Match the column I with column II and mark the appropriate choice .

Options

A. `(A) to (ii) , (B) to (iii) , (C) to (iv) , (D) to (i) `
B. `(A) to (iii) , (B) to (ii) , (C) to (i), (D) to (iv) `
C. `(A) to (i) , (B) to (iii) , (C) to (iv) , (D) to (ii)`
D. `(A) to (iv) , (B) to (iii) , (C) to (ii) , (D) to (i)`

Answer

D

Explanation

`Fe^(3+) : 3d^(5)` , since `CN^(-)` is a strong field ligand , pairing will take place .
`[Fe(CN)_(6)]^(3+) : (##NCERT_OBJ_FING_CHE_XII_C09_E01_073_S01##) " " [ because n = 1]`
`mu = sqrt(n(n+2)) = sqrt(1(1 + 2)) = sqrt3 = 1.732` B.M.
In `[CoF_(6)]^(3-)` , oxidation state of Co = + 3
`Co^(3+) : 3d^(6)` , since `F^(-)` is a weak field ligand , pairing will not occur .
`[CoF_(6)]^(3-) : (##NCERT_OBJ_FING_CHE_XII_C09_E01_073_S02##) " " [because n = 4]`
`mu = sqrt(4(4 + 2)) = sqrt(24) = 4.89` B.M.
In `[Fe(H_(2)O)_(6)]^(3+)` , oxidation state of Fe = +3
`Fe^(3+) : 3d^(5)` , since , `H_(2)O` is a weak field ligand , pairing will not take place .
`[Fe(H_(2)O)_(6)]^(3+) : (##NCERT_OBJ_FING_CHE_XII_C09_E01_073_S03##) " " [ because n = 5]`
`mu = sqrt(5(5 + 2)) = sqrt(35) = 5.92 B.M.`
In `[Co(NH_(3))_(6)]^(3+) ` , oxidation state of Co = + 3
`Co^(3+) : 3d^(6)` , since `NH_(3)` is a strong field ligand , pairing will take place .
`[Co(NH_(3))_(6)]^(3+) : (##NCERT_OBJ_FING_CHE_XII_C09_E01_073_S04##) " " [ because n = 0]`
`mu = sqrt(0(0 + 2)) = 0`