The number of equivalent contributing structures possible for XeO6^4- is m. The bond order of Xe - O bond is n.

Question

The number of equivalent contributing structures possible for  XeO^4-₆ is m. The bond order of Xe - O bond is n. Find (m x n) × 0.2. Round off your answer to the nearest integer.

Answer 1

Bond order of  XeO₆^4-  is 4/3 = n

It has 15 possible structures

∴ m = 15

Thus mn = 15 x 4/3 = 20

∴  0.2 mn = 0.2 x 20 = 4