JEE Main & Advanced
JEE Main Paper (Held On 10-Jan-2019 Morning)
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