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Contribution of atom at corners = 1/8 Contribution of atom at edges = 1/4 Contribution of atom at centres = 1 Contribution of atom at face = 1/2 So total no. of F = 8x1/4 + 2 = 4 so total no. of Xe = 8x1/8 +1 =2 So Xe2F4 = XeF2 Solution Contribution of atom at corners = 1/8 Contribution of atom at edges = 1/4 Contribution of atom at centres = 1 Contribution of atom at face = 1/2 So total no. of F = 8x1/4 + 2 = 4 so total no. of Xe = 8x1/8 +1 =2 So Xe2F4 = XeF2.

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Contribution of atom at corners = 1/8 Contribution of atom at edges = 1/4 Contribution of atom at centres = 1 Contribution of atom at face = 1/2 So total no. of F = 8x1/4 + 2 = 4 so total no. of Xe = 8x1/8 +1 =2 So Xe2F4 = XeF2 Solution Contribution of atom at corners = 1/8 Contribution of atom at edges = 1/4 Contribution of atom at centres = 1 Contribution of atom at face = 1/2 So total no. of F = 8x1/4 + 2 = 4 so total no. of Xe = 8x1/8 +1 =2 So Xe2F4 = XeF2

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  • 1. Contribution of atom at corners = 1/8 Contribution of atom at edges = 1/4 Contribution of atom at centres = 1 Contribution of atom at face = 1/2 So total no. of F = 8x1/4 + 2 = 4 so total no. of Xe = 8x1/8 +1 =2 So Xe2F4 = XeF2 Solution Contribution of atom at corners = 1/8 Contribution of atom at edges = 1/4 Contribution of atom at centres = 1 Contribution of atom at face = 1/2 So total no. of F = 8x1/4 + 2 = 4 so total no. of Xe = 8x1/8 +1 =2 So Xe2F4 = XeF2