What is the electron configuration of a sodium ion?

Dec 29, 2015

$1 {s}^{2} 2 {s}^{2} 2 {p}^{6}$

Explanation:

A neutral sodium (Na)atom has 11 electrons.
A sodium ion $\left(N {a}^{+}\right)$ has lost 1 electron and is left with 10 electrons, has a valency of $1 +$ and has the same electron configuration as the nearest noble gas, $N e$, with no unpaired electrons and filled outer shell, and in stable form.
Its electron configuration is hence, filling up from bottom most energy levels first as per the normal rules of filling energy levels and orbitals, 2 electrons in the 1s orbital, 2 electrons in the 2s orbital, and 6 electrons in the 2p oribital.
Hence the configuration $1 {s}^{2} 2 {s}^{2} 2 {p}^{6}$.

Dec 29, 2015

Hmmm, $N a$, $Z = 11$. The valence electron is in an $s$ shell.

Explanation:

$N a$: $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{1}$.

If the atomic number is $11$, there must be 11 electrons in the neutral atom. Why?

Electrons are distributed according to the $a u f b a u$ scheme.

So the outermost electron is removed from the electron configuration to give, $1 {s}^{2} 2 {s}^{2} 2 {p}^{6}$ .
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