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.