How many orbitals is/are possible with $n=3$, $l=1$ and $m_l=-1$ value?
How many orbitals is/are possible with $n=3$, $l=1$ and $m_l=-1$ value?
2
3
5
1
Solution
Symbol ' $n$ ' represent number of shells. Symbol ' $l$ ' represent value of subshell and symbol $m_l$ represent value of orientation of the orbital.
$\because$ Given, $n=3, l=1$ and $m_l=-1$
means an orbital is present in $p$-subshell of 3rd shell having orientation value $=-1$.
(i.e. $p_x$ or $p_y$ ).
Thus, only one orbital is possible for the given set of values of $n, l$ and $m_l$.
Hence, option (d) is the correct answer.