Animated Solution for Chemistry - Atomic Structure: Comprehension Passage

Welcome to a beautiful and comprehensive problem on atomic structure. We are presented with a matrix matching table that connects orbitals (Column 1) with their mathematical wave functions and nodal properties (Column 2), and their physical properties or graphs (Column 3).

Our mission is to decode this table step by step, establishing the correct links between the physics and the math, and then use our findings to answer three specific questions. Let's dive into the quantum realm!

Let's systematically break down the orbitals given in Column 1 and find their matches in Column 2.

1. The 1s Orbital (I): For the 1s orbital, the principal quantum number is $n=1$ and the azimuthal quantum number is $l=0$. The number of radial nodes is given by the formula $n-l-1$, which yields $1-0-1=0$. Because it is an s-orbital, it is spherically symmetric and has no angular dependence. Its wave function is purely exponential, decaying as $\Psi \propto e^{-Zr/a_0}$. This perfectly matches entry (i) in Column 2.

2. The 2s Orbital (II): For the 2s orbital, $n=2$ and $l=0$. Calculating the radial nodes gives $2-0-1=1$. This directly matches entry (ii), which states "One radial node".

3. The $2p_z$ Orbital (III): Here, $n=2$ and $l=1$. The wave function for a $p_z$ orbital depends on the angle $\theta$, specifically containing a $\cos \theta$ term. This matches entry (iii). Furthermore, when $\theta =90^{\circ }$ (which corresponds to the xy-plane), $\cos (90^{\circ })=0$. This means the probability of finding an electron on the xy-plane is zero, making it a nodal plane.

4. The $3d_{z^2}$ Orbital (IV): For this orbital, $n=3$ and $l=2$. The $3d_{z^2}$ orbital is unique; it has two conical nodes rather than planar nodes. Therefore, it does not have the xy-plane as a nodal plane, meaning it does not match entry (iv).

Now, let's look at the properties in Column 3.

Graph (P): The graph shows the radial wave function $\Psi (r)$ starting from a positive value, crossing the horizontal axis exactly once, and then asymptotically approaching zero. That single crossing is the physical manifestation of exactly one radial node. Which orbital did we find has one radial node? The 2s orbital (II)!

Statement (S): This statement claims that the energy needed to excite an electron from $n=2$ to $n=4$ is $\frac{27}{32}$ times the energy needed to excite it from $n=2$ to $n=6$. Let's verify this using the Bohr energy formula, $E_n=-13.6\frac{Z^2}{n^2}\text{ eV}$.

The energy difference for the first transition is:

The energy difference for the second transition is:

Taking the ratio of these two energies:

Notice how the $13.6Z^2$ terms completely cancel out! This means statement (S) is a universal truth for any hydrogen-like species, regardless of the specific orbital.

With our decoded matrix, answering the questions is a breeze.

Question 6: We need the only CORRECT combination. - Option (A) pairs $3d_{z^2}$ with an xy-nodal plane. False. - Option (B) pairs the 2s orbital with one radial node and Graph (P). Since Graph (P) also shows one radial node, this is a perfect, logically sound match! - Option (C) pairs $2p_z$ (0 radial nodes) with Graph (P) (1 radial node). False. - Option (D) pairs 1s (0 radial nodes) with 1 radial node. False. Answer: (B)

Question 7: We need the only INCORRECT combination for the ${\text{He}}^{+}$ ion. - Options (A), (B), and (C) all make correct statements about the 1s and 2s orbitals. - Look closely at Option (D): It pairs the 1s orbital (I) with equation (iii). Equation (iii) contains a $\cos \theta$ term, implying angular dependence. However, an s-orbital is spherically symmetric and has no angular dependence! This combination is fundamentally flawed. Answer: (D)

Question 8: We need the only CORRECT combination for the hydrogen atom. - Option (A) claims 1s has an xy-nodal plane. False. - Option (B) pairs 1s (0 radial nodes) with Graph (P) (1 radial node). False. - Option (C) pairs 2s with the 1s wave function equation (i). False. - Option (D) pairs the 1s orbital with its correct purely exponential wave function (i) and the universally true energy statement (S). This is absolutely correct. Answer: (D)