### 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

### 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

### Related Formula Specific metal sulfide precipitates exhibit characteristic colors used in qualitative inorganic analysis schemes. ### Core Logic Evaluating the colors of the specified inorganic sulfides: - (NH_4)_2S$(NH_4)_2S$ (Ammonium sulfide solution / yellow ammonium sulfide compound matrix) rightarrow$\rightarrow$ **Yellow** - PbS$PbS$ (Lead sulfide) rightarrow$\rightarrow$ **Black** - CuS$CuS$ (Copper sulfide) rightarrow$\rightarrow$ **Black** - As_2S_3$As_2S_3$ (Arsenic(III) sulfide) rightarrow$\rightarrow$ **Yellow** - As_2S_5$As_2S_5$ (Arsenic(V) sulfide) rightarrow$\rightarrow$ **Yellow** ### Step 1: Identifying Valid Items The sulfides matching the yellow color description are (A), (D), and (E). ### Pattern Recognition In qualitative salt analysis, arsenic belongs to Group IIB and precipitates as a bright yellow sulfide (As_2S_3$As_2S_3$). Transition metal sulfides like PbS$PbS$ and CuS$CuS$ typically form dark black precipitates. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: p-Block Elements