Which one of the following groupings represents a collection of isoelectronic species ?

Options

A. `Ca^(2+) , Cs^+, Br`
B. `Na^(+), Ca^(2+), Mg^(2+)`
C. `N^(3-), F^-, Na^+`
D. `Be, Al^(3+), CI^-`

Answer

To determine which grouping represents a collection of isoelectronic species, we need to identify the number of electrons in each species. Isoelectronic species are those that have the same number of electrons. Let's analyze the options step by step.

### Step 1: Identify the number of electrons in each species

1. **Calcium (Ca)**:
- Atomic number = 20
- Neutral calcium has 20 electrons. As a Ca²⁺ ion, it loses 2 electrons.
- Number of electrons = 20 - 2 = 18

2. **Cesium (Cs)**:
- Atomic number = 55
- Neutral cesium has 55 electrons. As a Cs⁺ ion, it loses 1 electron.
- Number of electrons = 55 - 1 = 54

3. **Bromine (Br)**:
- Atomic number = 35
- Neutral bromine has 35 electrons. As a Br⁻ ion, it gains 1 electron.
- Number of electrons = 35 + 1 = 36

### Conclusion for Option 1:
- The number of electrons in Ca²⁺, Cs⁺, and Br⁻ are 18, 54, and 36 respectively. **Not isoelectronic.**

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### Step 2: Analyze the next grouping

1. **Sodium (Na)**:
- Atomic number = 11
- Neutral sodium has 11 electrons. As a Na⁺ ion, it loses 1 electron.
- Number of electrons = 11 - 1 = 10

2. **Calcium (Ca)**:
- As calculated before, Ca²⁺ has 18 electrons.

3. **Magnesium (Mg)**:
- Atomic number = 12
- Neutral magnesium has 12 electrons. As a Mg²⁺ ion, it loses 2 electrons.
- Number of electrons = 12 - 2 = 10

4. **Nitrogen (N)**:
- Atomic number = 7
- Neutral nitrogen has 7 electrons. As an N³⁻ ion, it gains 3 electrons.
- Number of electrons = 7 + 3 = 10

5. **Fluorine (F)**:
- Atomic number = 9
- Neutral fluorine has 9 electrons. As an F⁻ ion, it gains 1 electron.
- Number of electrons = 9 + 1 = 10

### Conclusion for Option 2:
- The number of electrons in Na⁺, Mg²⁺, N³⁻, and F⁻ are 10, 10, 10, and 10 respectively. **These are isoelectronic.**

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### Step 3: Analyze the final grouping

1. **Beryllium (Be)**:
- Atomic number = 4
- Neutral beryllium has 4 electrons. As a Be²⁺ ion, it loses 2 electrons.
- Number of electrons = 4 - 2 = 2

2. **Aluminum (Al)**:
- Atomic number = 13
- Neutral aluminum has 13 electrons. As an Al³⁺ ion, it loses 3 electrons.
- Number of electrons = 13 - 3 = 10

3. **Chlorine (Cl)**:
- Atomic number = 17
- Neutral chlorine has 17 electrons. As a Cl⁻ ion, it gains 1 electron.
- Number of electrons = 17 + 1 = 18

### Conclusion for Option 3:
- The number of electrons in Be²⁺, Al³⁺, and Cl⁻ are 2, 10, and 18 respectively. **Not isoelectronic.**

### Final Answer:
The grouping that represents a collection of isoelectronic species is the one containing **Sodium (Na⁺), Magnesium (Mg²⁺), Nitrogen (N³⁻), and Fluorine (F⁻)**, all of which have **10 electrons**.

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