Given below are two statements: Statement I: H₂Se$\text{H}_2\text{Se}$ is more acidic than H₂Te$\text{H}_2\text{Te}$. Statement II: H₂Se$\text{H}_2\text{Se}$ has higher bond enthalpy for dissociation than H₂Te$\text{H}_2\text{Te}$. In the light of the above statements, choose the correct answer from the options given below:

Core Logic

Heating of Borax: Na₂B₄O₇ Δ 2NaBO₂ (X) + B₂O₃ (Y)$Na_2B_4O_7 \xrightarrow{\Delta} 2NaBO_2 \text{ (X)} + B_2O_3 \text{ (Y)}$

Borax bead test with copper sulfate (Non-Luminous flame, oxidizing): CuSO₄ + B₂O₃ Δ Cu(BO₂)₂ (Z) + SO₃$CuSO_4 + B_2O_3 \xrightarrow{\Delta} Cu(BO_2)_2 \text{ (Z)} + SO_3$ In Z, copper is present as Cu(BO₂)₂$Cu(BO_2)_2$, providing a blue bead. The oxidation state of Cu is +2.

In Luminous (reducing) flame: 2Cu(BO₂)₂ (Z) + 2NaBO₂ (X) + C Δ 2CuBO₂ (Q) + Na₂B₄O₇ + CO$2Cu(BO_2)_2 \text{ (Z)} + 2NaBO_2 \text{ (X)} + C \xrightarrow{\Delta} 2CuBO_2 \text{ (Q)} + Na_2B_4O_7 + CO$ The product Q is CuBO₂$CuBO_2$, a colorless/red opaque bead depending on conditions. The oxidation state of Cu in CuBO₂$CuBO_2$ is +1.

Step 1: Final Conclusion

Oxidation state of Cu in Z is +2, and in Q is +1.

Pattern Recognition

In Borax bead test, transition metals usually assume their higher stable oxidation state in the oxidizing flame, and are reduced to lower states (or native metal) in the reducing flame.

Chapter Mix

Class 11 Chemistry: The p-Block Elements Class 12 Chemistry: The d- and f-Block Elements

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