How does calcium obey the octet rule when reacting to form compounds?

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How does calcium obey the octet rule when reacting to form compounds?

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Answer 1 · 2016-02-01T00:31:06

It loses two electrons from its outermost shell.

Explanation:

Calcium, $Ca$ $"Ca"$ , is located in group 2 of the periodic table, which means that is has two electrons on its outermost shell.

In order to have a complete octet, calcium must lose these two outermost electrons, also called valence electrons.

Calcium will react with nonmetals to form ionic compounds.

The electron configuration for a neutral calcium atom looks like this

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

After the two outermost electrons are lost, which in a neutral atom occupy the fourth energy level, the calcium cation, ${Ca}^{2 +}$ $"Ca"^(2+)$ , is formed. Its electron configuration looks like this

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

By losing its valence electrons, calcium completes its octet. The outermost shell, which is now the third energy level, holds a total of eight electrons $\to$ $->$ a complete octet

For example, calcium will react with oxygen to form calcium oxide, $CaO$ $"CaO"$ . Calcium will donate its two valence electrons to oxygen, forming the ${Ca}^{2 +}$ $"Ca"^(2+)$ cation.

Oxygen will pick up these two electrons, forming the $O^{2 -}$ $"O"^(2-)$ anion. The resulting compound, $CaO$ $"CaO"$ , will thus be electrically neutral.

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How does calcium obey the octet rule when reacting to form compounds?

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