I"ve already read countless answers about the reason why $ceNF3$ has a smaller sized bond angle 보다 $ceNH3$ , but I can"t it seems to be ~ to understand them. Here"s my knowledge of the situation:

$ceNH3$: right here N is more electronegative 보다 H therefore a big electron cloud is overfilled over N. This will push the bond pairs $ceN-H$ far from the main atom. So, the edge of $ceH-N-H$ will certainly decrease.$ceNF3$: below F is much more electronegative 보다 N, therefore the lone pair cloud over N is scattered into the $ceN-F$ bonds. Thus the smaller electron cloud over main atom is unable to push the $ceN-F$ bonds away from itself as lot as the did previously. So, link angle must be greater than ahead case.
electronegativity vsepr-theory
re-publishing
enhance this concern
follow
edited Dec 29 "16 in ~ 23:53
*

jan
62.2k1111 yellow badges162162 silver badges345345 bronze badges
request Dec 28 "16 at 16:24
*

Gaurang TandonGaurang Tandon
8,9001010 gold badges5555 silver badges105105 bronze badges
$endgroup$
3
add a comment |

4 answers 4


energetic oldest Votes
9
$egingroup$
This can be argued on the communication of Bent"s rule; concisely stated

Atomic s character concentrates in orbitals directed toward electropositive substituents

What follows listed below is a crude explanation. Prior to that, I"ll keep in mind that we issue ourselves with the hybridisation the the orbitals at the central atom.

You are watching: What is the bond angle of nh3

Since s orbitals are lower in energy than p orbitals, lock are better at stabilising electrons. Which is why molecules choose to use them (s orbitals) in bonds whereby there is "more" electron thickness to stabilise.

Thus, Bent"s ascendancy can just as conveniently be formulated together follows: ns character concentrates in orbitals directed in the direction of electronegative substituents; (the main atom doesn"t have to waste its low energy s orbitals, the orbitals ~ above the electronegative atom, deserve to (possibly) take treatment of it.

What has all of this got to perform with bond angles? Well, an ext s character leader to larger bond angles. You can think of how s orbitals space spherical in shape, and a huge s character would certainly lead to more "spherical" hybrid orbitals, and also that would lead to bigger bond angles.

Going earlier to her question, us are supposed to to compare $ceNF3$ and also $ceNH3$.

See more: Player Points: What Are Player Points On Roblox Used For, (Pp): Roblox

Fluorine is an ext electronegative that hydrogen, and the $ceN-F$ bond would have better p character 보다 the $ceN-H$ bond, (or $ceN-H$ link has higher s character, whichever friend prefer). Hence $ceH-N-H$ angle is greater in $ceNH3$ than it is in $ceNF3$

P.S I"m sure you have the right to learn more about what I simply tried to describe (rather poorly) by reading the Wikipedia article, and/or a textbook and/or utilizing the search role on this website to discover related questions/answers.