Isoelectronic speices are :
`a.Na^(+) " " b.AI^(3+) " " c.Mg^(2+) " " d.Ca^(2+)`

Options

A. a,b and c
B. a,c and d
C. a,b and d
D. a,b,c and d

Explanation

To determine which of the given ions are isoelectronic species, we need to find out how many electrons each ion has after accounting for their charge.

### Step-by-Step Solution:

1. **Identify the number of electrons in each ion:**
- **Na⁺ (Sodium ion):**
- Atomic number of Na = 11
- Na⁺ has lost 1 electron: 11 - 1 = 10 electrons.

- **Al³⁺ (Aluminum ion):**
- Atomic number of Al = 13
- Al³⁺ has lost 3 electrons: 13 - 3 = 10 electrons.

- **Mg²⁺ (Magnesium ion):**
- Atomic number of Mg = 12
- Mg²⁺ has lost 2 electrons: 12 - 2 = 10 electrons.

- **Ca²⁺ (Calcium ion):**
- Atomic number of Ca = 20
- Ca²⁺ has lost 2 electrons: 20 - 2 = 18 electrons.

2. **Compare the number of electrons:**
- Na⁺ = 10 electrons
- Al³⁺ = 10 electrons
- Mg²⁺ = 10 electrons
- Ca²⁺ = 18 electrons

3. **Identify isoelectronic species:**
- Isoelectronic species are those that have the same number of electrons.
- From the calculations, Na⁺, Al³⁺, and Mg²⁺ all have 10 electrons, making them isoelectronic.
- Ca²⁺ has 18 electrons and is not isoelectronic with the others.

4. **Conclusion:**
- The isoelectronic species among the given options are Na⁺, Al³⁺, and Mg²⁺.

### Final Answer:
The isoelectronic species are: **a. Na⁺, b. Al³⁺, c. Mg²⁺.**