Q6E - Chemistry (OpenStax)

Question

Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds: (a) \({\rm{I}}\) (b) \({\rm{Sr}}\) (c) \({\rm{K}}\) (d) \({\rm{N}}\) (e) \({\rm{S}}\) (f) \({\rm{In}}\).

Step-by-Step Solution

Answer

1. The charge on monatomic ions is predicted as:\({{\rm{I}}^{\rm{ - }}}\).
2. The charge on monatomic ions is predicted as:\({\rm{S}}{{\rm{r}}^{{\rm{2 + }}}}\).
3. The charge on monatomic ions is predicted as:\({{\rm{K}}^{\rm{ + }}}\).
4. The charge on monatomic ions is predicted as:\({{\rm{N}}^{{\rm{3 - }}}}\).
5. The charge on monatomic ions is predicted as:\({{\rm{S}}^{{\rm{2 - }}}}\).
6. The charge on monatomic ions is predicted as:\({\rm{I}}{{\rm{n}}^{\rm{ + }}}\).

Step 1: Define Chemical Bonding

A chemical bond is a long-term attraction between atoms, ions, or molecules that allows chemical compounds to form.

Step 2: Predicting the charge on monatomic ions

The electronic configuration of \({\rm{I : -   }}\left( {{\rm{Kr}}} \right){\rm{4}}{{\rm{d}}^{{\rm{10}}}}{\rm{5}}{{\rm{s}}^{\rm{2}}}{\rm{5}}{{\rm{p}}^{\rm{5}}}\).

Iodine will receive one electron and form \({{\rm{I}}^{\rm{ - }}}\) in order to obtain the closest noble-gas electrical configuration (of Xenon).

Therefore, the charge on monatomic ions is:\({{\rm{I}}^{\rm{ - }}}\).

Step 3: Predicting the charge on monatomic ions

b.

The electronic configuration of \({\rm{Sr : -   }}\left( {{\rm{Kr}}} \right){\rm{5}}{{\rm{s}}^{\rm{2}}}\).

Strontium will loose two electrons and form \({\rm{S}}{{\rm{r}}^{{\rm{2 + }}}}\) in order to obtain the closest noble-gas electrical configuration (of Krypton).

Therefore, the charge on monatomic ions is:\({\rm{S}}{{\rm{r}}^{{\rm{2 + }}}}\).

Step 4: Predicting the charge on monatomic ions

c.

The electronic configuration of \({\rm{K : -   }}\left( {{\rm{Ar}}} \right){\rm{4}}{{\rm{s}}^{\rm{1}}}\).

Potassium will loose one electron and form \({{\rm{K}}^{\rm{ + }}}\) in order to obtain the closest noble-gas electrical configuration (of Argon).

Therefore, the charge on monatomic ions is:\({{\rm{K}}^{\rm{ + }}}\).

Step 5: Predicting the charge on monatomic ions

d.

The electronic configuration of \({\rm{N : -   }}\left( {{\rm{He}}} \right){\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{p}}^{\rm{3}}}\).

Nitrogen will receive three electrons and form \({{\rm{N}}^{{\rm{3 - }}}}\) in order to obtain the closest noble-gas electrical configuration (of Neon).

Step 6: Predicting the charge on monatomic ions

e.

The electronic configuration of \({\rm{S : -   }}\left( {{\rm{Ne}}} \right){\rm{3}}{{\rm{s}}^{\rm{2}}}{\rm{3}}{{\rm{p}}^{\rm{4}}}\).

Phosphorous will receive two electrons and form \({{\rm{S}}^{{\rm{2 - }}}}\) in order to obtain the closest noble-gas electrical configuration (of Argon).

Therefore, the charge on monatomic ions is:\({{\rm{S}}^{{\rm{2 - }}}}\).

Step 7: Predicting the charge on monatomic ions

f.

The electronic configuration of \({\rm{In : -   }}\left( {{\rm{Kr}}} \right){\rm{4}}{{\rm{d}}^{{\rm{10}}}}{\rm{5}}{{\rm{s}}^{\rm{2}}}{\rm{5}}{{\rm{p}}^{\rm{1}}}\).

\({\rm{I}}\)willloose one electron and form \({\rm{I}}{{\rm{n}}^{\rm{ + }}}\) in order to obtain the stability (having paired electron).

Therefore, the charge on monatomic ions is:\({\rm{I}}{{\rm{n}}^{\rm{ + }}}\).