Question

(A) \([\mathrm{MnBr_4}]^{2-}\)
(B) \([\mathrm{Cu(H_2O)_6}]^{2+}\)
(C) \([\mathrm{Ni(CN)_4}]^{2-}\)
(D) \([\mathrm{Ni(H_2O)_6}]^{2+}\)
Select correct order of spin-only magnetic moment among above complexes.

• A. A>D>B>C
• B. D>A>C>B
• C. D>B>A>C
• D. A>B>D>C

Hint:

Spin-only magnetic moment depends {only on unpaired electrons}. Strong-field ligands like \(\mathrm{CN^-}\) often cause pairing and reduce magnetism.

Answer 1

The Correct Option is A

Concept:
Spin-only magnetic moment depends on the number of unpaired electrons: \[ \mu = \sqrt{n(n+2)}\ \text{BM} \] where \(n\) is the number of unpaired electrons. Ligand field strength determines whether electrons pair (low spin) or remain unpaired (high spin).
Step 1: Determine Oxidation State, Geometry, and Spin
(A) \([\mathrm{MnBr_4}]^{2-}\):


Mn oxidation state: \(+2\) \(\Rightarrow d^5\)
\(\mathrm{Br^-}\) is a weak-field ligand \(\Rightarrow\) high spin
Geometry: tetrahedral
Unpaired electrons \(n = 5\)
(D) \([\mathrm{Ni(H_2O)_6
]^{2+}\):}

Ni oxidation state: \(+2\) \(\Rightarrow d^8\)
\(\mathrm{H_2O}\) is a weak-field ligand \(\Rightarrow\) high spin
Geometry: octahedral
Unpaired electrons \(n = 2\)
(B) \([\mathrm{Cu(H_2O)_6
]^{2+}\):}

Cu oxidation state: \(+2\) \(\Rightarrow d^9\)
One unpaired electron
Unpaired electrons \(n = 1\)
(C) \([\mathrm{Ni(CN)_4
]^{2-}\):}

Ni oxidation state: \(+2\) \(\Rightarrow d^8\)
\(\mathrm{CN^-}\) is a strong-field ligand \(\Rightarrow\) low spin
Geometry: square planar
All electrons paired \(\Rightarrow n = 0\)

Step 2: Compare Spin-only Magnetic Moments
\[ n:\quad A(5)>D(2)>B(1)>C(0) \] \[ \boxed{A>D>B>C} \]