Question

Calculate the oxidation number of the underlined atom in the following molecules:
\[{{H}_{2}}{{C}_{2}}{{O}_{4}}\], \[{{C}_{6}}{{H}_{12}}{{O}_{6}}\],\[P{{b}_{3}}{{O}_{4}}\], \[I{{F}_{7}}\], \[HClO\], \[O{{F}_{2}}\], \[Ni{{(CO)}_{4}}\], \[HAuC{{l}_{4}}\], \[Ba{{O}_{2}}\], \[M{{g}_{3}}{{N}_{2}}\], \[{{O}_{3}}\]

Answer 1

Hint:The oxidation state also known as oxidation number which describes the degree of oxidation i.e. loss of electrons of an atom in a chemical compound. Conceptually the oxidation state may be positive, negative or zero.

Complete step-by-step answer:Oxidation states are typically represented by integers which may be positive, zero, or negative. In some cases, the average oxidation state of an element is a fraction. The highest known oxidation state is reported to be +9 in the tetrox iridium (IX) cation \[Ir{{O}_{4}}^{+}\]. It is predicted that even a +10 oxidation state may be achievable by platinum in the tetrox platinum(X) cation \[Pt{{O}_{4}}^{2+}\]. The lowest oxidation state is −5, as for boron. Now let discuss the oxidation states of all the options given in question:
Oxidation state for hydrogen is +1 and oxygen is -2, then for carbon
\[{{H}_{2}}{{C}_{2}}{{O}_{4}}=2\times 1+2x+4\times -2=3\]
\[{{C}_{6}}{{H}_{12}}{{O}_{6}}=6x+12\times 1+6\times -2=0\]
Oxidation state for lead
\[P{{b}_{3}}{{O}_{4}}=3x+4\times (-2)=\frac{8}{3}\]
Oxidation state of halogens is -1, then oxidation state of iodine
\[I{{F}_{7}}=x-7=7\]
Oxidation state of chlorine is
\[HClO=1+x-2=-1\]
Oxidation state of oxygen
\[O{{F}_{2}}=x-2=2\]
Oxidation state of nickel, carbonyl group has +2 charge
\[Ni{{(CO)}_{4}}=x+8=-8\]
Oxidation state of gold is
\[HAuC{{l}_{4}}=1\times x+(-1\times 4)=3\]
Oxidation state of barium is
\[Ba{{O}_{2}}=x+4=-4\]
Oxidation state of magnesium
\[M{{g}_{3}}{{N}_{2}}=3x+10=-\frac{10}{3}\]
\[{{O}_{3}}=2\times 3=6\]

Note:The increase in oxidation state of an atom, through a chemical reaction, is known as an oxidation; a decrease in oxidation state is known as a reduction. Such reactions involve the formal transfer of electrons: a net gain in electrons being a reduction, and a net loss of electrons being an oxidation. For pure elements, the oxidation state is zero.