Question

Hybridisation state of Xe in $XeF_{2},XeF_{4}$and $XeF_{6}$respectively are

• A. $sp^{2},sp^{3},d,sp^{3}d^{2}$
• B. $sp^{3}d,sp^{3}d^{2},sp^{3}d^{3}$
• C. $sp^{3}d^{2},sp^{3}d,sp^{3}d^{3}$
• D. $sp^{2},sp^{3},sp^{3}d$

Answer 1

Answer: (B)

$sp^{3}d,sp^{3}d^{2},sp^{3}d^{3}$

Answer 2

: $XeF_{2} - sp^{3}d$

Total no. of valence electrons = 22

$\frac{22}{8} = 2(Q) + 6(R),\frac{6}{2} = 3(Q)$

X = 2 +3 = 5

Hybridisation is $sp^{3}d.$

$EeF_{4} -$

Hybridisation is $sp^{3}d^{2}$

Total no. of electrons in outermost shells = 8 +28 =36

$\frac{36}{8} = 4(Q) + 4(R),\frac{4}{2} = 2(Q) + 0(R)$

X = 4 + 2 +0 =6

Hybridisation is $Sp^{3}d^{2}$

$XeF_{6}$

Total no. of valence electrons = 8+ 42 = 50

$\frac{50}{8} = 6(Q) + 2(R),\frac{2}{2} = 1(Q)$

X= 6+ 1 = 7

Hybridisation is $Sp^{3}d^{3}$.