
Choose the element from the following that has the electronic configuration:
\[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}4{{s}^{1}}\]
A. $Fe$
B. $Co$
C. $Ni$
D. $Cu$
Answer
510.6k+ views
Hint: Consider the order in which the orbitals are filled according to their energies. Think about the energy stability of the orbitals when they are empty, partially-filled, half-filled, or completely filled.
Complete step by step solution:
We know that the electrons occupy the orbitals that have a lower energy, first. So, the orbitals with the lowest energy will be filled first and the orbitals with the highest energy will be filled up last. The order of the orbitals ranking from the lowest to the highest energy orbitals is:
\[1s<2s<2p<3s<3p<4s<3d\]
In the example given in the question, we can see that the $3d$ orbital has started getting filled up before the $4s$ orbital is completely filled despite it having a lower energy. The $4s$ is seen to have only 1 electron whereas the $3d$ orbital has 10 electrons.
We know that the order of stability for orbitals is:
Completely-filled orbitals $>$ half-filled orbitals $>$ partially filled orbitals
When the element has the configuration as $4{{s}^{2}}3{{d}^{9}}$, it has one completely filled and one partially filled orbital. But, when it is in the configuration $4{{s}^{1}}3{{d}^{10}}$, it has 1 completely filled and 1 half filled orbital. This makes the electronic configuration more stable.
The atomic number of this element as calculated from the number of electrons given in the configuration is 29. We know that the element copper has an atomic number of 29.
Hence, the answer to this question is D. $Cu$’
Note: Another element that shows a similar trend is Chromium. The atomic number of chromium is 24. It shows the electronic configuration of \[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{5}}4{{s}^{1}}\] instead of the configuration \[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{4}}4{{s}^{2}}\]. This again happened due to the stability patterns of the orbitals.
Complete step by step solution:
We know that the electrons occupy the orbitals that have a lower energy, first. So, the orbitals with the lowest energy will be filled first and the orbitals with the highest energy will be filled up last. The order of the orbitals ranking from the lowest to the highest energy orbitals is:
\[1s<2s<2p<3s<3p<4s<3d\]
In the example given in the question, we can see that the $3d$ orbital has started getting filled up before the $4s$ orbital is completely filled despite it having a lower energy. The $4s$ is seen to have only 1 electron whereas the $3d$ orbital has 10 electrons.
We know that the order of stability for orbitals is:
Completely-filled orbitals $>$ half-filled orbitals $>$ partially filled orbitals
When the element has the configuration as $4{{s}^{2}}3{{d}^{9}}$, it has one completely filled and one partially filled orbital. But, when it is in the configuration $4{{s}^{1}}3{{d}^{10}}$, it has 1 completely filled and 1 half filled orbital. This makes the electronic configuration more stable.
The atomic number of this element as calculated from the number of electrons given in the configuration is 29. We know that the element copper has an atomic number of 29.
Hence, the answer to this question is D. $Cu$’
Note: Another element that shows a similar trend is Chromium. The atomic number of chromium is 24. It shows the electronic configuration of \[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{5}}4{{s}^{1}}\] instead of the configuration \[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{4}}4{{s}^{2}}\]. This again happened due to the stability patterns of the orbitals.
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