Question

Atomic numbers of few elements are given below. Which of the pairs belongs to s-block?
A. 7, 14
B. 3, 20
C. 8, 15
D. 9, 17

Answer 1

As we know that s-block elements are those in which the last electron enters the ns energy shell. We can see that the electronic shell configuration of these are $n{{s}^{1}}$ o r$n{{s}^{2}}$.
- These are found to lie on the left side of the periodic table. Those elements that are having $n{{s}^{1}}$ configuration are called alkali metals. Whereas, those having $n{{s}^{2}}$ configuration are called alkaline earth metals.

Complete Solution :
- Let’s discuss about some of the general characteristics of alkali metals:
- As we know that on moving down the group the atomic number increases due to which atomic radius of alkali metals also increases.
- It is found that the size of the atoms increases down the group due to which the attraction between the nucleus and the electrons present in the outermost shell decreases. Due to this there is a decrease in ionization enthalpy takes place.
- It is also found that alkali metals have a low boiling and melting point.
- As we know that alkali metals have the electronic configuration of $n{{s}^{1}}$. Hence, the element Z=3 will have the electronic configuration of $1{{s}^{2}}2{{s}^{1}}$, and hence is an alkali metal.
- As we know that alkaline metals have the electronic configuration of $n{{s}^{2}}$. Hence, the element Z = 20 will have the electronic configuration of $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}$, and hence is an alkaline metal.
So, the correct answer is “Option B”.

Note: - We must note the difference in between the alkali and alkaline earth metals. As, alkali metals are those in which there is one valence shell present and have lower values of ionization enthalpy. Whereas, alkaline earth metals have two valence shell electrons and have higher values of ionization enthalpy.