What is the electron configuration (arrangement) by spdf notation of a sodium ion?

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What is the electron configuration (arrangement) by spdf notation of a sodium ion?

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Answer 1 · 2018-02-24T02:06:31

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

Explanation:

Sodium loses 1 electron from its outermost shell to get
the configuration of neon, or form the sodium ion.

The first shell has $1 s^{2}$ $1s^2$
The second shell has $2 s^{2}, 2 p^{6}$ $2s^2, 2p^6$

Answer 2 · 2018-02-24T03:56:05

$1 s^{2} 2 s^{2} 2 p^{6}$ $1s^2\2s^2\2p^6$ or just $[N e]$ $[Ne]$

Explanation:

A neutral sodium atom has an electron configuration of $[N e] 3 s^{1}$ $[Ne]3s^1$ .

Since it has $11$ $11$ electrons, it would want to lose one electron to remove its third shell, while keeping the other $2$ $2$ electron shells full. So, in a sodium ion, it becomes $N a^{+}$ $Na^+$ and has only $10$ $10$ electrons.

So, it will remove $1$ $1$ electron, and lose its $3 s$ $3s$ orbital. The electron configuration therefore just becomes $[N e]$ $[Ne]$ or if you want the full configuration, it is $1 s^{2} 2 s^{2} 2 p^{6}$ $1s^2\2s^2\2p^6$ .

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What is the electron configuration (arrangement) by spdf notation of a sodium ion?

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