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According to the Colby Chemistry Database, the bond angle of $\ce{PBr3}$ and $\ce{PCl3}$ are around $101$ degrees, but the bond angle of $\ce{PH3}$ is $92$ degrees. (Source: PCl3 PBr3 PH3)

There was a previous question on Stack Exchange about the bond angle difference in $\ce{PF3}$ and $\ce{PH3}$ here: PF3 and PH3, but the accepted answer mentions back bonding as a factor to increase the bond angle in $\ce{PF3}$. I was skeptical about this answer, because in $\ce{PCl3}$ and $\ce{PBr3}$, there should be less back bonding than in $\ce{PF3}$ because bromine and chlorine are both much larger than fluorine and there will be worse orbital overlap between the halogen. Less back bonding should make the bond angle of $\ce{PCl3}$ and $\ce{PBr3}$ smaller than that of $\ce{PF3}$, if back bonding were the major factor here? In reality, that is not the case. Why not?

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The larger angles can simply be explained as a result of repulsion between the larger atoms of $\ce{Br}$ and $\ce{Cl}$. Hydrogen atoms in $\ce{PH3},$ as the are so small, experience less repulsion as compared to $\ce{Br}$ atoms in $\ce{PBr3}$ or $\ce{Cl}$ atoms in $\ce{PCl3},$ therefore the larger bond angles in $\ce{PBr3}$ and $\ce{PCl3}.$

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It is because of the higher electronegativities of Br and Cl, than Hydrogen. So there is a higher repulsion of the bonding electrons.

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    $\begingroup$ You should be confirmed on ur answer! $\endgroup$ – Yaseen Khan Jan 1 at 9:08
  • $\begingroup$ oh! okay thank you for pointing me out, I forgot that. And I am kind of new to the community. $\endgroup$ – Sujee0_0 Jan 1 at 13:57

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