Question

Draw the structure for – XeOF4

Answer 1

Structure of $XeOF_4$.

• To draw the Lewis dot structure, the number of valence electrons present in the compound must be calculated.
• Accordingly number of valence electrons present in $XeOF_4$ is as follows:

Valence electrons

= Xe + O + 4(F)

= 8+6+4(7)

= 14+28

= 42

• Thus $XeOF_4$ has 42 valence electrons. These electrons must be allotted to the elements in such a way that each elements attain octet.
• In the given case, Xenon is exceptional as it can expand octet.
• Thus the structure of $XeOF_4$ comes out to be the structure which is given in the attachment.

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