Which of the following has largest ionic size

Options

A. `Li^+`
B. `K^+`
C. `Na^+`
D. `Cs^+`

Answer

To determine which ion has the largest ionic size, we need to analyze the given options based on their atomic numbers and the number of electrons remaining after ionization. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Ions
Let's assume the options provided are:
1. Li⁺ (Lithium ion)
2. Na⁺ (Sodium ion)
3. K⁺ (Potassium ion)
4. Rb⁺ (Rubidium ion)

### Step 2: Determine the Atomic Numbers
- Lithium (Li) has an atomic number of 3.
- Sodium (Na) has an atomic number of 11.
- Potassium (K) has an atomic number of 19.
- Rubidium (Rb) has an atomic number of 37.

### Step 3: Calculate the Electron Configuration After Ionization
- For Li⁺: Losing 1 electron from Li (3 electrons) leaves us with 2 electrons. The configuration is 1s².
- For Na⁺: Losing 1 electron from Na (11 electrons) leaves us with 10 electrons. The configuration is 1s² 2s² 2p⁶.
- For K⁺: Losing 1 electron from K (19 electrons) leaves us with 18 electrons. The configuration is 1s² 2s² 2p⁶ 3s² 3p⁶.
- For Rb⁺: Losing 1 electron from Rb (37 electrons) leaves us with 36 electrons. The configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 4p⁶.

### Step 4: Compare the Ionic Sizes
Ionic size generally increases as we move down a group in the periodic table because additional electron shells are added, which increases the distance between the nucleus and the outermost electrons.

- Li⁺ has 2 electrons in the first shell.
- Na⁺ has 10 electrons (2 in the first shell, 8 in the second shell).
- K⁺ has 18 electrons (2 in the first shell, 8 in the second shell, 8 in the third shell).
- Rb⁺ has 36 electrons (2 in the first shell, 8 in the second shell, 18 in the third shell, 8 in the fourth shell).

### Step 5: Conclusion
Since ionic size increases down the group, Rb⁺ will have the largest ionic size among the given ions.

**Final Answer: Rb⁺ has the largest ionic size.**