Animated Solution for Chemistry - s and p-Block Elements: The pair(s) of reagents that yield paramagnetic species is / are :

In the fascinating world of inorganic chemistry, magnetism is a direct window into the electronic structure of molecules. When a question asks us to identify paramagnetic species, it is essentially asking us to play detective and hunt for unpaired electrons.

If all electrons in a species are happily paired up in their orbitals, the substance is diamagnetic and repelled by magnetic fields. But if even a single electron is left without a partner, the species becomes paramagnetic and is drawn toward magnetic fields.

Let's break down each reaction in the options and analyze the electronic fate of their products.

The Blue Solution of Solvated Electrons

When sodium metal is dropped into ammonia, the outcome heavily depends on the physical state of the ammonia. If we assume ammonia is a gas, the reaction produces sodium amide and hydrogen gas:

Both ${\text{NaNH}}_2$ and ${\text{H}}_2$ are perfectly diamagnetic. However, in the context of alkali metals, "excess ${\text{NH}}_3$" almost universally implies liquid ammonia.

When sodium dissolves in liquid ammonia, it undergoes a beautiful ionization process:

The electron gets trapped in a cage of ammonia molecules, creating a solvated electron. This free, unpaired electron is highly paramagnetic and gives the solution its characteristic deep blue color. Therefore, Option A yields a paramagnetic species.

The Superoxide Anion

Next, we look at the combustion of potassium in excess oxygen. Alkali metals have a fascinating trend: as the cation gets larger, it can stabilize larger and more complex oxygen anions.

While lithium forms a normal oxide (${\text{Li}}_2\text{O}$) and sodium forms a peroxide (${\text{Na}}_2{\text{O}}_2$), the larger potassium atom forms a superoxide:

Potassium superoxide consists of ${\text{K}}^{+}$ and the superoxide anion, ${\text{O}}_2^{-}$. To determine its magnetic nature, we can simply count the valence electrons. Each oxygen atom has 6 valence electrons, and the negative charge adds one more, giving a total of 13 valence electrons.

An odd number of electrons mathematically guarantees that at least one electron must be unpaired. According to Molecular Orbital Theory, this unpaired electron resides in the ${\pi }^{\ast }$ antibonding orbital. Thus, ${\text{KO}}_2$ is paramagnetic, making Option B correct.

The Classic Nitric Acid Redox

Copper's reaction with nitric acid is a staple of redox chemistry. Because copper is below hydrogen in the reactivity series, it cannot displace hydrogen gas from acids. Instead, the powerful oxidizing nature of nitric acid takes over.

With dilute nitric acid, the reaction proceeds as follows:

Let's analyze the products. The copper(II) ion, ${\text{Cu}}^{2+}$, has an argon core with a $3d^9$ configuration. Nine electrons in the d-subshell mean there is exactly one unpaired electron, making it paramagnetic.

Furthermore, nitric oxide ($\text{NO}$) is a classic odd-electron molecule (15 total valence electrons), meaning it too has an unpaired electron and is paramagnetic in its gaseous state. Since both major products are paramagnetic, Option C is absolutely correct.

The Industrial Synthesis of Hydrogen Peroxide

Finally, we examine the reaction of 2-ethylanthraquinol with oxygen. This isn't just a random organic reaction; it is the cornerstone of the auto-oxidation process used industrially to manufacture hydrogen peroxide on a massive scale.

In this process, the hydroxyl groups (-OH) of the quinol are oxidized to form the double-bonded oxygens (=O) of the quinone, while the ${\text{O}}_2$ is reduced to ${\text{H}}_2{\text{O}}_2$.

If we draw the Lewis structures for both 2-ethylanthraquinone and hydrogen peroxide, we will find that every single electron is neatly paired in bonding or lone pairs. There are no free radicals. Therefore, all products are diamagnetic, and Option D is incorrect.

By carefully analyzing the electronic structures of the products, we have determined that the reactions in options A, B, and C produce species with unpaired electrons.