# Question #694f3

Dec 1, 2015

$s {p}^{2}$

#### Explanation:

carbon have 6 ${e}^{-}$

$\implies$

$1 {s}^{2} 2 {s}^{2} 2 {p}^{2}$ in normal state following klechkowski rule

$1 {s}^{2} 2 {s}^{2} 2 {p}_{x}^{1} 2 {p}_{y}^{1} 2 {p}_{z}^{0}$ you can see the orbital $2 {p}_{z}$ have no electron

So the orbital $2 {s}^{2}$ give one electron this is called the hybridization state

you have now $s {p}^{3}$ hybridization

BUT, to form double liaison you need $\pi$ electron

$\pi$ electron don't match with hybridization state, so the orbital $2 {p}_{z}$ get out

so finally you have $s {p}^{2}$

$- - - - - - - - - - - - - - - - - - - - -$

You can done this without any complication using the VSEPR Theory

Here the carbon is : $A {X}_{3}$ which means it's $s {p}^{2}$