I was trying to figure out if the bond angles in NF3 are larger than in NH3. My reasoning led me to the conclusion that they should be larger, though in reality the opposite is true (102 deg for NF3 and 106 deg for NH3). I can't pinpoint where I am mistaken.
Here is my reasoning:
According to VSEPR the repulsion for lone pair-bond e (electrons) is greater than bond e- bond e. Thus the bond angles will be determined by the strength of the lone-pair-bond e repulsion. F is more electronegative, thus it will pull closer to it the bond electrons, reduce the electron cloud density near N, reduce the repulsion lone-pair-bond e, thus decrease the lone pair-bond e angle and increase the bond e- bond e angle.
What am I missing?
2006-11-29
00:55:40
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3 answers
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asked by
bellerophon
6
in
Science & Mathematics
➔ Chemistry