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The d- and f- Block Elements Test - 38

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The d- and f- Block Elements Test - 38
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  • Question 1
    1 / -0
    For $$Zn^{2+}$$, $$Ni^{2+}$$, $$Cu$$, and $$Cr^{2+}$$ which of the following statements is correct?
    Solution
    Since $$Zn^{+2}$$ have fully filled $$d$$ orbital having no unpaired electrons,it does not undergo electronic transitions,hence colourless.However,rest of the elements given as $$Ni^{+2}$$ , $$Cu^{+2}$$ and $$Cr^{+2}$$ has unpaired electrons which can undergo electronic transitions and hence show colours.
  • Question 2
    1 / -0
    Transition elements form binary compounds with halogens. Which of the following elements will form $$MF_3$$ type compounds?
    Solution
    $$Cu\rightarrow { Cu }^{ + },{ Cu }^{ 2+ }\rightarrow CuCl,{ CuF }_{ 2 }$$
    generally transition metals form binary compounds with halogens but $$Cr$$ form $${ CrF }_{ 3 }$$
                                                                                                                                         $$\Rightarrow { MF }_{ 3 }$$ type compound
  • Question 3
    1 / -0
    Choose the correct answer from the alternative given.
    Amongst $$TiF_6^{2-}$$, $$CoF$$,   $$Cu_2$$ $$Cl_2$$ and $$NiCl$$, which are the colourless species? (atomic number of Ti = 22, Co = 27, Cu = 29, Ni = 28)
    Solution
    TiF$$_6^{2-}$$ -
    The oxidation state of Ti is +4.
    Electronic configuration of Ti$$^{4+}$$ is [Ar] 3d$$^0 4s^0$$.
    So, unpaired e$$^-$$ s is present in d-orbitals so it is colourless.
    Cu$$_2Cl_2$$-
    The oxidation state of Cu is +1.
    Electronic configuration of Cu$$^+$$ is [Ar] 3d$$^{10} 4s^0$$.
    So, no unpaired d e$$^-$$ s are present . Hence, the given compound is colourless.
  • Question 4
    1 / -0
    Colour of transition metal ions are due to absorption of some wavelength. This results in:
    Solution
    Colour of transition metal ions are due to absorption of the same wavelength. This results in $$d-d$$ transition.
  • Question 5
    1 / -0
    In the dichromate anion $$Cr_2O_7$$ :
    Solution
    There are $$6$$ $$Cr-O$$ terminal bonds which are equivalent due to resonance.
    (Refer to Image)

  • Question 6
    1 / -0
    Which of the following reactions do not result in the preparation of potassium dichromate from chromate?

    (I) 4$$FeCr_2O_4$$ + 8$$Na_2CO_3$$+ $$7O_2$$$$\rightarrow$$

    (II) $$Na_2CrO_4$$ + $$H_2SO_4$$ $$\rightarrow$$

    (III) $$Na_2Cr_2O_7$$+ $$2KCl$$$$\rightarrow$$
    Solution
    (i)  $${ 4FeCr }_{ 4 }{ O }_{ 4 }+8{ Na }_{ 2 }{ CO }_{ 3 }+7{ O }_{ 2 }\rightarrow 8{ Na }_{ 2 }{ CrO }_{ 4 }+2{ Fe }_{ 2 }{ O }_{ 3 }+8{ CO }_{ 2 }$$

    (ii) $$2{ Na }_{ 2 }{ CrO }_{ 4 }+{ H }_{ 2 }{ SO }_{ 4 }\rightarrow { Na }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }+{ Na }_{ 2 }{ SO }_{ 4 }+{ H }_{ 2 }O$$

    (iii) $${ Na }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }+2KCl\rightarrow { K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }+2NaCl$$
  • Question 7
    1 / -0
    Which of the following is not an actinide?
    Solution
    Erbium with atomic number $$68$$, electronic configuration $$[Xe]4f^{12}6s^2$$ belongs to lanthanides.
    $$\therefore$$, Erbium is not an actinide.
  • Question 8
    1 / -0
    Which of the following catalysts is not correctly matched with the reaction?
    Solution
    Wacker's process is the oxidation of ethylene to acetaldehyde in the presence of $$PdCl_2$$ as the catalyst.
  • Question 9
    1 / -0
    Match the column I with column II and mark the appropriate choice.
    Column IColumn II
    (A)Ni in the presence of hydrogen(i)Ziegler Natta catalyst
    (B)$$Cu_2Cl_2$$(ii)Contact process
    (C)$$V_2O_5$$(iii)Vegetable oil to ghee
    (D)Finely divided iron(iv)Sandmeyer reaction
    (E)$$TiCl_4 \, + \, Al (CH_3)_3$$(v)Haber's Process
    Solution
    A) $$Ni$$ in the presence of hydrogen is used in the reduction of vegetable oil to ghee.
    B) $$Cu_2Cl_2$$ is used in Sandmeyer Reaction for the synthesis of aryl halides.
    C) $$V_2O_5$$ is used in contact process for the oxidation of $$SO_2$$ to $$SO_3$$.
    D) Finely divided iron is used as catalyst in Haber's Process.
    E) $$TiCl_4+Al(CH_3)_3$$ is the Zieglar Natta Catalyst.
  • Question 10
    1 / -0
    Consider the following statements
    I. $$La(OH)_3$$ is least basic among hydroxides of lanthanides.
    II. $$Zr^{4+}$$ and $$Hf^{4+}$$ possess almost the same ionic radii.
    III. $$Ce^{4+}$$ can act as an oxidizing agent.
    Which of the above is/are true?
    Solution
    The ionic radii of $$Zr^{4+}$$ and $$Hf^{4+}$$ are almost similar due to the lanthanide contraction which means the steady decrease in the size of the atoms with the increase from lanthanum through Lutetium due to the poor shielding of the $$4f$$ $$e^-$$, the $$e^-$$ pull towards the nucleus increases thereby decreasing the size.
    $$Ce^{4+}$$ act as oxidizing agent as it oxidizes other elements by itself getting reduced as $$Ce^{3+}$$ is more stable than $$Ce^{4+}$$ oxidation state so in order to attain stable $$+3$$ oxidation it gets reduced.
    The basic characteristics of lanthanoid hydroxides decrease down the group.
    Thus $$La(OH)_3$$ is the most basic.
    Thus only statement II and III are correct.
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