Question

According to Valence Bond Theory, the hybridization of the central metal ion in [Ni(CN)$_4$]$^{2-}$ is:

• A. sp$^3$
• B. dsp$^2$
• C. d$^2$sp$^3$
• D. sp$^2$

Hint:

For complexes of coordination number 4:
If the metal is $\text{Ni}^{2+}$ ($d^8$ configuration) and the ligand is a strong-field ligand (like $\text{CN}^-$, CO), pairing occurs, resulting in $\text{dsp}^2$ (square planar, diamagnetic).
If the ligand is a weak-field ligand (like $\text{Cl}^-$, $\text{F}^-$), no pairing occurs, resulting in $\text{sp}^3$ (tetrahedral, paramagnetic).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question belongs to the topic "Coordination Compounds," specifically dealing with the Valence Bond Theory (VBT) to determine the hybridization, geometry, and magnetic behavior of a nickel coordination complex, $[\text{Ni}(\text{CN})_4]^{2-}$.

Step 2: Key Formula or Approach:
1. Identify the oxidation state of the central metal ion (Ni).
2. Write down the electronic configuration of the metal ion.
3. Analyze the field strength of the ligand ($\text{CN}^-$) to determine if pairing of electrons in the $d$-orbitals occurs.
4. Determine the vacant hybrid orbitals required to accommodate coordinate bonds from the ligands.

Step 3: Detailed Explanation:

• Let the oxidation state of Ni in the complex $[\text{Ni}(\text{CN})_4]^{2-}$ be $x$. Since CN has a charge of $-1$, we have:
\[ x + 4(-1) = -2 \implies x = +2 \]
Thus, nickel is in the $+2$ oxidation state ($\text{Ni}^{2+}$).

• The atomic number of Nickel is $28$. Its ground state electronic configuration is $[\text{Ar}] 3d^8 4s^2$.

• For the $\text{Ni}^{2+}$ ion, the configuration is $[\text{Ar}] 3d^8 4s^0$.

• In the presence of cyanide ($\text{CN}^-$), which is a strong-field ligand, spin-pairing of the $3d$ electrons occurs against Hund's rule.

• The eight $3d$ electrons, which occupied the five $3d$ orbitals leaving two unpaired, now pair up completely, occupying only four $3d$ orbitals.

• This leaves one $3d$ orbital empty, along with the vacant $4s$ and $4p$ orbitals:
Vacant orbitals available: one $3d$, one $4s$, and two $4p$ orbitals.

• These four vacant orbitals hybridize to form four equivalent $\text{dsp}^2$ hybrid orbitals.

• These four hybrid orbitals accept four lone pairs of electrons from the four cyanide ligands.

• The resulting geometry is square-planar, and since there are no unpaired electrons, the complex is diamagnetic.



Step 4: Final Answer:
The hybridization of the central metal ion in $[\text{Ni}(\text{CN})_4]^{2-}$ is $\text{dsp}^2$, which corresponds to option (B).