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 :

<p>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.</p> <p>Analyze each electron's quantum numbers:</p> <p><p><strong>A.</strong> $n=3, \ l=2 \quad \Rightarrow \quad n+l = 3+2 = 5$</p></p> <p><p><strong>B.</strong> $n=4, \ l=1 \quad \Rightarrow \quad n+l = 4+1 = 5$</p></p> <p><p><strong>C.</strong> $n=4, \ l=2 \quad \Rightarrow \quad n+l = 4+2 = 6$</p></p> <p><p><strong>D.</strong> $n=3, \ l=1 \quad \Rightarrow \quad n+l = 3+1 = 4$</p></p> <p>Comparing the values of $(n + l)$ and the principal quantum number $n$:</p> <p><p><strong>D</strong>: $n+l = 4$, and $n=3$.</p></p> <p><p><strong>A</strong> and <strong>B</strong>: $n+l = 5$, but <strong>A</strong> has $n=3$ while <strong>B</strong> has $n=4$. Therefore, <strong>A</strong> has a slightly lower energy than <strong>B</strong>.</p></p> <p><p><strong>C</strong>: $n+l = 6$.</p></p> <p>Thus, the correct order of increasing energy is:</p> <p>$D < A < B < C$</p> <p>Therefore, Option B is correct.</p>