The correct option is C SbCl2−5
By using formula,
Electron pairs=Number of valence electrons on central atom +Number of bond formation2
In SF4,
Electron pairs=6+42=5
Hence,
S has dsp3 hybridization
In I−3,
Electron pairs=7+2+12=5
Hence,
I has dsp3 hybridization
In PCl5,
Electron pairs=5+52=5
Hence,
P has dsp3 hybridization
In SbCl2−5,
Electron pairs=5+5+22=6
Sb has sp3d2 hybridization