Question

Electronic configurations of three elements, $A,B$ and $C$ are given below.

$A:1s^{2}2s^{2}2p^6$ ​

$B:1s^{2}2s^{2}2p^{6}3s^{2}3p^3$ ​

$C:1s^{2}2s^{2}2p^{6}3s^{2}3p^5$​

Stable form of $C$ may be represented by the formula:

• A. $C$
• B. $C_2$
• C. $C_4$
• D. $C_3$

Answer 1

The correct option is B $C_2$

Element	Electronic configuration
$C\left(Cl\right)$	$1s^{2}2s^{2}2p^{6}3s^{2}3p^5$

Electronic configuration suggests that $‘C^{\prime }$ is $Cl$ atom.

It is one electron short from the noble gas configuration, so to complete its octet and acquire stability, $Cl$ atoms share one electron with each other and form $C{l}_2$ or $C_2$ in the given case.

So, the stable form of $‘C^{\prime }$ is $C_2,$ hence, correct option is $\left(B\right).$