Consider the following elements In, Tl, Al, Pb, Sn and Ge. The most stable oxidation states of elements with highest and lowest first ionisation enthalpies, respectively, are

Related Formula

Core Logic

Step 1: Evaluate Statement-I:

• Most electronegative element = N (X$X$). Its oxide N₂O₃$\text{N}_2\text{O}_3$ is acidic.
• Least electronegative element = Sb (Y$Y$). Its oxide Sb₂O₃$\text{Sb}_2\text{O}_3$ is amphoteric.

Thus, Statement-I is TRUE.

Step 2: Evaluate Statement-II:

• BCl₃$\text{BCl}_3$ hydrolyses to form boric acid B(OH)₃$\text{B(OH)}_3$ and HCl$\text{HCl}$.
• Boron lacks d$d$-orbitals and cannot expand its coordination number to form hexaaqua species [B(H₂O)₆]³⁺$[\text{B}(\text{H}_2\text{O})_6]^{3+}$.

Thus, Statement-II is FALSE.

Pattern Recognition

Sees: [B(H₂O)₆]³⁺$[\text{B}(\text{H}_2\text{O})_6]^{3+}$ species for Boron. Shortcut: Boron can never show coordination number 6 due to absence of vacant d$d$-orbitals, making Statement-II immediately false.

Chapter Mix

Class 11 Chemistry: p-Block Elements Class 12 Chemistry: p-Block Elements

Core Logic

Evaluate statements based on Group 13 trends. (A) Ionic radii of trivalent cations generally increases down the group. (B⁺³ < Al⁺³ < Ga⁺³ < In⁺³ < Tl⁺³$B^{+3} < Al^{+3} < Ga^{+3} < In^{+3} < Tl^{+3}$). Statement A is incorrect. (B) Electronegativity in Group 13 drops from B to Al, but then increases slightly down to Tl due to poor shielding of d and f electrons. (B > Tl > In > Ga > Al$B > Tl > In > Ga > Al$). Statement B is incorrect. (C) Boron is the smallest and lacks d-orbitals, so its effective nuclear pull on valence electrons is strongest. It has the highest first IE. Statement C is correct. (D) Due to high charge density and Fajan's rules, trichlorides and triiodides of Group 13 (like BCl₃$BCl_3$, AlCl₃$AlCl_3$, etc.) exhibit significant covalent character. Statement D is correct.

Step 1: Final Selection

Only statements C and D are correct.

Pattern Recognition

Group 13 anomalies are extremely common: Electronegativity dips then rises (B > Tl > In > Ga > Al) and IE zig-zags (B > Tl > Ga > Al > In). Always be suspicious of 'smooth trends' in Group 13.

Chapter Mix

Class 11 Chemistry: The p-Block Elements

Related Formula

For a spontaneous reaction, Δ G° < 0$\Delta G^{\circ} < 0$.

Core Logic

Due to the inert pair effect, the stability of the lower oxidation state (+2$+2$) increases down the group in Group 14. Therefore, Pb²⁺$Pb^{2+}$ is more stable than Pb⁴⁺$Pb^{4+}$. Conversely, for lighter elements like Sn, the +4$+4$ oxidation state is more stable than +2$+2$.

Let's evaluate the first reaction:

Here, Pb⁴⁺$Pb^{4+}$ (PbO₂$PbO_2$) is being reduced to Pb²⁺$Pb^{2+}$ (PbO$PbO$), and Pb⁰$Pb^{0}$ is being oxidized to Pb²⁺$Pb^{2+}$. Since Pb²⁺$Pb^{2+}$ is the more stable state, this reaction is thermodynamically favorable (spontaneous). Thus, Δᵣ G°(1) < 0$\Delta_r G^{\circ}(1) < 0$.

Step 1: Evaluate Second Reaction

Now evaluate the second reaction:

Here, Sn⁴⁺$Sn^{4+}$ (SnO₂$SnO_2$) is being converted to Sn²⁺$Sn^{2+}$ (SnO$SnO$). However, we know that Sn⁴⁺$Sn^{4+}$ is more stable than Sn²⁺$Sn^{2+}$. This means the formation of Sn²⁺$Sn^{2+}$ from Sn⁴⁺$Sn^{4+}$ is not favored. The reaction is non-spontaneous. Thus, Δᵣ G°(2) > 0$\Delta_r G^{\circ}(2) > 0$.

