Which molecule has two lone pairs on the central atom?

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The discussion centers on identifying which molecule has two lone pairs on the central atom among PCl3, ICl4^+, CH3I, XeF4, and PCl6^-. Initially, PCl3 was considered, but it was clarified that it only has one lone pair. The correct answer is XeF4, which has two lone pairs, making it square planar in shape. Participants suggest drawing Lewis structures for better visualization of the molecular geometry. The conversation emphasizes the importance of understanding hybridization and the octet rule in determining molecular structures.
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I have to figure out which has two lone pairs on the central atom between PCl3, ICl4^+, CH3I, XeF4, PCl6^-.

Is it PCl3? I think I did it right but I am not sure.
Thank you.
 
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Think you are correct. The three 3p3 electrons of P should be bonded to the three Cls, each one forming single bonds. That leaves the pair of 3s2 electons unshared. Haven't even looked at the other choices, since I feel PCl3 is the one.

If incorrect, please someone let me know.

Best of luck.

Steve
 
hi Steve, thanks.
I re-checked my answer and realize I think I read the question wrong... Two lone pairs, meaning four all together (I was interpreting it like two, not two pairs) if that makes sense. Then I think it's XeF4.
 
Smith4046 said:
Think you are correct. The three 3p3 electrons of P should be bonded to the three Cls, each one forming single bonds. That leaves the pair of 3s2 electons unshared. Haven't even looked at the other choices, since I feel PCl3 is the one.

If incorrect, please someone let me know.

Best of luck.

Steve

The bonds are not strictly s and p bonds, but rather are spd hybridized.

PCl_{3} has only one lone pair. So, you can rule it out.

Carbon must obey the octet rule, so you can rule out iodomethane.

Try drawing Lewis structures for the other 3. One is square planar (octahedral), thanks to its two lone pairs.
 
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