Predict the total spin in ` Ni^(2+) ` ion :

Options

A. `+-(5)/(2)`
B. `+-(3)/(2)`
C. `+-(1)/(2)`
D. `+-1`

Answer

To predict the total spin in the Ni^(2+) ion, we can follow these steps:

### Step 1: Determine the electronic configuration of Ni
Nickel (Ni) has an atomic number of 28. The electronic configuration of neutral nickel is:
\[ \text{Ni: } [\text{Ar}] 3d^8 4s^2 \]

### Step 2: Account for the +2 oxidation state
When nickel loses 2 electrons to form Ni^(2+), the electrons are removed from the outermost shell first, which is the 4s orbital. Therefore, the electronic configuration for Ni^(2+) is:
\[ \text{Ni}^{2+}: [\text{Ar}] 3d^8 \]

### Step 3: Identify the number of unpaired electrons
In the 3d subshell, the distribution of electrons for 3d^8 is as follows:
- The 3d subshell can hold a maximum of 10 electrons and has 5 orbitals.
- For 8 electrons in the 3d subshell, the distribution will be:
- 2 electrons will pair up in two of the orbitals.
- 6 electrons will occupy the remaining 3 orbitals, with 2 of these orbitals containing 1 electron each (unpaired).

Thus, Ni^(2+) has **2 unpaired electrons** in the 3d subshell.

### Step 4: Calculate the total spin
The total spin (S) can be calculated using the formula:
\[ S = \frac{1}{2} \times n \]
where \( n \) is the number of unpaired electrons.

Substituting the value of unpaired electrons:
\[ S = \frac{1}{2} \times 2 = 1 \]

### Conclusion
The total spin for the Ni^(2+) ion is **1**.