Question

If an element A belongs to Period $3$ and Group $II$ then find the number of shells and valence electrons corresponding to the element A.
A.$3$ shells and $2$ valence electrons
B.$2$ shells and $3$ valence electrons
C.$3$ shells and $3$ valence electrons
D.$2$ shells and $2$ valence electrons

Answer 1

We know that the element that belong to the period $3$ as well as group II is the magnesium $\left( Mg \right)$ The number of shells an element has is equal to number of periods of periodic element and it belong to. The number of valence electrons of an element is equal to period of periodic elements then it belongs to.

Complete step-by-step answer: An element A which belongs to period $3$ as well as group $II$ of periodic elements which we have to determine that is number of shell and valence electron in element A.
Basically, the concept involved here is in a modern periodic table aligned with horizontal rows called as period and vertical columns called as groups. Whereas, period indicates number of shells whereas the group number indicates number of valence electrons.
Thus it's given that element A belong to period $3$ as well as group $II$
Now, Number of shells $=$ Period to which element and hence number of shells $=3$
Number of valence electrons $=$ Group to which the element and hence number of valence electrons $=2$
Therefore the given element is magnesium $\left( Mg \right)$
Hence, the correct answer is A i.e. the number of shell and valence electrons corresponding to the element A are $3$ shells and $2$ valence electrons.

Note: Note that in atomic physics, an electron shell may be of an orbit which is followed by electrons around the nucleus. Whereas the valence electron is outer shell electron that is associated with atom as well as that could only participate in formation of chemical bond if outer shell isn't closed in single covalent bond and both atoms are in bond in order to contribute one valence electron in order to form a shared pair.