Question

The correct order of first ionization enthalpy values of the following elements is :
(A) O (B) N (C) Be (D) F (E) B
Choose the correct answer from the options given below :

• A. B < D < C < E < A
• B. E < C < A < B < D
• C. C < E < A < B < D
• D. A < B < D < C < E

Answer 1

The Correct Option is B

The correct order of first ionization enthalpies among the given elements can be deduced based on periodic trends. In general, ionization enthalpy increases across a period from left to right due to increasing nuclear charge and decreases down a group due to an increase in atomic size. Additionally, elements with stable electronic configurations, such as half-filled or fully-filled orbitals, tend to have higher ionization enthalpies.

The general trend for ionization enthalpy in the periodic table is:

\( \text{Li} < \text{B} < \text{Be} < \text{C} < \text{O} < \text{N} < \text{F} < \text{Ne} \)

Among the given elements:

\( \text{E (B)} < \text{C (Be)} < \text{A (O)} < \text{B (N)} < \text{D (F)} \)

Hence, the correct order of first ionization enthalpy values is:

\( \text{E (B)} < \text{C (Be)} < \text{A (O)} < \text{B (N)} < \text{D (F)} \)

Answer 2

Step 1: Understanding Ionization Enthalpy Trend.
First ionization enthalpy is the energy required to remove one mole of electrons from one mole of atoms in the gaseous state.
It generally increases across a period (left to right) because of increasing nuclear charge and decreases down a group due to increasing atomic size and shielding.

Step 2: Identify the Elements and Their Groups.
- B (Boron) – Group 13
- C (Carbon) – Group 14
- N (Nitrogen) – Group 15
- O (Oxygen) – Group 16
- F (Fluorine) – Group 17
- Be (Beryllium) – Group 2

Step 3: Comparison and Explanation.
- Boron has lower ionization energy than Beryllium, despite being to the right, due to electron entering the 2p orbital which is easier to remove than from the full 2s orbital of Be.
- Oxygen has lower ionization energy than Nitrogen due to electron-electron repulsion in paired 2p electrons in oxygen.
- Fluorine has the highest ionization energy among these because of high nuclear charge and small atomic radius.

Step 4: Hence, the correct order is:
\[ \boxed{ \text{E (B)} < \text{C (Be)} < \text{A (O)} < \text{B (N)} < \text{D (F)} } \] or \[ B < Be < O < N < F \]