How many number of unpaired electrons are present in `Ca^(2+)` ?

Explanation

To determine the number of unpaired electrons in the ion \( \text{Ca}^{2+} \), we need to follow these steps:

### Step 1: Write the electronic configuration of neutral calcium (Ca).
Calcium (Ca) has an atomic number of 20. Therefore, its electronic configuration is:
\[ \text{Ca}: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 \]
This can also be abbreviated as:
\[ \text{Ca}: [\text{Ar}] 4s^2 \]
where [Ar] represents the electron configuration of Argon, which is \( 1s^2 2s^2 2p^6 3s^2 3p^6 \).

### Step 2: Determine the electronic configuration of \( \text{Ca}^{2+} \).
When calcium loses two electrons to form \( \text{Ca}^{2+} \), the two electrons are removed from the outermost shell (4s orbital). Thus, the electronic configuration for \( \text{Ca}^{2+} \) becomes:
\[ \text{Ca}^{2+}: [\text{Ar}] \]
This means that all the electrons in \( \text{Ca}^{2+} \) are now in the filled inner shells.

### Step 3: Analyze the electronic configuration for unpaired electrons.
In the configuration \( [\text{Ar}] \), all the electrons are paired. Since there are no electrons in the 4s orbital or any other higher energy orbitals, there are no unpaired electrons present.

### Final Answer
Therefore, the number of unpaired electrons in \( \text{Ca}^{2+} \) is:
\[ \text{Number of unpaired electrons} = 0 \]

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