question_answer

The type of hybridisation and number of lone pair (s) of electrons of Xe in \[XeO{{F}_{4}}\] respectively, are: [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

A) \[s{{p}^{3}}d\] and 2                          

B) \[s{{p}^{3}}{{d}^{2}}\] and 2

C) \[s{{p}^{3}}d\] and 1  

D)                  \[s{{p}^{3}}{{d}^{2}}\] and 1

Correct Answer: D

Solution :

Number of b.p. = 5 Number of l.p. = 1 Sp3d2 = hybridisation