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Some Basic Concepts of Chemistry question

2023 · 31 Jan · Shift 2 · Q20
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Some Basic Concepts of Chemistry question

2023 · 31 Jan · Shift 2 · Q20

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
A sample of a metal oxide has formula M0.83O1.00\mathrm{M}_{0.83} \mathrm{O}_{1.00}M0.83​O1.00​. The metal M\mathrm{M}M can exist in two oxidation states +2+2+2 and +3+3+3. In the sample of M0.83O1.00\mathrm{M}_{0.83} \mathrm{O}_{1.00}M0.83​O1.00​, the percentage of metal ions existing in +2+2+2 oxidation state is ‾\underline{\hspace{2cm}}​%\%%. (nearest integer)
Numerical answer
View written solutionFree

Correct answer: 59

  1. Given formula

    The oxide has composition: M0.83O1.00\mathrm{M}_{0.83}\mathrm{O}_{1.00}M0.83​O1.00​

    Oxygen is present as oxide ion, so each oxygen has oxidation state: −2-2−2

  2. Let the average oxidation state of metal be xxx

    Since the compound is electrically neutral: 0.83(x)+1(−2)=00.83(x) + 1(-2) = 00.83(x)+1(−2)=0

    0.83x=20.83x = 20.83x=2

    x=20.83≈2.4096x = \frac{2}{0.83} \approx 2.4096x=0.832​≈2.4096

    So the average oxidation state of metal is about: +2.41+2.41+2.41

  3. Let fraction of metal ions in +2 state be fff

    Then fraction in +3 state will be: 1−f1-f1−f

    Hence average oxidation state is: 2f+3(1−f)2f + 3(1-f)2f+3(1−f)

    This must equal 2.40962.40962.4096: 2f+3(1−f)=2.40962f + 3(1-f) = 2.40962f+3(1−f)=2.4096

  4. Solve for fff

    2f+3−3f=2.40962f + 3 - 3f = 2.40962f+3−3f=2.4096

    3−f=2.40963 - f = 2.40963−f=2.4096

    f=3−2.4096=0.5904f = 3 - 2.4096 = 0.5904f=3−2.4096=0.5904

    Therefore, percentage of metal ions in +2 state is: 0.5904×100=59.04%0.5904 \times 100 = 59.04\%0.5904×100=59.04%

  5. Nearest integer

    59%\boxed{59\%}59%​

  6. Comparison with stored answer

    Stored correct answer = 59

    Our derived answer = 59

    Hence, they agree.

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