Question

The orbital angular momentum quantum number of the state $\text{ }{{\text{S}}_{\text{2 }}}$ is:
A) 0
B) 1
C) 2
D) 3

Answer 1

Hint: The principal quantum number ‘n’ and orbital angular quantum number ‘l’ are related to each other as follows,
$\text{ No}\text{.of radial nodes = }n-l-1\text{ }$
Where n is the number of total nodes or principal quantum number, ‘l’ is the angular nodes or angular or azimuthal quantum number.

Complete answer:
The hydrogen-like species for example $\text{ L}{{\text{i}}^{\text{2+}}}\text{ }$have symmetrically spherical orbits. This species has one radial node. Node is defined as the point in the space at which the probability of finding the electrons has become zero.
On absorption of light radiation, this species or ions undergo the transition from the lower energy level let $\text{ }{{\text{S}}_{\text{1 }}}$ to the higher energy level$\text{ }{{\text{S}}_{\text{2 }}}$.
The $\text{ }{{\text{S}}_{\text{2 }}}$energy state also has one radial node. We are interested to find out the orbital quantum number for $\text{ }{{\text{S}}_{\text{2 }}}$the energy state.
It is given that the $\text{ }{{\text{S}}_{\text{2 }}}$has one radial node and its energy is equal to the ground state energy of a hydrogen atom.
Let's calculate the number of nodes ‘n’ for the state. We will use the following formula,
$\text{ }{{\text{E}}_{{{\text{S}}_{2}}}}\text{ = }\dfrac{-13.6\times {{Z}^{2}}}{{{n}^{2}}}\text{ = }{{\text{E}}_{\text{H}}}\text{ in ground state = }-13.6\text{ }$
Where Z is the atomic number, ‘n’ is the principal quantum number.
We are finding the principal quantum number of $\text{ }{{\text{S}}_{\text{2 }}}$the energy state of $\text{ L}{{\text{i}}^{\text{2+}}}\text{ }$species. Thus the atomic number is equal to 3.On substituting in the equation the ‘n’ is:
$\begin{align}
  & \text{ }\dfrac{-13.6\times {{(3)}^{2}}}{{{n}^{2}}}\text{ =}-13.6 \\
 & \Rightarrow {{n}^{2}}\text{ = }\dfrac{-13.6\times 9}{-13.6} \\
 & \therefore n\text{ = 3 } \\
\end{align}$
Thus, here the number of nodes is equal to 3.
 Let's calculate the orbital angular quantum number. The total radial nodes are given as follows,
$\text{ Number of radial nodes = }n-l-1\text{ }$
Then, the orbital angular quantum number is,
$\begin{align}
  & \text{ Number of radial nodes = }n-l-1\text{ } \\
 & \Rightarrow 1=\text{ 3}-l-1\text{ }\because n=3\text{ } \\
 & \therefore \text{ }l\text{ = 1 } \\
\end{align}$
Therefore, the $\text{ }{{\text{S}}_{\text{2}}}\text{ }$state has the orbital angular quantum number equal to 1.

Hence, (B) is the correct option.

Note:
Note that the principal quantum number is used to express the energy of an orbital or the shell. It also represents the number of nodes.
The orbital angular quantum number ‘l’ gives the idea about the subshell present in the shell. The value of ‘l’ is always one unit less than the ‘n’ $\text{ }\left( l=n-1 \right)$ .
For $\text{ }n=1\text{ , }l=0$ and have only s orbital
For$\text{ }n=2\text{ , }l=0,1\text{ }$, the shell has s and p as subshells.
For$\text{ }n=3\text{ , }l=0,1,2\text{ }$, the shell has s, p, and d subshells.