B has a smaller first ionization enthalpy than Be. Consider the following statements
(i) it is easier to remove 2p electron than 2s electron
(ii) 2p electron of B is more shielded from the nucleus by the inner core of electrons than the 2s electrons of Be
(iii) 2s electron has more penetration power than 2p electron
(iv) atomic radius of B is more than Be (atomic number B= 5, Be = 4) The correct statements are

Options

A. (i),(ii) and (iv)
B. (i),(iii) and (iv)
C. (i),(ii) and (iii)
D. (ii),(iii) and (iv)

Answer

To solve the question regarding the ionization enthalpy of Boron (B) and Beryllium (Be), we will analyze each statement provided in the question step by step.

### Step 1: Analyze the Electronic Configurations
- **Boron (B)** has an atomic number of 5, so its electronic configuration is:
\[ 1s^2 \, 2s^2 \, 2p^1 \]
- **Beryllium (Be)** has an atomic number of 4, so its electronic configuration is:
\[ 1s^2 \, 2s^2 \]

### Step 2: Evaluate Statement (i)
**Statement (i)**: It is easier to remove a 2p electron than a 2s electron.
- The 2s electrons in Beryllium are more tightly bound to the nucleus due to their lower energy and greater effective nuclear charge compared to the 2p electron in Boron. Thus, it is indeed easier to remove a 2p electron than a 2s electron.
- **Conclusion**: This statement is **correct**.

### Step 3: Evaluate Statement (ii)
**Statement (ii)**: The 2p electron of B is more shielded from the nucleus by the inner core of electrons than the 2s electron of Be.
- In Boron, the 2p electron experiences shielding from the 1s electrons, but the 2s electrons in Beryllium are also shielded by the same 1s electrons. However, since Beryllium has only 4 electrons compared to Boron's 5, the 2p electron in Boron is indeed more shielded.
- **Conclusion**: This statement is **correct**.

### Step 4: Evaluate Statement (iii)
**Statement (iii)**: The 2s electron has more penetration power than the 2p electron.
- The 2s orbital has a spherical shape which allows it to penetrate closer to the nucleus compared to the 2p orbital which has a dumbbell shape. Therefore, the 2s electron does have more penetration power than the 2p electron.
- **Conclusion**: This statement is **correct**.

### Step 5: Evaluate Statement (iv)
**Statement (iv)**: The atomic radius of B is more than Be.
- As we move across a period from left to right (from Beryllium to Boron), the effective nuclear charge increases, leading to a decrease in atomic radius. Therefore, Boron has a smaller atomic radius than Beryllium.
- **Conclusion**: This statement is **incorrect**.

### Final Conclusion
The correct statements are:
- Statement (i): Correct
- Statement (ii): Correct
- Statement (iii): Correct
- Statement (iv): Incorrect

Thus, the correct answer is that statements **(i), (ii), and (iii)** are correct.