Animated Solution for Chemistry - Atomic Structure: The 2s and the 2p orbital energies of hydrogen atom are $E_{2s}(\text{H})$ and $E_{2p}(\text{H})$, respectively. The 2s and the 2p orbital energies of lithium atom are $E_{2s}(\text{Li})$ and $E_{2p}(\text{Li})$, respectively. The correct option (s) about the orbital energies is (are)

Comparing Orbital Energies in Hydrogen and Lithium

When we study atomic structure, orbital energies are often treated as static numbers to be memorized. However, these energies are actually the result of a delicate, dynamic tug-of-war between the positively charged nucleus and the negatively charged electrons. In this problem, we are tasked with comparing the energies of the $2s$ and $2p$ orbitals, not just within a single atom, but across two very different systems: Hydrogen and Lithium.

Let's begin by analyzing the simplest atom in the universe: Hydrogen. Hydrogen is a single-electron system. Because there is only one electron, there are absolutely no inter-electronic repulsions to worry about.

In such pristine systems, the energy of an orbital is determined entirely by its principal quantum number, $n$. The azimuthal quantum number, $l$, which dictates the shape of the orbital, has no effect on the energy.

Since both the $2s$ and $2p$ orbitals in hydrogen share the same principal quantum number ($n=2$), they are what physicists call degenerate. This means their energies are exactly equal:

This fundamental rule immediately confirms that Option (B) is correct.

Now, let's shift our focus to the Lithium atom. Lithium has an atomic number of $Z=3$, meaning it possesses three electrons. This makes it a multi-electron system.

In a multi-electron atom, the electrons constantly repel each other. This repulsion, combined with the shielding effect of inner electrons, breaks the degeneracy we saw in hydrogen. The energy of a subshell is no longer dictated solely by $n$. Instead, we must rely on the $(n+l)$ rule.

Let's calculate the $(n+l)$ values for Lithium's subshells: - For the $2s$ orbital: $n=2,l=0⟹(n+l)=2$ - For the $2p$ orbital: $n=2,l=1⟹(n+l)=3$

Since $2<3$, the $2s$ orbital resides at a lower energy level than the $2p$ orbital. Therefore:

This confirms that Option (A) is also correct.

Hydrogen vs. Lithium

Here is where the problem gets truly fascinating. We must compare the $2s$ energy of Hydrogen directly with the $2s$ energy of Lithium. To do this, we need to understand Effective Nuclear Charge ($Z_{\text{eff}}$).

Imagine you are an electron residing in the $2s$ orbital. In Hydrogen, you feel the attractive pull of exactly one proton. However, in Lithium, the nucleus contains three protons. Even though the two inner $1s$ electrons try to shield you from the nucleus, their shielding is not perfect. The net positive charge—the $Z_{\text{eff}}$—pulling on the $2s$ electron in Lithium is significantly stronger than the pull of the single proton in Hydrogen.

What does a stronger pull mean for the energy? A higher $Z_{\text{eff}}$ pulls the electron closer to the nucleus, making it more tightly bound. In the realm of quantum mechanics, a more tightly bound state corresponds to a more negative (lower) energy level.

Because the $2s$ electron in Lithium is pulled harder than the $2s$ electron in Hydrogen, its energy drops lower:

This logical deduction confirms that Option (D) is correct, and naturally renders Option (C) incorrect.

By carefully distinguishing between single-electron and multi-electron systems, and by applying the concept of effective nuclear charge, we have successfully navigated this problem. The correct statements are (A), (B), and (D).