Question

What is the oxidation state of cr in CrPO4 and Cr3(PO4)2?

Answer 1

Answer:

Explanation:

Oxidation state of Cr in CrPO₄

As a general rule, the algebraic sum of all the oxidation numbers of all atoms in a neutral compound is zero.

The compound above is in its neutral state and we sum all the oxidation numbers and equate to zero:

The oxidation number of P is -2

                                           O is -2

Let the oxidation number of Cr be x:

                              x + (-2) + 4(-2) = 0

                              x -2-8 = 0

                               x -10 = 0

                                x = +10

For Cr in Cr₃(PO₄)₂

         Using the same rule:

        2(x) + 2[-2 + 4(-2)] = 0

           2x + 2(-2-8) = 0

            2x -20 = 0

                x = +10

Answer 2

Answer:

$CrPO_4: Cr +3\\Cr_3(PO_4)_2: Cr+2$

Explanation:

$CrPO_4$ and $Cr_3(PO_4)_2$ are compounds known as ternary salts

This means that they are formed by a metal or a non-metal and an anion

 Its formula is $M_x(X_yO_z)_w$, that is, the cation is written first and then the anion and the simplified charges are exchanged if possible.

The anion $PO_4$ has a load of -3: $PO_{4}^{-3}$

Let's look at the first compound $CrPO_4$ we observe that when exchanging the charges 3 of the$PO_4$ does not appear therefore the charges are simplified as the charges are completely simplified, it means that the chromium has the same valence (numerically but with opposite sign) that the anion

Therefore the oxidation state of Cr in $CrPO_4$ is +3

Let's look at the second compound $Cr_3(PO_4)_2$, it is observed that when exchanging the valences, the 3 of the $PO_4$ is with the chromium, and with the anion is  2

As valencia are not multiples, they cannot be simplified.

When exchanging the valences, the$PO_4$ has the valence corresponding to the chromium which in this case is + 2