### Core Logic Let us check the trends for the provided main group elements (mathrmAl, In, Tl$\mathrm{Al, In, Tl}$ from Group 13 and mathrmGe, Sn, Pb$\mathrm{Ge, Sn, Pb}$ from Group 14): - **Highest First Ionization Enthalpy (mathrmIE_1$\mathrm{IE}_1$):** Out of these options, Germanium (mathrmGe$\mathrm{Ge}$) sits highest and further right along its period layout, demonstrating the highest mathrmIE_1$\mathrm{IE}_1$ value among this set. Its most stable oxidation state is **+4$+4$**. - **Lowest First Ionization Enthalpy (mathrmIE_1$\mathrm{IE}_1$):** Indium (mathrmIn$\mathrm{In}$) lies lowest leftward among these relative coordinates, maintaining the lowest mathrmIE_1$\mathrm{IE}_1$. Its most stable group oxidation state is **+3$+3$** (as the inert pair effect is much more pronounced for the heavier element mathrmTl$\mathrm{Tl}$ which prefers +1$+1$). ### Pattern Recognition Sees: mathrmIE_1$\mathrm{IE}_1$ extrema vs stable oxidation state profiles. Trap: Forgetting that inert pair shifts display max stability values at +1$+1$ for mathrmTl$\mathrm{Tl}$ and +2$+2$ for mathrmPb$\mathrm{Pb}$, while lighter counterparts like mathrmIn$\mathrm{In}$ favor +3$+3$ and mathrmGe$\mathrm{Ge}$ favors +4$+4$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: The p-Block Elements Class 12 Chemistry: The p-Block Elements

### Core Logic Let us analyze both statements systematically: * **Statement I:** The textN-textN$\text{N}-\text{N}$ single bond is indeed weaker than the textP-textP$\text{P}-\text{P}$ single bond due to high inter-electronic repulsion between non-bonding lone pairs on the small nitrogen atoms. However, nitrogen has a smaller atomic size than phosphorus, making the textN-textN$\text{N}-\text{N}$ single bond shorter (140text pm$140\text{ \pm}$) compared to the textP-textP$\text{P}-\text{P}$ single bond (221text pm$221\text{ \pm}$). Thus, Statement I is false because it incorrectly claims it is longer. * **Statement II:** In Group 15, only nitrogen and phosphorus compounds in the +3$+3$ oxidation state readily undergo disproportionation. As we go down the group (As, Sb, Bi), the +3$+3$ oxidation state becomes increasingly stable due to the inert pair effect, meaning they do not readily undergo disproportionation. Hence, Statement II is false as a general trend across all group 15 elements. ### Step 1: Verification of Conclusions Since the textN-textN$\text{N}-\text{N}$ bond is shorter and heavier elements in the +3$+3$ state do not disproportionate readily, both statements are evaluated to be false. ### Pattern Recognition Sees: "textN-textN$\text{N}-\text{N}$ single bond longer" ightarrow$ ightarrow$ Absolute error. Small atoms form short bonds, always. Inert pair effect stabilizes +3$+3$ lower down the group, preventing disproportionation reactions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: p-Block Elements

### Core Logic - **Statement I is correct:** Moving left-to-right across a period from Group 13 to Group 14 increases the effective nuclear charge (Z_texteff$Z_{\text{eff}}$) and decreases the atomic radius. Consequently, more energy is required to extract an electron, so Group 14 elements exhibit higher first ionisation enthalpies. - **Statement II is incorrect:** Group 14 elements (like Carbon, Silicon, Germanium) build robust, highly stable three-dimensional covalent network crystal structures. As a result, the melting and boiling points of Group 14 elements are generally much higher than those of the corresponding Group 13 elements. ### Pattern Recognition Covalent network solids (Group 14) create huge upward steps in phase transition energy compared to Group 13 frameworks (e.g., Gallium, which melts at nearly room temperature). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: The p-Block Elements

### Related Formula textGroup 13 Ionisation Enthalpy Order: B > Tl > Ga > Al > In$\text{Group 13 Ionisation Enthalpy Order: } B > Tl > Ga > Al > In$ ### Core Logic The first ionisation enthalpy trend for Group 13 elements is irregular due to poor shielding by d and f electrons: - **Boron (B)** is the smallest atom in the group with no d-orbital shielding issues, so it has the **highest** ionisation enthalpy. - As we move down, poor shielding by 3d$3d$ electrons causes a slight increase at Gallium (Ga$Ga$). Poor shielding by 4f$4f$ electrons causes a sharp increase at Thallium (Tl$Tl$). - **Indium (In)** ends up with the weakest effective attraction for its outermost valence electron, giving it the **lowest** first ionisation enthalpy. Therefore, the highest and lowest elements are **B & In** respectively. ### Pattern Recognition Group 13 does not follow a linear downward trend. Remember the characteristic 'W' shape or zig-zag pattern of its ionisation energies. Indium sits at the absolute minimum point of this curve. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: The p-Block Elements

### Core Logic * **Statement I is true:** Nitrogen has a small atomic size and high electronegativity, allowing it to form strong multiple ppi - ppi$p\pi - p\pi$ bonds with oxygen atoms. This enables stable oxide forms across a wide range of oxidation numbers spanning +1$+1$ to +5$+5$ (e.g., N_2O_5$N_2O_5$). * **Statement II is true:** Nitrogen is a second-period element with a valence shell configuration of 2s^2 2p^3$2s^2 2p^3$. It completely lacks vacant 2d$2d$ orbitals. As a result, it cannot expand its octet beyond a covalency of 4, preventing the synthesis of pentahalides like NX_5$NX_5$. ### Pattern Recognition Second-period elements are structurally bounded by a maximum covalency of 4 due to the complete absence of d-orbitals. This explains why NF_5$NF_5$ is unstable/non-existent while PF_5$PF_5$ is easily synthesized. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The p-Block Elements