IBChemistry: HLDPExam Questions12. Atomic Structure (HL only)12.1 Electrons in Atoms

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First teaching 2014

Last exams 2024

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Electrons in Atoms (DP IB Chemistry: HL)

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12.1 Electrons in Atoms

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12.1 Electrons in Atoms

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1
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Which is true about the emission spectrum of hydrogen in the visible region?

  • The lines converge at longer wavelengths

  • The lines converge at higher frequency

  • The lines come from transitions between n=∞ and n=1

  • The lines are regularly spaced

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2
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The frequency of the point of convergence on a hydrogen emission spectrum is 32.883 x 1014 s-1.

What is the ionisation energy for one atom of hydrogen?

(h = 6.63 x 10-34 J s)

  • 32.883 x 1014 x 6.63 x 10-34 J

  • fraction numerator 6.63 cross times 10 to the power of negative 34 end exponent over denominator 32.883 cross times 10 to the power of 14 end fraction J

  • J

  • fraction numerator 32.883 cross times 10 to the power of 14 over denominator 6.63 cross times 10 to the power of negative 34 end exponent cross times 1000 end fraction spaceJ

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3
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The first five ionisation energies of an element are shown below.

What element could this ionisation energy graph belong to?

successive-ionisation-energy

  • N

  • P

  • Al

  • Na

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4
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Which are correct explanation(s) for the increase in ionisation energy across a period?

I. Nuclear charge increases

II. Atomic radius decreases

III. Shielding remains constant

  • I and II only

  • I and III only

  • I and III only

  • I,II and III

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5
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Which spectrum belongs to hydrogen?

eARGWkkX_hydrogen-spectrum

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    11 mark

    Which of the following calculations gives the correct calculation to find the energy, in kJ, for a photon of blue light given the wavelength ƛ = 550 nm.

    h = 6.626 x 10−34J s; c =  2.988 x 108 m s-1

    • begin mathsize 14px style fraction numerator 6.626 space cross times 10 to the power of negative 34 space end exponent cross times 2.988 cross times 10 to the power of 8 over denominator 550 cross times 10 to the power of negative 9 end exponent end fraction end style

    • begin mathsize 14px style fraction numerator 6.626 cross times 10 to the power of negative 34 end exponent cross times 2.988 cross times 10 to the power of 8 over denominator 550 cross times 1000 end fraction end style

    • begin mathsize 14px style fraction numerator 6.626 cross times 10 to the power of negative 34 end exponent cross times 2.988 cross times 10 to the power of 8 over denominator 550 cross times 10 to the power of negative 9 end exponent cross times 1000 end fraction end style

    • begin mathsize 14px style fraction numerator 6.626 cross times 10 to the power of negative 34 end exponent cross times 2.988 cross times 10 to the power of 8 over denominator 2.988 cross times 10 to the power of 8 cross times 1000 end fraction end style

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    21 mark

    Successive ionisation energies for an element, Y, are shown in the table below.

    Electrons removed

    1st

    2nd

    3rd

    4th

    5th
    Ionisation energy / kJ mol-1

    736

    1450

    7740

    10500

    13600

    What is the most likely formula for the ion of Y?

    • Y+

    • Y2+

    • Y3+

    • Y4+

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    31 mark

    Values for the successive ionisation energies for an unknown element are given in the table below.

    First ionisation energy / kJ mol-1

    Second ionisation energy / kJ mol-1

    Third ionisation energy / kJ mol-1

    Fourth ionisation energy / kJ mol-1

    420

    3600

    4400

    5900

    In which group of the periodic table would the unknown element be found?

    • 1

    • 2

    • 13

    • 14

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    41 mark

    The graph shows the first ionisation energies of some consecutive elements

    x-and-y-ionisation-energiesWhich statement is correct?

    • Y is in group 13

    • Y is in group 10

    • X is in group 15

    • X is in group 18

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    51 mark

    Which transition on the diagram corresponds to the ionisation of hydrogen in the ground state?

    ionisation-of-ground-state-hydrogen

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      1
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      Which statement is true about the spectra shown? hydrogen-spectrum-hl

      • All the lines in R have the same energy

      • Q and S could represent line emission spectra

      • Only S could represent a line emission spectrum for hydrogen

      • P indicates the element has 4 pairs of electrons at different energy levels 

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      2
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      The energy absorbed at the limit of convergence for helium is 19.6 x 10-18 J per atom.

      Which calculation would be used to calculate the wavelength, in m, for this electron transition?

      (c = 3.00 x 108 ms-1, h = 6.63 x 10-34 Js)

      • λ = fraction numerator 19.6 space cross times space 10 to the power of negative 18 end exponent space cross times space 3.00 space cross times space 10 to the power of 8 over denominator 6.63 space cross times space 10 to the power of negative 34 end exponent end fraction

      • λ = fraction numerator 3.00 space cross times 10 to the power of 8 space cross times 6.63 space cross times space 10 to the power of negative 34 end exponent space over denominator 19.6 space cross times space 10 to the power of negative 18 end exponent end fraction

      • λ =  fraction numerator 19.6 space cross times space 10 to the power of negative 18 end exponent space cross times space 6.63 space cross times space 10 to the power of negative 34 end exponent over denominator 3.00 space cross times space 10 to the power of 8 end fraction

      • λ = fraction numerator 3.00 space cross times space 10 to the power of 8 over denominator 19.6 space cross times space 10 to the power of negative 18 end exponent space cross times 6.63 space cross times space 10 to the power of negative 34 end exponent end fraction

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      3
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      The first ionisation energies (in kJ mol-1) of five successive elements are:

      2081, 496, 738, 578, 787

      What could these elements be?

      • First five elements in a period

      • Second to the sixth elements in a period

      • Last four elements of one period and the first one of the next period

      • Last element of one period and the first four elements of the next period

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      4
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      A period 3 element forms an oxide M2O3.

      Which represents the first four successive ionisation energies of M?

        Ionisation energy / kJmol-1
        First Second Third Fourth
      A. 496 4560 6940 9540
      B. 578 1820 2740 11600
      C. 1012 1907 2914 4964
      D. 736 1450 7740 10500

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        5
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        Between which ionisation energies of silicon will there be the greatest difference?

        • Between the second and fourth ionisation energies

        • Between the first and third ionisation energies

        • Between the fourth and fifth ionisation energies

        • Between the fifth and sixth ionisation energies

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