How can I calculate the valence electrons of ions?

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How can I calculate the valence electrons of ions?

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Answer 1 · 2014-05-07T01:28:07

Valence electrons are the electrons contained in the outermost or valence shell of an atom or ion.

Explanation:

You determine the valence electrons by using the Periodic Table.

Let's assume you're using a table that has the groups numbered 1-18.

The electron configurations of ions are those of the neutral atoms plus or minus a number of electrons equal to the charge on the ion.

Groups 15 to 17

Write the electronic structure for the neutral atom. Then add electrons to get a total of eight.

E.g., for ${Cl}^{-}$ $"Cl"^-$ , we get
$Cl : 1 s^{2} l 2 s^{2} 2 p^{6} l 3 s^{2} 3 p^{5}$ $"Cl": 1s^2 color(white)(l)2s^2 2p^6color(white)(l) 3s^2 3p^5$ but ${Cl}^{-}$ $"Cl"^-$ has one more electron.
∴ ${Cl}^{-} : 1 s^{2} l 2 s^{2} 2 p^{6} l 3 s^{2} 3 p^{6}$ $"Cl"^"-": 1s^2 color(white)(l)2s^2 2p^6 color(white)(l)3s^2 3p^6$

Groups 1, 2, and 13

Write the electronic structure for the neutral atom. Then remove the outermost electrons.

For ${Na}^{+}$ $"Na"^+$ :
$Na : 1 s^{2} l 2 s^{2} 2 p^{6} l 3 s$ $"Na": 1s^2 color(white)(l)2s^2 2p^6 color(white)(l)3s$ , but ${Na}^{+}$ $"Na"^+$ has one less electron
∴ ${Na}^{+} : 1 s^{2} l 2 s^{2} 2 p^{6}$ $"Na"^+: 1s^2 color(white)(l)2s^2 2p^6$

Groups 3 to 12

Remove $s$ $s$ electrons before $d$ $d$ electrons.

E.g., for ${Cr}^{3+}$ $"Cr"^"3+"$
$Cr : 1 s^{2} l 2 s^{2} 2 p^{6} l 3 s^{2} 3 p^{6} l 4 s 3 d^{5}$ $"Cr": 1s^2 color(white)(l)2s^2 2p^6 color(white)(l)3s^2 3p^6 color(white)(l)4s 3d^5$
∴ ${Cr}^{3+} : 1 s^{2} l 2 s^{2} 2 p^{6} l 3 s^{2} 3 p^{6} l 3 d^{3}$ $"Cr"^"3+": 1s^2 color(white)(l)2s^2 2p^6 color(white)(l)3s^2 3p^6 color(white)(l)3d^3$

You remove the $4 s$ $4s$ electron first, followed by two of the $3 d$ $3d$ electrons.

EXAMPLES

Write the electron configurations for $O^{2-}$ $"O"^"2-"$ , ${Ca}^{2+}$ $"Ca"^"2+"$ , and ${Zn}^{2+}$ $"Zn"^"2+"$ .

**Solutions **

$O : 1 s^{2} l 2 s^{2} 2 p^{4}$ $"O": 1s^2 color(white)(l)2s^2 2p^4$
$O^{2-} : 1 s^{2} l 2 s^{2} 2 p^{6}$ $"O"^"2-": 1s^2 color(white)(l)2s^2 2p^6$

$Ca : 1 s^{2} l 2 s^{2} 2 p^{6} l 3 s^{2} 3 p^{6} l 4 s^{2}$ $"Ca": 1s^2 color(white)(l)2s^2 2p^6 color(white)(l)3s^2 3p^6 color(white)(l)4s^2$
${Ca}^{2+} : 1 s^{2} l 2 s^{2} 2 p^{6} l 3 s^{2} 3 p^{6}$ $"Ca"^"2+": 1s^2color(white)(l) 2s^2 2p^6 color(white)(l)3s^2 3p^6$

$Zn : 1 s^{2} l 2 s^{2} 2 p^{6} l 3 s^{2} 3 p^{6} l 4 s^{2} 3 d^{10}$ $"Zn": 1s^2 color(white)(l)2s^2 2p^6 color(white)(l)3s^2 3p^6 color(white)(l)4s^2 3d^10$
${Zn}^{2+} : 1 s^{2} l 2 s^{2} 2 p^{6} l 3 s^{2} 3 p^{6} l 3 d^{10}$ $"Zn"^"2+": 1s^2 color(white)(l)2s^2 2p^6 color(white)(l)3s^2 3p^6 color(white)(l)3d^10$

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How can I calculate the valence electrons of ions?

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