Question

write the distribotion of electrons in carbon and sodium​

Answer 1

Answer:

Carbon = (2,4)

Sodium = (2,8,1)

Answer 2

The distribution of electrons in Carbon(C) = 2,4 and in Sodium (Na) = 2,8,1.

Explanation:

Carbon is a non-metal that is placed in Group 14 (IVa) of the periodic table.

C has 6 electrons, which occupy 2 shells and for filling of orbitals electrons follow the

$2n^{2} rule\\where\ n = Shell\ number$ .

The electronic configuration of C is $1s^{2} 2s^{2} 2p^{2}$.

The distribution of electrons can be written as 2,4.

Sodium (Na) is an alkali metal present in Group 1 of the periodic table.

Na has 11 electrons, which occupy 3 shells. The electrons are filled according to the same  $2n^{2} rule.$

So the electronic configuration of Na is $1s^{2} 2s^{2} 2p^{6} 3s^{1}$.

The distribution of electrons can be written as 2,8,1.