
The number of orbitals in n = 3 is:
A) 1
B) 4
C) 9
D) 16
Answer
576.3k+ views
Hint: The principal quantum number for the shell: $n = 3$
We know that for a given principal quantum number n, the number of orbitals is given by ${n^2}$.
Complete step by step answer:
We have four quantum numbers, the set of which is unique for an electron and can be used to represent the same. However, there are some rules that restrict the permissible values of these quantum numbers. Let’s have a brief look at these quantum numbers one by one:
> Principal quantum number $\left( n \right)$: It represents the shell and is a deciding factor in determining the size of an orbital. It also has an impact on the energy of the orbitals.
> Azimuthal quantum number $\left( l \right)$: For a given value of n, the permissible values of l are 0,1,2,...n - 1. l can be used to identify the subshell and its value can be used to deduce the shape of orbitals.
> Magnetic quantum number $\left( {{m_l}} \right)$: It gives the orbital orientation and the number of permissible values is $\left( {2l + 1} \right)$ with values ranging from - l to + l.
> Spin quantum number ${m_s}$: it gives the spin orientation and have only two permissible values $\left( { - \dfrac{1}{2}, + \dfrac{1}{2}} \right)$.
These restrictions on the permissible values of quantum numbers also put restrictions to the number of orbitals present in a given shell. Let’s calculate it for n = 3 by using the above rules as follows:
For n = 3, the permissible sub-shells would have $l = 0,\;l = 1$ and $l = 2$. Now, we will calculate the number of orbitals in each subshell:
For l = 0 the only possible orbital would have ${m_l} = 0$
For l = 1 the permissible values are: ${m_l} = - 1, 0, + 1$. So, we have 3 orbitals in this sub-shell.
For l = 2 the permissible values are: ${m_l} = - 2, - 1, 0, + 1, + 2$. So, we have 5 orbitals in this sub-shell.
Now we can add the number from each to find out that there are a total 9 orbitals for $n = 3$.
We can also use the formula ${n^2}$ to find the same as ${3^2} = 9$
Hence, the correct option is C.
Note: We have to be careful with the notations of shell, sub-shell, orbits and orbitals. An orbital is the space or the region where probability of finding the electron is maximum on the other hand orbits are defined as the path that gets established in a circular motion by revolving the electron around the nucleus.
We know that for a given principal quantum number n, the number of orbitals is given by ${n^2}$.
Complete step by step answer:
We have four quantum numbers, the set of which is unique for an electron and can be used to represent the same. However, there are some rules that restrict the permissible values of these quantum numbers. Let’s have a brief look at these quantum numbers one by one:
> Principal quantum number $\left( n \right)$: It represents the shell and is a deciding factor in determining the size of an orbital. It also has an impact on the energy of the orbitals.
> Azimuthal quantum number $\left( l \right)$: For a given value of n, the permissible values of l are 0,1,2,...n - 1. l can be used to identify the subshell and its value can be used to deduce the shape of orbitals.
> Magnetic quantum number $\left( {{m_l}} \right)$: It gives the orbital orientation and the number of permissible values is $\left( {2l + 1} \right)$ with values ranging from - l to + l.
> Spin quantum number ${m_s}$: it gives the spin orientation and have only two permissible values $\left( { - \dfrac{1}{2}, + \dfrac{1}{2}} \right)$.
These restrictions on the permissible values of quantum numbers also put restrictions to the number of orbitals present in a given shell. Let’s calculate it for n = 3 by using the above rules as follows:
For n = 3, the permissible sub-shells would have $l = 0,\;l = 1$ and $l = 2$. Now, we will calculate the number of orbitals in each subshell:
For l = 0 the only possible orbital would have ${m_l} = 0$
For l = 1 the permissible values are: ${m_l} = - 1, 0, + 1$. So, we have 3 orbitals in this sub-shell.
For l = 2 the permissible values are: ${m_l} = - 2, - 1, 0, + 1, + 2$. So, we have 5 orbitals in this sub-shell.
Now we can add the number from each to find out that there are a total 9 orbitals for $n = 3$.
We can also use the formula ${n^2}$ to find the same as ${3^2} = 9$
Hence, the correct option is C.
Note: We have to be careful with the notations of shell, sub-shell, orbits and orbitals. An orbital is the space or the region where probability of finding the electron is maximum on the other hand orbits are defined as the path that gets established in a circular motion by revolving the electron around the nucleus.
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