Question

Which of the following sets of elements have the same number of electrons in the outermost shell?
A. Elements with atomic numbers 30,48,80
B. Elements with atomic numbers 14,15,16
C. Elements with atomic numbers 20,30,50
D. Elements with atomic numbers 10,18,26

Answer 1

Hint: There are many concepts of outermost shells of an element, such as if the elements are in the same group then they have the same number of electrons in their outermost shell. Also, we can use noble gas for writing the electronic configuration which simplifies it.

Complete step by step answer:
So here we have to write the electronic configuration of elements given here in order to find the outermost shells. Let us go through the option one by one.

(A) We have elements with atomic numbers 30, 48 and 80.
Before that let us write the noble gas group, so that we can write the electronic configuration more easily as it has all 6 electrons in its outermost shell. Therefore, we take it as general notation to represent the rest of them.
Noble groups have Helium (Z=2), Neon(Z=10), Argon(Z=18), Krypton(Z=36), Xenon(Z=54) and Radon(Z=86).
First, we have atomic number 30
i.e., $\left[ {Ar} \right]3{d^{10}}4{s^2}$, we took Argon since the noble gas below 30 is Ar. Hence we just have to write the remaining electronic configuration.
Next, we have atomic number 48
i.e., $\left[ {Kr} \right]4{d^{10}}5{s^2}$, here Kr has atomic number 36 which is less than 48.
Next, we have atomic number 80.
i.e., $\left[ {{\text{Xe}}} \right]4{f^{14}}5{d^{10}}4{s^2}$
Hence, we can see all of them have valence shells as s and the 2 electrons in its valence shell.

(B) We have elements with atomic numbers 14, 15 and 16.
We just have to apply logic here, as the atomic number is increasing by 1, which means the electrons in the valence shell also increase by 1, hence this won’t have the same number of electrons.

(C) We have elements with atomic numbers 20,30 and 50.
20 $ \Rightarrow \left[ {Ar} \right]4{s^2}$
30 $ \Rightarrow \left[ {Ar} \right]3{d^{10}}4{s^2}$
50 $ \Rightarrow \left[ {Kr} \right]4{d^{10}}5{s^2}5{p^2}$
These do not have the same electrons in its outermost shell.

(D) We have elements with atomic numbers 10, 18 and 26.Here we know that 10 and 18 noble gases with completely filled electrons in its outermost shell but element with atomic number 26 is not a noble atom as the next noble atom is Krypton (Z=36). Hence they don’t have the same number of electrons in the outermost shell.

Hence option (A) is correct.

Note: We can use the 2, 8, 8, 18, 18, 32 rule to find out the number of electrons in its outer shell very quickly as all the element which is in same group will have same number of electrons in its outermost shell, hence they all follow this 2-8-8-18-18-32 rule. For quick reference we can check noble gases itself you will get a more clear idea.