Question

The probability density plots of 1s and 2s orbitals are given in the diagram. The density of dots in a region represents the probability density of finding electrons in the region. On the basis of the above diagram which of the following statement(s) is/are correct?



(A) 1s and 2s orbitals are spherical in shape.
(B) The probability density of finding the electron is maximum near the nucleus
(C) The probability of finding the electron at a given distance is equal in all directions
(D) The probability density of electrons for 2s orbital decreases uniformly as distance from the nucleus increases.

Answer 1

Hint: First, we will explain the accuracy of the first option. We will follow this procedure for the rest of the options as well which will provide us with our required answer.

Complete step by step solution:
> We have to note that we can have multiple correct options. Refer to the solution below.
From the figure it is clear that the orbitals 1s and 2s will be spherical in shape. It doesn’t matter if the principal quantum number is 1 or 2. Since it is mentioned that it is the s-subshell, it will always be spherical. So, it is clear that option A stands correct.
Since we already know that the nodes are always present near the nucleus. So, the probability of finding the electrons will also be maximum as we move towards the nucleus. So, we can say that option B also stands correct.
> The probability of finding any electron at any given distance is always equal in all the directions. But it is so possible only in s-subshells. Which means that the statement stands correct only for orbitals having spherical shapes.
The probability of having an electron close to the nucleus is maximum, as the number of angular nodes = $l = 0\left( {l = s} \right)$. And, complete node count = $n - l$
The number of nodes = 0 for 1s-orbital. The probability density at the nucleus is therefore maximum and decreases dramatically with the distance from the nucleus.
Number of nodes = $2 - 1 = 1$. In 2s-orbital.Therefore, it has two maxima separated by a nodal surface area. The probability density is at the maximum point close to the nucleus and decreases rapidly to zero and then rises again and decreases suddenly after the limit hits.
The probability density of electrons present in 2s orbitals first increases then decreases. Thus, this fact simply renders our option D incorrect.

Hence, we can simply say that option A, option B and option C are correct options.

Note: The shell of an electron can be considered as an orbit followed by electrons around the nucleus of an atom. The shell nearest to a nucleus is the "1 shell (1s)" (also known as the "K shell"), after which comes the 2 shell "2s" (or "L shell") and so it goes on as we move away from the nucleus.