Question

The shape of \( [Ni(CN)_4]^{2- \) is:}

• A. Tetrahedral
• B. Square planar
• C. Octahedral
• D. Trigonal bipyramidal

Hint:

Most \( d^8 \) metal ions with strong field ligands form square planar complexes.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The geometry of a coordination complex depends on the coordination number and the hybridization of the central metal atom.
For a coordination number of 4, the hybridization can be \(sp^3\) (tetrahedral) or \(dsp^2\) (square planar).
Step 2: Detailed Explanation:
In the complex ion \([Ni(CN)_4]^{2-}\):
1. The oxidation state of Nickel is \(+2\).
2. The electronic configuration of \(Ni\) is \([Ar] 3d^8 4s^2\), so \(Ni^{2+}\) is \([Ar] 3d^8\).
3. The ligands are cyanide ions (\(CN^-\)), which are very strong field ligands.
4. According to Valence Bond Theory (VBT), strong field ligands cause the electrons in the \(3d\) orbitals to pair up if possible.
5. The 8 electrons in the \(3d\) subshell usually occupy all five orbitals (three pairs and two singles). Under the influence of \(CN^-\), the two single electrons pair up, leaving one \(3d\) orbital empty.
6. The vacant \(3d\) orbital, the \(4s\) orbital, and two \(4p\) orbitals hybridize to form four \(dsp^2\) hybrid orbitals.
7. These four hybrid orbitals point toward the corners of a square.
Therefore, the geometry is square planar.
If the ligands were weak (like \(Cl^-\)), pairing wouldn't happen, hybridization would be \(sp^3\), and the shape would be tetrahedral.
Step 3: Final Answer:
The complex \([Ni(CN)_4]^{2-}\) has a square planar shape.