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# How would you identify the period, block and group of the element with the electron configuration $\left[ {Ar} \right]3{d^7}4{s^2}$?

Last updated date: 28th Feb 2024
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Hint: Try to draw the electronic configuration of the element as indicated by Aufbau principle and you will get the group number, period and block of that element from the configuration itself.
$n\left( {period} \right){\text{ }} = 4\;,\;d - block\;,\;Group{\text{ }}9{\text{ }}or{\text{ }}identically{\text{ }}group{\text{ }}VIII{\text{ }}\left( B \right)$

Complete step by step solution:
Aufbau principle expresses that, in the ground condition of a particle or particle, electrons are to be first filled in the nuclear orbitals having lowest accessible energy levels prior to involving higher levels. For instance, the $1s1s$ subshell is to be filled before the $2s$ subshell is involved.
According to the given inquiry, the nuclear number of the element with the electron configuration $\left[ {Ar} \right]3{d^7}4{s^2}$ is $27$.
Along these lines, the electronic configuration of the element with nuclear number $27$ will be,
$1{s^2},2{s^2},2{p^6},3{s^2},3{p^6},4{s^2},3{d^7}$
The symbol $\left[ {Ar} \right]$ in the consolidated electron configuration of this element takes after the electron configuration of a ground-state argon $Ar$ particle. It speaks to all internal shell electrons of this particle. (see this issue for insights regarding dense electron configurations.)
On the highest point of that, a ground-state iota of this element would contain valence electrons $3{d^7}4{s^2}$. It would lie in the period directly under the one containing argon. Argon is the end element of the ${3^{rd}}$ period. Thus, this element is in the fourth period of the periodic table. (What's more, subsequently $n = 4$)
The block on the periodic table has a place which is reliant on the sort of the involved electron orbital of most noteworthy possible energy.
Mentioning the Aufbau Diagram over, the electron of the most elevated possible energy in a ground-state particle of this element lies in a $4d$ orbital (the one out of an orbital that is reached after any remaining involved nuclear orbitals.) That element, accordingly, is situated in the $d$ block of the periodic table.
A particle of this element contains $9$ valence electrons in the ground state. It would along these lines be in $IUPAC$ Group $9$ of the periodic table, which relates to old $IUPAC{\text{ }}Group{\text{ }}VIII$.

Note:
According to Aufbau’s principle, the order in which the orbitals fill up is as follows:$1s,2s,2p,3s,3p,4s,3d,4p,5s,4d,5p,6s,4f,5d,6p,7s,5f,6d,7p$ and so on.