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Symmetrical tetrahedral molecules (like $\ce{CH4}$) have a bond angle of $109.5$. Those with lone pairs in place of one atom (like $\ce{NH3}$) have bond angles less than $109.5$.

The standard explanation for why is that the lone pairs are "larger" than the hydrogens, creating repulsive forces that push the hydrogens down, decreasing the bond angle.

However, this explanation fails when confronted with $\ce{CH3Cl}$. According to the above logic, the $\ce{Cl}$, which is much bigger than the hydrogens, should repel the hydrogens just like a lone pair, making the $\ce{H-C-H}$ bond angle smaller than $109.5$. However, the $\ce{H-C-H}$ bond angle is actually $110.5$ (see here, I could not find a more authoritative source).

Why is this (or is my source wrong)?

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Cl is an electronegative than carbon which is further more electronegative than hydrogen and thus Chlorine pulls the bond pairs towards itself, polarizing it. In this it pulls the electrons towards it increasing the bond angle to110.5.

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Cl is more electronegative than C and H. Hence Cl will pull the electron toward itself. Again the C is also more electronegative than H. Hence C will pull the electron towards itself. Hence repulsion from the Cl side will less. The repulsion between bonds of C-H will be more and hence the angle enter image description here

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  • $\begingroup$ Please add more detail. This aim of this site is quality, not short answers. I think your close but please consider revising. $\endgroup$ – A.K. Aug 23 '18 at 18:09

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