Why is the Br-C-H bond angle in CH3Br smaller than the Cl-C-H bond angle in CH3Cl?

Can someone explain for me why the Br-C-H bond angle in CH3Br is smaller than the Cl-C-H bond angle in CH3Cl? From what I know, F-C-H in CH3F has a smaller bond angle than Cl-C-H in CH3Cl because F is more electronegative than Cl, which then attracts the electron in the bonding pair closer to F, resulting in a smaller bond angle. I therefore expected that Br-C-H will have a larger bond angle than Cl-C-H because Br is less electronegative than Cl. However, the data show that Br-C-H has a smaller bond angle. Can someone please help!

Data (taken from CCCBDB) CH3Br Br-C-H = 107.7 degree H-C-H = 111.2 degree

CH3Cl Cl-C-H = 110.8 degree H-C-H = 108.2 degree

• These are small differences so who really knows why. That said, the $\ce{C-X}$ bond length is 1.939 A in methyl bromide, but only 1.781 A in methyl chloride (sterics?) (ref), a significant difference. The longer $\ce{C-Br}$ bond implies that it contains more p-character than the corresponding $\ce{C-Cl}$ bond. Consequently, there is more s-character in the methyl bromide $\ce{C-H}$ bonds than in the methyl chloride $\ce{C-H}$ bonds. More s-character in the $\ce{C-H}$ bonds means a larger $\ce{H-C-H}$ angle.
– ron
Nov 11, 2021 at 18:44