Animated Solution for Chemistry - Atomic Structure: The number of orbitals associated with quantum numbers $n=5$, $m_s=+\frac{1}{2}$ is

Imagine you are trying to locate a specific person in a massive, sprawling city. You would need their zip code, street name, building number, and finally, their exact apartment. In the microscopic universe of an atom, electrons are located using a very similar system called Quantum Numbers.

These numbers act as the ultimate cosmic address system. The Principal Quantum Number ($n$) tells us the main shell or the "zip code." The Azimuthal Quantum Number ($l$) tells us the subshell or the "street." The Magnetic Quantum Number ($m_l$) specifies the exact orbital or the "building." Finally, the Spin Quantum Number ($m_s$) tells us which specific "bed" the electron occupies inside that orbital.

In this problem, we are given a partial address: $n=5$ and $m_s=+\frac{1}{2}$. Our mission is to find out how many "buildings" (orbitals) are associated with this specific address.

Let's start with the broadest piece of information: $n=5$. This tells us we are looking at the fifth principal shell of the atom.

How many orbitals exist in an entire principal shell? Quantum mechanics provides us with a beautifully elegant rule: for any given principal quantum number $n$, the total number of orbitals in that shell is exactly $n^2$.

Why is this the case? Let's break it down. For $n=5$, the possible values for the azimuthal quantum number $l$ range from $0$ to $n-1$. So, $l$ can be $0,1,2,3,$ or $4$. These correspond to the $5s,5p,5d,5f,$ and $5g$ subshells.

Each subshell contains $2l+1$ orbitals: - For $l=0$ ($5s$): $2(0)+1=1$ orbital - For $l=1$ ($5p$): $2(1)+1=3$ orbitals - For $l=2$ ($5d$): $2(2)+1=5$ orbitals - For $l=3$ ($5f$): $2(3)+1=7$ orbitals - For $l=4$ ($5g$): $2(4)+1=9$ orbitals

If we sum these up: $1+3+5+7+9=25$.

So, there are exactly 25 orbitals in the $n=5$ shell.

Now, we encounter the trap set by the examiners. The question doesn't just say $n=5$; it adds the condition $m_s=+\frac{1}{2}$.

Many students see this and think, "Ah! The spin is restricted to only positive half. Since half the electrons are spin-up and half are spin-down, I must divide the number of orbitals by two!"

This is a classic conceptual pitfall. Let's revisit our hotel analogy. According to Pauli's Exclusion Principle, an orbital is like a hotel room with exactly two beds. One bed is for an electron with spin-up ($m_s=+\frac{1}{2}$), and the other bed is for an electron with spin-down ($m_s=-\frac{1}{2}$).

The question asks: How many orbitals are associated with the quantum number $m_s=+\frac{1}{2}$?

Think about it. Does restricting the spin eliminate any of the hotel rooms? No! Every single one of those 25 rooms has a bed for a spin-up electron. Therefore, every single one of the 25 orbitals is capable of hosting an electron with $m_s=+\frac{1}{2}$.

The spin condition $m_s=+\frac{1}{2}$ does not reduce the number of available orbitals; it merely specifies which "seat" inside the orbital we are talking about. Since every orbital has exactly one seat for a spin-up electron, the number of orbitals associated with this quantum state remains equal to the total number of orbitals in the shell.

Thus, the total number of orbitals is simply $n^2=5^2=25$.