Question

Calculate the oxidation number of underlined atoms in the following compounds
And ions $C{{H}_{4,}}S{{b}_{2}}{{0}_{5}},{{C}_{6}}{{H}_{12}}{{O}_{6}}$ ?

Answer 1

Hint : Oxidation state of a compound is equal to the sum of charge over the compound. The oxidation state of group $1$ metals is considered to be positive one $+1$ and of group $2$ metals is considered to be positive two $+2$ .

Complete Step By Step Answer:
The oxidation state of oxygen is generally considered negative two $-2$ but changes to negative one $1$ in case of peroxide and positive one $+1$ in case of ${{F}_{2}}O$ .
Hydrogen oxidation number is generally taken as positive one $1$ but in case of metal hydrides it is taken as negative one $1$ .
The overall sum of oxidation number of a compound or ion is equal to the charge on the compound or ion.
For the given $3$ compounds let’s check them one by one
$\Rightarrow 1-C{{H}_{4}}$
In this compound oxidation state of H is $+1$ and $4$ hydrogen atoms are their let oxidation state of C be n than
$\Rightarrow n+4 \times 1=0$
$\Rightarrow n=-4$
As total charge on $C{{H}_{4}}$ is zero
$\Rightarrow 2-\text{ }S{{b}_{2}}{{O}_{5}}$
In this compound oxidation state of O will be $-2$ and $5$ O atoms are present let the oxidation state of Sb atom be n than
$\Rightarrow 2n+5 \times (-2)=0$
$\Rightarrow 2n=10$
$\Rightarrow n=+5$
As the total charge on the compound is zero.
$\Rightarrow 3\text{ }-\text{ }{{C}_{6}}{{H}_{12}}{{O}_{6}}$
In this compound oxidation state of H is $+1$ and $6$ H atoms are there and of O is $-2$ and $6$ O atoms, let the oxidation state of C atom be n than
$\Rightarrow 6n+12 \times (1)+6 \times (-2)=0$
$\Rightarrow 6n+12-12=0$
$\Rightarrow 6n=0$
$\Rightarrow n=0$

Note :
Oxidation state can also be calculated using the structures of compounds. The more electronegative elements have negative oxidation number and less electronegative number have negative oxidation state.