Question

Elements with atomic number ${\text{56}}$ belong to which block?
A) s
B) p
C) d
D) f

Answer 1

Hint: The atomic number of the element has been given which means one can write the electronic configuration based on that. The electronic configuration gives the idea of electron presence in the orbitals and the last orbital shows the block of that element.

Complete step by step answer:
1) First of all we will try to write the electronic configuration of the given element based on the given atomic number of an element. The atomic number is ${\text{56}}$ and the electronic configuration is as below,
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}4{s^2}4{p^6}4{d^{10}}5{s^2}5{p^6}6{s^2}$
In the above electronic configuration if we calculate the total electrons by doing the addition of electrons that are present in the superscript then the total will be fifty-six electrons.
2) Now if we analyze the above electronic configuration we can say that the last electrons that are present in the orbital of the element are ${\text{6s}}$ orbital.
3) The ${\text{6s}}$ orbital shows that the outermost electrons are present in s orbital of the element having the atomic number as ${\text{56}}$.
Now from the above observations, we can say that the element with atomic number ${\text{56}}$ belongs to s block which shows option A as the correct choice of answer.

Note:
The element with the atomic number ${\text{56}}$ is Barium which is placed in the s block of the periodic table in group second and period sixth. We can even write the electronic configuration as $\left[ {Xe} \right]6{s^2}$ with respect to a noble gas. One must remember while writing the electronic configuration to fill the electrons in an orbital according to the order of the Aufbau principle and the Pauli exclusion principle.