How do valence electrons affect ionic bonding?

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How do valence electrons affect ionic bonding?

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Answer 1 · 2014-04-22T06:53:41

Lets take the ionic formula for Calcium Chloride is $C a C l_{2}$ $CaCl_2$

Calcium is an Alkaline Earth Metal in the second column of the periodic table. This means that calcium $s^{2}$ $s^2$ has 2 valence electrons it readily gives away in order to seek the stability of the octet. This makes calcium a Ca+2 cation.

Chlorine is a Halogen in the 17th column or $s^{2} p^{5}$ $s^2p^5$ group.
Chlorine has 7 valence electrons. It needs one electron to make it stable at 8 electrons in its valence shells. This makes chlorine a $C l^{- 1}$ $Cl^(−1)$ anion.

Ionic bonds form when the charges between the metal cation and non-metal anion are equal and opposite. This means that two $C l^{- 1}$ $Cl^(−1)$ anions will balance with one $C a^{+ 2}$ $Ca^(+2)$ cation.

This makes the formula for calcium chloride, $C a C l_{2}$ $CaCl_2$ .

For the example Aluminum Oxide $A l_{2} O_{3}$ $Al_2O_3$

Aluminum $s^{2} p^{1}$ $s^2p^1$ has 3 valence electrons and an oxidation state of +3 or $A l^{+ 3}$ $Al^(+3)$
Oxygen $s^{2} p^{4}$ $s^2p^4$ has 6 valence electrons and an oxidation state of -2 or $O^{- 2}$ $O^(−2)$

The common multiple of 2 and 3 is 6.
We will need 2 aluminum atoms to get a +6 charge and 3 oxygen atoms to get a -6 charge. When the charges are equal and opposite the atoms will bond as $A l_{2} O_{3}$ $Al_2O_3$ .

I hope this is helpful.
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How do valence electrons affect ionic bonding?

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