[`PdCl_(2)(PM e_(3))_(2)]` is a diamagnetic complex of Pd (II). How many unpaired electrons are presennt in analogous complex of Ni (II)?

Options

A. Zero
B. 1
C. 2
D. 3

Answer

To solve the question regarding the number of unpaired electrons in the analogous complex of Ni(II) based on the given diamagnetic complex of Pd(II), we can follow these steps:

### Step 1: Understand the given complex
The complex given is \[PdCl_2(PMe_3)_2\]. Palladium (Pd) is in the +2 oxidation state, which means it has lost two electrons.

### Step 2: Determine the electron configuration of Pd(II)
Palladium has an atomic number of 46. The electron configuration of neutral Pd is \[ [Kr] 4d^{10} 5s^0 \]. For Pd(II), we remove two electrons from the 4d subshell:
- Pd(II) configuration: \[ [Kr] 4d^8 \]

### Step 3: Analyze the geometry and hybridization
The complex \[PdCl_2(PMe_3)_2\] is square planar due to the presence of weak field ligands (Cl and PMe3). In a square planar arrangement, the d-orbitals split into two sets: one set of lower energy (dxy, dxz, dyz) and one set of higher energy (dx2-y2, dz2).

### Step 4: Determine the number of unpaired electrons in Pd(II)
Since the complex is diamagnetic, all electrons in the d-orbitals are paired. Therefore, for Pd(II) in this complex, there are **0 unpaired electrons**.

### Step 5: Determine the electron configuration of Ni(II)
Nickel has an atomic number of 28. The electron configuration of neutral Ni is \[ [Ar] 3d^8 4s^2 \]. For Ni(II), we remove two electrons from the 4s subshell:
- Ni(II) configuration: \[ [Ar] 3d^8 \]

### Step 6: Analyze the geometry and hybridization of Ni(II) complex
In the analogous complex of Ni(II), we can assume a similar ligand environment as the Pd complex, which is likely to be square planar as well. However, since Ni(II) is a 3d element, the splitting of d-orbitals may differ from that of Pd.

### Step 7: Determine the number of unpaired electrons in Ni(II)
For Ni(II) with a configuration of \[3d^8\]:
- The d-orbitals will fill as follows:
- Each of the five d-orbitals will receive one electron first (5 electrons), and then pairing will begin.
- After filling 5 orbitals with 5 electrons, we have 3 remaining electrons to fill the orbitals, resulting in:
- 2 orbitals will contain paired electrons, and 2 orbitals will contain one unpaired electron each.

Thus, for Ni(II), there are **2 unpaired electrons**.

### Final Answer
The analogous complex of Ni(II) has **2 unpaired electrons**.

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