Solveeit Logo
Login

Question

Question: If an atom has electronic configuration \(1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^3}4{s^2}\), it will be p...

If an atom has electronic configuration 1s22s22p63s23p63d34s21{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^3}4{s^2}, it will be placed in
A. Second group
B. Third group
C. Fifth group
D. Sixth group

Explanation

Solution

All the elements in the group have the same outer shell configuration. The periodic table can also be divided into blocks, such as the s block, p block, d block and f block. Elements are divided into these blocks on the basis of in which shell does the incoming electron go.

Complete step by step answer:
The periodic table is organised in such a way, that elements with the same outer shell configuration will fall in the same group.
For example, all the elements in group one will have the outer shell electronic configuration as ns1,n=period numbern{s^1},n = period\,number Elements in first group are Na,KNa,K Na=1s12s22p63s1Na = 1{s^1}2{s^2}2{p^6}3{s^1}
K=1s22s22p63s23p64s1K = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1}
Notice that the last electron in both the elements goes into the s shell and the s shell is half filled with only one electron. The periodic table is divided as: s block=1−2groups\,block = 1 - 2group
d block=3−12blockd\,block = 3 - 12block
p block=13−18blockp\,block = 13 - 18block
According to the Aufbau principle the of 4s4sorbital is less than that of 3d3d orbital and hence the 4s4s orbital is filled first, but while writing the electronic configuration 3d3d is mentioned before 4s4s . Hence, The outermost electronic configuration for elements in the d block goes like, (n−1)d1−10ns2(n - 1){d^{1 - 10}}n{s^2}. The d electrons have shell number one less than the s shell number. So, in the configuration 1s22s22p63s23p63d34s21{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^3}4{s^2}, the last shell to be filled is the d shell and the its configuration is 3d33{d^3}. This means that the incoming electron goes to the d block and hence, the elements must belong to the d block in the periodic table. The general configuration of the elements is, (n−1)d1ns2(n - 1){d^1}n{s^2}, we will add the value of the outermost electron to get the value of shell number. ∴1+2=3\therefore 1 + 2 = 3 Hence, Elements with outer shell electronic configuration as (n−1)d1ns2(n - 1){d^1}n{s^2} will be in the third group. In our question, the outer shell electronic configuration is 3d34s23{d^3}4{s^2}. By adding the value of the outermost shell electrons we get 3+2=53 + 2 = 5 Hence, this element belongs to the fifth group.

So, the correct answer is “Option C”.

Note: The d block elements are also known as transition metal, and they have a separate group called as inner transition metals or the f block elements. These elements are mostly the radioactive materials found in the periodic table.
The metal with electronic configuration: 1s22s22p63s23p63d34s21{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^3}4{s^2} is Chromium with atomic number 2323 .
Transition materials have the most wide range applications out of any other group of elements.If the outermost electrons are in the p orbital, for example if the element has three electrons in the p orbital and ten electrons in the d orbital, then it goes in the thirteenth group. But if it only has electrons in the d orbital and not the p, it goes in the group number which will be the sum of s and d orbital.