Answer
a.
Work Step by Step
(a) In manganese (II) phosphate, we see that this particular atom of manganese (Mn) has an oxidation number of $2+$ by looking at the Roman numeral next to it. We know that phosphate ($PO_4$) has an oxidation number of $3-$. In this compound, manganese and phosphate combine in a three-to-two ratio, which means we have three manganese atoms for every two phosphate ions. The formula for this compound should be $Mn_3(PO_4)_2$; therefore, the formula given is incorrect.
(b) For nickel (II) bromide, we see that this particular atom of nickel (Ni) has an oxidation number of $2+$, and bromine (Br) has an oxidation number of $1-$. This means that we need to have one nickel atom for every two bromine atoms. Therefore, the formula for this compound should be $NiBr_2$; therefore, the formula given is correct.
(c) In chromium (III) sulfide, this particular atom of chromium (Cr) has an oxidation number of $3+$, and sulfur (S) has an oxidation number of $2-$. Chromium and sulfur combine in a two-to-three ratio, which means we will need two atoms of chromium for every three atoms of sulfur. Therefore, the formula for this compound should be $Cr_2S_3$; therefore, the formula given is correct.
(d) For copper (II) sulfate hexahydrate, we see that this particular atom of copper (Cu) has an oxidation number of $2+$, and sulfate ($SO_4$) has an oxidation number of $2-$. This particular atom of copper combines with sulfate in a two-to-two ratio; we reduce this ratio to its lowest whole number equivalent, meaning we need one copper atom for every one sulfate ion. Therefore, the formula for this compound should be $CuSO_4 · 6H_2O$; therefore, the formula given is correct.
