Electronic configuration of four elements A, B, C and D are given below : (A) 1mathrms^2 2mathrms^2 2mathrmp^3$1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^3$ (B) 1mathrms^2 2mathrms^2 2mathrmp^4$1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^4$ (C) 1mathrms^2 2mathrms^2 2mathrmp^5$1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^5$ (D) 1mathrms^2 2mathrms^2 2mathrmp^2$1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^2$ Which of the following is the correct order of increasing electronegativity (Pauling's scale)?

### Core Logic Let's list the identity of the elements given by their atomic numbers (Z$Z$): * Z = 19 implies textPotassium (K)$Z = 19 \implies \text{Potassium (K)}$, an alkali metal in Period 4. * Z = 32 implies textGermanium (Ge)$Z = 32 \implies \text{Germanium (Ge)}$, a metalloid in Group 14, Period 4. * Z = 35 implies textBromine (Br)$Z = 35 \implies \text{Bromine (Br)}$, a halogen in Group 17, Period 4. * Z = 87 implies textFrancium (Fr)$Z = 87 \implies \text{Francium (Fr)}$, an alkali metal in Period 7. **Ionization Enthalpy Trends**: 1. Ionization enthalpy increases across a period from left to right due to an increase in effective nuclear charge. 2. Ionization enthalpy decreases down a group due to increasing atomic radius and increasing screening effects. Comparing alkali metals (K$K$ and Fr$Fr$), Francium (Fr, [Rn]7s^1$Fr, [Rn]7s^1$) is located far lower in Group 1, possessing an immense atomic size and the highest shielding. This makes its outermost valence electron exceptionally loose and effortless to remove. ### Pattern Recognition Alkali metals always define the absolute minimum first ionization energy in their respective horizontal rows. Among alkali metals, values monotonically decrease downwards, validating Francium (Z=87$Z=87$) immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

### Core Logic Let's analyze the periodic trend among the listed elements: Li, Na, Be, Mg, B, Al. * Atomic radius decreases across a period due to increasing effective nuclear charge (Z_texteff$Z_{\text{eff}}$). * Atomic radius increases down a group due to addition of electron shells. Comparing Period 2 elements (Li, Be, B): Boron (B$B$) has the highest atomic number here and thus the smallest atomic radius. Boron forms an oxide where its oxidation state is +3$+3$, which gives B_2O_3$B_2O_3$. This matches the structural template A_2O_3$A_2O_3$. ### Pattern Recognition Smallest element in Period 2 (excluding noble gases) is on the far right. Boron belongs to Group 13, so it forms traditional trivalent acidic oxides (A_2O_3$A_2O_3$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

### Related Formula Atomic radius decreases across a period due to increase in effective nuclear charge (Z_texteff$Z_{\text{eff}}$) and increases down a group due to addition of new electronic shells. ### Core Logic Let us analyze the elements in option (3): - mathrmBe$\mathrm{Be}$ and mathrmMg$\mathrm{Mg}$ belong to Group 2. Down the group, atomic radius increases: mathrmMg > mathrmBe$\mathrm{Mg} > \mathrm{Be}$. - mathrmMg$\mathrm{Mg}$, mathrmAl$\mathrm{Al}$, and mathrmSi$\mathrm{Si}$ belong to Period 3. Across the period, atomic radius decreases: mathrmMg > mathrmAl > mathrmSi$\mathrm{Mg} > \mathrm{Al} > \mathrm{Si}$. Combining these trends, the correct decreasing order is: mathrmMg > mathrmBe > mathrmAl > mathrmSi$$\mathrm{Mg} > \mathrm{Be} > \mathrm{Al} > \mathrm{Si}$$ Therefore, the order given in option (3) mathrmBe > mathrmMg > mathrmAl > mathrmSi$\mathrm{Be} > \mathrm{Mg} > \mathrm{Al} > \mathrm{Si}$ is incorrect. ### Pattern Recognition Sees: Group 2 and Period 3 comparison. Trap: Assuming atomic radius always increases down a group regardless of cross-period shifts; mathrmMg$\mathrm{Mg}$ is larger than mathrmBe$\mathrm{Be}$, making mathrmBe > mathrmMg$\mathrm{Be} > \mathrm{Mg}$ fundamentally incorrect. Shortcut: Check adjacent group/period boundaries to immediately isolate inversions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

### Core Logic Let us check each criteria statement: * **A is incorrect:** Electron gain can be endothermic for stable configurations like noble gases or alkaline earth metals. * **B is correct:** Removing an electron from a stable atomic nucleus always requires input energy, hence Delta H > 0$\Delta H > 0$ (endothermic). * **C is correct:** textBe \ (1s^2 2s^2)$\text{Be} \ (1s^2 2s^2)$ has a stable, fully-filled subshell configuration, making its first ionization energy higher than textB \ (1s^2 2s^2 2p^1)$\text{B} \ (1s^2 2s^2 2p^1)$ where the electron is removed from a higher energy p$p$-orbital. * **D is incorrect:** Due to inductive withdrawal and shifting effective charge distribution, electronegativity alters slightly contextually across different molecular systems (CCl_4 > CH_4$CCl_4 > CH_4$). * **E is incorrect:** Cesium (textCs$\text{Cs}$) is the most electropositive Group 1 element. ### Step 1: Match with Choices Statements B and C are definitively evaluated to be correct, corresponding to option (1). ### Pattern Recognition Shortcut: Ionization energy is strictly endothermic (+ Delta H$+ \Delta H$). Beryllium versus Boron is a classic fully-filled subshell anomaly (textIE_1 text Be > textB$\text{IE}_1 \text{ Be} > \text{B}$). Knowing these isolates option (1) immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

### Core Logic Let's evaluate the anomalies within Group 13 trends: * **Atomic Radius Order:** Due to the poor shielding effect of the filled 3d$3d$ electron subshell in Gallium (transition contraction), its outer valence shell experiences a stronger nuclear pull. Consequently, the atomic radius sequence exhibits an inversion: textCorrect Atomic Size: B < Ga < Al < In < Tl$$\text{Correct Atomic Size: } B < Ga < Al < In < Tl$$ Therefore, the pair (Al < Ga)$(Al < Ga)$ provided in the problem statement is incorrect. * **Ionic Radius Order (M^3+$M^{3+}$):** In the fully ionized +3$+3$ configuration state, the transition contraction anomaly is overridden by standard shell count physics. The ionic radius follows the uniform down-the-group trend: Al^3+ < Ga^3+$$Al^{3+} < Ga^{3+}$$ Hence, Gallium (Ga$Ga$) has the larger ionic radius between the two elements. Its atomic number is 31. ### Pattern Recognition Transition contraction heavily distorts the *atomic* radius profile of Gallium, but standard descending size progression rules are fully restored when evaluating the *ionic* M^3+$M^{3+}$ radius profile. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties Class 12 Chemistry: The p-Block Elements