Match the complex ions given in Column I with the hybridization and number of unpaired electrons given in Column II and assign the correct code. 

Column l 

Column II

(Hybridisation, number of unpaired electrons)

A. Cr(H2O)63+

1. dsp2, 1

B.Co(CN)42-

2. sp3d2, 5

C.Ni(NH3)62+

3. d2sp3, 3

D.MnF64-

4. sp3d2, 2

Codes

          A         B          C           D

1.       2          3           4           1

2.       3          1           4           2

3.       5          4           3           2

4.       4          5           3           2

Hint: In the presence of strong field ligand pairing of electron takes place in metal d orbital.

Formation of inner orbital complex and outer orbital complex determines hybridization of the molecule which depends upon field strength of ligand and number of vacant d orbitals,

(i) Strong field ligand forms an inner orbital complex with hybridization d2sp3

(il) Weak field ligand forms an outer orbital complex with hybridization sp3d2

According to VBT, hybridization, and the number of unpaired electrons of coordination compounds can be calculated as

A. Cr(H2O)63+

The orbital diagram of Cr3+ in Cr(H2O63+ is as follows:

Hybridisation = d2sp3

n (number of unpaired electrons) =3

B.

 Co(CN)42-Orbital diagram of Co2+ in Co(CN)42- is                            

  

Hybridisation = dsp2

n=1

C.

Ni(NH3)62+

Orbital diagram of Ni2+ in Ni(NH3)62+ is 

Hybridisation = sp3d2

 n=2

D.

 MnF64-Orbital diagram of Mn2+ in MnF64 is 

Hybridization = sp3d2

  n= b

Hence, the correct choice is option 2.