Match each set of hybrid orbitals from LIST-I with complex(es) given is LIST-II LIST-I LIST-II (P) dsp2 (1) [FeF₆]4– (Q) sp3 (2) [Ti(H₂O)₃Cl₃] (R) sp3d2 (3) [Cr(NH₃)₆]3+ (S) d2sp3 (4) [FeCl₄]2– (5) Ni(CO)₄ (6) [Ni(CN)₄]2– The correct option is
A.P - 5; Q - 4, 6; R - 2, 3; S - 1
B.P - 5, 6; Q - 4; R - 3; S - 1, 2
C.P - 6; Q - 4, 5; R - 1; S - 2, 3✓ Correct
D.P - 4, 6; Q - 5, 6; R - 1, 2; S - 3
Explanation
4- 6 2+ 6 2 3 3 +3 1 3+ 3 6 +3 3 2- 4 2+ 6 4 8 2 2- 4 2+ 8 (1)[FeF ] Fe =[Ar]3d (2)[Ti(H O) Cl ] Ti =[Ar]3d (3)[Cr(NH ) ] Cr =[Ar]3d (4)[FeCl ] Fe =[Ar]3d (5)[Ni(CO) ] Ni=[Ar]3d 4s (6)[Ni(CN) ] Ni =[Ar]3d Hence, the correct answer is option (c)