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Do Mg and S Form an Ionic Bond?

Would Mg and S form an ionic bond, and if so what ions form and what is the empirical formula of the compound?

Subject: General Chemistry Chapter: Chemical Bonds Topic: Lewis Diagrams for Atoms and Simple Ions Answer included
would mg and s form an ionic bond ionic bond ionic compound electron transfer valence electrons magnesium sulfide MgS formula Mg2+ ion
Accepted answer Answer included

Answer to “would mg and s form an ionic bond”

Magnesium (Mg) and sulfur (S) form an ionic bond because Mg is a metal that readily forms a cation by losing electrons, while S is a nonmetal that forms an anion by gaining electrons; the electrostatic attraction between these oppositely charged ions yields the ionic compound magnesium sulfide, MgS.

Key idea: Ionic bonding is driven by electron transfer that produces a cation and an anion, followed by strong Coulombic attraction between the ions in a crystal lattice.

Step 1: Identify valence electrons and typical ion charges

Magnesium is in Group 2, so it has 2 valence electrons. Sulfur is in Group 16, so it has 6 valence electrons. A stable noble-gas-like configuration is reached when Mg loses 2 electrons and S gains 2 electrons.

Element Valence electrons (group trend) Most common ion formed How stability is reached
Mg 2 \(\mathrm{Mg^{2+}}\) Loses 2 electrons to achieve a filled outer shell in the preceding noble-gas core.
S 6 \(\mathrm{S^{2-}}\) Gains 2 electrons to complete an octet in the valence shell.

Step 2: Write the electron-transfer (half-reaction) viewpoint

  1. Magnesium oxidation (electron loss): \[ \mathrm{Mg \rightarrow Mg^{2+} + 2e^-} \]
  2. Sulfur reduction (electron gain): \[ \mathrm{S + 2e^- \rightarrow S^{2-}} \]

The transferred electrons cancel when the processes occur together, leaving oppositely charged ions that attract.

Step 3: Combine ions to obtain a neutral empirical formula

Ionic compounds are electrically neutral overall. A single \(\mathrm{Mg^{2+}}\) balances a single \(\mathrm{S^{2-}}\):

\[ (+2) + (-2) = 0 \]

Therefore, the simplest ratio is \(1:1\), giving the empirical formula \(\mathrm{MgS}\). The lattice can be summarized as:

\[ \mathrm{Mg^{2+} + S^{2-} \rightarrow MgS(s)} \]

Visualization: Electron transfer and ion pairing

Electron transfer from Mg to S forming Mg2+ and S2- A schematic showing Mg with 2 valence electrons transferring two electrons to S with 6 valence electrons, resulting in Mg2+ and S2- and the 1:1 formula MgS. Mg 2 valence e⁻ S 6 valence e⁻ 2 e⁻ Mg²⁺ cation S²⁻ anion ion formation Empirical formula: MgS (1:1)
The diagram shows Mg donating two electrons to S, producing \(\mathrm{Mg^{2+}}\) and \(\mathrm{S^{2-}}\); their charges balance in a 1:1 ratio to form \(\mathrm{MgS}\).

Common checks and pitfalls

  • Charge balance: ionic formulas must satisfy total charge \(= 0\); for \(\mathrm{Mg^{2+}}\) and \(\mathrm{S^{2-}}\), the smallest neutral ratio is \(1:1\).
  • Octet logic: sulfur gains 2 electrons (not 1) to complete an octet, so \(\mathrm{S^{2-}}\) is expected in simple ionic bonding with Group 2 metals.
  • Name vs formula: “magnesium sulfide” corresponds to \(\mathrm{MgS}\); subscripts come from charge balancing, not from valence electron counts alone.
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