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**Hint:**In order to find the period of an element we must know the electronic configuration of that element. And the highest number in the electronic configuration, which denotes the valence electron shell, is the period to which the element belongs to.

**Complete step by step answer:**

We know that by saying the “tenth” element, we are actually referring to the element’s atomic number. Hence, we have to find out the period to which the element having atomic number $10$ belongs to. The number of electrons in any element is equal to its atomic number. Hence, this element has a total of $10$ electrons. We will now try to devise the electronic configuration of this element by using the Aufbau principle. As we know, the first orbital in any electronic configuration is $1s$, which can hold a total of $2$ electrons. The next is $2s$, which can also hold $2$ electrons. Next comes $2p$, which can hold a total of six electrons. As we can see, from these first three orbitals itself fill up our element’s $10$ electrons and there are no more electrons to be added. Hence, the electronic configuration is:

$1{s^2}2{s^2}2{p^6}$

From this, we can see that the highest shell being filled is the second shell and it is the valence (outermost) shell for this element. Hence, the element belongs to period two.

**So, the correct answer is Option B.**

**Note:**The highest shell, or the valence shell, also denotes the element’s principal quantum number, denoted by $n$. The Aufbau principle states that electrons are filled into atomic orbitals in the increasing order of orbital energy level. That is, $1s$ is filled before $2s$, which is filled before $2p$ and so on. Note that we can also approach this question like this: We know the first shell can accommodate $2$ electrons and the second shell can accommodate a total of $8$ electrons. Therefore, the total of $10$ electrons are filled in the first two shells, and hence, as the period number is equal to the total number of shells, this element belongs to the second period.

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