JEE MAIN - Chemistry (2022 - 29th July Evening Shift - No. 2)
Given below are the quantum numbers for 4 electrons.
A. $$\mathrm{n}=3,l=2, \mathrm{~m}_{1}=1, \mathrm{~m}_{\mathrm{s}}=+1 / 2$$
B. $$\mathrm{n}=4,l=1, \mathrm{~m}_{1}=0, \mathrm{~m}_{\mathrm{s}}=+1 / 2$$
C. $$\mathrm{n}=4,l=2, \mathrm{~m}_{1}=-2, \mathrm{~m}_{\mathrm{s}}=-1 / 2$$
D. $$\mathrm{n}=3,l=1, \mathrm{~m}_{1}=-1, \mathrm{~m}_{\mathrm{s}}=+1 / 2$$
The correct order of increasing energy is :
Explanation
The order of increasing energy of electrons in an atom can generally be determined using the principal quantum number ($n$) and the azimuthal quantum number ($l$). The energy can be compared using the rule: the higher the $(n + l)$ value, the higher the energy. If two electrons have the same $(n + l)$ value, the electron with the higher $n$ value has the higher energy.
Analyze each electron's quantum numbers:
A. $$n=3, \ l=2 \quad \Rightarrow \quad n+l = 3+2 = 5$$
B. $$n=4, \ l=1 \quad \Rightarrow \quad n+l = 4+1 = 5$$
C. $$n=4, \ l=2 \quad \Rightarrow \quad n+l = 4+2 = 6$$
D. $$n=3, \ l=1 \quad \Rightarrow \quad n+l = 3+1 = 4$$
Comparing the values of $(n + l)$ and the principal quantum number $n$:
D: $n+l = 4$, and $n=3$.
A and B: $n+l = 5$, but A has $n=3$ while B has $n=4$. Therefore, A has a slightly lower energy than B.
C: $n+l = 6$.
Thus, the correct order of increasing energy is:
$$ D < A < B < C $$
Therefore, Option B is correct.
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