Answer
424.5k+ views
Hint: According to Hund’s principle which states that, when electron starts filling up subshells, they do it such that the electrons of the same spin must solely occupy the orbitals within the subshell first and then the electrons of opposite spin will start filling up the remaining space in the orbitals.
Complete answer:
We know that the order of stability is defined as:
Fully-filled orbital $>$ half-filled orbital $>$ partially-filled orbital
We also know that s-orbital can accommodate a maximum of 2 electrons, p-orbital can accommodate a maximum of 6 electrons, d-orbital can accommodate a maximum of 10 electrons and f-orbital can accommodate a maximum of 14 electrons.
So, in $4{{s}^{1}}\text{ }3{{d}^{5}}$we see that both the s and p orbitals are half-filled. In case of chromium, the expected electronic configuration is $4{{s}^{2}}\text{ }3{{d}^{4}}$, yet actually we see that one electron from 4s gets transferred to 3d orbital making it $4{{s}^{1}}\text{ }3{{d}^{5}}$. This happens because $4{{s}^{1}}\text{ }3{{d}^{5}}$ being a more stable configuration.
Electron orbitals are most stable when fully-filled or half-filled, hence the most stable configuration of electrons for 3d subshells is either $3{{d}^{10}}$or $3{{d}^{5}}$. In the case of chromium, after $4{{s}^{2}}\text{ }3{{d}^{4}}$ configuration is attained, and electron from s-orbital gets transferred to 3d subshell because $3{{d}^{5}}$ is much more stable configuration than $3{{d}^{4}}$. This is why the configuration for chromium is $4{{s}^{1}}\text{ }3{{d}^{5}}$.
So, the correct answer is “Option A”.
Note: In the case of chromium, it is an exception of the Aufbau principle and the systematic rule of the principle does not comply with its electron configuration. The classical rule that there are orbitals with different energy levels state that
$1s\text{ }<\text{ }2s\text{ }<\text{ }2p\text{ }<\text{ }3s\text{ }<\text{ }3p\text{ }<\text{ }4s\text{ }<\text{ }3d\text{ }<\text{ }4p$ and so on. Same results are seen in $Cu$ where the expected configuration is $[Ar]4{{s}^{2}}3{{d}^{9}}$ but the observed configuration is $[Ar]4{{s}^{1}}3{{d}^{10}}$
Complete answer:
We know that the order of stability is defined as:
Fully-filled orbital $>$ half-filled orbital $>$ partially-filled orbital
We also know that s-orbital can accommodate a maximum of 2 electrons, p-orbital can accommodate a maximum of 6 electrons, d-orbital can accommodate a maximum of 10 electrons and f-orbital can accommodate a maximum of 14 electrons.
So, in $4{{s}^{1}}\text{ }3{{d}^{5}}$we see that both the s and p orbitals are half-filled. In case of chromium, the expected electronic configuration is $4{{s}^{2}}\text{ }3{{d}^{4}}$, yet actually we see that one electron from 4s gets transferred to 3d orbital making it $4{{s}^{1}}\text{ }3{{d}^{5}}$. This happens because $4{{s}^{1}}\text{ }3{{d}^{5}}$ being a more stable configuration.
![seo images](https://www.vedantu.com/question-sets/17f572ba-35f0-4b6a-8119-979cdd44c2722797076504277655382.png)
Electron orbitals are most stable when fully-filled or half-filled, hence the most stable configuration of electrons for 3d subshells is either $3{{d}^{10}}$or $3{{d}^{5}}$. In the case of chromium, after $4{{s}^{2}}\text{ }3{{d}^{4}}$ configuration is attained, and electron from s-orbital gets transferred to 3d subshell because $3{{d}^{5}}$ is much more stable configuration than $3{{d}^{4}}$. This is why the configuration for chromium is $4{{s}^{1}}\text{ }3{{d}^{5}}$.
So, the correct answer is “Option A”.
Note: In the case of chromium, it is an exception of the Aufbau principle and the systematic rule of the principle does not comply with its electron configuration. The classical rule that there are orbitals with different energy levels state that
$1s\text{ }<\text{ }2s\text{ }<\text{ }2p\text{ }<\text{ }3s\text{ }<\text{ }3p\text{ }<\text{ }4s\text{ }<\text{ }3d\text{ }<\text{ }4p$ and so on. Same results are seen in $Cu$ where the expected configuration is $[Ar]4{{s}^{2}}3{{d}^{9}}$ but the observed configuration is $[Ar]4{{s}^{1}}3{{d}^{10}}$
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