Consider the following reactions. Na₂B₄O₇ Δ 2X + Y$Na_{2}B_{4}O_{7} \xrightarrow{\Delta} 2X + Y$ CuSO₄ + Y Non-Luminous flame Z + SO₃$CuSO_{4} + Y \xrightarrow{\text{Non-Luminous flame}} Z + SO_{3}$ 2Z + 2X + Carbon Luminous flame 2Q + Na₂B₄O₇ + CO$2Z + 2X + \text{Carbon} \xrightarrow{\text{Luminous flame}} 2Q + Na_{2}B_{4}O_{7} + CO$ The oxidation states of Cu in Z and Q, respectively are:

Related Formula

Core Logic

Let's analyze the properties of Tellurium compounds:

• Tellurium Dioxide (TeO₂$\mathrm{TeO}_2$):
• 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), TeO₂$\mathrm{TeO}_2$ is oxidising because the lower oxidation states (like element Tellurium or +2$+2$) are chemically accessible. Thus, TeO₂$\mathrm{TeO}_2$ acts as an oxidising agent.
• Tellurium Hydride (TeH₂$\mathrm{TeH}_2$):
• Tellurium is a very large atom. The orbital overlap between Tellurium and Hydrogen is extremely poor. Hence, the Te-H$\mathrm{Te-H}$ bond is very long and has very low bond dissociation energy.
• This allows TeH₂$\mathrm{TeH}_2$ to easily release H^+$\mathrm{H^+}$ in solution, making it highly acidic.

Step 1: Final Verification

Therefore, the nature of TeO₂$\mathrm{TeO}_2$ is oxidising, and the nature of TeH₂$\mathrm{TeH}_2$ is acidic.

Pattern Recognition

Periodic Trend: As we go down Group 16:

• Acidic strength of hydrides increases: H₂O < H₂S < H₂Se < H₂Te$\mathrm{H_2O < H_2S < H_2Se < H_2Te}$.
• Reducing character of hydrides also increases.
• Reducing power of dioxides decreases: SO₂$\mathrm{SO_2}$ (reducing) arrow TeO₂$\rightarrow \mathrm{TeO_2}$ (oxidising).

Chapter Mix

Class 12 Chemistry: p-Block Elements

Related Formula

Group 13 elements (B, Al, Ga, In, Tl$\mathrm{B}, \mathrm{Al}, \mathrm{Ga}, \mathrm{In}, \mathrm{Tl}$) show highly anomalous periodic trends due to the intervention of filled d-orbitals (d-block contraction in Ga$\mathrm{Ga}$) and f-orbitals (lanthanoid contraction in Tl$\mathrm{Tl}$).

Core Logic

Evaluate each specified trend against official physical constants:

• Atomic radius: Due to d-block contraction, gallium (Ga$\mathrm{Ga}$) is smaller than aluminum (Al$\mathrm{Al}$):

Hence, the given order is Incorrect.

• Electronegativity: Electronegativity first decreases from B$\mathrm{B}$ to Al$\mathrm{Al}$, then increases down the group due to poor shielding of d and f electrons:

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:

Hence, the given order is Incorrect (it is completely reversed).

• 1st$1^{\mathrm{st}}$ Ionisation Energy: Shows an irregular trend due to ineffective shielding by d and f electrons:

Hence, this order is Correct.

Step 2: Conclusion

Only the Electronegativity (B) and 1st$1^{\mathrm{st}}$ Ionisation Energy (D) orders are correct, matching Option (1).

Pattern Recognition

Group 13 elements do not follow monotonic trends. The poor shielding of 3d¹⁰$3d^{10}$ and 4f¹⁴$4f^{14}$ electrons increases the effective nuclear charge on valence electrons, causing anomalies in atomic radius (Ga < Al$\mathrm{Ga} < \mathrm{Al}$) and pulling electronegativities and ionization energies upward as you go further down.

Chapter Mix

Class 11 Chemistry: The p-Block Elements Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Core Logic

For Group 14 (C, Si, Ge, Sn, Pb$\text{C}, \text{Si}, \text{Ge}, \text{Sn}, \text{Pb}$):

• The ionisation energies generally decrease down the group, but there is an anomaly between Sn$\text{Sn}$ and Pb$\text{Pb}$ due to relativistic contraction / poor shielding of 4f electrons in Pb$\text{Pb}$.
• Thus, the first ionisation enthalpy of Tin (Sn$\mathrm{Sn}$) is 708 kJ mol⁻¹$708 \text{ kJ mol}^{-1}$ and Lead (Pb$\mathrm{Pb}$) is 715 kJ mol⁻¹$715 \text{ kJ mol}^{-1}$. These are indeed the lowest in the group.
• Hence, element A is Sn$\mathrm{Sn}$ and B is Pb$\mathrm{Pb}$.

Nature of their ions:

• A²⁺ = Sn²⁺$\mathrm{A}^{2+} = \mathrm{Sn}^{2+}$: Since Sn⁴⁺$\mathrm{Sn}^{4+}$ is more stable than Sn²⁺$\mathrm{Sn}^{2+}$, Sn²⁺$\mathrm{Sn}^{2+}$ readily undergoes oxidation to +4$+4$, acting as a strong reducing agent.
• B⁴⁺ = Pb⁴⁺$\mathrm{B}^{4+} = \mathrm{Pb}^{4+}$: Due to the strong inert pair effect, Pb²⁺$\mathrm{Pb}^{2+}$ is highly stable compared to Pb⁴⁺$\mathrm{Pb}^{4+}$. Thus, Pb⁴⁺$\mathrm{Pb}^{4+}$ 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 Pb⁴⁺$\mathrm{Pb}^{4+}$ oxidising. Tin's stable state is +4$+4$, making Sn²⁺$\mathrm{Sn}^{2+}$ reducing.

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 (Group 16$\text{Group 16}$):

• Bond Dissociation Enthalpy (ΔdisH$\Delta_{\text{dis}}H$): As we descend the group from Selenium to Tellurium, the size of the central atom increases significantly (rTe > rSe$r_{\text{Te}} > r_{\text{Se}}$). This increase in size leads to poorer orbital overlap with the small 1s$1s$ orbital of hydrogen, resulting in a longer and weaker M-H$\text{M-H}$ bond. Consequently, the bond dissociation enthalpy decreases:

Thus, Statement II is true.

• Acidic Strength: A weaker bond dissociates more easily in aqueous solution to release H^+$\text{H}^+$ ions. Since the Te-H$\text{Te-H}$ bond is weaker than the Se-H$\text{Se-H}$ bond, H₂Te$\text{H}_2\text{Te}$ releases protons much more readily than H₂Se$\text{H}_2\text{Se}$, making it a stronger acid:

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.

Chapter Mix

Class 12 Chemistry: p-Block Elements