## Textbook 4.37

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

Luis Torres 1C
Posts: 30
Joined: Fri Apr 06, 2018 11:05 am

### Textbook 4.37

4.37
Identify the hybrid orbitals used by the phosphorus atom in each of the following species: (d) PCl3

I have been using the following drawing method to identify hybrid orbitals, and it has been successful so far. However, when I get to part d, my answer is sp3d, but the correct answer is sp3. Could someone please explain where I am going wrong? Is there a rule that I am forgetting about, or a better method I should be using?
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Nicole Shak 1L
Posts: 35
Joined: Wed Nov 22, 2017 3:03 am

### Re: Textbook 4.37

For PCl3 when drawing out the Lewis structures, you can see that P has 4 regions of electron density surrounding it, so this is why it is sp3. I think that drawing out the structures for these problems also helps.

ErinKim1I
Posts: 31
Joined: Fri Apr 06, 2018 11:03 am

### Re: Textbook 4.37

P has 4 areas of electron density because P has 3 bonding pairs AND a lone pair. Lone pairs are represented by both spin up and spin down.
so the answer is sp3 with the 3s orbital containing both a spin up and spin down, and the 3p orbital having 3 unpaired e-.

Myles Chang 1B
Posts: 29
Joined: Fri Apr 06, 2018 11:02 am

### Re: Textbook 4.37

I think your confusion stems from the fact that the 2 lone electrons are found in one orbital, meaning that the P atom only has 4 regions of electron density.