Question

The correct structure of the complex \([Ni(CN)_4]^{2-}\) and its magnetic property are

• A. Tetrahedral, Paramagnetic
• B. Square planar, Diamagnetic
• C. Tetrahedral, Diamagnetic
• D. Square planar, Paramagnetic

Hint:

A very important exception in coordination chemistry is: \[ [Ni(CN)_4]^{2-} \] which is a square planar, diamagnetic complex due to the strong-field nature of \(CN^{-}\). In contrast, \[ [NiCl_4]^{2-} \] contains the weak-field ligand \(Cl^{-}\), adopts tetrahedral geometry and is paramagnetic.

Answer 1

The Correct Option is B

Concept: The geometry and magnetic behaviour of a coordination compound depend upon:

• Oxidation state of the central metal ion.

• Electronic configuration of the metal ion.

• Nature of the ligand (strong field or weak field).

• Hybridization adopted by the central metal atom.
Strong field ligands such as \(CN^{-}\) produce a large crystal field splitting energy and tend to pair the electrons present in the d-orbitals. This often results in low-spin complexes with fewer or no unpaired electrons.

Step 1: Determine the oxidation state of nickel. Let the oxidation state of nickel be \(x\). For the complex \[ [Ni(CN)_4]^{2-} \] we have \[ x + 4(-1) = -2 \] \[ x - 4 = -2 \] \[ x = +2 \] Therefore, nickel is present as \[ Ni^{2+} \]

Step 2: Determine the electronic configuration of \(Ni^{2+}\). The atomic number of nickel is 28. Its ground state electronic configuration is \[ Ni=[Ar]\,3d^8\,4s^2 \] Removing two electrons to form \(Ni^{2+}\), \[ Ni^{2+}=[Ar]\,3d^8 \] Thus, the metal ion possesses a \(d^8\) electronic configuration.

Step 3: Examine the effect of the ligand \(CN^{-}\). The cyanide ion is a strong field ligand and occupies a high position in the spectrochemical series. Because of the strong ligand field produced by \(CN^{-}\), electrons in the d-orbitals pair up before occupying higher energy orbitals. For a \(d^8\) metal ion in the presence of a strong field ligand, pairing occurs and the complex prefers \(dsp^2\) hybridization.

Step 4: Determine the geometry of the complex. The \(dsp^2\) hybridization involves: \[ 1d + 1s + 2p \] orbitals. This hybridization produces a \[ \boxed{\text{Square Planar Geometry}} \] around the central metal ion.

Step 5: Determine the magnetic behaviour. In the square planar \(d^8\) configuration produced by a strong field ligand, all electrons become paired. Hence the number of unpaired electrons is \[ n=0 \] Since there are no unpaired electrons, the complex is \[ \boxed{\text{Diamagnetic}} \]

Final Answer Thus, the complex \[ [Ni(CN)_4]^{2-} \] is \[ \boxed{\text{Square Planar and Diamagnetic}} \] Hence, the correct option is \[ \boxed{\text{(B)}} \]