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P Block Elements question

2007 · Shift 1 · Q16
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P Block Elements question

2007 · Shift 1 · Q16

JEE AdvancedChemistryP Block ElementsMCQ+3 / −1
The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions. The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers +2, +4 and +6. XeF 4{}_44​ reacts violently with water to give XeO 3{}_33​. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.XeF 4{}_44​ and XeF 6{}_66​ are expected to be
  1. A
    oxidizing
  2. B
    reducing
  3. C
    unreactive
  4. D
    strongly basic
View written solutionFree

Correct answer: A

  1. Analyze the oxidation state of Xenon:

    • In any compound, fluorine, being the most electronegative element, always exhibits an oxidation state of -1 (except in F₂).
    • In XeF₄, let the oxidation state of Xenon (Xe) be x. The sum of oxidation states in a neutral molecule is zero. Therefore, x + 4(-1) = 0, which gives x = +4.
    • In XeF₆, let the oxidation state of Xenon (Xe) be y. Similarly, y + 6(-1) = 0, which gives y = +6.

  2. Evaluate the stability and chemical nature:

    • Xenon is a noble gas with a stable electronic configuration. Its most stable oxidation state is 0.
    • The oxidation states +4 and +6 are very high positive oxidation states for Xenon. Compounds with elements in such high oxidation states are generally unstable and have a strong tendency to gain electrons to return to a more stable, lower oxidation state.
    • A substance that gains electrons in a chemical reaction is called an oxidizing agent (as it causes the other substance to be oxidized).
    • Therefore, both XeF₄ and XeF₆ have a strong tendency to accept electrons and be reduced (e.g., to Xe(0) or a compound with Xe in a +2 state). This makes them powerful oxidizing agents.

  3. Consider chemical reactions as evidence:

    • XeF₄ and XeF₆ are excellent fluorinating agents, which is a form of oxidation. For example, they can oxidize platinum: XeF4+Pt→Xe+PtF4\text{XeF}_4 + \text{Pt} \rightarrow \text{Xe} + \text{PtF}_4XeF4​+Pt→Xe+PtF4​ Here, Xe is reduced from +4 to 0, and Pt is oxidized from 0 to +4.
    • Similarly, XeF₆ can oxidize hydrogen: XeF6+3H2→Xe+6HF\text{XeF}_6 + 3\text{H}_2 \rightarrow \text{Xe} + 6\text{HF}XeF6​+3H2​→Xe+6HF Here, Xe is reduced from +6 to 0, and H is oxidized from 0 to +1.

  4. Evaluate the given options:

    • A: oxidizing: This is consistent with our analysis. XeF₄ and XeF₆ are strong oxidizing agents due to the high oxidation state of Xenon.
    • B: reducing: A reducing agent gets oxidized. This would require Xenon to increase its oxidation state further, which is energetically unfavorable. So, this is incorrect.
    • C: unreactive: The provided text itself states that XeF₄ reacts violently with water. These compounds are known to be highly reactive. So, this is incorrect.
    • D: strongly basic: These are covalent compounds of non-metals. Their hydrolysis products include hydrofluoric acid (HF) and xenon trioxide (XeO₃), which is an acidic oxide. Thus, they are not basic. So, this is incorrect.

  5. Conclusion: Based on the high positive oxidation states of Xenon in XeF₄ and XeF₆, they are expected to be strong oxidizing agents.

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