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Answer 1 · 2015-12-01T23:50:46

$sp^2$

carbon have 6 $e^-$

$=>$

$1s^(2)2s^(2)2p^2$ in normal state following klechkowski rule

$1s^(2)2s^(2)2p_x^(1)2p_y^(1)2p_z^0$ you can see the orbital $2p_z$ have no electron

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

you have now $sp^3$ hybridization

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

$pi$ electron don't match with hybridization state, so the orbital $2p_z$ get out

so finally you have $sp^2$

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

You can done this without any complication using the VSEPR Theory

Here the carbon is : $AX_3$ which means it's $sp^2$