Question

Which one of the following is a correct electronic configuration of sodium?

• A. 2,8
• B. 8,2,1
• C. 2,1,8
• D. 2,8,1

Answer 1

The Correct Option is D

The question asks for the correct electronic configuration of sodium. Let's first understand the basics of electronic configuration.

Each element in the periodic table has a specific number of electrons arranged in shells around the nucleus. These shells follow specific rules: each shell can hold a maximum number of electrons given by the formula \(2n^2\), where \(n\) is the shell number (1, 2, 3, etc.). 

Sodium (Na) has an atomic number of 11, which means it has 11 electrons. We need to distribute these electrons across the available shells:

1. The first shell (K-shell) can hold up to 2 electrons: \(2\cdot1^2 = 2\).
2. The second shell (L-shell) can hold up to 8 electrons: \(2\cdot2^2 = 8\).
3. The third shell (M-shell) can hold up to 18 electrons, but we only distribute the leftover electrons into this shell after filling the previous shells.

Putting it all together for sodium (11 electrons):

• First shell (K): 2 electrons
• Second shell (L): 8 electrons
• The remaining electrons go to the third shell (M): 1 electron

Thus, the correct electronic configuration of sodium is 2,8,1. This corresponds to option 2,8,1.

Let's rule out the incorrect options:

• Option 2,8 is incorrect because it only accounts for 10 electrons, but sodium has 11 electrons.
• Option 8,2,1 is incorrect because the distribution does not follow the filling order of shells based on energy levels.
• Option 2,1,8 is also incorrect, as it suggests electrons are present in shells that are not filled in the usual order observed in periodic elements.

Therefore, the correct choice is 2,8,1, which accurately represents sodium's electronic configuration.