Animated Solution for Chemistry - s and p-Block Elements: In which of the following arrangements, the sequence is not strictly according to the property written against it?

This problem tests your mastery over multiple groups and periods simultaneously. We are presented with four sequences and their corresponding properties. Our objective is to identify the sequence that contradicts the property written next to it. Let's dissect them one by one.

The Inert Pair Effect Let's evaluate option (a): $C{O}_2<Si{O}_2<Sn{O}_2<Pb{O}_2$ for increasing oxidising power.

As we descend Group 14, the atomic size increases, and the shielding effect of the inner $d$ and $f$ electrons becomes poor. This poor shielding means the nucleus holds onto the outermost $s$-electrons very tightly, making them reluctant to participate in bonding. This phenomenon is known as the inert pair effect.

Because of this, the $+2$ oxidation state becomes significantly more stable than the $+4$ state for heavier elements like Lead ($Pb$). Therefore, $P{b}^{4+}$ in $Pb{O}_2$ is highly unstable and desperately wants to gain two electrons to become $P{b}^{2+}$. This strong tendency to get reduced makes $Pb{O}_2$ a powerful oxidising agent. The sequence correctly represents increasing oxidising power.

The Size Factor Next, let's look at option (b): $HF<HCl<HBr<HI$ for increasing acid strength.

Acidic strength in hydrogen halides is primarily governed by the bond dissociation energy. As we move down Group 17 from Fluorine to Iodine, the atomic radius of the halogen increases dramatically. A larger atomic radius results in a longer $H-X$ bond length.

We know that a longer bond is a weaker bond. Consequently, the $H-I$ bond requires the least amount of energy to break, allowing it to release an $H^{+}$ ion most readily in an aqueous solution. Thus, $HI$ is the strongest acid, and the given sequence is perfectly correct.

The Half-Filled Anomaly Now, let's examine option (d): $B<C<O<N$ for increasing first ionisation enthalpy.

The general trend across a period is that ionisation enthalpy increases due to the increasing effective nuclear charge. However, Nitrogen ($N$) presents a classic exception.

The electronic configuration of Nitrogen is $1s^{2}2s^{2}2p^3$. The $2p$ subshell is exactly half-filled, which grants it extra exchange energy and symmetrical stability. Because of this enhanced stability, it requires more energy to remove an electron from Nitrogen than from Oxygen ($2p^4$). Therefore, $I{E}_1(N)>I{E}_1(O)$, making the sequence $B<C<O<N$ absolutely correct.

The Electron Density Trap Finally, we arrive at option (c): $N{H}_3>P{H}_3>As{H}_3>Sb{H}_3$ for increasing basic strength.

According to Lewis theory, a base is an electron pair donor. In Group 15 hydrides, the central atom possesses one lone pair of electrons. The basicity depends on how easily this lone pair can be donated.

Nitrogen is a very small atom. Its lone pair is concentrated in a tiny volume, resulting in a very high electron density. This makes Ammonia ($N{H}_3$) an excellent electron donor and a strong base.

Conversely, as we move down the group to Antimony ($Sb$), the atomic size increases massively. The lone pair is now diffused over a much larger volume, causing the electron density to drop drastically. A diffused lone pair is poorly available for donation, making Stibine ($Sb{H}_3$) a very weak base.

The actual trend for basic strength is a decreasing order: $N{H}_3>P{H}_3>As{H}_3>Sb{H}_3$.

However, the property written against this sequence in the question claims it is an "increasing basic strength". This is a direct contradiction! The sequence shows decreasing basicity, while the label says increasing. Therefore, option (c) is the incorrect arrangement and the right answer to our problem.