} = 1$.$Ne_2$ $(20 \ e^-)$: $	ext{B.O.} = 0$.$N_2$ $(14 \ e^-)$: $	ext{B.O.} = 3$.$He_2$ $(4 \ e^-)$: $	ext{B.O.} = 0$.Thus,only $C_2$ and $O_2$ have a bond order of $2$. The total count is $2$.

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

Number of molecules having bond order $2$ from the following molecules is: $C_2, O_2, Be_2, Li_2, Ne_2, N_2, He_2$

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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

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