91Ó°ÊÓ

Begin by writing the chemical symbol and charge for each ion involved in the pairing. These ion charges will help balance the overall charge of the compound.For part (a), ions are \(\mathrm{Fe}^{3+}\) and \(\mathrm{OH}^{-}\).For part (b), ions are \(\mathrm{Cs}^{+}\) and \(\mathrm{NO}_{3}^{-}\).For part (c), ions are \(\mathrm{V}^{2+}\) and \(\mathrm{CH}_{3}\mathrm{COO}^{-}\).For part (d), ions are \(\mathrm{Li}^{+}\) and \(\mathrm{PO}_{4}^{3-}\).For part (e), ions are \(\mathrm{In}^{3+}\) and \(\mathrm{O}^{2-}\).

Create a neutral compound by balancing the total positive and negative charges for each pair of ions.(a) \(\mathrm{Fe}^{3+}\) and \(\mathrm{OH}^{-}\): Since 1 \(\mathrm{Fe}^{3+}\) requires 3 \(\mathrm{OH}^{-}\) ions to balance the charge, the formula is \(\mathrm{Fe(OH)}_{3}\).(b) \(\mathrm{Cs}^{+}\) and \(\mathrm{NO}_{3}^{-}\): Both ions have charges that cancel out directly, so the formula is \(\mathrm{CsNO}_{3}\).(c) \(\mathrm{V}^{2+}\) and \(\mathrm{CH}_{3}\mathrm{COO}^{-}\): 1 \(\mathrm{V}^{2+}\) matches with 2 \(\mathrm{CH}_{3}\mathrm{COO}^{-}\) ions, giving the formula \(\mathrm{V(CH}_{3}\mathrm{COO})_{2}\).(d) \(\mathrm{Li}^{+}\) and \(\mathrm{PO}_{4}^{3-}\): 3 \(\mathrm{Li}^{+}\) ions are needed per \(\mathrm{PO}_{4}^{3-}\) ion, resulting in the formula \(\mathrm{Li}_{3}\mathrm{PO}_{4}\).(e) \(\mathrm{In}^{3+}\) and \(\mathrm{O}^{2-}\): To balance the charges, 2 \(\mathrm{In}^{3+}\) ions are needed per 3 \(\mathrm{O}^{2-}\) ions, resulting in \(\mathrm{In}_{2}\mathrm{O}_{3}\).

Check each formula to ensure charge neutrality and correct combination:- \(\mathrm{Fe}^{3+} + 3 \times \mathrm{OH}^{-} = 0\)- \(\mathrm{Cs}^{+} + \mathrm{NO}_{3}^{-} = 0\)- \(\mathrm{V}^{2+} + 2 \times \mathrm{CH}_{3}\mathrm{COO}^{-} = 0\)- \(3 \times \mathrm{Li}^{+} + \mathrm{PO}_{4}^{3-} = 0\)- \(2 \times \mathrm{In}^{3+} + 3 \times \mathrm{O}^{2-} = 0\)If all calculations hold true without errors, the formulas are correct!