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

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.