Question

write the chemical formula
namechemical formula
nickel (iii) iodide
copper (ii) sulfide
vanadium (iv) chloride
vanadium (v) chloride
gold (iii) oxide
tin (iv) oxide

Explanation:

Step1: Determine the ions

• For Copper (II) sulfide: Copper ion is \(Cu^{2+}\), sulfide ion is \(S^{2 -}\).
• For Vanadium (IV) chloride: Vanadium ion is \(V^{4+}\), chloride ion is \(Cl^{-}\).
• For Vanadium (V) chloride: Vanadium ion is \(V^{5+}\), chloride ion is \(Cl^{-}\).
• For Gold (III) oxide: Gold ion is \(Au^{3+}\), oxide ion is \(O^{2 -}\).
• For Tin (IV) oxide: Tin ion is \(Sn^{4+}\), oxide ion is \(O^{2 -}\).

Step2: Use charge balance

• Copper (II) sulfide: To balance charges (\(+ 2\) and \(-2\)), formula is \(CuS\).
• Vanadium (IV) chloride: \(V^{4+}\) and \(Cl^{-}\), so \(VCl_{4}\) (\(4\times(- 1)+4 = 0\)).
• Vanadium (V) chloride: \(V^{5+}\) and \(Cl^{-}\), so \(VCl_{5}\) (\(5\times(-1)+5 = 0\)).
• Gold (III) oxide: \(Au^{3+}\) and \(O^{2 -}\), cross - multiply charges: \(Au_{2}O_{3}\) (\(2\times3+3\times(-2)=0\)).
• Tin (IV) oxide: \(Sn^{4+}\) and \(O^{2 -}\), cross - multiply charges: \(SnO_{2}\) (\(4 + 2\times(-2)=0\)).

Answer:

Copper (II) sulfide: \(CuS\)
Vanadium (IV) chloride: \(VCl_{4}\)
Vanadium (V) chloride: \(VCl_{5}\)
Gold (III) oxide: \(Au_{2}O_{3}\)
Tin (IV) oxide: \(SnO_{2}\)