Question

(a) Draw the geometrical isomers of the complex \([ \text{Pt(NH}_3)_2\text{Cl}_2 ]\).

Hint:

Presence of unpaired electrons and allowed d-d transitions cause colour in complexes. CN\(^{-}\) induces strong pairing.

Answer 1

(a) This is a square planar complex of type MA\(_2\)B\(_2\). Two geometrical isomers are possible: - **Cis-isomer:** both Cl ligands adjacent - **Trans-isomer:** Cl ligands opposite \[ \boxed{ \text{Cis-}\ce{[Pt(NH3)2Cl2]} \quad \text{and} \quad \text{Trans-}\ce{[Pt(NH3)2Cl2]} } \] (b) Give the electronic configuration of \( d^4 \) ion when \(\Delta_0>P\).
Solution:
(b) When \(\Delta_0>P\), the ligand field is strong → **low-spin configuration**
\[ d^4 \Rightarrow \text{All electrons pair in } t_{2g} \text{ first: } t_{2g}^4e_g^0 \] \[ \boxed{\text{Low-spin: } t_{2g}^4 e_g^0} \] (c) Solution of \([ \text{Ni(H}_2\text{O)}_6 ]^{2+}\) is green in colour, whereas \([ \text{Ni(CN)}_4 ]^{2-}\) is colourless. Give reason.
(Ni = 28) Solution:
(c) - In \([ \text{Ni(H}_2\text{O)}_6 ]^{2+}\): Weak field ligand → octahedral, \( d^8 \) → partially filled d-orbitals → d-d transition allowed → green color. - In \([ \text{Ni(CN)}_4 ]^{2-}\): Strong field ligand (CN\(^{-}\)) → low-spin, square planar → all electrons paired → d-d transition not allowed → colourless. \[ \boxed{\text{Colour arises due to d-d transition. Strong field ligand prevents it.}} \]