**69.** From the magnetic behaviour of \( [\ce{NiCl4}]^{2-} \) (paramagnetic) and \( [\ce{Ni(CO)4}] \) (diamagnetic), choose the correct geometry and oxidation state. **Options:** (1) \( [\ce{NiCl4}]^{2-} \) : Ni\(^{2+} \), square planar \( [\ce{Ni(CO)4}] \) : Ni(0), square planar (2) \( [\ce{NiCl4}]^{2-} \) : Ni\(^{2+} \), tetrahedral \( [\ce{Ni(CO)4}] \) : Ni(0), tetrahedral (3) \( [\ce{NiCl4}]^{2-} \) : Ni\(^{2+} \), tetrahedral \( [\ce{Ni(CO)4}] \) : Ni\(^{2+} \), square planar (4) \( [\ce{NiCl4}]^{2-} \) : Ni(0), tetrahedral \( [\ce{Ni(CO)4}] \) : Ni(0), square planar

Key Concepts Needed

1. Oxidation State

   • Chloride is (-1), the whole complex is (-2). So Ni must be (+2) in $[\text{NiCl}_4]^{2-}$.
   • Carbonyl (CO) is a neutral ligand (0 charge). $[\text{Ni(CO)}_4]$ is neutral, so Ni is in the 0 oxidation state.

2. Ligand Field Strength & Magnetic Behaviour

   • Weak-field ligands (e.g. Cl$^-$, F$^-$, OH$^-$) usually cause small splitting between $e$ and $t_2$ (tetrahedral) or $t_{2g}$ and $e_g$ (octahedral) sets. Electrons remain unpaired ⇒ paramagnetism.
   • Strong-field ligands (e.g. CO, CN$^-$) cause a large crystal field splitting, favouring electron pairing ⇒ diamagnetism.

3. Geometry Preference of $d^8$ Systems

   • Ni$^{2+}$ is $d^8$. In tetrahedral fields, splitting ($\Delta_t$) is smaller, so remaining unpaired (paramagnetic) is common. In square planar fields, splitting is very large; electrons pair, making the complex diamagnetic.
   • Experimental clue: $[\text{NiCl}_4]^{2-}$ is paramagnetic, so the splitting cannot be large ⇒ it is tetrahedral, not square planar.

4. Electronic Configuration of Ni(0)

   • Ni(0) is $d^{10}$. A completely filled $d$ shell is already paired, giving diamagnetism regardless of geometry. $[\text{Ni(CO)}_4]$ is known from experiments and textbooks to be tetrahedral.

Logical Chain a Student Should Use

1. Determine oxidation state from charges.
2. Convert oxidation state to $d$ electron count (Ni is atomic number 28).
3. Use ligand field strength to predict whether the electrons stay paired or unpaired.
4. Match paramagnetic/diamagnetic data with possible geometries.
5. Check which option matches every condition.

What is the question?

We have two nickel compounds. One behaves like a magnet (paramagnetic) and the other does not (diamagnetic). From this information, we must guess two things for each compound:

1. How many positive charges are really on the nickel atom (oxidation state).
2. What shape the four surrounding ligands make around nickel (square‐planar like a flat square or tetrahedral like a four-corner pyramid).

How to think like a 10-year-old scientist

• Imagine an octopus (the nickel ion) with 8 arms (its $d$-electrons).
• If the arms are forced into a small space by strong ropes (strong-field ligands like CO), they pair up and stay calm—no magnetic antics (diamagnetic).
• If the ropes are loose (weak-field ligands like Cl$^-$), the arms wiggle freely and you can feel a magnetic pull (paramagnetic).
• A calm octopus can happily sit in either a flat square or a pyramid, but the magnetic, wiggly octopus usually prefers the pyramid (tetrahedral) because it needs elbow room.
• CO does not add or remove charge from Ni, so the octopus stays neutral (Ni(0)).
• Each Cl$^-$ steals one positive charge from Ni, so Ni becomes Ni$^{2+}$ in $[\text{NiCl}_4]^{2-}$.

Putting it together:

• Wiggly octopus $[\text{NiCl}_4]^{2-}$ → Ni$^{2+}$, tetrahedral.
• Calm octopus $[\text{Ni(CO)}_4]$ → Ni(0), also tetrahedral but without magnetism.

Hence Option (2) is correct.

Step 1: Oxidation States

For $[\text{NiCl}_4]^{2-}$:
Let oxidation state of Ni be $x$.
Each Cl$^-$ contributes (-1).
$x + 4(-1) = -2 \Rightarrow x = +2$

For $[\text{Ni(CO)}_4]$:
CO is neutral. Overall charge 0.
$x + 4(0) = 0 \Rightarrow x = 0$

Step 2: $d$-Electron Count

Ni atomic number 28 ⇒ ground-state $[Ar]3d^84s^2$.
Ni$^{2+}$ ⇒ $d^8$ (lost two 4s electrons).
Ni(0) ⇒ $d^{10}$ (keeps all $d$ electrons).

Step 3: Use Magnetic Data

• $[\text{NiCl}_4]^{2-}$ is paramagnetic ⇒ unpaired electrons are present.
• $[\text{Ni(CO)}_4]$ is diamagnetic ⇒ all electrons are paired.

Step 4: Geometry Decision for Ni$^{2+}$ ($d^8$)

• Square planar $d^8$ complexes (e.g. with CN$^-$, CO) are usually diamagnetic because large splitting forces pairing.
• Tetrahedral $d^8$ complexes formed with weak-field ligands (Cl$^-$) have smaller splitting, leaving two unpaired electrons ⇒ paramagnetic. Hence $[\text{NiCl}_4]^{2-}$ must be tetrahedral.

Step 5: Geometry for Ni(0) ($d^{10}$)

$d^{10}$ configuration is inherently paired, so both tetrahedral and square planar would be diamagnetic.
Experimentally and per 18-electron rule, $[\text{Ni(CO)}_4]$ is tetrahedral.

Conclusion

$[\text{NiCl}_4]^{2-}:\; \text{Ni}^{2+},\; \text{tetrahedral}$ $[\text{Ni(CO)}_4]:\; \text{Ni(0)},\; \text{tetrahedral}$

Therefore the correct option is Option (2).