Question

Magnesium, atomic number 12, has the electron configuration $ [Ne]3{s^2} $ . To what group does magnesium belong?

Answer 1

Hint: For this question, you have to know about periodic tables. Modern periodic law tells us about the physical and chemical properties of the elements which are the periodic function of their atomic number. With the help of electronic configuration, we can find the group of the element.

Complete Step By Step Answer:
The modern periodic table is based on the atomic number of the elements. The atomic number of the element is defined as the total number of protons present in the nucleus present in that element. As we know, the modern periodic table consists of $ 18 $ vertical columns which are known as a group. The elements present in the same group have the same characteristic feature which is that they have the same number of valence shell electrons. For $ s $ block; the group of the element is equal to number of valence shell electron, for $ p $ block; the group of the element is equal to number of valence shell electron $ + 10 $ , for $ d $ block; the group of the element is equal to number of electrons in $ number\;of\,electron\;in\;\left( {n - 1} \right)d\;subshell $ + $ \left( {number\;of\;electrons\,in\;ns\,subshell} \right) $ and for $ f $ block, the group of elements is always $ 3 $ .
In the question, the atomic number of magnesium is 12 and its electronic configuration is $ \left[ {Ne} \right]3{s^2} $ so it belongs to the $ s $ block. As we know the $ s $ block contains only two groups. To find the group number we will see the valence shell electron. Magnesium has $ 2 $ valence shell electrons and magnesium belongs to the group $ {2^{nd}} $ of the periodic table.

Note:
In $ s $ block, there are two groups; $ {1^{st}} $ group and the $ {2^{nd}} $ group. To find in which group the element is, we have to check the valence shell electron. If the element has one valence shell electron then it is in the $ {1^{st}} $ group whereas if the element has two valence electrons then it is in the $ {2^{nd}} $ group.