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 $5(3 p)<2(3 d)=4(3 d)<3(4 p)=6(4 p)<1(4 d)$.
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