I am reading a book: Concise Inorganic Chemistry by J.D.Lee (Fifth Edition)
In the chapter-The Covalent Bond, the author says:

$\ce{NF3}$ and $\ce{NH3}$ both have structures based on a tetrahedron with one corner occupied by a lone pair. The high electronegativity of $\ce{F}$ pulls the bonding electrons further away from $\ce{N}$ than in $\ce{NH3}.$ Thus repulsion between bond pairs is less in $\ce{NF3}$ than in $\ce{NH3}$. Hence the lone pair in $\ce{NF3}$ causes a greater distortion from tetrahedral and gives an $\ce{F-N-F}$ Bond angle of 102°30', compared with 107°48' in $\ce{NH3}$. The same effect is found in $\ce{H2O}$ (bond angle 104°27') and $\ce{F2O}$ (bond angle 102°)

I am not able to understand why the pulling of the bonding electrons (in $\ce{NF3}$) by F results in reduced repulsion between the bond pairs and how does this ultimately lead to a smaller bond angle ($\ce{F-N-F}$) in $\ce{NF3}$?

  • $\begingroup$ If the electrons are more pulled towards $\ce{F}$, the distance between the three bond pairs increase from one another and hence electrostatic repulsion decreases, so they can go a little bit closer to achieve the new equilibrium distance , which results in shrinking of bond angles. $\endgroup$
    – Soumik Das
    Jul 19 '18 at 11:10
  • $\begingroup$ It would be helpful if you could give a pictorial representation of what you are trying to say, your comment isn't quite what I am looking for. $\endgroup$ Jul 19 '18 at 11:20
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    $\begingroup$ Related: Why does NF3 have a smaller bond angle than NH3? $\endgroup$ Jul 19 '18 at 13:50
  • $\begingroup$ Thanks @Gaurang, that was helpful. I actually wanted to know how the author has reached the conclusion, which is still not clear to me. Also the posts mention 'Bent's rule' which I am not aware of. $\endgroup$ Jul 19 '18 at 14:07
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    $\begingroup$ chemistry.stackexchange.com/questions/15620/what-is-bents-rule $\endgroup$
    – Mithoron
    Jul 19 '18 at 15:20