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Question: An element with the atomic number \[19\] will most likely combine chemically with the element whose ...

An element with the atomic number 1919 will most likely combine chemically with the element whose atomic number is :
A. 1717
B. 1111
C. 1818
D. 2020

Explanation

Solution

At first, we will write the electronic configurations of all the given atomic numbers.
Then, we will try to match those electronic configurations with the electronic configuration of nearest Noble gas atom.
Then we can easily find out the element which will combine with the element having atomic number 19.19.

Complete step by step answer:
Electronic configuration of atomic number - 19 - $$$${\text{1}}{{\text{s}}^2}{\text{2}}{{\text{s}}^2}{\text{2}}{{\text{p}}^6}{\text{3}}{{\text{s}}^2}{\text{3}}{{\text{p}}^6}{\text{4}}{{\text{s}}^1}
Electronic configuration of atomic number - 17{\text{ }}\left( A \right) - $$$${\text{1}}{{\text{s}}^2}{\text{2}}{{\text{s}}^2}{\text{2}}{{\text{p}}^6}{\text{3}}{{\text{s}}^2}{\text{3}}{{\text{p}}^5}
Electronic configuration of atomic number - 11{\text{ }}\left( B \right) - $$$${\text{1}}{{\text{s}}^2}{\text{2}}{{\text{s}}^2}{\text{2}}{{\text{p}}^6}{\text{3}}{{\text{s}}^1}
Electronic configuration of atomic number - 18{\text{ }}\left( C \right) - $$$${\text{1}}{{\text{s}}^2}{\text{2}}{{\text{s}}^2}{\text{2}}{{\text{p}}^6}{\text{3}}{{\text{s}}^2}{\text{3}}{{\text{p}}^6}
Electronic configuration of atomic number−20 (D)−- 20{\text{ }}\left( D \right) - 1s22s22p63s23p64s2{\text{1}}{{\text{s}}^2}{\text{2}}{{\text{s}}^2}{\text{2}}{{\text{p}}^6}{\text{3}}{{\text{s}}^2}{\text{3}}{{\text{p}}^6}{\text{4}}{{\text{s}}^2}

So, from the electronic configurations, we can see that the atomic number−18 (option −C) - 18{\text{ }}\left( {option{\text{ }} - C} \right) is the electronic configuration of a noble gas. So, this is a stable electronic configuration.
This noble gas is the nearest noble gas of both the atomicnumbers−17 (option −A) and 20 (option−D) - 17{\text{ }}\left( {option{\text{ }} - A} \right){\text{ }}and{\text{ }}20{\text{ }}\left( {option - D} \right) and also the atomic number−19. - 19.
So, all the atoms having atomic numbers - 17$$$$\left( {option{\text{ }} - A} \right){\text{ }}and{\text{ }}20{\text{ }}\left( {option - D} \right)andatomic number-1919 will try to attain the stable electronic configuration of atomic number−18 (option−C) - 18{\text{ }}\left( {option - C} \right)
Now, the atomic number−19 - 19 has one excess electron than the atomic number−18 (option −C) - 18{\text{ }}\left( {option{\text{ }} - C} \right) and the atomic number−17(option −A) - 17\left( {option{\text{ }} - A} \right) has one less electron than the atomic number 18 (option −C).18{\text{ }}\left( {option{\text{ }} - C} \right).
So, these two (atomic number 19 andatomic number 17)\left( {atomic{\text{ }}number{\text{ }}19{\text{ }}andatomic{\text{ }}number{\text{ }}17} \right) will combine most likely.

So, option-AA is the correct option.

Note:
1. We should remember that all the elements try to attain a stable electronic configuration of it’s nearest noble gas.
2. If the situation was given for atomic number−16, - 16, it will most likely combine with the atomic number−20. - 20.
3. We have to remember that all the atoms try to fulfil its valence shell with electrons.