Molecules having the same number of hybrid orbitals, have same hybridisation and number of hybrid orbitals, H=‌
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[V+X−C+A] where, V= no. of valence electrons of central atom X= no. monovalent atoms C= charge on cation A= Charge on anion. (1) ln‌SF4, H=‌
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[6+4−0+0)=5 (2) ln‌l3H=‌
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[7+2+1]=5 (3) ln‌SbCl52−,H=‌
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[5+5+2)=6 (4) ln‌PCl5, H=‌
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[5+5+0−0]=5 Since, only SbCl52− has different number of hybrid orbitals (ie,6) from the other given species, its hybrisation is different from the others, ie, sp3d2. (The hybridsation of other species is sp3d ).