1subhomoy, hybridisation of XeOF4 is sp3d2 as it contains 5 bonded pairs and 1 lone
pair i.e. 4 single bonds with flourine and one double bond with oxygen and remains 1 lone pair with
Xe.thus its structure is square pyramidal
3 Answers
1@mentor.....hyn=sp3d2....
reason:5bp+1lp=6
@me.....hyn=sp3d3
reason:6bp+1lp=7
now i think that i am correct because in first case no. of e- in Xe atom is 5+2=7 but in my explanation it is 6+2=8 which is correct......
so am i correct???????if i am wrong where am i wrong??????
1
1Hey hows is going. I had a question. I understand that O is double bonded so wouldn't that fulfill a pair per bond? If it is double bonded due to its oxidization of -2, and single bonded with the 4 Fs. Wouldn't be 6 bonds and one lone pair around Xe making it 7. Therefore s,p3,d3 would be the formula. If i have a discrepency in my understanding of Xe's hybridization please help me rectify it.