`IE_(1), IE_(2)` and `IE_(3)` values are `100, 150` and `1500 eV` respectively. The element can be

Options

A. `Na`
B. `B`
C. `Be`
D. `F`

Explanation

To determine the element based on the given ionization energy values (IE1 = 100 eV, IE2 = 150 eV, and IE3 = 1500 eV), we can follow these steps:

### Step 1: Analyze the Ionization Energy Values
The first ionization energy (IE1) is 100 eV, the second ionization energy (IE2) is 150 eV, and the third ionization energy (IE3) is significantly higher at 1500 eV. This large jump between IE2 and IE3 suggests that removing the third electron is much more difficult than removing the first two.

**Hint:** Look for a significant increase in ionization energy values, as this indicates a change in the stability of the electron configuration.

### Step 2: Understand the Implications of Ionization Energies
The first two ionization energies are relatively low compared to the third. This indicates that after the removal of the first two electrons, the resulting ion (M²⁺) has achieved a stable electron configuration, likely resembling that of a noble gas.

**Hint:** Consider the electron configuration of common elements and how they relate to noble gas configurations.

### Step 3: Identify Possible Elements
Elements that typically have low first and second ionization energies and a high third ionization energy are usually found in Group 2 of the periodic table (alkaline earth metals). For example, magnesium (Mg) loses two electrons easily to form Mg²⁺, which has the same electron configuration as neon, a noble gas.

**Hint:** Recall the groups in the periodic table and their typical ionization energy trends.

### Step 4: Conclusion
Based on the analysis, the element in question is likely to be an alkaline earth metal, specifically one that forms a stable M²⁺ ion. The large jump to IE3 indicates that the element is likely to be magnesium (Mg), as it forms Mg²⁺ with a stable noble gas configuration.

**Hint:** Confirm your answer by checking the periodic table for elements in Group 2 and their ionization energy trends.

### Final Answer
The element can be identified as magnesium (Mg), which is an alkaline earth metal.