## 2F.1 7th Edition

$sp, sp^{2}, sp^{3}, dsp^{3}, d^{2}sp^{3}$

Mark 1D
Posts: 63
Joined: Fri Sep 28, 2018 12:18 am

### 2F.1 7th Edition

Can someone explain the solution for 2F.1 in the 7th Edition? "State the relative orientation of each of the following hybrid orbitals: (a)sp^3; (b) sp; (c)sp^3d^2; (d) sp^2.

Manya Bali 4E
Posts: 66
Joined: Fri Sep 28, 2018 12:23 am

### Re: 2F.1 7th Edition

From class, we know that the number of hybrid orbitals equals the regions of electron density.

(a) One s orbital and 3 p orbitals hybridize to create 4 sp^3 orbitals. This means that there are 4 regions of electron density around the central atom so there must be a tetrahedral arrangement.

(b) One s orbital and 1 p orbital hybridize to create 2 sp orbitals. This means there are 2 regions of electron density so there must be a linear arrangement.

Try the others and feel free to comment if it still doesn't make sense!