### Related Formula textBond Strength propto frac1textSize difference$\text{Bond Strength} \propto \frac{1}{\text{Size difference}}$ textAcidic Strength propto frac1textM-H Bond Dissociation Energy$\text{Acidic Strength} \propto \frac{1}{\text{M-H Bond Dissociation Energy}}$ ### Core Logic Let's analyze the properties of Tellurium compounds: 1. **Tellurium Dioxide** (mathrmTeO_2$\mathrm{TeO}_2$): - Due to the **inert pair effect**, the +6$+6$ oxidation state of Tellurium is less stable, whereas its +4$+4$ state is relatively stable. However, in comparison to sulphur dioxide (which is a strong reducing agent), mathrmTeO_2$\mathrm{TeO}_2$ is oxidising because the lower oxidation states (like element Tellurium or +2$+2$) are chemically accessible. Thus, mathrmTeO_2$\mathrm{TeO}_2$ acts as an **oxidising agent**. 2. **Tellurium Hydride** (mathrmTeH_2$\mathrm{TeH}_2$): - Tellurium is a very large atom. The orbital overlap between Tellurium and Hydrogen is extremely poor. Hence, the mathrmTe-H$\mathrm{Te-H}$ bond is very long and has very **low bond dissociation energy**. - This allows mathrmTeH_2$\mathrm{TeH}_2$ to easily release mathrmH^+$\mathrm{H^+}$ in solution, making it highly **acidic**. ### Step 1: Final Verification Therefore, the nature of mathrmTeO_2$\mathrm{TeO}_2$ is oxidising, and the nature of mathrmTeH_2$\mathrm{TeH}_2$ is acidic. ### Pattern Recognition Periodic Trend: As we go down Group 16: - Acidic strength of hydrides increases: mathrmH_2O < H_2S < H_2Se < H_2Te$\mathrm{H_2O < H_2S < H_2Se < H_2Te}$. - Reducing character of hydrides also increases. - Reducing power of dioxides decreases: mathrmSO_2$\mathrm{SO_2}$ (reducing) rightarrow mathrmTeO_2$\rightarrow \mathrm{TeO_2}$ (oxidising). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: p-Block Elements

### Related Formula Group 13 elements (mathrmB, mathrmAl, mathrmGa, mathrmIn, mathrmTl$\mathrm{B}, \mathrm{Al}, \mathrm{Ga}, \mathrm{In}, \mathrm{Tl}$) show highly anomalous periodic trends due to the intervention of filled d-orbitals (d-block contraction in mathrmGa$\mathrm{Ga}$) and f-orbitals (lanthanoid contraction in mathrmTl$\mathrm{Tl}$). ### Core Logic Evaluate each specified trend against official physical constants: - **Atomic radius**: Due to d-block contraction, gallium (mathrmGa$\mathrm{Ga}$) is smaller than aluminum (mathrmAl$\mathrm{Al}$): textRadius (pm): mathrmB(88) < mathrmGa(135) < mathrmAl(143) < mathrmIn(167) < mathrmTl(170)$\text{Radius (pm): } \mathrm{B}(88) < \mathrm{Ga}(135) < \mathrm{Al}(143) < \mathrm{In}(167) < \mathrm{Tl}(170)$ Hence, the given order is *Incorrect*. - **Electronegativity**: Electronegativity first decreases from mathrmB$\mathrm{B}$ to mathrmAl$\mathrm{Al}$, then increases down the group due to poor shielding of d and f electrons: textElectronegativity: mathrmAl(1.5) < mathrmGa(1.6) < mathrmIn(1.7) < mathrmTl(1.8) < mathrmB(2.0)$\text{Electronegativity: } \mathrm{Al}(1.5) < \mathrm{Ga}(1.6) < \mathrm{In}(1.7) < \mathrm{Tl}(1.8) < \mathrm{B}(2.0)$ Hence, this order is *Correct*. ### Step 1: Analyze density and ionization energy trends - **Density**: Increases down the group as atomic mass increases much faster than atomic volume: textDensity (g/cm^3text): mathrmB(2.35) < mathrmAl(2.70) < mathrmGa(5.90) < mathrmIn(7.31) < mathrmTl(11.85)$\text{Density (g/cm}^3\text{): } \mathrm{B}(2.35) < \mathrm{Al}(2.70) < \mathrm{Ga}(5.90) < \mathrm{In}(7.31) < \mathrm{Tl}(11.85)$ Hence, the given order is *Incorrect* (it is completely reversed). - **1^mathrmst$1^{\mathrm{st}}$ Ionisation Energy**: Shows an irregular trend due to ineffective shielding by d and f electrons: textIE_1mathrm~(kJ/mol): mathrmIn(558) < mathrmAl(577) < mathrmGa(579) < mathrmTl(589) < mathrmB(801)$\text{IE}_1\mathrm{~(kJ/mol): } \mathrm{In}(558) < \mathrm{Al}(577) < \mathrm{Ga}(579) < \mathrm{Tl}(589) < \mathrm{B}(801)$ Hence, this order is *Correct*. ### Step 2: Conclusion Only the Electronegativity (B) and 1^mathrmst$1^{\mathrm{st}}$ Ionisation Energy (D) orders are correct, matching Option (1). ### Pattern Recognition Group 13 elements do not follow monotonic trends. The poor shielding of 3d^10$3d^{10}$ and 4f^14$4f^{14}$ electrons increases the effective nuclear charge on valence electrons, causing anomalies in atomic radius (mathrmGa < mathrmAl$\mathrm{Ga} < \mathrm{Al}$) and pulling electronegativities and ionization energies upward as you go further down. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: The p-Block Elements Class 11 Chemistry: Classification of Elements and Periodicity in Properties

