Considering the following precipitation reaction: Pb (NO3) 2 (aq) + 2KI (aq) → PbI2 (s) + 2KNO3 (aq) Which ion would NOT be present in the complete ionic equation?

The question is incomplete and confusing.

In the complete ionic equation you write all the ions that are formed. Those are: Pb²⁺, NO₃⁻, K⁺, and I⁻. They all are present in the complete ionic equation.

In the net ionic equation, the spectator ions do not appear. They are: NO₃⁻ and K⁺. They would not be present in the net ionic equation, but they do in the complete ionic equation.

See below the details.

Explanation:

Which compound will not form ions?

1. Write the balanced molecular equation:

Pb (NO₃) ₂ (aq) + 2KI (aq) → PbI₂ (s) + 2KNO₃ (aq)

2. Write the ionizations for the ionic aqueous compounds:

Pb (NO₃) ₂ (aq) → Pb⁺² (aq) + 2NO₃⁻ (aq)

2KI (aq) → 2K⁺ (aq) + 2I⁻ (aq)

2KNO₃ (aq) → 2K⁺ (aq) + 2NO₃⁻ (aq)

3. Write the complete ionic equation:

Pb⁺² (aq) + 2NO₃⁻ (aq) + 2K⁺ (aq) + 2I⁻ (aq) → PbI₂ (s) + 2K⁺ (aq) + 2NO₃⁻ (aq)

Hence, since PbI₂ (s) does not ionize, but stays in solid form, it will not form ions.

All, Pb⁺², NO₃⁻, K⁺, and I⁻ will be present in the total ionic equation.

It is in the net ionic equation that the spectator ions are removed. Those, are NO₃⁻ and K⁺, because they are on both sides of the complete ionic equation.