Question

The number of unpaired electrons calculated in \(\{Co(NH_3)_6\}^{3+}\) and \(\{CoF_6\}^{3-}\) are

• A. 4 and 4
• B. 0 and 2
• C. 2 and 4
• D. 0 and 4

Hint:

Strong field ligands (NH\(_3\), CN\(^-\)) form low spin complexes; weak field ligands (F\(^-\), Cl\(^-\)) form high spin complexes.

Answer 1

The Correct Option is D

Step 1: Find oxidation state and electronic configuration of Co.
Cobalt (Z = 27):
\[ Co: [Ar]\,3d^7\,4s^2 \]
Step 2: For \([Co(NH_3)_6]^{3+\).}
Oxidation state of Co is \(+3\).
\[ Co^{3+}: [Ar]\,3d^6 \]
NH\(_3\) is a strong field ligand, so it causes pairing (low spin).
So \(d^6\) in octahedral low spin:
\[ t_{2g}^6 e_g^0 \]
Unpaired electrons = \(0\).
Step 3: For \([CoF_6]^{3-}\).
Let oxidation state of Co be \(x\).
\[ x + 6(-1) = -3 \Rightarrow x = +3 \]
So again \(Co^{3+}: 3d^6\).
F\(^-\) is a weak field ligand, so complex is high spin.
In high spin octahedral \(d^6\):
\[ t_{2g}^4 e_g^2 \]
Unpaired electrons = \(4\).
Step 4: Final conclusion.
So unpaired electrons are \(0\) and \(4\).
Final Answer:
\[ \boxed{\text{(D) 0 and 4}} \]