The quantum number of four electrons are given below:
I. n=4, l=2, ml=-2, ms=-1/2
II. n=3, l=2, ml=1

Q: Why do we completely ignore $m_l$ and $m_s$ when comparing energies?

In a multi-electron atom, the energy of an orbital is primarily determined by its size (given by $n$) and its shape (given by $l$). The magnetic quantum number $m_l$ and spin quantum number $m_s$ only describe the spatial orientation and spin direction, which do not affect the energy unless the atom is placed in an external magnetic or electric field.

Q: What if this was a hydrogen atom instead of a multi-electron atom?

Great question! For a single-electron system like the hydrogen atom, there is no inter-electronic repulsion. Therefore, the energy depends ONLY on the principal quantum number $n$. In that case, all subshells within the same shell (like 3s, 3p, 3d) are degenerate, meaning they have the exact same energy.

Q: Why does the orbital with lower $n$ have lower energy when $(n+l)$ is the same?

When $(n+l)$ is the same, the orbital with the lower $n$ value is physically closer to the nucleus on average. Because it is closer, it experiences a stronger electrostatic pull from the positively charged nucleus (higher effective nuclear charge), making it more stable and thus lower in energy.

Q: What do the $l$ values 0, 1, 2, 3 actually correspond to?

The azimuthal quantum number $l$ determines the subshell type and its shape. $l=0$ is a spherical s-orbital, $l=1$ is a dumbbell-shaped p-orbital, $l=2$ is a double-dumbbell d-orbital, and $l=3$ is a complex f-orbital.

Q: Can the $(n+l)$ rule ever be violated?

Yes, there are exceptions to the Aufbau principle, especially for transition metals like Chromium and Copper, where half-filled or fully-filled d-subshells offer extra stability. However, for predicting the general energy order of isolated orbitals, the $(n+l)$ rule works perfectly.

Question

The quantum number of four electrons are given below:
I. $n=4, l=2, m_l=-2, m_s=-\frac{1}{2}$
II. $n=3, l=2, m_l=1, m_s=+\frac{1}{2}$
III. $n=4, l=1, m_l=0, m_s=+\frac{1}{2}$
IV. $n=3, l=1, m_l=1, m_s=-\frac{1}{2}$
The correct order of their increasing energies will be

Options:

(A) IV < III < II < I

(B) I < II < III < IV

(C) IV < II < III < I

(D) I < III < II < IV

Sigma Percentile · Accepted Answer

Imagine you are trying to find a specific person in a massive, bustling city. You would need their country, city, street, and house number. In the quantum world, finding an electron around a nucleus requires a similar address system, known as **quantum numbers**.

Every electron in an atom is uniquely identified by a set of four quantum numbers: the principal quantum number ($n$), the azimuthal quantum number ($l$), the magnetic quantum number ($m_l$), and the spin quantum number ($m_s$). However, when it comes to determining the **energy** of an electron in a multi-electron atom, not all quantum numbers are created equal.

### The Energy of an Electron

In a simple hydrogen atom, which only has one electron, the energy of that electron depends entirely on the principal quantum number, $n$. The larger the $n$, the further the electron is from the nucleus, and the higher its energy.

But in a multi-electron atom, things get a bit more complicated. Electrons don't just interact with the nucleus; they also repel each other. This inter-electronic repulsion leads to phenomena known as **shielding** and **penetration**. Because of these effects, the energy of an orbital depends on both its size (determined by $n$) and its shape (determined by $l$).

### The $(n+l)$ Rule

To predict the order in which electrons fill the orbitals, we use the **Aufbau Principle**, which states that electrons fill lower-energy atomic orbitals before filling higher-energy ones. The relative energies of these orbitals are governed by the **$(n+l)$ rule** (also known as the Madelung energy ordering rule).

The rule is beautifully simple:
1. The energy of an orbital increases as the sum of $(n+l)$ increases.
2. If two orbitals have the same $(n+l)$ value, the orbital with the lower $n$ value has lower energy.

Let's apply this rule to our four contestants to find their correct energy order.

### Analyzing the Contestants

**Electron I:** 
We are given $n=4$ and $l=2$. 
The sum is $(n+l) = 4 + 2 = 6$. 
*(Note: $l=2$ corresponds to a d-orbital, so this is the $4d$ subshell).*

**Electron II:** 
We are given $n=3$ and $l=2$. 
The sum is $(n+l) = 3 + 2 = 5$. 
*(This is the $3d$ subshell).*

**Electron III:** 
We are given $n=4$ and $l=1$. 
The sum is $(n+l) = 4 + 1 = 5$. 
*(This is the $4p$ subshell).*

**Electron IV:** 
We are given $n=3$ and $l=1$. 
The sum is $(n+l) = 3 + 1 = 4$. 
*(This is the $3p$ subshell).*

### Breaking the Tie

Right away, we can see that **Electron IV** has the lowest $(n+l)$ value of $4$, making it the lowest in energy. Conversely, **Electron I** has the highest $(n+l)$ value of $6$, making it the highest in energy.

But what about **Electron II** and **Electron III**? Both have an $(n+l)$ value of $5$. This is where the second part of our rule comes into play. When there is a tie in the $(n+l)$ value, we look at the principal quantum number, $n$. The orbital with the lower $n$ value is closer to the nucleus on average, experiences a stronger effective nuclear charge, and is therefore more stable (lower in energy).

For Electron II ($3d$), $n=3$. 
For Electron III ($4p$), $n=4$.

Since $3 < 4$, the energy of Electron II is strictly less than the energy of Electron III.

### The Final Verdict

Putting it all together, we arrange the electrons in increasing order of their energies:

$$E_{IV} < E_{II} < E_{III} < E_{I}$$

This perfectly matches the sequence **IV < II < III < I**. Notice how we completely ignored the magnetic quantum number ($m_l$) and the spin quantum number ($m_s$)? In the absence of an external magnetic or electric field, these quantum numbers do not affect the energy of the orbital. They merely describe the spatial orientation and the spin state of the electron.

Mastering the $(n+l)$ rule is like having a master key to the architecture of the periodic table. It dictates the electronic configuration of elements and, by extension, their chemical behavior!

Select Answer:

Visualized Solution

$(n + l)$ Rule

Electron I

Electron II

Electron III

Electron IV

Resolving the Tie

Final Energy Order

The Sigma Insight: Quantum Mechanical Model

The Energy of an Electron

The $(n + l)$ Rule

Analyzing the Contestants

Breaking the Tie

The Final Verdict

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