The first `(IE_(1))` and second `(E_(2))` ionization energies (kJ/mol) of few elements designated by Roman numerals are given below. Which of these would be an alkali metal?

Options

A. `{:(,IE_(1),IE_(2)),(I,2372,5251):}`
B. `{:(,IE_(1),IE_(2)),(II,520,7300):}`
C. `{:(,IE_(1),IE_(2)),(III,900,1760):}`
D. `{:(,IE_(1),IE_(2)),(IV,1680,3380):}`

Answer

To determine which of the given elements is an alkali metal based on their ionization energies (IE1 and IE2), we can follow these steps:

### Step 1: Understand Ionization Energies
Ionization energy is the energy required to remove an electron from an atom. The first ionization energy (IE1) is the energy needed to remove the first electron, while the second ionization energy (IE2) is the energy needed to remove a second electron after the first has been removed.

### Step 2: Identify Alkali Metals
Alkali metals belong to Group 1 of the periodic table. They have one electron in their outermost shell (ns1 configuration). When this outer electron is removed, the atom achieves a stable noble gas configuration, which results in a significant increase in the energy required to remove a second electron.

### Step 3: Analyze the Given Ionization Energies
We need to look for a significant jump between the values of IE1 and IE2. A large jump indicates that the first electron removed was from the outer shell, while the second electron is being removed from a much more stable inner shell.

### Step 4: Compare the Options
- If we have the following hypothetical values for ionization energies:
- Option I: IE1 = 500 kJ/mol, IE2 = 1000 kJ/mol
- Option II: IE1 = 520 kJ/mol, IE2 = 7300 kJ/mol
- Option III: IE1 = 600 kJ/mol, IE2 = 1200 kJ/mol
- Option IV: IE1 = 550 kJ/mol, IE2 = 1100 kJ/mol

From the above options, we can see that:
- **Option II** shows a significant jump from 520 kJ/mol to 7300 kJ/mol. This large increase suggests that the first electron is being removed from the outer shell, while the second electron is being removed from a much more stable inner shell.

### Step 5: Conclusion
Based on the analysis, **Option II** is the alkali metal because it shows the largest jump in ionization energy between IE1 and IE2.