Question

Match List - I with List - II:

List - I:

(A) \([ \text{MnBr}_4]^{2-}\) 
(B) \([ \text{FeF}_6]^{3-}\) 
(C) \([ \text{Co(C}_2\text{O}_4)_3]^{3-}\) 
(D) \([ \text{Ni(CO)}_4]\) 

List - II: 

(I) d²sp³ diamagnetic 
(II) sp²d² paramagnetic 
(III) sp³ diamagnetic 
(IV) sp³ paramagnetic

• A. (A)-(III),(B)-(II),(C)-(I),(D)-(IV)
• B. (A)-(IV),(B)-(I),(C)-(II),(D)-(III)
• C. (A)-(I),(B)-(II),(C)-(III),(D)-(IV)
• D. (A)-(IV),(B)-(I),(C)-(III),(D)-(II)

Hint:

To determine the hybridization and magnetic properties, consider the oxidation state and geometry of the metal center in the complex.

Answer 1

The Correct Option is A

To solve the problem of matching List-I with List-II, we need to determine the hybridization of each complex and their magnetic properties:

1. For \([ \text{MnBr}_4]^{2-}\):
   Manganese in \([ \text{MnBr}_4]^{2-}\) has a coordination number of 4 and contains no unpaired electrons; thus, it is diamagnetic. The hybridization of this complex is sp³.
2. For \([ \text{FeF}_6]^{3-}\):
   The complex \([ \text{FeF}_6]^{3-}\) involves \(\text{Fe}^{3+}\), which has a coordination number of 6. Fluoride is a weak field ligand, resulting in high spin, thus with unpaired electrons and is paramagnetic. The hybridization is sp²d².
3. For \([ \text{Co(C}_2\text{O}_4)_3]^{3-}\):
   In this case, \(\text{Co}^{3+}\) forms low-spin complexes with oxalate as a ligand. It has no unpaired electrons hence is diamagnetic and features d²sp³ hybridization.
4. For \([ \text{Ni(CO)}_4]\):
   This complex contains \(\text{Ni}^{0}\) which forms a tetrahedral complex with carbonyl as a ligand, resulting in unpaired electrons. Therefore, it is paramagnetic with sp³ hybridization.

Thus, the correct matches are:

List-I	List-II
(A) \([ \text{MnBr}_4]^{2-}\)	(III) sp³ diamagnetic
(B) \([ \text{FeF}_6]^{3-}\)	(II) sp²d² paramagnetic
(C) \([ \text{Co(C}_2\text{O}_4)_3]^{3-}\)	(I) d²sp³ diamagnetic
(D) \([ \text{Ni(CO)}_4]\)	(IV) sp³ paramagnetic

The correct answer is:

(A)-(III),(B)-(II),(C)-(I),(D)-(IV)