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The d- and f- Block Elements Test - 57

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The d- and f- Block Elements Test - 57
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  • Question 1
    1 / -0
    What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?
    Solution

    Correct Answer: Option C

    The oxidation state of chromium in K2Cr2O7K_2Cr_2O_7 ​is +6+6, whereas the oxidation state of nitrogen in HNO3HNO_3 is +5+5.

    Since the maximum oxidation state of nitrogen is +5+5, so it won’t get oxidize further. Rather it will react to form dichromate ion and a water molecule ( Cr2O72Cr_2O_{7}^{2-} and  H2OH_2O ).

    The equation involved is:

    2K2CrO4+2HNO3K2Cr2O7+2KNO3+H2O2K_2CrO_4+2HNO_3\rightarrow K_2Cr_2O_7+2KNO_3+H_2O

    Therefore, when potassium chromate solution is treated with an excess of dilute nitric acid, Cr2O72Cr_2O_{7}^{2-} and  H2OH_2O are formed.

  • Question 2
    1 / -0
    In Cr2O72{Cr_{2}O_{7}}^{2-} ion, the maximum number of bonds having the same Cr - O bond length is :
    Solution
    In Cr2O72Cr_{2}O_{7}^{2-} ion, 6 (CrO)(Cr - O) bonds have the same bond length 163 pm because of the resonance and 2 (CrO)(Cr - O) bonds have a bond length of 179 pm.

  • Question 3
    1 / -0

    Which of the following pairs has almost same colour?

    Solution
    The color of ions depends upon no of unpair electrons for transition.

    Electronic configuration of 63Eu2+,is[Xe]4f7_{63}Eu^{2+}, is [Xe] 4f^7 and 65Tb3+is[Xe]4f8_{65}Tb^{3+} is [Xe]4f^8.

    They both have almost same number of electrons for the transition, so they show the almost same color.

    Hence, option C is correct.
  • Question 4
    1 / -0

    The increasing order of paramagnetism of the following compound is:

    (I)MnSO4.4H2O MnSO_{4}.4H_{2}O 

    (II)FeSO4.7H2OFeSO_{4}.7H_{2}O

    (III)  NiSO4.6H2O   NiSO_{4}.6H_{2}O

    (IV)CuSO4.5H2OCuSO_{4}.5H_{2}O

    Solution
    Mn2+,Fe2+,Ni2+,Cu2+{ Mn }^{ 2+ },{ Fe }^{ 2+ },{ Ni }^{ 2+ },{ Cu }^{ 2+ } - central metal atoms
    3d53d63d83d9{ 3d }^{ 5 }\quad { 3d }^{ 6 }\quad { 3d }^{ 8 }\quad { 3d }^{ 9 }
    n=5n=4n=2n=1n=5\quad n=4\quad n=2\quad n=1
    I>II>III>IV\Rightarrow I>II>III>IV
    decreasing order of paramagnetism.
  • Question 5
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    Directions For Questions

    The presence of incompletely filled d- and f-orbitals is mainly responsible for colour of transition metals and inner transition metals. The para and diamagnetic behaviour of the metals is again due to the presence or absence of unpaired electrons in their d or f levels. In some cases, colour may appear due to charge transfer rather than d-d transition.
    The energy gap between 4f and 5d subshells is large, while the same between 5f, 6d and 7s levels is relatively small.

    ...view full instructions

    Statements that are true of d-and f-block elements are:

    1. The colour of K2Cr2O7K_{2}Cr_{2}O_{7} is due to d-d transition.

    2.  Actinides have larger number of oxidation states than lanthanides.

    3.  Tripositive ions of Pr(At.No.=59)Pr(At. No. =59) and Tm(At.No.=69)Tm(At. No. =69) have the same (green) colour.

    4.  Actinides have paramagnetic moments less than theoretically predicted values.
    Solution
    In K2Cr2O7K_{2}Cr_{2}O_{7} the color is due to charge transfer and not d-d transition.

    Actinides exhibit a larger number of oxidation states than lanthanides because, there is a very small energy gap between 5f, 6d, and 7s orbitals.

    Pr3+Pr^{3+} has two 4f electrons (both unpaired). Tm3+Tm^{3+} has twelve 4f electrons (5 pairs and 2 unpaired). Both Pr3+Pr^{3+} and Tm3+Tm^{3+}, with 2 unpaired electrons each, have the same colour.

    5f electrons in actinides are less effectively shielded resulting in quenching of orbital contribution of magnetic moments. So their magnetic moments are less than theoretically predicted values.
  • Question 6
    1 / -0

    Directions For Questions

    Lanthanides and actinides are strongly paramagnetic. Lanthanide ions have high charge (+3) but have large size resulting in small charge to size ratio (low charge density). In actinides 5f electrons are far too diffused than 4f electrons in lanthanides. There is a very small energy gap between 5f, 6d and 7s electrons. All these factors influence the physical and chemical properties of lanthanides and actinides.

    ...view full instructions

    Regarding the magnetic properties of lanthanides and actinides the correct statement is:
    Solution
    Magnetic moments of actinides are significantly less than their theoretically predicted values. This is because electrons of actinides are less effectively shielded which results in quenching of the orbital contribution. 

    Hence, option C is the correct statement.
  • Question 7
    1 / -0

    The actinoids exhibit more number of oxidation states in general than lanthanoids because: 

    Solution

    Correct Answer: Option D

    Explanation:

    The common oxidation number of lanthanoid is +3+3 but +2+2 and +4+4 ions are also found in the solution.

    Actinoids show +3+3 as oxidation state but some Actinoids are stable till +7+7 oxidation states.

    Actinoids have more oxidation states than the Lanthanoids as 5f5f orbitals are at a greater distance from the nucleus than 4f4f orbitals.

    Therefore lanthanoids are larger in size than in comparison to actinoids.

  • Question 8
    1 / -0
    Dipositive and tripositive ions of A and B possess same number of electron pairs, but different number of unpaired electrons. Both are paramagnetic. Identify the atomic numbers of A and B.
    Solution
    Let us assume the atomic number of A and B be 22 and 24 respectively.
    A+2(20):1s22s22p63s23p64s03d2{ A }^{ +2 } (20): { 1s }^{ 2 }{ 2s }^{ 2 }2p^{ 6 }{ 3s }^{ 2 }{ 3p }^{ 6 }{ 4s }^{ 0 }{ 3d }^{ 2 } (2 unpaired electron, hence paramagnetic)

    B+3(21):1s22s22p63s23p64s03d3{ B }^{ +3 } (21): { 1s }^{ 2 }{ 2s }^{ 2 }2p^{ 6 }{ 3s }^{ 2 }{ 3p }^{ 6 }{ 4s }^{ 0 }{ 3d }^{ 3 } (3 unpaired electron, hence paramagnetic)

    Hence, D is the answer.
  • Question 9
    1 / -0
    Atoms of the transition element are smaller than those of the s-block elements. This is because of  :
    Solution
    Atoms of the transition element are smaller than those of the s-block elements due to contraction in size in periods and also new elctrons  are added to the penultimate d-shell rather than to the outermost shell of the atom.
  • Question 10
    1 / -0
    The value of the magnetic moment of a particular ion is 2.83 Bohr magneton. The ion is :
    Solution
    The value of μ=\mu= 2.83 BM (=n(n+2)=\sqrt{n(n+2)}) corresponds to the presence of two unpaired electrons. 

    So, the ion is Ni2+(3d8)Ni^{2+}(3d^8) since it has two unpaired electrons in d-subshell.
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