Question

question 15 which of the following sets of quantum numbers is for the highest energy orbital a.n = 4,l= 3,m= 3 b.n = 4,l= 2,m= 2 c.n = 4,l = 2,m= 1 d.n = 4,l= 1,m= 0 question 16 which set of quantum numbers is for the lowest energy orbital? a.n = 3,l= 2, m = 2 b.n = 3,l= 1, m = 1 c.n = 2,l= 0, m = 0 d.n = 2,l= 1, m = 0

Explanation:

Step1: Recall energy - orbital rule

The energy of an orbital is mainly determined by the principal quantum number $n$ and the angular - momentum quantum number $l$. The higher the value of $n + l$, the higher the energy of the orbital. When $n + l$ values are the same, the orbital with the higher $n$ value has higher energy.

Step2: Calculate $n + l$ for each option in Question 15

• Option a: $n = 4$, $l=3$, so $n + l=4 + 3=7$.
• Option b: $n = 4$, $l = 2$, so $n + l=4 + 2=6$.
• Option c: $n = 4$, $l = 2$, so $n + l=4 + 2=6$.
• Option d: $n = 4$, $l = 1$, so $n + l=4 + 1=5$.

Step3: Calculate $n + l$ for each option in Question 16

• Option a: $n = 3$, $l = 2$, so $n + l=3 + 2=5$.
• Option b: $n = 3$, $l = 1$, so $n + l=3 + 1=4$.
• Option c: $n = 2$, $l = 0$, so $n + l=2+0 = 2$.
• Option d: $n = 2$, $l = 1$, so $n + l=2 + 1=3$.

Answer:

Question 15: A. $n = 4,l = 3,m = 3$
Question 16: C. $n = 2,l = 0,m = 0$