### Core Logic For Group 14 (textC, textSi, textGe, textSn, textPb$\text{C}, \text{Si}, \text{Ge}, \text{Sn}, \text{Pb}$): - The ionisation energies generally decrease down the group, but there is an anomaly between textSn$\text{Sn}$ and textPb$\text{Pb}$ due to relativistic contraction / poor shielding of 4f electrons in textPb$\text{Pb}$. - Thus, the first ionisation enthalpy of Tin (mathrmSn$\mathrm{Sn}$) is 708 text kJ mol^-1$708 \text{ kJ mol}^{-1}$ and Lead (mathrmPb$\mathrm{Pb}$) is 715 text kJ mol^-1$715 \text{ kJ mol}^{-1}$. These are indeed the lowest in the group. - Hence, element **A** is mathrmSn$\mathrm{Sn}$ and **B** is mathrmPb$\mathrm{Pb}$. Nature of their ions: - mathrmA^2+ = mathrmSn^2+$\mathrm{A}^{2+} = \mathrm{Sn}^{2+}$: Since mathrmSn^4+$\mathrm{Sn}^{4+}$ is more stable than mathrmSn^2+$\mathrm{Sn}^{2+}$, mathrmSn^2+$\mathrm{Sn}^{2+}$ readily undergoes oxidation to +4$+4$, acting as a strong **reducing agent**. - mathrmB^4+ = mathrmPb^4+$\mathrm{B}^{4+} = \mathrm{Pb}^{4+}$: Due to the strong **inert pair effect**, mathrmPb^2+$\mathrm{Pb}^{2+}$ is highly stable compared to mathrmPb^4+$\mathrm{Pb}^{4+}$. Thus, mathrmPb^4+$\mathrm{Pb}^{4+}$ is eager to reduce to +2$+2$, acting as a strong **oxidising agent**. ### Pattern Recognition Inert pair effect becomes extremely prominent at the bottom of the group. Lead's most stable state is +2$+2$, making mathrmPb^4+$\mathrm{Pb}^{4+}$ oxidising. Tin's stable state is +4$+4$, making mathrmSn^2+$\mathrm{Sn}^{2+}$ reducing. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: p-Block Elements Class 11 Chemistry: Periodic Classification of Elements

### Core Logic Let us analyze the periodic properties of chalcogen hydrides (textGroup 16$\text{Group 16}$): 1. **Bond Dissociation Enthalpy (Delta_textdisH$\Delta_{\text{dis}}H$)**: As we descend the group from Selenium to Tellurium, the size of the central atom increases significantly (r_textTe > r_textSe$r_{\text{Te}} > r_{\text{Se}}$). This increase in size leads to poorer orbital overlap with the small 1s$1s$ orbital of hydrogen, resulting in a longer and weaker textM-H$\text{M-H}$ bond. Consequently, the bond dissociation enthalpy decreases: Delta_textdisH: textH_2textSe (276 text kJ mol^-1) > textH_2textTe (238 text kJ mol^-1)$\Delta_{\text{dis}}H: \text{H}_2\text{Se } (276 \text{ kJ mol}^{-1}) > \text{H}_2\text{Te } (238 \text{ kJ mol}^{-1})$ **Thus, Statement II is true.** 2. **Acidic Strength**: A weaker bond dissociates more easily in aqueous solution to release textH^+$\text{H}^+$ ions. Since the textTe-H$\text{Te-H}$ bond is weaker than the textSe-H$\text{Se-H}$ bond, textH_2textTe$\text{H}_2\text{Te}$ releases protons much more readily than textH_2textSe$\text{H}_2\text{Se}$, making it a stronger acid: textAcidic Strength: textH_2textSe < textH_2textTe$\text{Acidic Strength: } \text{H}_2\text{Se} < \text{H}_2\text{Te}$ **Thus, Statement I is false.** ### Pattern Recognition For binary hydrides down any group (like Group 15, 16, or 17), atomic size increase weakens the covalent bond. A weaker bond releases protons more effectively, meaning that both **acidic strength and reducing character increase down the group**, while thermal stability decreases. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: p-Block Elements

### Core Logic The main element of the apatite mineral family (e.g., fluorapatite Ca_5(PO_4)_3F$Ca_{5}(PO_{4})_{3}F$) is Phosphorus (P$P$). The first four elements of Group 15 are N$N$, P$P$, As$As$, Sb$Sb$. First ionization enthalpy decreases regularly down the group: IE_1(N) > IE_1(P) > IE_1(As) > IE_1(Sb)$IE_1(N) > IE_1(P) > IE_1(As) > IE_1(Sb)$ Sorting the given enthalpy data values in decreasing order: 1402 > 1012 > 947 > 834$1402 > 1012 > 947 > 834$ Assigning these to the elements: * N = 1402text kJ mol^-1$N = 1402\text{ kJ mol}^{-1}$ * P = 1012text kJ mol^-1$P = 1012\text{ kJ mol}^{-1}$ * As = 947text kJ mol^-1$As = 947\text{ kJ mol}^{-1}$ * Sb = 834text kJ mol^-1$Sb = 834\text{ kJ mol}^{-1}$ ### Pattern Recognition Apatite family = Phosphorus reference. Match the elements down a column directly to a monotonic numerical array. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: p-Block Elements