69. From the magnetic behaviour of [NiCl4]2− (paramagnetic) and [Ni(CO)4] (diamagnetic), choose the correct geometry and oxidation state. Option 1: [NiCl4]2− : Ni2+, square planar; [Ni(CO)4] : Ni(0), square planar. Option 2: [NiCl4]2− : Ni2+, tetrahedral; [Ni(CO)4] : Ni(0), tetrahedral. Option 3: [NiCl4]2− : Ni2+, tetrahedral; [Ni(CO)4] : Ni2+, square planar. Option 4: [NiCl4]2− : Ni(0), tetrahedral; [Ni(CO)4] : Ni(0), square planar.

Key Concepts Needed

1. Oxidation State
   • For a complex $[ML_n]^{q}$, add ligand charges and overall charge to deduce metal oxidation state.

2. Crystal Field Theory (CFT)
   • Octahedral vs. tetrahedral vs. square-planar splitting patterns.
   • Weak-field ligands (Cl⁻, Br⁻, I⁻, H₂O) cause small splitting ($\Delta$) → high-spin → more unpaired electrons → paramagnetic.
   • Strong-field ligands (CN⁻, CO, phosphines) cause large splitting → low-spin → electrons pair → diamagnetic if all electrons paired.

3. Magnetic Behavior
   • Paramagnetic: at least one unpaired electron (measured by $\mu = \sqrt{n(n+2)}$ BM).
   • Diamagnetic: all electrons paired, $\mu \approx 0$.

4. Ni Electronic Configuration
   • Ground state Ni atom: $[Ar]\,3d^8\,4s^2$.
   • $Ni^{2+}$: loses two 4s electrons → $3d^8$.
   • $Ni(0)$: keeps 4s electrons but often promoted/hybridized for bonding.

Reasoning Path

1. Find oxidation states:
   • In $[NiCl_4]^{2-}$: each $Cl^-$ is $-1$. Total ligand charge = $-4$. Overall complex charge = $-2$ → $\text{Oxidation state of Ni} = x = -2 +4 = +2$
   • In $[Ni(CO)_4]$: CO is neutral. Complex is neutral → $x = 0$
   So, $[NiCl_4]^{2-}$ has $Ni^{2+}$, $[Ni(CO)_4]$ has $Ni(0)$.

2. Predict geometry using ligand strength and magnetism:
   • $Ni^{2+}\,(3d^8)$ in a tetrahedral field ($Cl^-:$ weak-field) gives configuration $e^4 t_2^4$ with two unpaired electrons → paramagnetic.
   • Square-planar $Ni^{2+}$ would require strong-field ligands and yield diamagnetic low-spin ($d_{x^2-y^2}$ empty). That contradicts the observed paramagnetism. Therefore $[NiCl_4]^{2-}$ must be tetrahedral.

   • For $Ni(0)$ with strong CO ligands, hybridisation $sp^3$ gives a tetrahedral molecule. All 3d orbitals are filled (10 electrons) and 4s/4p electrons donate into CO π* back-bonding, leaving no unpaired electrons → diamagnetic.

   Square-planar $Ni(0)$ would have unusual electron count (16e) and is less stable than tetrahedral 18-electron rule satisfaction.

3. Match with options:
   Only Option 2 states:
   • $[NiCl_4]^{2-}$: $Ni^{2+}$, tetrahedral
   • $[Ni(CO)_4]$: $Ni(0)$, tetrahedral
   which fits the oxidation states and magnetic data.

What’s going on here?

Imagine nickel (Ni) is like a house with rooms (orbitals).
• Strong guests (strong-field ligands such as CO) make the original residents pair up in the lower rooms so everyone fits downstairs. No one is left roaming, so the house looks very calm (diamagnetic).
• Weak guests (weak-field ligands such as Cl⁻) don’t insist on pairing; residents stay spread out in both upstairs and downstairs rooms, some remain unpaired and run around (paramagnetic).

Because of the way the rooms are shaped, if the house is tetrahedral (4 corners like a pyramid) the splitting between rooms is small, guests are weak, and people don’t bother pairing. If the house is square-planar (flat like a board), splitting is large, so people pair up when the guest is strong.

So:

• [NiCl₄]²⁻ has weak guests → small splitting → unpaired electrons → paramagnetic → tetrahedral.
• [Ni(CO)₄] has strong guests → big splitting → all electrons paired → diamagnetic → square-planar? Wait! In the special case of Ni(0) with CO, it actually forms tetrahedral but remains diamagnetic because Ni uses its 4s and 4p orbitals for bonding, leaving its 3d full and paired.

Put the clues together and you’ll get the right choice!

Step-by-Step Solution

1. Oxidation State of Ni in each complex

   [NiCl_4]^{2-}:\;&4(Cl^{-}) = -4\\ \text{Overall charge} = -2\\ x + (-4) = -2 \Rightarrow x = +2\\[5pt] [Ni(CO)_4]:\;&CO\text{ is neutral}\\ \text{Overall charge} = 0 \Rightarrow x = 0 \end{aligned}$$ So, $Ni^{2+}$ in the first, $Ni(0)$ in the second.

2. Determine geometry and magnetic behavior For $Ni^{2+}$ ($3d^8$): • Tetrahedral (weak field Cl⁻) configuration: e^{4}_{(\uparrow\downarrow\uparrow\downarrow)}\,t_2^{4}_{(\uparrow\downarrow\uparrow\downarrow)} → 2 unpaired electrons → paramagnetic (observed).
   • Square-planar (strong field) would give all paired → diamagnetic (contradicts observation).

   For $Ni(0)$ ($3d^{10}$): • CO is strong field. Hybridisation $sp^3$ gives tetrahedral $[Ni(CO)_4]$ with filled 3d and donated 4s/4p electrons. All electrons paired → diamagnetic (observed).

3. Compare with given options Option 2 exactly matches: • $[NiCl_4]^{2-}$: $Ni^{2+}$, tetrahedral, paramagnetic.
   • $[Ni(CO)_4]$: $Ni(0)$, tetrahedral, diamagnetic.

Therefore, the correct answer is Option 2.