some chemical reactants are listed in the table below. complete the table by filling in the oxidation state…

Explanation:

Step1: Oxidation state of O in (H_3O^+)

Let the oxidation state of O be (x). For (H_3O^+), we know that the oxidation state of H is (+ 1). Using the formula: (3\times(+1)+x=+1) (since the overall charge of (H_3O^+) is (+1)) (3 + x=+1) (x=+1 - 3=-2)

Step2: Oxidation state of C in (CO_3^{2-})

Let the oxidation state of C be (y). The oxidation state of O is (-2). Using the formula: (y+3\times(-2)=-2) (since the overall charge of (CO_3^{2-}) is (-2)) (y-6=-2) (y=-2 + 6=+4)

Step3: Oxidation state of Cr in (CrO_4^{2-})

Let the oxidation state of Cr be (z). The oxidation state of O is (-2). Using the formula: (z+4\times(-2)=-2) (since the overall charge of (CrO_4^{2-}) is (-2)) (z-8=-2) (z=-2 + 8=+6)

Step4: Oxidation state of C in (HCO_3^{-})

Let the oxidation state of C be (w). The oxidation state of H is (+1) and of O is (-2). Using the formula: (+1+w+3\times(-2)=-1) (since the overall charge of (HCO_3^{-}) is (-1)) (1+w - 6=-1) (w=-1+6 - 1=+4)

Answer:

species	oxidation state of highlighted atom
(H_3O^+(aq))	(-2)
(CO_3^{2-}(aq))	(+4)
(CrO_4^{2-}(aq))	(+6)
(HCO_3^{-}(aq))	(+4)