In `NI(CO)_4`, electron configuration and the oxidation state of Ni are

Options

A. `3d^8 4s^2 , 0`
B. `3d^10 , 0`
C. `3d^8, + 2`
D. none of these

Explanation

To solve the question regarding the electron configuration and oxidation state of Nickel (Ni) in the coordination compound Ni(CO)₄, we can follow these steps:

### Step 1: Determine the oxidation state of Nickel (Ni)
In the coordination compound Ni(CO)₄, we need to find the oxidation state of Nickel.

- The carbonyl (CO) is a neutral ligand, meaning it has no charge.
- Since there are four CO ligands, the total charge contributed by them is \(4 \times 0 = 0\).
- The overall charge of the coordination compound is also neutral (0).

Let the oxidation state of Nickel be \(X\). The equation can be set up as follows:

\[ X + (4 \times 0) = 0 \]

This simplifies to:

\[ X = 0 \]

Thus, the oxidation state of Nickel in Ni(CO)₄ is 0.

### Step 2: Determine the electron configuration of Nickel (Ni)
Nickel has an atomic number of 28. The electron configuration of Nickel in its ground state is determined by filling the orbitals according to the Aufbau principle.

- The electron configuration can be written as follows:
- The first 18 electrons fill the orbitals up to Argon (Ar): \(1s^2 2s^2 2p^6 3s^2 3p^6\).
- The next 10 electrons fill the 4s and 3d orbitals: \(4s^2 3d^8\).

Thus, the complete electron configuration of Nickel is:

\[ \text{Ni: } [\text{Ar}] 4s^2 3d^8 \]

### Final Answer
- The electron configuration of Ni in Ni(CO)₄ is: \([Ar] 4s^2 3d^8\)
- The oxidation state of Ni in Ni(CO)₄ is: 0

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