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the total number of lone pair present in XeF4 is
a)10
b)12
c)14
d)16
plz answer asap and explain.

There are 3 lone pairs present on each F atom and two lone pairs present on Xe. So, the total number of  lone pairs are: 4 x 3 + 2 = 14 (c).
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total no. of lone pairs=4*3+2=14 (C)
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By the help of single linkage concept,we draw the structure of this compound.

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14 is the answer

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Lone pair present in Xef4
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