Which of the following is true for ${\mathrm{O}}_2$and ${\mathrm{C}}_2$ molecules according to M.O.T.?

Since oxygen contains 8 electrons. Consequently, $O_2$'s electronegative configuration

​the result of having 16 electrons

${\sigma }_{1\mathrm{s}}^2{\sigma }_{1\mathrm{s}}^{\ast 2}{\sigma }_{2\mathrm{s}}^2{\sigma }_{2\mathrm{s}}^{\ast 2}{\sigma }_{2\mathrm{pz}}^2{\pi }_{2\mathrm{py}}^2={\pi }_{2\mathrm{pr}}^2{\pi }_{2\mathrm{py}}^{\ast 1}={\pi }_{2\mathrm{p}}^{\ast }$

Bond order :- $\frac{\left|N_a-N_b\right|}{2}=\frac{|6-10|}{2}=2$

$O_2$ has a bond order of 2.

Given that carbon contains six electrons, the electrical configuration of carbon with twelve electrons is

${\sigma }_{1\mathrm{s}}^2{\sigma }_{1\mathrm{s}}^{\ast 2}{\sigma }_{2\mathrm{s}}^2{\sigma }_{2\mathrm{s}}^{\ast }{\pi }_{2\mathrm{p}y}^2={\pi }_{2\mathrm{p}}^2$ 

Bond order:-  $\frac{|4-8|}{2}=2$

So, the bond order of  $C_2$ is 2 .

${\mathrm{C}}_2$and ${\mathrm{O}}_2$ have same  B.O = 2.0  but $C_2$ has $2\mathrm{\pi }$ bonds while $O_2$ has 10 and  $1\mathrm{\pi }$.

Hence correct option is C.