Question

1. write the formula for lead (ii) oxide (0.5 points) *

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1. write the formula for manganese (ii) oxide (0.5 points) *

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1. write the formula for tin (ii) chloride (0.5 points) *

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1. write the formula for iron (iii) oxide (0.5 points) *

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Explanation:

Step1: Determine the ions

• Lead (II) has a charge of \(Pb^{2+}\), oxide has a charge of \(O^{2 -}\).
• Manganese (II) has a charge of \(Mn^{2+}\), oxide has a charge of \(O^{2 -}\).
• Tin (II) has a charge of \(Sn^{2+}\), chloride has a charge of \(Cl^{-}\).
• Iron (III) has a charge of \(Fe^{3+}\), oxide has a charge of \(O^{2 -}\).

Step2: Use the criss - cross method

• For Lead (II) Oxide: \(Pb^{2+}\) and \(O^{2 -}\), the formula is \(PbO\) (since \(\frac{2}{2}=1\) for both ions).
• For Manganese (II) Oxide: \(Mn^{2+}\) and \(O^{2 -}\), the formula is \(MnO\) (since \(\frac{2}{2}=1\) for both ions).
• For Tin (II) Chloride: \(Sn^{2+}\) and \(Cl^{-}\), using criss - cross (swap the charges as sub - scripts), the formula is \(SnCl_{2}\) (because \(2\) from \(Sn^{2+}\) goes to \(Cl\) and \(1\) from \(Cl^{-}\) goes to \(Sn\)).
• For Iron (III) Oxide: \(Fe^{3+}\) and \(O^{2 -}\), using criss - cross (swap the charges as sub - scripts), the formula is \(Fe_{2}O_{3}\) (because \(3\) from \(Fe^{3+}\) goes to \(O\) and \(2\) from \(O^{2 -}\) goes to \(Fe\)).

Answer:

1. \(PbO\)
2. \(MnO\)
3. \(SnCl_{2}\)
4. \(Fe_{2}O_{3}\)