Question

The oxidation number of Mn is maximum in

1 point

(a) MnO2

(b) K2MnO4

(c) Mn3O4

(d) KMnO4.​

Answer 1

Option D KMnO4 is the correct answer.

• The Oxidation number of an element is defined as the total number of elements an element has either lost or gained during the bond formation process.
• The positive value of the oxidation number implies the donation of electrons.
• The negative value of the oxidation number implies receiving electrons from other atoms.
• From the given options, (assume the oxidation state of Mn as x before calculating the final value).

a) MnO2:

=> x + 2(-2) = 0,

=> x = +4. (Oxidation state of Mn in MnO2 is +4).

b) K2MnO4:

=> 2(1) + x + 4(-2) = 0,

=> x = +6. (Oxidation state of Mn in K2MnO4 is +6).

c) Mn3O4:

=> 3x + 4(-2) = 0,

=> x = +8/3. (Oxidation state of Mn in Mn3O4 is +8/3, fractional oxidation state is also possible).

d) KMnO4:

=> 1(1) + x + 4(-2) = 0,

=> x = +7. (Oxidation state of Mn in KMnO4 is +7).

Hence, the oxidation state of Mn is highest in KMnO4. Option D is the correct answer.

Answer 2

To Find:

The maximum oxidation number of Mn $=?$

Step-by-step explanation:

(a) Oxidation number of Mn in $MnO_2$

The oxidation number of oxygen is $-2$

Let x be the oxidation number of Mn in $MnO_2$

$x+2\left(-2\right)=0$

$x-4=0$

$x=+4$

Hence, the oxidation number of Mn in $MnO_2$ is $+4$.

(b) Oxidation number of Mn in $K_2MnO_4$

The oxidation number of potassium is $+1$

Let y be the oxidation number of Mn in $K_2MnO_4$

$2\left(+1\right)+y+4\left(-2\right)=0$

$2+y-8=0$

$y-6=0$

$y=+6$

Hence, the oxidation number of Mn in $K_2MnO_4$ is $+6$.

(c) Oxidation number of Mn in $Mn_3O_4$

The oxidation number of oxygen is $-2$

Let z be the oxidation number of Mn in  $Mn_3O_4$

$3\left(z\right)+4\left(-2\right)=0$

$3z-8=0$

$3z=+8$

$z=\frac{+8}3$

Hence, the oxidation number of Mn in $Mn_3O_4$ is $\frac{+8}3$.

(d) Oxidation number of Mn in $KMnO_4$

The oxidation numbers of potassium and oxygen are $+1$ and $-2$

Let a be the oxidation number of Mn in $KMnO_4$

$1+a+4\left(-2\right)=0$

$1+a-8=0$

$a-7=0$

$a=+7$

Hence, the oxidation number of Mn in $KMnO_4$ is $+7$.

Therefore, the oxidation number of Mn is maximum in (d) $\boldsymbol K\boldsymbol M\boldsymbol n{\boldsymbol O}_{\mathbf4}$