Question

A coordination compound has the formula \[ [Co(NH_3)_4Cl_2]Cl \] The magnetic moment of the complex is found to be approximately \(4.9\ BM\). The hybridization and geometry of the complex ion are respectively:

• A. \(dsp^2\), square planar
• B. \(sp^3\), tetrahedral
• C. \(d^2sp^3\), octahedral
• D. \(sp^3d^2\), octahedral

Hint:

Remember: High-spin octahedral complex \[ \rightarrow sp^3d^2 \] Low-spin octahedral complex \[ \rightarrow d^2sp^3 \] Magnetic moment is often the key to distinguishing between them.

Answer 1

The Correct Option is D

Concept: This question integrates:
• Coordination compounds
• Oxidation state
• Crystal Field Theory
• Magnetic moment
• Hybridization Such integrated questions are extremely common in CUET.

Step 1: Determine the oxidation state of cobalt. The complex is: \[ [Co(NH_3)_4Cl_2]^+ \] because one chloride ion is present outside the coordination sphere. Let oxidation state of Co be \(x\). \[ x+4(0)+2(-1)=+1 \] \[ x-2=1 \] \[ x=+3 \] Therefore: \[ Co^{3+} \]

Step 2: Determine the electronic configuration. Atomic number of cobalt: \[ 27 \] Electronic configuration: \[ [Ar]\,3d^7\,4s^2 \] For: \[ Co^{3+} \] remove three electrons. \[ 3d^6 \] Thus: \[ Co^{3+} = d^6 \]

Step 3: Use magnetic moment data. Given: \[ \mu \approx4.9BM \] Using: \[ \mu = \sqrt{n(n+2)} \] where \(n\) is the number of unpaired electrons. Checking: \[ \sqrt{4(4+2)} = \sqrt{24} = 4.90 \] Thus: \[ n=4 \] Hence the complex contains four unpaired electrons.

Step 4: Determine field strength. Presence of four unpaired electrons indicates a weak-field situation. No pairing occurs. Therefore: \[ t_{2g}^{4}e_g^{2} \] High-spin arrangement.

Step 5: Determine hybridization. For high-spin octahedral complexes: \[ sp^3d^2 \] hybridization occurs using outer orbitals. Geometry remains: \[ \text{Octahedral} \]

Step 6: Eliminate remaining options. Option (A): Square planar complexes usually show \(dsp^2\). Not applicable. Option (B): Coordination number is 6, not 4. Incorrect. Option (C): Would represent low-spin inner orbital complex. Magnetic moment does not support this.

Step 7: Final conclusion. Hybridization: \[ sp^3d^2 \] Geometry: \[ \text{Octahedral} \] Hence: \[ \boxed{\text{Option (D)}} \]