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### HW Q 4.11

Posted: Thu Nov 15, 2018 3:23 pm
Initially when I did part a for SCl4, I thought it's shape was supposed to be tetrahedral. But the answer says it is see saw. Can someone explain why its shape is see saw and not tetrahedral and what is a key way of differentiating the two.

### Re: HW Q 4.11

Posted: Thu Nov 15, 2018 3:29 pm
SCl4 has a seesaw molecular geometry because you must take into account the effect that the lone pair on S has on shape; if there was no lone pair on SCl4, the shape would be tetrahedral.

### Re: HW Q 4.11

Posted: Thu Nov 15, 2018 3:30 pm
laurenho-4c wrote:SCl4 has a seesaw molecular geometry because you must take into account the effect that the lone pair on S has on shape; if there was no lone pair on SCl4, the shape would be tetrahedral.

If there was no lone pair it would just be SCl4. Would it not be square planar?

### Re: HW Q 4.11

Posted: Thu Nov 15, 2018 3:44 pm
Phoebe Chen 4I wrote:
laurenho-4c wrote:SCl4 has a seesaw molecular geometry because you must take into account the effect that the lone pair on S has on shape; if there was no lone pair on SCl4, the shape would be tetrahedral.

If there was no lone pair it would just be SCl4. Would it not be square planar?

To be square planar, a molecule would need the VSEPR formula AX4E2 (2 lone pairs). These lone pairs would situate themselves 180 degrees away from one another, leaving the other 4 atoms to arrange themselves in a square around the central atom and between the lone pairs.

For a hypothetical SCl4 without lone pairs, the VSEPR formula would be AX4, which corresponds with a tetrahedral molecular geometry.

### Re: HW Q 4.11

Posted: Thu Nov 15, 2018 3:54 pm
I understand that the lone pair causes the shape to be seesaw, but can we assume that the lone pairs will be on the top? Would the shape be different if the lone pair was elsewhere?