Question

Which is the most non-metallic among the following ?
A. $1{s^2}2{s^2}2{p^6}3{s^1}$
B. $1{s^2}2{s^2}2{p^5}$
C. $1{s^2}2{s^2}2{p^6}3{s^2}$
D. $1{s^2}2{s^2}2{p^3}$

Answer 1

Hint: We will solve this problem with the help of a trend of periodic table that is in a period on moving from left to right metallic character decreases and non-metallic character increases. So first of all we have to find out the elements from their given electronic configurations.

Complete step by step solution:
> Non-metals are poor conductors of electricity and heat. It is malleable or ductile.
> Non-metals are kept at the right side of the periodic table and possess high ionization energies and electron affinities. On moving right to left in a period non-metallic character decreases due to increase in atomic size and decrease in the force of attraction between the nucleus and the electrons in the outermost shell.
> We know that element with electronic configuration $1{s^2}2{s^2}2{p^6}3{s^1}$ has atomic number eleven which is sodium $(Na)$ in periodic table.
> Then next to it is magnesium with electronic configuration $1{s^2}2{s^2}2{p^6}3{s^2}$ and atomic number twelve.
> The element with electronic configuration $1{s^2}2{s^2}2{p^6}3{s^2}$is fluorine with atomic number nine.
> The last element is nitrogen with $1{s^2}2{s^2}2{p^3}$and atomic number seven. Fluorine is most non-metallic among all the elements given in the option because the non-metallic character increases as we go from left to right. Fluorine atoms have the maximum tendency to accept an electron characteristic of non-metal. Hence option B is correct.

Note: We have approached this problem with the trend of periodic table. Fluorine has the highest ionisation energy among all the given elements and it is also last in the periodic table hence it has the most non-metallic character among sodium, magnesium and nitrogen.