Question

The number of complexes among the following having exactly four unpaired electrons is \[ [Cr(H_2O)_6]^{2+}, \; [Mn(H_2O)_6]^{2+}, \; [Fe(H_2O)_6]^{2+}, \; [Co(H_2O)_6]^{3+}, \] \[ [Cu(H_2O)_6]^{2+}, \; [CoF_6]^{3-}, \; [Cr(CN)_6]^{4-}, \; [MnCl_4]^{2-} \]

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

For coordination compounds, always remember: \[ CN^- > NH_3 > H_2O > F^- > Cl^- \] in ligand field strength. Strong field ligands cause pairing, whereas weak field ligands generally produce high-spin complexes.

Answer 1

The Correct Option is C

Concept: The magnetic behavior of coordination compounds depends upon the number of unpaired electrons present in the d-orbitals of the central metal ion. To determine the number of unpaired electrons, we must:

• Determine the oxidation state of the metal.

• Determine the electronic configuration of the metal ion.

• Identify whether the ligand is weak field or strong field.

• Determine whether electron pairing occurs.

• Count the number of unpaired electrons.
Weak field ligands such as \[ H_2O,\;F^-,\;Cl^- \] usually produce high-spin complexes. Strong field ligands such as \[ CN^- \] usually produce low-spin complexes.

Step 1: Analyse \([Cr(H_2O)_6]^{2+}\) Oxidation state of chromium: \[ x+6(0)=+2 \] \[ x=+2 \] Electronic configuration: \[ Cr=[Ar]3d^54s^1 \] \[ Cr^{2+}=[Ar]3d^4 \] Water is a weak field ligand. Hence high-spin \(d^4\). Number of unpaired electrons: \[ 4 \] Thus, \[ [Cr(H_2O)_6]^{2+} \] has exactly four unpaired electrons.

Step 2: Analyse \([Mn(H_2O)_6]^{2+}\) \[ Mn^{2+} = 3d^5 \] Water is weak field. High-spin \(d^5\). Number of unpaired electrons: \[ 5 \] Therefore this complex is not counted.

Step 3: Analyse \([Fe(H_2O)_6]^{2+}\) \[ Fe^{2+} = 3d^6 \] Water is weak field. High-spin configuration. Number of unpaired electrons: \[ 4 \] Hence this complex contributes one count.

Step 4: Analyse \([Co(H_2O)_6]^{3+}\) \[ Co^{3+} = 3d^6 \] Because of the high oxidation state, pairing becomes significant. Number of unpaired electrons: \[ 4 \] Therefore this complex also contributes.

Step 5: Analyse \([Cu(H_2O)_6]^{2+}\) \[ Cu^{2+} = 3d^9 \] Number of unpaired electrons: \[ 1 \] Not counted.

Step 6: Analyse \([CoF_6]^{3-}\) \[ Co^{3+} = 3d^6 \] \(F^-\) is a weak field ligand. Therefore high-spin complex. Number of unpaired electrons: \[ 4 \] This complex is counted.

Step 7: Analyse \([Cr(CN)_6]^{4-}\) \[ Cr^{2+} = 3d^4 \] \(CN^-\) is a strong field ligand. Pairing occurs. Number of unpaired electrons: \[ 2 \] Not counted.

Step 8: Analyse \([MnCl_4]^{2-}\) \[ Mn^{2+} = 3d^5 \] Tetrahedral complexes are generally high spin. Number of unpaired electrons: \[ 5 \] Not counted.

Step 9: Count all complexes with four unpaired electrons \[ [Cr(H_2O)_6]^{2+} \] \[ [Fe(H_2O)_6]^{2+} \] \[ [Co(H_2O)_6]^{3+} \] \[ [CoF_6]^{3-} \] Total: \[ 4 \] \[ \boxed{4} \] Hence the correct answer is \[ \boxed{\text{Option (C)}} \]