Question

A coordination compound \([CoF_6]^{3-}\) is converted into \([Co(CN)_6]^{3-\). Which of the following correctly represents the hybridization and number of unpaired electrons in the final complex?}

• A. \(sp^3d^2\), \(4\) unpaired electrons
• B. \(d^2sp^3\), \(0\) unpaired electrons
• C. \(sp^3d^2\), \(2\) unpaired electrons
• D. \(d^2sp^3\), \(2\) unpaired electrons

Hint:

For octahedral \(d^6\) complexes, strong field ligands such as \(CN^-\) usually produce low-spin diamagnetic complexes.

Answer 1

The Correct Option is B

Concept: The magnetic behaviour and hybridization of coordination compounds depend upon the electronic configuration of the metal ion and the field strength of the ligands. \(CN^-\) is a strong field ligand and produces electron pairing in the \(3d\) orbitals.

Step 1: Determine the oxidation state of cobalt. \[ x+6(-1)=-3 \] \[ x=+3 \] Therefore, \[ Co^{3+} \] Electronic configuration: \[ Co=[Ar]3d^74s^2 \] \[ Co^{3+}=[Ar]3d^6 \]

Step 2: Effect of cyanide ligand. Since \(CN^-\) is a strong field ligand, electrons pair up in the lower-energy \(t_{2g}\) orbitals. Configuration becomes: \[ t_{2g}^{6}e_g^{0} \]

Step 3: Determine magnetic nature. All six electrons become paired. \[ \text{Number of unpaired electrons}=0 \] Hence the complex is diamagnetic.

Step 4: Determine hybridization. Inner \(d\)-orbitals participate in bonding. \[ d^2sp^3 \] Thus the complex is an inner orbital octahedral complex. \[ \boxed{d^2sp^3,\;0\text{ unpaired electrons}} \]