Animated Solution for Chemistry - d and f-Block Elements: The incorrect statement(s) among (1)-(3) is (are) 1. W(VI) is more stable than Cr(VI). 2. in the presence of HCl, permanganate titrations provide satisfactory results. 3. some lanthanoid oxides can be used as phosphors.

The beauty of inorganic chemistry lies not in rote memorization, but in uncovering the logical patterns that govern the elements. Today, we are going to play the role of a chemical detective. We have three statements—three suspects, if you will—and our job is to interrogate each one using the fundamental laws of the d and f-block elements to find the imposter.

Let's dive into the investigation!

The Heavyweight Champion

Our first statement claims: W(VI) is more stable than Cr(VI).

To verify this, we need to look at Group 6 of the periodic table, which houses Chromium (Cr), Molybdenum (Mo), and Tungsten (W). If you recall the trends in the p-block, you might remember the "inert pair effect," where heavier elements prefer lower oxidation states (like Lead preferring +2 over +4).

However, the d-block plays by a completely different set of rules!

In the transition metals, as we move down a group from the 3d series to the 4d and 5d series, the higher oxidation states become increasingly stable. Why does this happen?

The 4d and 5d orbitals are much larger and more spatially extended than the compact 3d orbitals. This allows the heavier metals like Tungsten to form much stronger covalent bonds with electronegative atoms like oxygen or halogens. The immense energy released from forming these strong bonds more than compensates for the high ionization energies required to strip away six electrons.

Because of this, Tungsten in the +6 state (like in ${\text{WO}}_3$) is incredibly stable and unreactive. On the other hand, Chromium in the +6 state (like in the dichromate ion, ${\text{Cr}}_2{\text{O}}_7^{2-}$) is highly unstable and acts as a ferocious oxidizing agent, desperately trying to gain electrons to reach the more stable +3 state.

Verdict: Statement 1 is absolutely correct.

The Titration Saboteur

Our second statement claims: In the presence of HCl, permanganate titrations provide satisfactory results.

Imagine you are in the laboratory setting up a redox titration. You have your beautiful, deep purple Potassium Permanganate (${\text{KMnO}}_4$) in the burette. You need an acidic medium for the reaction to proceed, so you reach for a bottle of Hydrochloric Acid ($\text{HCl}$).

Stop right there! This is a classic laboratory disaster.

Potassium Permanganate is one of the strongest oxidizing agents we have. Its job in a titration is to oxidize the analyte (the substance you are trying to measure). However, if you introduce $\text{HCl}$ into the flask, the permanganate will look at the chloride ions (${\text{Cl}}^{-}$) and say, "I can oxidize those too!"

Instead of reacting solely with your analyte, a significant portion of your precious ${\text{KMnO}}_4$ is wasted oxidizing the chloride ions into greenish-yellow chlorine gas (${\text{Cl}}_2$). Because the permanganate is consumed in this side reaction, your burette reading will be artificially high, completely ruining the stoichiometry and accuracy of your titration.

This is exactly why we use dilute Sulfuric Acid (${\text{H}}_2{\text{SO}}_4$) instead. The sulfate ion (${\text{SO}}_4^{2-}$) is already in its maximum +6 oxidation state and cannot be oxidized any further by the permanganate.

Verdict: Statement 2 is a blatant lie. It is incorrect.

The Glowing Rare Earths

Our final statement claims: Some lanthanoid oxides can be used as phosphors.

Let's shift our focus to the f-block, specifically the lanthanides. Have you ever wondered what makes the vibrant red, green, and blue colors on an old CRT television screen or inside modern fluorescent lamps? The secret lies in the unique quantum mechanics of the f-orbitals.

Many lanthanide ions, such as Europium (${\text{Eu}}^{3+}$) and Terbium (${\text{Tb}}^{3+}$), possess unpaired electrons buried deep within their 4f subshells. Because these f-orbitals are shielded from the outside environment by the filled 5s and 5p shells, their energy levels remain incredibly sharp and well-defined.

When these ions absorb energy, their electrons jump to higher f-orbital energy levels. When they relax back down to their ground state, they release that energy as photons of visible light. This phenomenon is known as an f-f transition.

Because the energy levels are so sharp, the emitted light is incredibly pure in color. Europium oxides give us brilliant reds, while Terbium compounds give us piercing greens. Therefore, lanthanoid oxides are indeed heavily utilized as phosphors in various display and lighting technologies.

Verdict: Statement 3 is perfectly correct.

We have successfully interrogated all three statements. We found that Statement 1 and Statement 3 are scientifically sound, while Statement 2 describes a flawed laboratory procedure.

The question specifically asks us to identify the incorrect statement.

Therefore, the only incorrect statement is the second one, making our final answer option (b). Always remember to read the question carefully—finding the truth is only half the battle; answering what is actually asked is the key to victory!