Question

The electronic configuration of carbon is $1{s^2}2{s^2}2{p^2}$ . There are 12 electrons in ${C_2}$ . The correct electronic configuration of ${C_2}$ molecule is:
A.$(\sigma 1{s^2})(\sigma \times 1{s^2})(\sigma 2{s^2})(\sigma \times 2{s^2})(\sigma 2{p_z}^2)(\pi 2{p_x}^2)$
B.$(\sigma 1{s^2})(\sigma \times 1{s^2})(\sigma 2{s^2})\left( {\sigma \times 2{s^2}} \right)\left( {\pi 2p{x^2} = \pi 2p{y^2}} \right)$
C.$(\sigma 1{s^2})(\sigma \times 1{s^2})(\sigma 2{s^2})(\sigma \times 2{s^2})(\sigma 2p{z^2})\left( {\pi 2p{x^1} = \pi 2p{y^1}} \right)$
D.$(\sigma 1{s^2})(\sigma \times 1{s^2})(\sigma 2{s^2})(\sigma \times 2{s^2})\left( {\pi 2p{x^2} = \pi 2p{y^1}} \right)$

Answer 1

There are two types of Molecular orbitals one is called Bonding Molecular orbital and the other one is called Anti Bonding Molecular orbital. BMO is formed by the addition of atomic orbitals, Anti Bonding Molecular Orbitals are formed by the subtraction of atomic orbitals.

Complete step by step answer:
Carbon has an atomic number 6. Its electronic configuration is $1{s^2}2{s^2}2{p^2}$ . Thus, a Carbon atom has a total of 12 electrons. There are several rules for adding electrons in the molecular orbitals. We were using the same way as the electrons adding to the atomic orbitals. First, we should fill the electron in the molecular orbital with lowest energy. The maximum number of electrons in a molecular orbital should not exceed two. And the two electrons must have opposite spin. If two or more molecular orbitals have the same energy the pairing of electrons will occur only after each orbital of the same has one electron. By using those rules, we can fill electrons in the molecular orbital
Carbon molecules are formed by the combination of two carbon molecules. Each carbon has 6 electrons. There are 12 electrons to be accommodate in the $(\sigma 1s)(\sigma \times 1s)(\sigma 2s)\left( {\sigma \times 2s} \right)\left( {\pi 2{p_x} = \pi 2{p_y}} \right)$ Bonding Molecular orbitals.
Molecular orbital electronic configuration up to 14 electrons is
$(\sigma 1s)(\sigma \times 1s)(\sigma 2s)\left( {\sigma \times 2s} \right)\left( {\pi 2{p_x} = \pi 2{p_y}} \right)(\sigma 2{p_z})\left( {\pi \times 2{p_x} = \pi \times 2{p_y}} \right)$

Molecular orbital electronic configuration of more than 14 ${e^ - }$ is

$(\sigma 1{s^2})(\sigma \times 1{s^2})(\sigma 2{s^2})(\sigma \times 2{s^2})(\sigma 2{p_z})\left( {\pi 2{p_x} = \pi 2{p_y}} \right)\left( {\pi \times 2{p_x} = \pi \times 2{p_y}} \right)(\sigma \times 2{p_z})$
Carbon molecule having less than 14 electrons. So, we can use the first electronic configuration.

$(\sigma 1{s^2})(\sigma \times 1{s^2})(\sigma 2{s^2})\left( {\sigma \times 2{s^2}} \right)\left( {\pi 2{p_x}^2 = \pi 2{p_y}^2} \right)$
Hence, the correct answer is option (B) i.e $(\sigma 1{s_2})(\sigma \times 1{s_2})(\sigma 2{s_2})\left( {\sigma 2{s^2}} \right)\left( {\pi 2{p_x}^2 = \pi 2{p_y}^2} \right)$

Note: We can find the Bond order of molecules by filling electrons in the Molecular orbitals.
Bond order= $\;\dfrac{1}{2}$ [Number of electrons in Bonding Molecular orbital- Number of electrons in Anti bonding Molecular orbitals]. As the Bond order increases Bond strength increases and Bond length decreases.