Xenon forms three binary fluorides by direct combination of the elements under controlled conditions : Xe + F₂ (Xe in excess, 673 K) → (A) ; Xe + F₂ (1 : 5, 673 K, 6-7 bar) → (B) ; Xe + F₂ (1 : 20, 57

Question

Xenon forms three binary fluorides by direct combination of the elements under controlled conditions : Xe + F₂ (Xe in excess, 673 K) → (A) ; Xe + F₂ (1 : 5, 673 K, 6-7 bar) → (B) ; Xe + F₂ (1 : 20, 573 K, 60-70 bar) → (C). Total number of lone pairs on the central atom in (A), (B) and (C) taken together is :

Answer:

6

Solution:

A = XeF₂ (3 lone pairs on Xe), B = XeF₄ (2 lone pairs), C = XeF₆ (1 lone pair). Total = 3 + 2 + 1 = 6.

TestHub · Accepted Answer

A = XeF₂ (3 lone pairs on Xe), B = XeF₄ (2 lone pairs), C = XeF₆ (1 lone pair). Total = 3 + 2 + 1 = 6.

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