Let's determine the oxidation states of Mn in the reactants and products:
In MnO$_4^{2-}$: Let $x$ be the oxidation state of Mn.
Then $x + 4(-2) = -2$, so $x = +6$.
In MnO$_4^-$: $x + 4(-2) = -1$, so $x = +7$.
In MnO$_2$: $x + 2(-2) = 0$, so $x = +4$.
The oxidation states of Mn in the reaction are +6, +7, and +4. Therefore, the oxidation states not shown are +2 and +3.
Given below are two statements:
Statement (I): The first ionization energy of Pb is greater than that of Sn.
Statement (II): The first ionization energy of Ge is greater than that of Si.
In light of the above statements, choose the correct answer from the options given below:
List I | List II | ||
A | Down’s syndrome | I | 11th chormosome |
B | α-Thalassemia | II | ‘X’ chromosome |
C | β-Thalassemia | III | 21st chromosome |
D | Klinefelter’s syndrome | IV | 16th chromosome |
The velocity (v) - time (t) plot of the motion of a body is shown below :
The acceleration (a) - time(t) graph that best suits this motion is :