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The period of elements with atomic number \[20\] in the periodic table is:
A. \[1\]
B. \[2\]
C. \[3\]
D. \[4\]

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Last updated date: 20th Jun 2024
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Answer
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Hint: We know that the period of the periodic table probably was originally associated with the new row number but it actually corresponds to the outermost electron shell in which the electron resides for an atom in the ground state.

Step by step answer: Each row is a period. The first period consists only of hydrogen and lithium, through neon, the third sodium through argon. In the periodic table, elements are arranged in the period through the shell; if an element is present in the third period, it means it has \[3\] shells. Like (K, L and M).
According to the question, elements have an atomic number \[20\] , when we arrange it by bohr bury rule, K,L and M shell we can also find the period.
When we do electronic configuration of atomic number \[20\] we get,
$\begin{array}{c}
{\rm{Atomic}}\;{\rm{number}}\;\left( {20} \right) = \begin{array}{*{20}{c}}
K&L&M&N
\end{array}\\
 = \begin{array}{*{20}{c}}
2&8&8&2
\end{array}
\end{array}$
Here we see that, there are four shells present, so the atomic number $20$ is placed in ${{\rm{4}}^{{\rm{th}}}}$ period.
It is well known that the period of an element corresponds to the principal quantum number of the valence shell.
\[{\rm{The}}\;{\rm{electronic}}\;{\rm{configuration}} = 1{s^2}1{s^2}2{p^6}3{s^2}3{p^6}4{s^2}\]
The last \[2\] electron are in the s-orbital of the having the principal quantum number \[4\] since it is an s-block element having \[2\] and since the outermost shell’s principal quantum number is\[4\], the element is present in period \[4\].
Therefore, the correct answer is D.
;
Note: There are two ways by which we can determine the period of the element, by bohr bury rule and by the principal quantum number. Principal quantum numbers represent the period of the element.