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<h3>Ions</h3>
The atoms in a stable compound have a <u>noble gas</u> election configuration.
Ionic compounds are assumed to be in the solid state unless otherwise noted.
When <u>two nonmetals</u> react to form a <u>covalent bond</u>, they <u>share</u> elections in a way that completes the valence election configuration of both atoms.
When a nonmetal and a main group metal react to form a binary ionic compound, the ions form so that the <u>nonmetal</u> achieves the electron configuration of the <u>next noble gas atom</u>, and the <u>valence orbitals</u> of <u>the metal are emptied</u>.
- Nonmetals will gain $e^-$
- Metals will lose $e^-$

What ionic compound will form between calcium and oxygen?
- Let's consider their election configurations:
$$
\begin{align}
Ca: [AR]\;4s^2 \hspace{25pt} Ca^{2+}:[AR] \\
O: [He]\;2s^22p^4 \hspace{25pt} O^{2-}:[HE]\;2s^22p^6
\end{align}
$$
- What would the chemical equation of election transfer be?
$$Ca + O \rightarrow Ca^{2+}+O^{2-}$$
- What is the ionic compound formula?
$$Ca^{2+}\;O^{2-} \rightarrow CaO$$

Groups 1A, 2A and 3A cation charge:
- Positive charge equal to their group number except for $H$.
Groups 5A, 6A, and 7A anion charge:
- Negative charge equal to $8$ - Group #.
There are some exceptions though:
- Tin: $Sn^{2+}$ and $Sn^{4+}$
- Lead: $Pb^{2+}$ and $Pb^{4+}$
- Bismuth: $B^{3+}$ and $B^{5+}$
- Thallium: $Tl^{+}$ and $Tl^{3+}$

<h3>Ions</h3>
The atoms in a stable compound have a <u>noble gas</u> election configuration.
Ionic compounds are assumed to be in the solid state unless otherwise noted.
When <u>two nonmetals</u> react to form a <u>covalent bond</u>, they <u>share</u> elections in a way that completes the valence election configuration of both atoms.
When a nonmetal and a main group metal react to form a binary ionic compound, the ions form so that the <u>nonmetal</u> achieves the electron configuration of the <u>next noble gas atom</u>, and the <u>valence orbitals</u> of <u>the metal are emptied</u>.
- Nonmetals will gain $e^-$
- Metals will lose $e^-$

What ionic compound will form between calcium and oxygen?
- Let's consider their election configurations:
$$
\begin{align}
Ca: [AR]\;4s^2 \hspace{25pt} Ca^{2+}:[AR] \\
O: [He]\;2s^22p^4 \hspace{25pt} O^{2-}:[HE]\;2s^22p^6
\end{align}
$$
- What would the chemical equation of election transfer be?
$$Ca + O \rightarrow Ca^{2+}+O^{2-}$$
- What is the ionic compound formula?
$$Ca^{2+}\;O^{2-} \rightarrow CaO$$

Groups 1A, 2A and 3A cation charge:
- Positive charge equal to their group number except for $H$.
Groups 5A, 6A, and 7A anion charge:
- Negative charge equal to $8$ - Group #.
There are some exceptions though:
- Tin: $Sn^{2+}$ and $Sn^{4+}$
- Lead: $Pb^{2+}$ and $Pb^{4+}$
- Bismuth: $B^{3+}$ and $B^{5+}$
- Thallium: $Tl^{+}$ and $Tl^{3+}$