Number of electrons that vanadium (Z=23) has in p-orbitals is equal to .........

Q: Why does the 4s orbital fill before the 3d orbital?

According to the $n+l$ rule, the $4s$ orbital has a lower energy level ($4+0=4$) compared to the $3d$ orbital ($3+2=5$). Electrons always prefer to occupy the lowest available energy state first, which is why $4s$ fills before $3d$.

Q: Are there any p-electrons in the 4th shell for Vanadium?

No. Vanadium only has $23$ electrons. After filling the $4s$ orbital with $2$ electrons, the remaining $3$ electrons go into the $3d$ orbital. The $4p$ orbital remains completely empty.

Q: What if the question asked for the electronic configuration of the $V^{2+}$ ion?

When transition metals form positive ions, they lose electrons from their outermost shell first. For Vanadium, the outermost shell is the 4th shell ($4s$). So, $V^{2+}$ would lose its two $4s$ electrons, resulting in a configuration ending in $3d^3$. The number of $p$-electrons would remain unchanged at $12$.

Q: How many unpaired electrons does Vanadium have?

Looking at the $3d^3$ subshell, Hund's rule tells us that these three electrons will occupy separate degenerate orbitals with parallel spins before any pairing occurs. Therefore, Vanadium has exactly $3$ unpaired electrons.

Q: Does the 3p subshell fill before the 4s?

Yes. The $3p$ subshell has an $n+l$ value of $3+1=4$, while the $4s$ subshell also has an $n+l$ value of $4+0=4$. When the $n+l$ values are the same, the orbital with the lower principal quantum number ($n$) fills first. Thus, $3p$ fills before $4s$.

Question

Number of electrons that vanadium ($Z=23$) has in p-orbitals is equal to ……… .

Sigma Percentile · Accepted Answer

### Unveiling the Electrons of Vanadium

When we look at the periodic table, transition metals like Vanadium often present fascinating puzzles regarding their electronic structure. In this problem, we are tasked with acting like atomic detectives to find out exactly how many electrons reside in the $p$-orbitals of a neutral Vanadium atom.

To solve this, we must first recall the atomic number of Vanadium, which is $Z = 23$. This fundamental piece of information tells us that a neutral atom of Vanadium contains exactly $23$ electrons. Our mission is to distribute these electrons into their respective energy levels and subshells to see where they all end up.

### The Rules of the Quantum World

We cannot simply throw electrons into an atom randomly. We must follow the **Aufbau Principle**, which dictates that electrons will always occupy the lowest available energy orbitals before moving on to higher ones.

The sequence of filling is determined by the $(n+l)$ rule, giving us the following strict order:

$$1s < 2s < 2p < 3s < 3p < 4s < 3d$$

By following this sequence, we ensure that the atom remains in its most stable, ground-state configuration.

### Writing the Electronic Configuration

Let's start distributing our $23$ electrons step-by-step:
1. The $1s$ orbital takes $2$ electrons: $1s^2$
2. The $2s$ orbital takes $2$ electrons: $2s^2$
3. The $2p$ subshell takes $6$ electrons: $2p^6$
4. The $3s$ orbital takes $2$ electrons: $3s^2$
5. The $3p$ subshell takes $6$ electrons: $3p^6$
6. The $4s$ orbital takes $2$ electrons: $4s^2$

At this point, we have placed $2 + 2 + 6 + 2 + 6 + 2 = 20$ electrons. We have exactly $3$ electrons remaining. According to our sequence, these last three electrons will enter the $3d$ subshell.

Thus, the complete ground-state electronic configuration for Vanadium is:

$$V_{23} : 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3$$

### Zeroing in on the p-Electrons

With the full configuration laid out, we can now answer the specific question asked: How many electrons are in the $p$-orbitals?

We simply scan our configuration and identify all the $p$-subshells:
- We have a fully filled **$2p$ subshell**, which contains $6$ electrons.
- We also have a fully filled **$3p$ subshell**, which contains another $6$ electrons.

There are no electrons in the $4p$ subshell or any higher $p$-orbitals.

### The Final Calculation

To find the total number of $p$-electrons, we just add them up:

$$	ext{Total } p	ext{-electrons} = 6 	ext{ (from } 2p) + 6 	ext{ (from } 3p) = 12$$

And there we have it! The total number of electrons that Vanadium has in its $p$-orbitals is **$12$**. Always remember that writing out the full electronic configuration is the most foolproof way to avoid silly mistakes in these types of questions.

Animated Solution for Chemistry - d and f-Block Elements: Number of electrons that vanadium ( $Z = 23$ ) has in p-orbitals is equal to ……… .

Enter Numerical Value:

Visualized Solution

$Z = 23$

$Aufbau Principle$

$Electronic Configuration$

$Counting p -electrons$

$Total p -electrons$

$What’s Next?$

The Sigma Insight: d-block Elements

Unveiling the Electrons of Vanadium

The Rules of the Quantum World

Writing the Electronic Configuration

Zeroing in on the p-Electrons

The Final Calculation

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Animated Solution for Chemistry - d and f-Block Elements: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Number of electrons that vanadium (Z=23) has in p-orbitals is equal to ……… .