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  4. Hybridisation state of Xe in XeF2, XeF4 and XeF6 respectively are

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Q. Hybridisation state of $Xe$ in $XeF_2, XeF_4$ and $XeF_6$ respectively are

Chemical Bonding and Molecular Structure

Solution:

$XeF_{2} - sp^{3}d;$
Total no. of valence electrons $= 22$
$\frac{22}{8}= 2\left(Q\right)+6\left(R\right), \frac{6}{2}= 3\left(Q\right) $
$ X= 2+3=5$
Hybridisation is $sp^{3}d$.
$XeF_{4} -sp^{3}d^{2} $
Total no. of electrons in outermost shells $= 8 + 28 = 36 $
$ \frac{36}{8}= 4\left(Q\right)+4\left(R\right), \frac{4}{2} = 2\left(Q\right)+0\left(R\right) $
$ X=4+2+0=6$
Hybridisation is $ sp^{3}d^{2}$.
$XeF_{6}-sp^{3}d^{3} $
Total no. of valence electrons $= 8 + 42 = 50$
$ \frac{50}{8}= 6\left(Q\right)+2\left(R\right), \frac{2}{2} $
$= 1\left(Q\right), X = 6+1 = 7 $
Hybridisation is $sp^{3}d^{3}.$