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B
(i) and (iv)
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C
(ii) and(iii)
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D
(i) and (iii)
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Solution
The correct option is B (i) and (iv) Both SbF5 and SF4 have all the p orbitals ocuppied. Hence they involve vacant d-orbitals to form [SbF6]− and [SF5]−.
H3NBF3 here,the lone pair of N is donated to vacant p orbital of B
AlCl−4 here, the lone pair of Cl is donated to vacant p orbital of Al
Hence (i) and (iv) are correct