Question

1st attempt
see periodic table see hint
select the orbital with the highest energy in a multielectron atom:
choose one:
n = 5, l = 1
n = 2, l = 1
n = 5, l = 2
n = 4, l = 3

Explanation:

Step1: Recall energy - order rule

In a multielectron atom, the energy of an orbital is determined by the $(n + l)$ rule. The higher the value of $(n + l)$, the higher the energy of the orbital. If two orbitals have the same $(n + l)$ value, the orbital with the higher $n$ - value has higher energy.

Step2: Calculate $(n + l)$ for each option

For $n = 5,l = 1$: $(n + l)=5 + 1=6$.
For $n = 2,l = 1$: $(n + l)=2 + 1=3$.
For $n = 5,l = 2$: $(n + l)=5+2 = 7$.
For $n = 4,l = 3$: $(n + l)=4 + 3=7$.

Step3: Compare the values

Since the $(n + l)$ values of $n = 5,l = 2$ and $n = 4,l = 3$ are equal ($(n + l)=7$), and for equal $(n + l)$ values, the orbital with the higher $n$ - value has higher energy. So the orbital with $n = 5,l = 2$ has higher energy than the orbital with $n = 4,l = 3$. Also, $(n + l)=7$ is higher than $(n + l)=6$ and $(n + l)=3$.

Answer:

$n = 5,l = 2$