Assertion: KMnO_4 is purple in colour due to | vedclass

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(D) In $KMnO_4$,Manganese is in the $+7$ oxidation state.The electronic configuration of $Mn^{+7}$ is $[Ar] 3d^0 4s^0$.Since there are no $d$-electron

Vedclass · Accepted Answer

Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Vedclass · Answer

(D) In $KMnO_4$,Manganese is in the $+7$ oxidation state.The electronic configuration of $Mn^{+7}$ is $[Ar] 3d^0 4s^0$.Since there are no $d$-electrons,the purple colour of $KMnO_4$ cannot be due to $d-d$ transitions.Instead,the colour arises due to ligand-to-metal charge transfer $(LMCT)$ from oxygen to manganese.Thus,both the Assertion and the Reason are true,and the Reason correctly explains the Assertion.

Column $A$	Column $B$
$(1)$ Brass	$(a)$ $Ni$ $(60\%)$,$Cr$ $(40\%)$
$(2)$ Bronze	$(b)$ $Cu$ $(80\%)$,$Sn$ $(20\%)$
$(3)$ Cupronickel	$(c)$ $Cu$ $(90\%)$,$Sn$ $(10\%)$
$(4)$ Nichrome	$(d)$ $Cu$ $(70\%)$,$Zn$ $(30\%)$
	$(e)$ $Cu$ $(75-85\%)$,$Ni$ $(15-25\%)$

Assertion: $KMnO_4$ is purple in colour due to charge transfer.
Reason: There is no electron present in $d$-orbitals of Manganese in $MnO_4^-$.

Similar Questions

Match Column $A$ with Column $B$:

Column $A$ Column $B$

$(1)$ Brass $(a)$ $Ni$ $(60\%)$,$Cr$ $(40\%)$

$(2)$ Bronze $(b)$ $Cu$ $(80\%)$,$Sn$ $(20\%)$

$(3)$ Cupronickel $(c)$ $Cu$ $(90\%)$,$Sn$ $(10\%)$

$(4)$ Nichrome $(d)$ $Cu$ $(70\%)$,$Zn$ $(30\%)$

$(e)$ $Cu$ $(75-85\%)$,$Ni$ $(15-25\%)$

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Assertion: $KMnO_4$ is purple in colour due to charge transfer.Reason: There is no electron present in $d$-orbitals of Manganese in $MnO_4^-$.