
What is the oxidation number of $Mn$ in $Mn{{O}_{2}}$?
A.$+2$
B.$+4$
C.$-2$
D.$-4$
Answer
572.7k+ views
Hint: The aggregate number of electrons that an atom either absorbs or loses in order to form a chemical bond with another atom, also called oxidation state. By summing up the oxidation of the element present here in the molecule $Mn{{O}_{2}}$ and assuming the oxidation number of manganese as $x$ we will find its oxidation number.
Complete answer:
To calculate the oxidation number there are some general rules that we need to know first which are given as follows:
When we see a neutral compound we say that all oxidation numbers of that compound must add up to zero. To identify the neutral compound is that they don’t have any charge on them for example ${{H}_{2}}O$(water).
If we take about ions so all the oxidation number of it must add up to the charge on the ion.
Now if we talk about the free elements like $Na,Fe$ and ${{H}_{2}}$ so they have an oxidation number of zero.
One important fact about oxygen is that its oxidation number is $-2$ except that with fluorine or peroxide its oxidation is different.
Now we will calculate the oxidation number by considering the rules we have discuss till now,
We will consider here that the total oxidation number is going to add up into zero.
And we will assume the oxidation number of Manganese as $x$.
And as we know the oxidation number of oxygen is $-2$.
Now by putting the data in the form of equation we get:
$x+(-4)=0$
By rearranging the equation we get
$x=+4$
So we get the oxidation number of manganese as $+4$.
So Option B is the correct answer.
Note:
Here manganese oxide is the neutral compound which has its oxidation number equal to zero so as when we calculate the oxidation number of their individual element, we will get their sum as zero.
Transition metal oxides with +1, +2 and +3 oxidation numbers are ionic complexes composed of metal ions and oxide ions. As covalent compounds with covalent metal-oxygen bonds, these transition metal oxides with oxidation numbers +4, +5, +6, and +7 act.
Complete answer:
To calculate the oxidation number there are some general rules that we need to know first which are given as follows:
When we see a neutral compound we say that all oxidation numbers of that compound must add up to zero. To identify the neutral compound is that they don’t have any charge on them for example ${{H}_{2}}O$(water).
If we take about ions so all the oxidation number of it must add up to the charge on the ion.
Now if we talk about the free elements like $Na,Fe$ and ${{H}_{2}}$ so they have an oxidation number of zero.
One important fact about oxygen is that its oxidation number is $-2$ except that with fluorine or peroxide its oxidation is different.
Now we will calculate the oxidation number by considering the rules we have discuss till now,
We will consider here that the total oxidation number is going to add up into zero.
And we will assume the oxidation number of Manganese as $x$.
And as we know the oxidation number of oxygen is $-2$.
Now by putting the data in the form of equation we get:
$x+(-4)=0$
By rearranging the equation we get
$x=+4$
So we get the oxidation number of manganese as $+4$.
So Option B is the correct answer.
Note:
Here manganese oxide is the neutral compound which has its oxidation number equal to zero so as when we calculate the oxidation number of their individual element, we will get their sum as zero.
Transition metal oxides with +1, +2 and +3 oxidation numbers are ionic complexes composed of metal ions and oxide ions. As covalent compounds with covalent metal-oxygen bonds, these transition metal oxides with oxidation numbers +4, +5, +6, and +7 act.
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