enhance this concern
edited Dec 29 "16 in ~ 23:53
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request Dec 28 "16 at 16:24
Gaurang TandonGaurang Tandon
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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.
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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 $ceNH3$.
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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.