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How many valence electrons does Cr have ?

Last updated date: 13th Jun 2024
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Hint: Chromium belongs to the group \[6\] in periodic table. \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^5}4{s^1}\]is electronic configuration of Chromium in ground state. Atomic number of Chromium is \[24\]. In its outermost shell, six electrons are existing.

Complete step by step answer:
-Atomic number of Chromium is \[24\] .
-It goes to the group \[6\] of the periodic table. Cr is a transition metal in Group \[6\] on the Periodic Table of .
 - The Electronic configuration of Cr is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^5}4{s^1}\].
-The valence electron of Cr grip its \[4s\] and \[3d\] electrons, because they are nearby sufficient in energy that more than one electron can be recycled to bond.
Its electronic configuration as an atom is \[\left[ {Ar} \right]3{d^5}4{s^1}\], so it has 6 valence electrons .
There most definitely cannot be \[4\] or\[\;2\;\] (that's just impossible... you can't neglect some \[3d\] electrons and not the rest and quiet account for all valence electrons).For chromium, it is sufficient of a steadying to maximize its entire spin state by having altogether unpaired electrons in a \[3{d^5}4{s^1}\] configuration as a substitute of\[\;3{d^4}4{s^2}\]
The 3d electrons are not strongly believed by the nucleus. Hence they are also likely to lose. So we can say the no. of valence electrons in Cr are
\[1{\text{ }} + {\text{ }}5{\text{ }} = {\text{ }}6\]
For the record of the d-block elements, the electrons in the d subshell are calculated as valence electrons.
-Valence electrons can be intended by resolute using following steps
While atomic number of elements is known, electronic configuration should be written giving to Aufbau principle, then valence electrons are the number of electrons currently in the outermost shell.
As in Chromium, six electrons are current in the outermost shell, so Chromium has six valence electrons.

Atomic number of Chromium is \[24\]. So the electronic configuration given to the Aufbau principle is \[\left[ {Ar} \right]3{d^5}4{s^1}\] .
Note that after writing the electron configuration for an atom like Cr, the \[3d\] is generally written before the \[4s\]. Together the configurations have the right numbers of electrons in every orbital, it is just a material of how the electronic configuration note is written (why).
Then we have (quiet incorrect) \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^4}4{s^2}\]