SF4 has trigonal bipyramidal geometry and see-saw shape as shown in below structure:Here, axial bonds are longer than equatorial bonds. Thus, SF4 has unequal bond length.
BF−4,SiF4 have tetrahedral geometry where all bonds are equal in respective compounds.
XeF4 has square planar shape where all four Xe−F bonds are equal.
Hence, option (c) is the correct answer.