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:
The output (Y) of the given logic gate is similar to the output of an/a :
A | B | Y |
0 | 0 | 1 |
0 | 1 | 0 |
1 | 0 | 1 |
1 | 1 | 0 |
List I (Spectral Lines of Hydrogen for transitions from) | List II (Wavelength (nm)) | ||
A. | n2 = 3 to n1 = 2 | I. | 410.2 |
B. | n2 = 4 to n1 = 2 | II. | 434.1 |
C. | n2 = 5 to n1 = 2 | III. | 656.3 |
D. | n2 = 6 to n1 = 2 | IV. | 486.1 |