Question

How many orbitals are present in an energy level$?$
(i) $2n$
(ii) $n + 2$
(iii) ${n^2}$
(iv) $2{n^2}$

Answer 1

Hint: For a given energy level $\left( n \right)$, we know that ${l_{\max \,}}\, = \,n - 1$ and there exits $\left( {2l + 1} \right)$ orbitals per subshell where $l$ represents azimuthal quantum number or the angular momentum and $n$represent the energy level or the principal quantum number. Using these calculate the number of orbitals present in each energy level.

Complete step-by-step answer:
We know that the quantum number representing energy level is the Principal Quantum number, \[n\].
The quantum number corresponding to angular momentum is the azimuthal quantum number, $l$.
$l$ can range from $0\,\,to\,\,\left( {n - 1} \right)$, hence ${l_{\max \,}}\, = \,n - 1$.
Also the number of orbitals per subshell $ = \,\,\left( {2l + 1} \right)$.
Therefore we can say,
For each energy level $n$there exists $\sum\limits_{l\, = \,0}^{n - 1} {\left( {2l + 1} \right)} $ orbitals in each energy level.
$ = \,\left\{ {\left[ {2\left( 0 \right) + 1} \right] + \left[ {2\left( 1 \right) + 1} \right] + ........ + \left( {2\left( {{l_{\max }}} \right) + 1} \right)} \right\}$ orbitals in each energy level.
$ = \,\left\{ {2\left( 0 \right) + 2\left( 1 \right) + .....2\left( {{l_{\max }}} \right)} \right\} + \left\{ {1 + 1 + 1...... + n} \right\}$orbitals in each energy level.
$ = \,2\left( {0 + 1 + 2..... + {l_{\max }}} \right) + n$ orbitals in each energy level.
$ = \,2\left\{ {\dfrac{{{l_{\max }} \times \left( {{l_{\max }} + 1} \right)}}{2}} \right\}\, + \,n$ orbitals in each energy level.
$ = \,2\left\{ {\dfrac{{n\left( {n - 1} \right)}}{2}} \right\} + n$ orbitals in each energy level. $\left( {\because \,{l_{\max }}\, = \,n - 1} \right)$
$ = \,n\left( {n - 1} \right) + n$ orbitals in each energy level.
$ = \,{n^2} + n - n$ orbitals in each energy level.
$ = \,{n^2}$ orbitals in each energy level.

Hence the correct answer is (iii) ${n^2}$.

Additional information:
Orbital is the three dimensional region where probability of finding the electron is maximum. An orbital is characterized by three quantum numbers principal quantum number $\left( n \right)$, azimuthal quantum number $\left( l \right)$ and the magnetic quantum number $\left( {{m_l}} \right)$ which correspond to energy level, angular momentum and vector component of angular momentum respectively.From the above calculations we have seen in each energy level total number of orbitals present $ = \,{n^2}$

Now, for the first energy level i.e. $n\, = \,1$
-The number of orbitals present $ = \,{1^2}\, = \,1$ which represent the $1s$orbital.
-For the second energy level i.e. $n\, = \,2$
-The number of orbitals present $ = \,{2^2}\, = \,4$ which represent the $1s,\,2{p_x},\,2{p_y},\,2{p_z}$orbitals.
-For the third energy level i.e. $n\, = \,3$
-The number of orbitals present $ = \,{3^2}\, = \,9$ which represent the $3s,\,3{p_x},\,3{p_y},\,3{p_z},\,3{d_{xz}},\,3{d_{yz}},\,3{d_{xy}},\,3{d_{{x^2} - {y^2}}},\,3{d_{{z^2}}}$orbitals.

Note: The problem can be approached in a different way also. If you know the maximum number of electrons an energy level can hold, you can use the fact that each individual orbital can hold no more than 2 electrons to determine how many orbitals are present.