Match the complex given in List-I with type of hybridisation given in…

The correct answer is Option (1) → (A)-(IV), (B)-(II), (C)-(III), (D)-(I)

In valence bond theory the hybridisation of a complex depends on the number of d electrons of the metal and on the strength of the ligand. A strong field ligand (CO, \( \ce{CN-} \), \( \ce{NH3} \) with \( \ce{Co^{3+}} \)) pairs up the d electrons and frees inner 3d orbitals, giving dsp² or d²sp³. A weak field ligand (\( \ce{F-} \)) leaves the electrons unpaired, so outer 4d orbitals are used, giving sp³d².

• (A) \( \ce{[Ni(CO)4]} \) — This matches (IV) sp³. CO is neutral, so Ni is 0; Ni (Z = 28) is \( \mathrm{[Ar]\,3d^{8}\,4s^{2}} \). CO pushes the two 4s electrons into 3d (\( \mathrm{3d^{10}} \)), and the empty 4s and three 4p orbitals give a tetrahedral, diamagnetic complex.

• (B) \( \ce{[Ni(CN)4]^2-} \) — This matches (II) dsp². x + 4(−1) = −2, so x = +2; \( \ce{Ni^{2+}} \) is \( \mathrm{3d^{8}} \) (both 4s electrons lost). \( \ce{CN-} \) pairs the eight electrons into four 3d orbitals; one 3d, the 4s and two 4p orbitals give a square planar complex.

• (C) \( \ce{[Co(NH3)6]^3+} \) — This matches (III) d²sp³. \( \ce{NH3} \) is neutral, so Co is +3; Co (Z = 27) is \( \mathrm{[Ar]\,3d^{7}\,4s^{2}} \), and \( \ce{Co^{3+}} \) is \( \mathrm{3d^{6}} \). \( \ce{NH3} \) pairs the six electrons into three 3d orbitals; the two free 3d, the 4s and three 4p orbitals give an inner orbital octahedral complex (diamagnetic).

• (D) \( \ce{[CoF6]^3-} \) — This matches (I) sp³d². x + 6(−1) = −3, so x = +3, again \( \mathrm{3d^{6}} \). \( \ce{F-} \) is a weak field ligand, so four electrons stay unpaired; the 4s, three 4p and two 4d orbitals give an outer orbital, high spin octahedral complex.

Shortcut: \( \ce{[Ni(CN)4]^2-} \) being square planar gives (B)-(II), found only in Option (1).

NCERT (2026-27 edition):
Valence Bond Theory

Table 5.2: Number of Orbitals and Types of Hybridisations