A) from \[s{{p}^{2}}\]to \[ds{{p}^{2}}\]
B) from \[s{{p}^{2}}\]to \[s{{p}^{3}}\]
C) from \[s{{p}^{3}}\] to \[s{{p}^{2}}\]
D) from \[s{{p}^{3}}\] to \[s{{p}^{3}}d\]
Correct Answer: B
Solution :
[b] In \[B{{F}_{3}}\], B is \[s{{p}^{2}}\] hybridized with one empty \[{{p}_{z}}\] orbital. The empty \[{{p}_{z}}\] orbital of \[B{{F}_{3}}\] can be filled by lone pair of molecules such as \[N{{H}_{3}}\]. When this occurs a tetrahedral molecule or ion is formed which is \[s{{p}^{3}}\] hybridized.
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