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In $NH_4^+$, the $/_H-N-H$ bond angles are $109.5$ $""^@$, as we would expect for a regular tetrahedron. $F_2O$, and $H_3O^+$ share this electronic configuration, but here the central $O$ atom bears 2 and 1 lone pairs of electrons respectively. The lone pairs would tend to compress the $/_F-O-F$ and $/_H-O-H$ bond angles somewhat to give approx. $104-5^@$.

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