IE1 of Cu (3d10 4s1) Is Less Than That Of Zn (3d10 4s2) Because It Is Easy To Remove An Electron From A Half-filled s-orbital As Compared To Fully-filled s-orbital. However, In Case Of IE2, The Electron Is Lost From 3d orbital Of Cu And 4s orbital Of Zn And Loss Of Electron From 4s orbital Is Easier As Compared To Fully-filled 3d orbital.

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Question

IE1 of Cu (3d10 4s1) is less than that of Zn (3d10 4s2) because it is easy to remove an electron from a half-filled s-orbital as compared to fully-filled s-orbital. However, in case of IE2, the electron is lost from 3d orbital of Cu and 4s orbital of Zn and loss of electron from 4s orbital is easier as compared to fully-filled 3d orbital.

Solution

Correct option is

IE1 of Cu is less than that of Zn and IE2 of Cu is more than that of Zn

IE1 of Cu (3d10 4s1) is less than that of Zn (3d10 4s2) because it is easy to remove an electron from a half-filled s-orbital as compared to fully-filled s-orbital. However, in case of IE2, the electron is lost from 3d orbital of Cu and 4s orbital of Zn and loss of electron from 4s orbital is easier as compared to fully-filled 3d orbital.

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