Animated Solution for Chemistry - Atomic Structure: The correct statement about probability density (except at infinite distance from nucleus) is

Imagine trying to find a hyperactive firefly in a massive, dark stadium. You can't pinpoint its exact location, but you can predict where it is most likely to be. This is exactly how we treat electrons in the quantum mechanical model of the atom. We abandon the idea of fixed orbits (like planets around the sun) and instead embrace the concept of probability density.

In quantum mechanics, the behavior of an electron is described by a mathematical function called the wave function, denoted by the Greek letter $\psi$ (psi). However, $\psi$ itself doesn't have a direct physical meaning. It was the brilliant physicist Max Born who proposed that the square of the absolute value of the wave function, $|\psi {|}^2$, represents the probability density—the probability of finding an electron in a specific tiny volume of space around the nucleus.

Let's tackle the options in our problem one by one. Option (b) suggests that probability density can be negative for a 2p orbital.

Think about this logically: can a probability ever be less than zero? You can have a $0\%$ chance of winning the lottery, but you can't have a $-10\%$ chance! Mathematically, since probability density is defined as $|\psi {|}^2$, and the square of any real (or complex magnitude) number is always non-negative, probability density can never be negative. It is strictly $|\psi {|}^2\ge 0$. Thus, option (b) is fundamentally flawed and immediately eliminated.

The Concept of Nodes

If probability density is the chance of finding an electron, what happens when that chance drops exactly to zero? We call these regions of zero probability nodes.

There are two types of nodes in an atom: 1. Angular Nodes: These are flat planes or cones where the probability is zero. The number of angular nodes is simply equal to the azimuthal quantum number, $l$. 2. Radial Nodes: These are spherical shells (like the empty spaces between the layers of an onion) where the probability is zero. The number of radial nodes is calculated using the formula:

Where $n$ is the principal quantum number and $l$ is the azimuthal quantum number.

Now, let's apply our radial node formula to the remaining options to see which orbital has a probability density that drops to zero at a finite distance from the nucleus.

Analyzing the 1s Orbital (Option a): For a 1s orbital, $n=1$ and $l=0$. Plugging this into our formula: The 1s orbital has zero radial nodes. Its probability density is highest right at the nucleus and smoothly exponentially decays as you move outward. It only truly reaches zero at an infinite distance. Therefore, option (a) is incorrect.

Analyzing the 2s Orbital (Option d): For a 2s orbital, $n=2$ and $l=0$. Let's calculate the nodes: The 2s orbital has exactly one radial node. This means there is a specific spherical boundary around the nucleus where the probability density is exactly zero. Option (d) claims it can never be zero, which is completely false.

Analyzing the 3p Orbital (Option c): For a 3p orbital, $n=3$ and $l=1$ (since $s=0,p=1,d=2,f=3$). Let's calculate the nodes: The 3p orbital possesses exactly one radial node. If you were to plot the radial probability density $|R(r){|}^2$ against the distance $r$ from the nucleus, the curve would start at zero, rise to a small peak, drop back down to exactly zero (this is the node!), and then rise to a larger peak before finally tapering off towards infinity.

Because the 3p orbital has a radial node, its probability density can be zero at a specific, finite distance from the nucleus.

By systematically applying the rules of quantum mechanics and the formula for radial nodes, we have proven that the probability density of a 3p orbital drops to zero at a finite distance. This makes Option (c) the undeniably correct statement.

Understanding nodes isn't just about memorizing a formula; it's about visualizing the beautiful, complex, and layered architecture of the quantum atom!