In the isoelectronic species the ionic radii (Å ) of `N^(3-)` , Ne and `Al^(+3)` are respectively given by :

Options

A. 1.36, 1.40, 1.71
B. ) 1.36, 1.71, 1.40
C. 1.71, 1.40, 1.36
D. 1.71, 1.36, 1.40

Answer

To solve the question regarding the ionic radii of the isoelectronic species \( N^{3-} \), Ne, and \( Al^{3+} \), we need to follow these steps:

### Step 1: Identify the number of electrons in each species.
- **For \( N^{3-} \)**: Nitrogen (N) has an atomic number of 7, meaning it has 7 protons and normally 7 electrons. Since it has a \( 3- \) charge, it gains 3 additional electrons. Therefore, \( N^{3-} \) has \( 7 + 3 = 10 \) electrons.

- **For Ne**: Neon (Ne) has an atomic number of 10, meaning it has 10 protons and 10 electrons.

- **For \( Al^{3+} \)**: Aluminum (Al) has an atomic number of 13, meaning it has 13 protons and normally 13 electrons. Since it has a \( 3+ \) charge, it loses 3 electrons. Therefore, \( Al^{3+} \) has \( 13 - 3 = 10 \) electrons.

### Step 2: Confirm that all species are isoelectronic.
Since all three species have 10 electrons, they are indeed isoelectronic.

### Step 3: Analyze the effect of charge on ionic size.
- **Negative Charge**: For anions (like \( N^{3-} \)), the addition of electrons increases electron-electron repulsion, leading to a larger ionic radius.

- **Neutral Species**: For neutral species (like Ne), the size is relatively stable and is determined by the balance of protons and electrons.

- **Positive Charge**: For cations (like \( Al^{3+} \)), the loss of electrons results in a higher effective nuclear charge acting on the remaining electrons, which pulls them closer to the nucleus, resulting in a smaller ionic radius.

### Step 4: Rank the ionic radii based on the charge.
- \( N^{3-} \) will have the largest radius because it has the most electrons and a negative charge.
- Ne will have an intermediate radius as it is neutral.
- \( Al^{3+} \) will have the smallest radius due to its positive charge.

### Step 5: Conclusion on the order of ionic radii.
The order of ionic radii from largest to smallest is:
\[ N^{3-} > Ne > Al^{3+} \]

### Step 6: Assign the given ionic radii values.
If we were given specific values for the ionic radii, we would assign them based on our analysis:
- \( N^{3-} \) would correspond to the largest value,
- Ne would correspond to the middle value,
- \( Al^{3+} \) would correspond to the smallest value.

### Final Answer:
Thus, the ionic radii (in Å) of \( N^{3-} \), Ne, and \( Al^{3+} \) are respectively in the order of increasing size.

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