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Arrange the electrons specified by the quantum numbers (i) $n=4,\, l=1$ (ii) $n=4,\, l=0$ (iii) $n=3,\, l=2$ (iv) $n=3,\, l=1$ in order of increasing energy, from lowest to highest.

1. (iv) < (ii) < (i) < (iii)
2. (ii) < (iv) < (i) < (iii)
3. (i) < (iii) < (iv) < (ii)
4. (iii) < (i) < (iv) < (ii)

Correct answer: (ii) < (iv) < (i) < (iii)

Solution

Using the $(n+l)$ rule, lower $n+l$ means lower energy. Here, (ii) is $4s$ with $n+l=4$, (iv) is $3p$ with $n+l=4$, (i) is $4p$ with $n+l=5$, and (iii) is $3d$ with $n+l=5$. For ties, lower $n$ is lower in energy, so the order is $4s < 3p < 4p < 3d$.

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