Pattern Recognition

Inert pair effect drives heavier elements to lower oxidation states (spontaneous, Δ G < 0$\Delta G < 0$) and lighter elements to higher oxidation states (making reduction to lower state non-spontaneous, Δ G > 0$\Delta G > 0$).

Chapter Mix

Class 11 Chemistry: p-Block Elements Class 12 Chemistry: Thermodynamics

Core Logic

Let's examine the Pauling electronegativity values for Group 15 elements: N = 3.0$\text{N} = 3.0$ P = 2.1$\text{P} = 2.1$ As = 2.0$\text{As} = 2.0$ Sb = 1.9$\text{Sb} = 1.9$ Bi = 1.9$\text{Bi} = 1.9$

The question states that the difference in EN between 'X' and P (2.1$2.1$) is strictly greater than the difference between P (2.1$2.1$) and 'Y'.

Let's test Option 1: X = N, Y = As. Difference between X (N) and P = |3.0 - 2.1| = 0.9$|3.0 - 2.1| = 0.9$ Difference between P and Y (As) = |2.1 - 2.0| = 0.1$|2.1 - 2.0| = 0.1$ Since 0.9 > 0.1$0.9 > 0.1$, this set perfectly satisfies the given condition.

Pattern Recognition

The electronegativity drop from the 2nd period (N) to the 3rd period (P) is very sharp compared to the gradual decrease down the rest of the group (P to As to Sb to Bi). This makes N an extreme outlier in EN differences.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity Class 12 Chemistry: p-Block Elements

Related Formula

Steric Number = Number of σ-bonds + Number of lone pairs on central atom$\text{Steric Number} = \text{Number of } \sigma\text{-bonds} + \text{Number of lone pairs on central atom}$

Core Logic

Evaluate the VSEPR structure of XeO₂F₂$XeO_2F_2$: Central atom Xe has 8 valence electrons. It forms 2 double bonds with Oxygen (2 × 2 = 4$2 \times 2 = 4$ electrons used) and 2 single bonds with Fluorine (2 × 1 = 2$2 \times 1 = 2$ electrons used). Total electrons used in bonding = 6$6$. Remaining valence electrons on Xe = 8 - 6 = 2$8 - 6 = 2$, which forms 1$1$ lone pair. Steric Number = 4 (bond pairs) + 1 (lone pair) = 5$\text{Steric Number} = 4 (\text{bond pairs}) + 1 (\text{lone pair}) = 5$ electron pairs. Hybridization is sp³d$sp^3d$. Geometry is trigonal bipyramidal. According to Bent's rule, the highly electronegative F atoms occupy axial positions, while the double-bonded O atoms and the lone pair occupy equatorial positions. This results in a "see-saw" shape.

Step 1: Check Statements

A. True. The molecule has a see-saw shape. B. False. Xe is surrounded by 4 σ$\sigma$-bonds, 2 π$\pi$-bonds, and 1 lone pair. In total, counting individual bonds, Xe has 7 electron pairs in its valence shell (4 σ$\sigma$, 2 π$\pi$, 1 lp) = 14 valence electrons. C. False. The O atoms occupy equatorial positions, so the O-Xe-O bond angle is close to 120°$120^{\circ}$, not 180°$180^{\circ}$. D. True. The F atoms are axial, so the F-Xe-F bond angle is close to 180°$180^{\circ}$. E. False. The central Xe atom has 14 valence electrons involved (8 from octet, plus sharing from ligands), not 16.

We need to find the INCORRECT statements, which are B, C, and E.

Pattern Recognition

In sp³d$sp^3d$ hybridization, bulky groups (double bonds, lone pairs) always occupy equatorial positions to minimize 90°$90^{\circ}$ repulsions. Highly electronegative groups (F) are forced into axial positions.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: p-Block Elements