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Answer 1 · 2015-12-01T23:50:46

$s p^{2}$ $sp^2$

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

$\Rightarrow$ $=>$

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

$1 s^{2} 2 s^{2} 2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{0}$ $1s^(2)2s^(2)2p_x^(1)2p_y^(1)2p_z^0$ you can see the orbital $2 p_{z}$ $2p_z$ have no electron

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

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

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

$π$ $pi$ electron don't match with hybridization state, so the orbital $2 p_{z}$ $2p_z$ get out

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

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

You can done this without any complication using the VSEPR Theory

Here the carbon is : $A X_{3}$ $AX_3$ which means it's $s p^{2}$ $sp^2$