Determine the oxidation state of the metals in pbcro4. 1. pb = + 1, cr = + 4 2. pb = + 2, cr = + 4 3. pb = + 2, cr = + 6 4. pb = + 1, cr = + 1 5. pb = + 1, cr = + 6

PbCrO₄ is made of a cation and anion

anion is CrO₄²⁻

we need to then find the oxidation number of Pb which is the same as its charge

when we make compounds, the charges of the cations and anions are exchanged.

Charge of Pb is then Pb²⁺

since charge of monoatomic ion is equal to oxidation number of the ion

oxidation number of Pb = + 2

net charge of CrO₄²⁻ is - 2

oxidation number of Cr + (4 x oxidation number of O) = - 2

oxidation number of O is usually - 2

oxidation number of Cr = y

y + (4 x - 2) = - 2

y - 8 = - 2

y = + 6

oxidation number of Cr is + 6

answer is

3. pb = + 2, cr = + 6