Question

The probability density plots of 1s and 2s orbitals are given in Fig.
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 statements is incorrect? 1s2s

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

Answer 1

The correct option is C The probability density of electrons for 2s orbital decreases uniformly as distance from the nucleus increases.

The probability density of electrons for 2s orbital decreases uniformly as distance from the nucleus increases

Option (D) is the correct answer

Solution:
Analyzing the options:Option (A): and 2s orbitals are spherical in shape.
From the figure it is corroborated that 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, option (A) stands correct.

Option (B):
The probability of finding the
electron is maximum near the nucleus.

For both 1s and 2s orbitals, the probability of
finding the electrons will increase, as we move
towards the nucleus. So, we can say that option
(B) also stands correct.

Option (C):
The probability of finding the
electron at a given distance is equal in all
directions.

The probability of finding any electron at any
given distance is always equal in all the
directions. But it is only possible in s-subshells.
Which means that the statement stands correct
only for orbitals having spherical shapes.

Option D:

The probability density of
electron for 2s orbital decrease uniformaly as distance from the nucleus increase.

The probability of having an electron close to the nucles is maxium, as the number of angular nodes, $l=0(l=s)$. And Complete node count = $n-1$

The number of nodes is 0 for 1s- orbital . The probability density at the nucleus is therefore maximum and decrease dramatically with the distance from the nucleus. Number of nodes in 2s- Orbital = $2-1=1,$ therefore it has two maxima separated by a nodal surface area.

The probability density of maximum point close to the nucleus and decrease rapidly to zero and then rises again and decrease suddenly after the limits hits.

The probability density of electrons present in 2s orbitals first increases then decreases then again increases as moving towards the nucleus. Thus, this fact simply renders option (d) incorrect.