The electron configuration of some neutral atoms are given below :
`(i) 1s^(2) 2s^(2) (ii) 1s^(2) 2s^(2) 2p^(1) (iii) 1s^(2) 2s^(2) 2p^(4) (iv) 1s^(2) 2s^(2) 2p^(3)`
Which of these electronic configuration would be expected th have the highest
`(a) Delta_(i) H_(1) ( b) Delta_(i) H_(2) (c ) Delta_(i) H_(3) (d ) Delta_(i) H_(4)`

Explanation

To determine which electronic configuration has the highest ionization energy, we need to analyze the given configurations step by step.

### Step 1: Identify the Elements
The electronic configurations provided are:
1. (i) 1s² 2s² - This corresponds to Beryllium (Be).
2. (ii) 1s² 2s² 2p¹ - This corresponds to Boron (B).
3. (iii) 1s² 2s² 2p⁴ - This corresponds to Oxygen (O).
4. (iv) 1s² 2s² 2p³ - This corresponds to Nitrogen (N).

### Step 2: Understand Ionization Energy
Ionization energy is the energy required to remove an electron from an atom in the gaseous state. Several factors affect ionization energy, including:
- The effective nuclear charge (Z_eff) experienced by the outermost electrons.
- The distance of the outermost electrons from the nucleus.
- The electron configuration, particularly whether the outer shell is full or half-full.

### Step 3: Analyze Trends in Ionization Energy
- As we move across a period from left to right, the ionization energy generally increases due to the increasing nuclear charge, which pulls the electrons closer to the nucleus.
- However, fully filled and half-filled orbitals provide extra stability, making it harder to remove an electron.

### Step 4: Compare the Configurations
Now, let's analyze the configurations based on the above points:
- **Beryllium (Be)**: 1s² 2s² - Fully filled 2s subshell, relatively low ionization energy.
- **Boron (B)**: 1s² 2s² 2p¹ - Less stable than Be, higher ionization energy than Be.
- **Nitrogen (N)**: 1s² 2s² 2p³ - Half-filled 2p subshell, more stable than Boron and Oxygen, thus has a higher ionization energy.
- **Oxygen (O)**: 1s² 2s² 2p⁴ - Less stable than Nitrogen due to increased electron-electron repulsion in the 2p subshell, leading to lower ionization energy than Nitrogen.

### Conclusion
Based on the analysis, the order of ionization energies from highest to lowest is:
1. Nitrogen (N) - Highest ionization energy due to half-filled stability.
2. Oxygen (O)
3. Boron (B)
4. Beryllium (Be) - Lowest ionization energy.

Thus, the configuration expected to have the highest ionization energy is **(iv) 1s² 2s² 2p³ (Nitrogen)**.

### Final Answer
**(d) Delta_(i) H_(4)** corresponds to the configuration with the highest ionization energy.