Answers: Coordination Compounds Worksheet

 

Answers: Coordination Compounds Worksheet

Section A: Objective Questions

1. Hexaamminecobalt(III) ion
2. Linkage isomerism
3. Oxidation state of Cr = +3
4. Ligand = Cyanide (CN⁻)
5. Coordination number = 6

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Section B: Short Answer Questions

6. Structures of geometrical isomers of [Pt(NH₃)₂Cl₂]:

   • Cis-isomer: Both Cl atoms adjacent to each other.
   • Trans-isomer: Cl atoms opposite to each other.

7. $\text{[Fe(CN}_6\text{]}^{3-}$ is low spin because CN⁻ is a strong field ligand, causing pairing of electrons.
   $\text{[FeF}_6\text{]}^{3-}$ is high spin because F⁻ is a weak field ligand, so no pairing occurs.

8. Formula: $\text{[Cu(NH₃)}_4\text{]SO₄}$

9. In $\text{[Ni(CO)}_4\text{]}$, CO is a strong field ligand, causing pairing of 3d electrons. Ni forms dsp² hybrid orbitals, making the complex diamagnetic.

10. Significance of CFSE:

• Determines the stability of complexes.
• Influences properties like color, magnetic behavior, and reactivity.

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Section C: Long Answer Questions

(a) Crystal Field Splitting in Octahedral Field:

• Splits d-orbitals into $t_{2g}$ (lower energy) and $e_g$ (higher energy).
• Energy difference = $\Delta_o$.

(b) Weak vs. Strong Field Ligands:

• Weak field ligands: Small $\Delta_o$, high-spin complexes.
• Strong field ligands: Large $\Delta_o$, low-spin complexes.

• Preparation: $\text{[CoF}_6\text{]}^{3-}$ forms by reacting Co³⁺ with F⁻ ligands.
• Structure: Octahedral.
• Magnetic Properties: High spin due to F⁻ being a weak field ligand, paramagnetic.

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Section D: Assertion-Reason Questions

13. (a) Both A and R are true, and R is the correct explanation of A.

14. (a) Both A and R are true, and R is the correct explanation of A.