The quantum numbers of six electrons are given below. Arrange them in order of increasing energies
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:
1. $n=4, I=2, m_{i}=-2, m_{s}=-1 / 2$
2. $n=3, l=2, m_{i}=1, m_{s}=+1 / 2$
3. $n=4, l=1, m_{i}=0, m_{s}=+1 / 2$
4. $n=3,1=2, m_{i}=-2, m_{s}=-1 / 2$
5. $n=3, I=1, m_{i}=-1, m_{s}=+1 / 2$
6. $n=4, l=1, m_{i}=0, m_{s}=+1 / 2$
For $n=4$ and $/=2$, the orbital occupied is $4 d$.
For $n=3$ and $/=2$, the orbital occupied is $3 d .$
For $n=4$ and $/=1$, the orbital occupied is $4 p$.
Hence, the six electrons i.e., $1,2,3,4,5$, and 6 are present in the $4 d, 3 d, 4 p, 3 d, 3 p$, and $4 p$ orbitals respectively.
Therefore, the increasing order of energies is $5(3 p)<2(3 d)=4(3 d)<3(4 p)=6(4 p)<1(4 d)$.