The isoelectronic species are
1. `Na^(+)" " 2. Al^(3+)`
3. `Mg^(2+) " " 4. Ca^(2+)`

Options

A. 1,2 and 3
B. 1,3 and 4
C. 1,2 and 4
D. 1,2,3 and 4

Explanation

To determine which of the given species are isoelectronic, we need to find the number of electrons in each ion. Isoelectronic species are those that have the same number of electrons. Let's analyze each ion step by step.

### Step 1: Determine the number of electrons in Na⁺
- Sodium (Na) has an atomic number of 11, which means it has 11 protons and 11 electrons in its neutral state.
- Since Na⁺ has lost one electron, it will have:
\[ \text{Number of electrons in Na⁺} = 11 - 1 = 10 \]

### Step 2: Determine the number of electrons in Al³⁺
- Aluminium (Al) has an atomic number of 13, which means it has 13 protons and 13 electrons in its neutral state.
- Since Al³⁺ has lost three electrons, it will have:
\[ \text{Number of electrons in Al³⁺} = 13 - 3 = 10 \]

### Step 3: Determine the number of electrons in Mg²⁺
- Magnesium (Mg) has an atomic number of 12, which means it has 12 protons and 12 electrons in its neutral state.
- Since Mg²⁺ has lost two electrons, it will have:
\[ \text{Number of electrons in Mg²⁺} = 12 - 2 = 10 \]

### Step 4: Determine the number of electrons in Ca²⁺
- Calcium (Ca) has an atomic number of 20, which means it has 20 protons and 20 electrons in its neutral state.
- Since Ca²⁺ has lost two electrons, it will have:
\[ \text{Number of electrons in Ca²⁺} = 20 - 2 = 18 \]

### Conclusion
Now, let's summarize the number of electrons in each ion:
- Na⁺: 10 electrons
- Al³⁺: 10 electrons
- Mg²⁺: 10 electrons
- Ca²⁺: 18 electrons

The isoelectronic species are Na⁺, Al³⁺, and Mg²⁺, as they all have 10 electrons.

### Final Answer
The isoelectronic species are: **Na⁺, Al³⁺, and Mg²⁺.**

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