Animated Solution for Chemistry - Periodicity in Properties: Within each pair of elements F and Cl, S and Se, and Li and Na, respectively, the elements that release more energy upon an electron gain are

Imagine an atom floating in the vast emptiness of space. Suddenly, a rogue electron zips by. The atom's nucleus, with its positive charge, reaches out and pulls that electron in. This cosmic capture isn't just a silent event; it's a violent collision that releases a burst of energy. In the world of chemistry, we call this burst of energy the Electron Gain Enthalpy (${\Delta }_{eg}H$).

The more an atom "wants" that electron, the more energy it releases when it finally catches it. In this problem, we are playing a game of cosmic tug-of-war. We have three pairs of elements, and we need to figure out which element in each pair is the most eager to grab an extra electron.

Before we dive into the specific pairs, let's establish a fundamental rule of the periodic table. As you move down a group, atoms get larger. They add more and more electron shells, like layers of an onion.

Because the atom is larger, the positively charged nucleus is buried deep inside. When a new electron approaches, it doesn't feel a very strong pull from that distant nucleus. Furthermore, the existing inner electrons act like a shield, blocking the nucleus's attractive force.

Therefore, the general rule is simple: Smaller atoms exert a stronger pull on incoming electrons and release more energy upon gaining them. Mathematically, we can say that the magnitude of electron gain enthalpy is inversely proportional to atomic size: $|{\Delta }_{eg}H|\propto \frac{1}{\text{Atomic Size}}$.

Group 16 and Group 1

Let's test our rule on the first two pairs.

First, we look at Group 16: Sulfur ($\text{S}$) and Selenium ($\text{Se}$). If you glance at the periodic table, Sulfur sits right above Selenium. This means Sulfur has fewer electron shells and is significantly smaller than Selenium. Following our golden rule, the incoming electron can get much closer to Sulfur's nucleus. Thus, Sulfur releases more energy than Selenium.

Next, we jump over to Group 1: Lithium ($\text{Li}$) and Sodium ($\text{Na}$). The logic here is identical. Lithium is at the top of the group, making it smaller than Sodium. The nucleus of Lithium has a tighter grip on any passing electron. Therefore, Lithium releases more energy than Sodium.

So far, so good. The rule works perfectly. But chemistry is never without its surprises.

The Fluorine Anomaly

Now we arrive at Group 17, the halogens: Fluorine ($\text{F}$) and Chlorine ($\text{Cl}$).

If we blindly follow our rule, Fluorine is above Chlorine, so it is smaller. It is also the most electronegative element in the entire periodic table. It absolutely loves electrons. So, Fluorine should release the most energy, right?

Wrong. Here lies one of the most famous traps in all of chemistry.

Fluorine is indeed very small. In fact, it is too small. Its outermost shell is the $2p$ subshell, which is extremely compact. When you try to force a brand new electron into this tiny, crowded space, the existing seven valence electrons push back fiercely. This is known as inter-electronic repulsion.

Imagine trying to shove a 10th person into a tiny elevator that already has 9 people in it. They are going to push back!

Chlorine, on the other hand, uses the much larger $3p$ subshell. It has plenty of room. When an extra electron arrives, Chlorine welcomes it with open arms, without the intense repulsion seen in Fluorine. Because the incoming electron faces less resistance, Chlorine actually releases more energy than Fluorine.

Let's bring it all together. We analyzed our three pairs and found our winners: - From Group 17, Chlorine ($\text{Cl}$) beats Fluorine due to less inter-electronic repulsion. - From Group 16, Sulfur ($\text{S}$) beats Selenium due to its smaller size. - From Group 1, Lithium ($\text{Li}$) beats Sodium due to its smaller size.

Our winning trio is $\text{Cl}$, $\text{S}$, and $\text{Li}$. This perfectly matches option (c).

The beauty of this problem lies in how it tests both your knowledge of general periodic trends and your awareness of the critical exceptions that make chemistry so fascinating. Always remember: rules are important, but understanding why the rules exist is what truly makes you a master of the subject.