
In which of the following compounds, manganese has an oxidation number equal to that of iodine in $KI{O_4}$ ?
(A) Potassium manganate
(B) Potassium permanganate
(C) Manganous chloride
(D) Manganese chloride
Answer
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Hint: You can find the oxidation number of any atom in the compound by use of the equation that the overall charge on the compound is the sum of the oxidation numbers of all the elements.
Complete step by step solution:
Here, we will first find the oxidation number of iodine in $KI{O_4}$ and then find the compound having the same oxidation number of manganese as iodine.
- So, to find the oxidation number of iodine, we will use the following equation.
Overall charge on $KI{O_4}$ = Oxidation number of K + Oxidation number of I + 4(Oxidation number of O)
0 = +1 + Oxidation number of I + 4(-2)
Oxidation number of I = 8-1 = +7
Now, we will find the oxidation number of Mn in the given compounds.
A) Potassium manganate (${K_2}Mn{O_4}$)
We know that
Overall charge on ${K_2}Mn{O_4}$ = 2(Oxidation number of K) + Oxidation number of Mn + 4(Oxidation number of O)
0 = 2(+1) + Oxidation number of Mn + 4(-2)
Oxidation number of Mn = 8-2 = +6
B) Potassium permanganate ($KMn{O_4}$)
We know that
Overall charge on $KMn{O_4}$ = Oxidation number of K + Oxidation number of Mn + 4(Oxidation number of O)
0 = 1 + Oxidation number of Mn + 4(-2)
Oxidation number of Mn = 8-1 =+7
C) Manganous chloride and D) Manganese chloride
- Both of these compounds are the same and have a molecular formula of $MnC{l_2}$.
We know that
Overall charge on $MnC{l_2}$ = Oxidation number of Mn + 2(Oxidation number of Cl)
0 = Oxidation number of Mn + 2(-1)$Mn{O_4}^ - $
Oxidation number of Mn = +2
Thus, from the above calculations, we obtained that the oxidation number of Mn in potassium permanganate is the same as of iodine in $KI{O_4}$.
Thus, the correct answer is (B).
Note: The Mn has an oxidation number of +6 in manganate ions ($Mn{O_4}^{2 - }$). Mn has an oxidation number of +7 in all permanganate ions. Manganous refers to the lowest positive value of oxidation state of manganese which is +2.
Complete step by step solution:
Here, we will first find the oxidation number of iodine in $KI{O_4}$ and then find the compound having the same oxidation number of manganese as iodine.
- So, to find the oxidation number of iodine, we will use the following equation.
Overall charge on $KI{O_4}$ = Oxidation number of K + Oxidation number of I + 4(Oxidation number of O)
0 = +1 + Oxidation number of I + 4(-2)
Oxidation number of I = 8-1 = +7
Now, we will find the oxidation number of Mn in the given compounds.
A) Potassium manganate (${K_2}Mn{O_4}$)
We know that
Overall charge on ${K_2}Mn{O_4}$ = 2(Oxidation number of K) + Oxidation number of Mn + 4(Oxidation number of O)
0 = 2(+1) + Oxidation number of Mn + 4(-2)
Oxidation number of Mn = 8-2 = +6
B) Potassium permanganate ($KMn{O_4}$)
We know that
Overall charge on $KMn{O_4}$ = Oxidation number of K + Oxidation number of Mn + 4(Oxidation number of O)
0 = 1 + Oxidation number of Mn + 4(-2)
Oxidation number of Mn = 8-1 =+7
C) Manganous chloride and D) Manganese chloride
- Both of these compounds are the same and have a molecular formula of $MnC{l_2}$.
We know that
Overall charge on $MnC{l_2}$ = Oxidation number of Mn + 2(Oxidation number of Cl)
0 = Oxidation number of Mn + 2(-1)$Mn{O_4}^ - $
Oxidation number of Mn = +2
Thus, from the above calculations, we obtained that the oxidation number of Mn in potassium permanganate is the same as of iodine in $KI{O_4}$.
Thus, the correct answer is (B).
Note: The Mn has an oxidation number of +6 in manganate ions ($Mn{O_4}^{2 - }$). Mn has an oxidation number of +7 in all permanganate ions. Manganous refers to the lowest positive value of oxidation state of manganese which is +2.
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