# lp = ½[ (valence e⁻ of A )- n(A:X σ bonds;2 perO) ±charge ]
ICl2-: ½[ (7 valence e⁻ of I )- 2(I:Cl σ bonds;) +1(-ve charge) ] =3lp = AX2E3 sp3d hybridization
H2O: ½[ (6 valence e⁻ of O)- n(O:H σ bonds ] = 2lp AX2E2 sp^3
CO2: O=C=O C sp
NO2-: ½[ (5 valence e⁻ of N )- 4(2 O) +1] =1lp AX2E sp^2 hybridization
The figure you have in your pic is of sp^3 hybridization (four lobes) so answer is H2O
