Assertion: $KMn{O_4}$ is purple in colour due to charge transfer.
Reason: There is no electron present in d-orbitals of manganese in $MnO_4^ - .$
(A) Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
(B) Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
(C) Assertion is correct but Reason is incorrect
(D) Assertion is incorrect but Reason is correct
Answer
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Hint: $Mn$ has 5 electrons in its d-orbital. So it can take one electron. This transfer of electrons takes place in the presence of visible light. The oxidation state of $Mn$ in $KMn{O_4}$ is $ + 7$. So, the d-orbital will have no electron present it for d-d transition.
Complete Step by step answer: In $KMn{O_4}$, let the oxidation state of $Mn$ be $x$
The oxidation state of $K$ is 1.
The oxidation state of $O$ is $ - 2$
Thus the oxidation state of ${O_4}$ is $ - 8$
Since the compound together is neutral, we can write
$1 + x - 8 = 0$
$ \Rightarrow x = 7$
Thus the oxidation state of $Mn$ in $KMn{O_4}$ is $ + 7$
Therefore, there is no electron present in the d-orbital.
Now, transition metals show different colors due to the d-d transition. That is, the transfer of electrons from d-orbitals. But in this case, the color in $Mn$ is not because of d-d transition.
The color in $Mn$ is because of the charge transfer from oxygen to $Mn$ in visible light. So the color is due to the charge transfer spectra. And that color is purple.
Therefore, the assertion is correct.
Now, let the oxidation state of $Mn$ in $MnO_4^ - $ be $x$
Oxidation state of $O$ is $ - 2$
So, oxidation state of ${O_4}$ is $ - 8$
Hence, oxidation state of $O_4^ - $ is $ - 7$
Thus, $x - 7 = 0$
Therefore, oxidation state of $Mn$ is $ + 7$
Now, the electronic configuration of $Mn$ is $\left[ {Ar} \right]3{d^5}4{s^2}$
So for $ + 7$ oxidation state, there will be no electron in the d orbital of $Mn$. Therefore, reason is also correct.
But since, the color of $KMn{O_4}$ is not due to d-d transition, reason is not the correct explanation of assertion.
Therefore, from the above explanation the correct option is (B) both assertion and reason are correct but reason is not the correct explanation for Assertion.
Note: The color of $KMn{O_4}$ is due to the charge transfer transition by absorption of light. To understand this question, you need to know how to find oxidation state and how oxidation state affects the properties of different elements. In this question, finding out the oxidation state of $Mn$ was the key concept to solve it.
Complete Step by step answer: In $KMn{O_4}$, let the oxidation state of $Mn$ be $x$
The oxidation state of $K$ is 1.
The oxidation state of $O$ is $ - 2$
Thus the oxidation state of ${O_4}$ is $ - 8$
Since the compound together is neutral, we can write
$1 + x - 8 = 0$
$ \Rightarrow x = 7$
Thus the oxidation state of $Mn$ in $KMn{O_4}$ is $ + 7$
Therefore, there is no electron present in the d-orbital.
Now, transition metals show different colors due to the d-d transition. That is, the transfer of electrons from d-orbitals. But in this case, the color in $Mn$ is not because of d-d transition.
The color in $Mn$ is because of the charge transfer from oxygen to $Mn$ in visible light. So the color is due to the charge transfer spectra. And that color is purple.
Therefore, the assertion is correct.
Now, let the oxidation state of $Mn$ in $MnO_4^ - $ be $x$
Oxidation state of $O$ is $ - 2$
So, oxidation state of ${O_4}$ is $ - 8$
Hence, oxidation state of $O_4^ - $ is $ - 7$
Thus, $x - 7 = 0$
Therefore, oxidation state of $Mn$ is $ + 7$
Now, the electronic configuration of $Mn$ is $\left[ {Ar} \right]3{d^5}4{s^2}$
So for $ + 7$ oxidation state, there will be no electron in the d orbital of $Mn$. Therefore, reason is also correct.
But since, the color of $KMn{O_4}$ is not due to d-d transition, reason is not the correct explanation of assertion.
Therefore, from the above explanation the correct option is (B) both assertion and reason are correct but reason is not the correct explanation for Assertion.
Note: The color of $KMn{O_4}$ is due to the charge transfer transition by absorption of light. To understand this question, you need to know how to find oxidation state and how oxidation state affects the properties of different elements. In this question, finding out the oxidation state of $Mn$ was the key concept to solve it.
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