Answer to Question #161560 in General Chemistry for mike

Question

 Determine the oxidation number of the element underlined in each of the following.

a). Zn(OH)42- b). CrO2Cl2 c). NO2- d). Cr2O3

Answer 1

The oxidation number of oxygen O is -2 in the majority of its compounds. Similarly, the oxidation number of H is almost always equal to +1.

The oxidation number of an element in a given compound can be calculated as the total formal charge of the ion (equal to zero for neutral molecules) minus the sum of the formal charges of the other elements, this difference divided by the number of the atoms of the element in the compound.

a) Zn(OH)₄^2-: the oxidation number of Zn is: $-2 -(-2)·4 - (+1)·4 = +2$ .

b) CrO₂Cl₂: the oxidation number of Cl is -1, so the oxidation number of Cr is : $0-(-1)·2-(-2)·2 = +6$ .

c) NO₂^-: in this nitrite anion, the oxidation number of N is: $-1 - (-2)·2 = +5$ .

d) Cr₂O₃: the oxidation number of Cr in this compound is: $(0 - (-2)·3)/2 = +3$ .