Bond order of C_2 vapour is : | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The total number of electrons in a $C_2$ molecule is $12$ $(6 + 6)$.According to Molecular Orbital Theory,the electronic configuration of $C_2$ is

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The total number of electrons in a $C_2$ molecule is $12$ $(6 + 6)$.According to Molecular Orbital Theory,the electronic configuration of $C_2$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2$.Bond order is calculated as: $	ext{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.Here,$N_b = 8$ and $N_a = 4$.$	ext{Bond Order} = \frac{8 - 4}{2} = \frac{4}{2} = 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

Bond order of $C_2$ vapour is :

Similar Questions

Which one of the following is paramagnetic?

The correct bond order in the following species is

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 of $He_2^+$ molecule ion is

Which of the following concepts is $NOT$ part of the valence bond theory $(VBT)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module