Question

The atomic numbers of $Ni$ and $Cu$ are $28$ and $29$ respectively. The electron configuration $Is^2 2s^2 2p^6 3s^2 3p^6 3d^{10}$ represents

• A. $Cu^{+}$
• B. $Cu^{2+}$
• C. $Ni^{2+}$
• D. $Ni$

Answer 1

The Correct Option is A

Electronic configuration of

$_{28}Ni = 1s^2, \,2s^2 \,2p^6, \, 3s^2\,3p^6\,3d^8, \,4s^2$
$ Ni^{2+} = 1s^2, \,2s^2\,2p^6, \,3s^2 \,3p^6\, 3d^8, \,4s^0$
$_{29}Cu = 1s^2 ,\,2s^2\,2p^6 ,3s^2\,3p^63d^{10} ,\,4s^1$
$Cu^+ = Is^2,\, 2s^2\,2p^6, 3s^2\,3p^6\,3d^{10}, \,4s^0$

So, the given configuration is of $ Cu^+$

Answer 2

Ans: The distribution of electrons in the atomic orbitals can be known by looking at their electronic configuration. It helps in determining the valency of the element and the properties of the elements. 

The maximum number of electrons that can be added to a shell is because of th =e principal quantum number that is represented by n. The formula for finding the maximum number of electrons is 2n² . The maximum number of electron that can be added in the shells are as follows: 

• K shell: 2 
• L shell: 8
• M shell: 18 
• N shell: 32 

The order of filing of the electrons are 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹º4p⁶ 5s² …..

In the question, Ni has the atomic number of 28. 

So the electronic configuration will be 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁸

Now, in Ni²⁺, two electrons  has been removed. According to the Aufbau principle the after filling the electrons the electronic configuration will be 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸

The atomic number of Cu is 29. Its electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹º. This is so because half filled and fully filled shells are more stable. 

Now for Cu⁺, there  is a loss of 1 electron thus the electronic configuration will be 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹º. 

In Cu²⁺, there is loss of two electrons, the electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁹

Thus the answer is Cu⁺