The electronic configuration of element `A, B`, and `C`are `[He]2s^(1),[Ne]3s^(1)`, and `[Ar]4s^(1)`, respectively. Which one of the following order is correct for the `IE_(1) ("in" kJ mol^(-1))` of `A`,`B,` and `C`?

Options

A. `AgtBgtC`
B. `CgtBgtA`
C. `BgtCgtA`
D. `CgtAgtB`

Explanation

To determine the correct order of ionization energies (IE) for elements A, B, and C based on their electronic configurations, we can follow these steps:

### Step 1: Identify the Elements
- Element A has the electronic configuration `[He] 2s^1`, which corresponds to Lithium (Li).
- Element B has the electronic configuration `[Ne] 3s^1`, which corresponds to Sodium (Na).
- Element C has the electronic configuration `[Ar] 4s^1`, which corresponds to Potassium (K).

### Step 2: Understand the Trend in Ionization Energy
Ionization energy refers to the energy required to remove an electron from an atom in the gaseous state. The general trend in the periodic table is that ionization energy decreases as you move down a group. This is due to:
- Increasing atomic size: As you move down a group, the outermost electron is farther from the nucleus.
- Increased shielding effect: Inner electrons shield the outermost electron from the nuclear charge, making it easier to remove.

### Step 3: Apply the Trend to Elements A, B, and C
- Element A (Li) is at the top of the group (alkali metals), so it will have the highest ionization energy.
- Element B (Na) is below Li, so it will have a lower ionization energy than Li.
- Element C (K) is below Na, so it will have the lowest ionization energy among the three.

### Step 4: Write the Order of Ionization Energies
Based on the above analysis, the order of ionization energies from highest to lowest is:
- IE1(A) > IE1(B) > IE1(C)
- Therefore, the correct order is: A > B > C.

### Final Answer
The correct order for the ionization energies of elements A, B, and C is:
**A > B > C**

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