Question

The hybridisation states of Ni in the 3 complexes Ni(CO)\(_4\), [Ni(NH\(_3\))\(_6\)]\(^{2+}\), [Ni(CN)\(_4\)]\(^{2-}\) are

• A. \( \text{dsp}^2, \text{sp}^3\text{d}^2, \text{sp}^3 \)
• B. \( \text{sp}^3, \text{sp}^3\text{d}^2, \text{sp}^3 \)
• C. \( \text{sp}^3, \text{sp}^3\text{d}^2, \text{sp}^3 \)
• D. \( \text{sp}^3, \text{d}^2\text{sp}^3, \text{dsp}^2 \)

Answer 1

The Correct Option is C

Step 1: Hybridisation of Ni in Ni(CO)\(_4\).
In \( \text{Ni(CO)}_4 \), Nickel forms four bonds with four CO molecules. This is a tetrahedral structure, and Nickel uses \( \text{sp}^3 \) hybridisation to form the bonds. Step 2: Hybridisation of Ni in [Ni(NH\(_3\))\(_6\)]\(^{2+}\).
In \( [\text{Ni(NH}_3\text{)}_6]^{2+} \), Nickel is in a +2 oxidation state. With six ligands, the geometry is octahedral, and Nickel undergoes \( \text{sp}^3\text{d}^2 \) hybridisation. Step 3: Hybridisation of Ni in [Ni(CN)\(_4\)]\(^{2-}\).
In \( [\text{Ni(CN)}_4]^{2-} \), Nickel is in a +2 oxidation state with four cyanide ligands. The geometry is square planar, so Nickel undergoes \( \text{sp}^3 \) hybridisation. Step 4: Conclusion.
Therefore, the hybridisation states are \( \text{sp}^3 \), \( \text{sp}^3\text{d}^2 \), and \( \text{sp}^3 \), corresponding to option (C).
Final Answer: (C) \( \text{sp}^3, \text{sp}^3\text{d}^2, \text{sp}^3 \)