| (I) \[{{(C{{H}_{3}})}_{2}}O+B{{F}_{3}}\to \,\,{{(C{{H}_{3}})}_{2}}O\,\,\to \,\,B{{F}_{3}}\] |
| (II) \[{{(Si{{H}_{3}})}_{2}}O+B{{F}_{3}}\to {{(Si{{H}_{3}})}_{2}}O\,\,\to \,\,B{{F}_{3}}\] |
| (III) \[{{H}_{3}}N+B{{F}_{3}}\to \,\,{{H}_{3}}N\to B{{F}_{3}}\] |
| (IV) \[{{(C{{H}_{3}})}_{2}}O+BF_{4}^{\Theta }\,\,\to \,\,{{(C{{H}_{3}})}_{2}}O\to BF_{4}^{\Theta }\] |
A) I, II
B) II, III
C) II, IV
D) Only IV
Correct Answer: C
Solution :
(II) \[Si{{H}_{3}}\] has vacant d-orbital So back bonding possible (IV) Octet of \[BF_{4}^{\Theta }\] is complete.
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