Certainly! Let's determine the empirical formulas for binary ionic compounds that can be formed from the given ions: [tex]\( Mg^{2+} \)[/tex], [tex]\( V^{5+} \)[/tex], [tex]\( Br^{-} \)[/tex], and [tex]\( S^{2-} \)[/tex].
1. Magnesium Bromide (MgBr₂):
- Magnesium ion: [tex]\( Mg^{2+} \)[/tex]
- Bromide ion: [tex]\( Br^{-} \)[/tex]
- To balance the charges, we need 2 Bromide ions to balance the 2+ charge of one Magnesium ion.
- Therefore, the empirical formula is [tex]\( \text{MgBr}_2 \)[/tex].
2. Magnesium Sulfide (MgS):
- Magnesium ion: [tex]\( Mg^{2+} \)[/tex]
- Sulfide ion: [tex]\( S^{2-} \)[/tex]
- Both ions have charges that are equal in magnitude but opposite in sign, thus one Magnesium ion will combine with one Sulfide ion.
- Therefore, the empirical formula is [tex]\( \text{MgS} \)[/tex].
3. Vanadium(III) Bromide (VBr₃):
- Vanadium ion: [tex]\( V^{3+} \)[/tex]
- Bromide ion: [tex]\( Br^{-} \)[/tex]
- To balance the charges, we need 3 Bromide ions to balance the 3+ charge of one Vanadium ion.
- Therefore, the empirical formula is [tex]\( \text{VBr}_3 \)[/tex].
4. Vanadium Sulfide (V₂S₅):
- Vanadium ion: [tex]\( V^{5+} \)[/tex]
- Sulfide ion: [tex]\( S^{2-} \)[/tex]
- To balance the charges, we need multiple Vanadium and Sulfide ions such that the total positive and negative charges are the same. We can use 2 Vanadium ions and 5 Sulfide ions:
- [tex]\( 2 \times 5^+ = 10^+ \)[/tex]
- [tex]\( 5 \times 2^- = 10^- \)[/tex]
- Therefore, the empirical formula is [tex]\( \text{V}_2\text{S}_5 \)[/tex].
So, the empirical formulas of the four binary ionic compounds formed are:
[tex]\[
\begin{aligned}
&\text{1. Magnesium Bromide: MgBr}_2 \\
&\text{2. Magnesium Sulfide: MgS} \\
&\text{3. Vanadium(III) Bromide: VBr}_3 \\
&\text{4. Vanadium Sulfide: V}_2\text{S}_5 \\
\end{aligned}
\][/tex]
