Question

An element 'A' when combines with four F-atoms, forms $A{F}_4$ which has square planar shape and also has lone pair(s). Select correct statement with respect to compounds of A.

• A. $A{F}_4$ is polar in nature
• B. $A{F}_3^{+}$ has all identical bond
• C. $A{F}_5^{+}$ has all identical bond
• D. $A{F}_2$ has non-zero net dipole moment

Answer 1

The correct option is C $A{F}_5^{+}$ has all identical bond

2 lone pair and 4 bond pair so A belongs to ${18}^{th}$ group.
From the shape we can say that A has 8 valence electron.
So $A{F}_3^{+}$ has T-shape with axial and equatorial bond but $A{F}_5^{+}$ has octahedral geometry with one lone pair so all bonds are identical.