An atomic orbital has $n=3$. What are the possible values of $/$ and $m_{l}$ ?

Solution:

(i) $n=3$ (Given)

For a given value of $n, I$ can have values from 0 to $(n-1)$.

$\therefore$ For $n=3$

$I=0,1,2$

For a given value of $I, m_{1}$ can have $(2 I+1)$ values.

For $I=0, m=0 I=1, m=-1,0,1 I=2, m=-$

$2,-1,0,1,2$

$\therefore$ For $n=3$

$I=0,1,2$

$m_{0}=0$

$m_{1}=-1,0,1 m_{2}=-$

$2,-1,0,1,2$

(ii) For $3 d$ orbital, $I=2$.

For a given value of $I, m_{i}$ can have $(2 I+1)$ values i.e., 5 values.

$\therefore$ For $l=2 m_{2}=-2$,

$-1,0,1,2$

(iii) Among the given orbitals only $2 s$ and $2 p$ are possible. $1 p$ and $3 f$ cannot exist.

For $p$-orbital, $I=1$.

For a given value of $n, I$ can have values from zero to $(n-1)$.

$\therefore$ For $/$ is equal to 1 , the minimum value of $n$ is 2 .

Similarly,

For $f$-orbital, $I=4$.

For $I=4$, the minimum value of $n$ is $5 .$

Hence, $1 p$ and $3 f$ do not exist.