Question

Which of the following can react with ${{K}_{2}}C{{r}_{2}}{{O}_{7}}?$
(A) $S{{O}_{3}}^{2-}$
(B) $C{{O}_{3}}^{2-}$
(C) $S{{O}_{4}}^{2-}$
(D) $N{{O}_{3}}^{-}$

Answer 1

Hint: To solve this question we have to find the oxidation state of each element given in the option. Oxidation number also known as oxidation state is defined as the total number of electrons which an atom either gain or loses to form a chemical bond with other atom or we can say that the oxidation number is the charge an atom would get if the compound were made up of ions.

Complete step by step solution:
Let’s find the oxidation state in each of the given option:
Oxidation state of S in $S{{O}_{3}}^{2-}$
Oxidation state of O in the given compound= -2
Oxidation state of the sulphur=\[x+3(-2)=-2\]
$x = +4$
Oxidation state of C in $C{{O}_{3}}^{2-}$
Oxidation state of O in the given compound= -2
Oxidation state of the C=\[x+3(-2)=-2\]
$x = +4$
Oxidation state of S in $S{{O}_{4}}^{2-}$
Oxidation state of O in the given compound= -2
Oxidation state of the sulphur=\[x+4(-2)=-2\]
$x = +6$
Oxidation state of N in $N{{O}_{3}}^{-}$
Oxidation state of O in the given compound= -2
Oxidation state of the sulphur=\[x+3(-2)=-1\]
$x = +5$
Except $S{{O}_{3}}^{2-}$all other compounds are already present in their maximum oxidation state hence $S{{O}_{3}}^{2-}$ can easily react with ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$

Hence the correct answer is option (A) i.e. $S{{O}_{3}}^{2-}$.

Note: The oxidation number of an atom in the neutral material containing atoms of only one type of element is zero. Which means that the oxidation number of ${{O}_{2}},{{O}_{3}},{{P}_{4}}$ etc. is zero. The number of simple ions oxidizing is equal to the charge on the ion, and in a neutral compound the sum of the oxidation state of all the elements is zero.