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 match List-I with List-II, we analyze the hybridization and magnetic properties of each complex:

1. For \([ \text{MnBr}_4]^{2-}\):
   Manganese in \([ \text{MnBr}_4]^{2-}\) has a coordination number of 4 and no unpaired electrons, making it diamagnetic with sp³ hybridization.
2. For \([ \text{FeF}_6]^{3-}\):
   The \(\text{Fe}^{3+}\) complex \([ \text{FeF}_6]^{3-}\) has a coordination number of 6. Fluoride is a weak field ligand, leading to a high-spin configuration with unpaired electrons, thus it is paramagnetic. The hybridization is sp²d².
3. For \([ \text{Co(C}_2\text{O}_4)_3]^{3-}\):
   \(\text{Co}^{3+}\) forms low-spin complexes with oxalate. This complex has no unpaired electrons, is diamagnetic, and exhibits d²sp³ hybridization.
4. For \([ \text{Ni(CO)}_4]\):
   This complex features \(\text{Ni}^{0}\) forming a tetrahedral structure with carbonyl ligands. It possesses unpaired electrons, rendering it paramagnetic with sp³ hybridization.

The corresponding 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 pairings are:

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