Question: 55. Consider the following compounds:
\underline{\mathrm{K O}}_{2}, \mathrm{H}_{2} \underline{\mathrm{O}}_{2} and \mathrm{H}_{2} \underline{\mathrm{S O}}_{4}
The oxidation states of the underlined elements in them are, respectively,
(1) +1,-1, and +6
(2) +2,-2, and +6
(3) +1,-2, and +4
(4) +4,-4, and +6
Answer: Option (1)
Explanation:
We determine the oxidation state of the underlined element in each compound.
In \mathrm{KO_2}, potassium is an alkali metal and has oxidation state +1. Therefore, the underlined potassium has oxidation state +1.
In \mathrm{H_2O_2} (hydrogen peroxide),
oxygen is present in a peroxide.
The oxidation state of oxygen in peroxides is -1.
Hence, the underlined oxygen has oxidation state -1.
In \mathrm{H_2SO_4}, let the oxidation state of sulfur be x.
Hydrogen has oxidation state +1 and oxygen has oxidation state -2.
2(+1)+x+4(-2)=0 2+x-8=0 x=+6Thus, the oxidation states are +1, -1, and +6, respectively.
Therefore, the correct answer is Option (1).