If I.E. of Na, Mg and Si are respectively 496, 737 and 786 kJ `"mol"^(-)` The I.E. of Al in KJ `"mol"^(-)` is :-

Options

A. 575
B. 760
C. 390
D. 1120

Explanation

To find the ionization energy (I.E.) of aluminum (Al) based on the given ionization energies of sodium (Na), magnesium (Mg), and silicon (Si), we can follow these steps:

### Step 1: Understand the Ionization Energies
The ionization energy is the energy required to remove an electron from an atom in its gaseous state. The given ionization energies are:
- I.E. of Na = 496 kJ/mol
- I.E. of Mg = 737 kJ/mol
- I.E. of Si = 786 kJ/mol

### Step 2: Write the Electronic Configurations
To understand the trends in ionization energy, we can write the electronic configurations of the elements:
- Sodium (Na, atomic number 11): 1s² 2s² 2p⁶ 3s¹
- Magnesium (Mg, atomic number 12): 1s² 2s² 2p⁶ 3s²
- Aluminum (Al, atomic number 13): 1s² 2s² 2p⁶ 3s² 3p¹
- Silicon (Si, atomic number 14): 1s² 2s² 2p⁶ 3s² 3p²

### Step 3: Analyze the Trends
From the electronic configurations, we can observe the following trends:
- Ionization energy generally increases across a period due to increasing nuclear charge.
- Sodium has the lowest I.E. because it has one electron in its outermost shell (3s¹).
- Magnesium has a higher I.E. than sodium because it has a completely filled 3s² subshell.
- Aluminum has a lower I.E. than magnesium because it has one electron in the 3p subshell (3s² 3p¹), which is easier to remove than an electron from a filled subshell.
- Silicon has a higher I.E. than aluminum because it has two electrons in the 3p subshell (3s² 3p²), making it harder to remove an electron.

### Step 4: Determine the Position of Aluminum
Since aluminum's I.E. is between magnesium and silicon, we can conclude:
- I.E. of Mg (737 kJ/mol) < I.E. of Al < I.E. of Si (786 kJ/mol)

### Step 5: Estimate the Ionization Energy of Aluminum
Given that aluminum's I.E. is between 737 kJ/mol and 786 kJ/mol, we can estimate a reasonable value. The options provided are:
1. 575 kJ/mol
2. 760 kJ/mol
3. 800 kJ/mol
4. 700 kJ/mol

Since 760 kJ/mol falls between 737 kJ/mol and 786 kJ/mol, it is a plausible value for the I.E. of aluminum.

### Conclusion
Thus, the ionization energy of aluminum is **760 kJ/mol**.

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