Arrange the following in the order of increasing ionization enthalpy :
`(i) 1 s^(2) 2s^(2) 2p^(6) 3s^(2) (ii) 1s^(2) 2s^(2) 2p^(6) 3s^(1) (iii) 1s^(2) 2s^(2) 2p^(6) (iv) 1s^(2) 2s^(2) 2p^(2) (v) 1s^(2) 2s^(2) 2p^(3)`

Explanation

To arrange the given electron configurations in the order of increasing ionization enthalpy, we need to consider the factors that influence ionization energy, such as atomic size and electron configuration stability. Here’s a step-by-step breakdown:

### Step 1: Identify the Electron Configurations
The given electron configurations are:
1. (i) \(1s^2 2s^2 2p^6 3s^2\) - Argon (Ar) with 3rd shell filled
2. (ii) \(1s^2 2s^2 2p^6 3s^1\) - Chlorine (Cl) with 1 electron in the 3rd shell
3. (iii) \(1s^2 2s^2 2p^6\) - Neon (Ne) with a complete outer shell
4. (iv) \(1s^2 2s^2 2p^2\) - Carbon (C) with 2 electrons in the 2p subshell
5. (v) \(1s^2 2s^2 2p^3\) - Nitrogen (N) with 3 electrons in the 2p subshell

### Step 2: Analyze the Stability of the Configurations
- **Stable Configurations**: Atoms with completely filled or half-filled subshells are more stable and require more energy to remove an electron.
- **Less Stable Configurations**: Atoms with unfilled subshells are less stable and require less energy to remove an electron.

### Step 3: Compare the Configurations
1. **(i) \(1s^2 2s^2 2p^6 3s^2\)**: This has a filled outer shell (n=3) and is stable. It will have a higher ionization enthalpy.
2. **(ii) \(1s^2 2s^2 2p^6 3s^1\)**: This has one electron in the outer shell (n=3), making it less stable than (i) but more stable than (iv) and (v).
3. **(iii) \(1s^2 2s^2 2p^6\)**: This is a noble gas configuration (Ne) and is very stable, thus has a high ionization enthalpy.
4. **(iv) \(1s^2 2s^2 2p^2\)**: This has 2 electrons in the 2p subshell, making it less stable than (iii) and (v).
5. **(v) \(1s^2 2s^2 2p^3\)**: This has a half-filled 2p subshell, which is relatively stable but less than the noble gas configuration.

### Step 4: Order the Configurations Based on Ionization Enthalpy
Based on the analysis:
- (iv) \(1s^2 2s^2 2p^2\) - least stable, lowest ionization enthalpy
- (v) \(1s^2 2s^2 2p^3\) - more stable than (iv)
- (ii) \(1s^2 2s^2 2p^6 3s^1\) - less stable than (i)
- (i) \(1s^2 2s^2 2p^6 3s^2\) - more stable than (ii)
- (iii) \(1s^2 2s^2 2p^6\) - highest stability, highest ionization enthalpy

### Final Order of Increasing Ionization Enthalpy
The order of increasing ionization enthalpy is:
**(iv) < (v) < (ii) < (i) < (iii)**