The fluoride for which the dipole moment is not equal to zero, is :

Step-by-step solution

As per VSEPR Theory XeF 4 ⇒ H ⇒ 1 2 8 + 4 = sp 3 d 2 6 Geometry square planar with 2 L.P. μ → = 0 (symmetry) CF 4 ⇒ H ⇒ 1 2 4 + 4 = sp 3 4 Tetrahedral geometry μ → = 0 (symmetry) PF 5 ⇒ H = 1 2 5 + 5 = sp 3 d 5 Trigonal bipyramidal μ → = 0 (Symmetry) SF 4 ⇒ H = 1 2 6 + 4 = sp 3 d 5 With one L.P see saw geometry μ → ≠ 0 has a value μ > 0 due to two S - F bonds in equatorial plane.