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The d- and f- Block Elements Test - 75

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The d- and f- Block Elements Test - 75
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  • Question 1
    1 / -0
    Which of the following pair will have effective magnetic moment equal?
    Solution
    $$ Cr^{2+} $$ and $$ Fe^{2+ }$$ will have an effective magnetic moment equal since they have the same number of unpaired electrons.

    $$ Cr^{2+} - [Ar]3d^4 \to$$ 4 unpaired electrons

    $$ Fe^{2+} - [Ar]3d^6\to $$ 4 unpaired electrons
  • Question 2
    1 / -0
    Which of the following weighs less when weighed in magnetic field?
    Solution
    $$ ScCl_3 \rightarrow Sc^{3+}  + 3Cl^- $$

    $$ScCl_3$$ has no unpaired electron so will show diamagnetic character and will be repelled, so will weigh less. 

  • Question 3
    1 / -0
    Which of the following is more paramagnetic? 
    Solution
    $$ Fe^{3+} \to [Ar]3d^5\to $$ having 5 unpaired electrons have the highest number of unpaired electrons so it will be more paramagnetic.
  • Question 4
    1 / -0
    Effective magnetic moment of $$ Sc^{3+} $$ ion is 
    Solution
    Magnetic moment, $$\mu=\sqrt{n(n+2)}$$

    Where $$n =$$ number of unpaired electrons

    For $$ Sc^{3+} = [Ar]3d^0 , \quad n = 0 , \quad \therefore  \mu = 0 $$
  • Question 5
    1 / -0
    $$ V_2O_5 $$ is useful as catalyst in 
    Solution
    $$ V_2O_5 $$ is useful as catalyst in manufacturing of $$ H_2SO_4 $$
    Hene, Option "A" is the correct answer.
  • Question 6
    1 / -0
    Which of the following will show increase in weight when kept in magnetic field? 
    Solution
    $$ Fe_2(SO_4)_3 \to Fe^{3+} -[Ar]3d^5 - 5$$ electrons are unpaired. So, Fe will be attracted in the magnetic field so it will show an increase in weight.
  • Question 7
    1 / -0
    Which one of the following is expected to be paramagnetic complex?
    Solution
    The configuration of $$Ni^{2+}$$ in $$[Ni(H_2O)_6]^{2+}$$ has two unpaired electrons so it is paramagnetic.

    $$[Ni(H_2O)_6]^{2+}\to Ni^{2+}\to [Ar]3d^8\to$$ two unpaired electrons in the complex.
  • Question 8
    1 / -0
    $$ [Sc(H_2O)_6]^{3+} $$ ion is 
    Solution
    $$ [Sc(H_2O)_6]^{3+} \to\ _{21}Sc = [Ar] 3d^1 4s^2 $$

    $$ Sc^{3+}= [Ar]3d^0 4s^0 \to $$ no unpaired electrons in d subshell, so it is diamagnetic and colourless and because of the six ligand it is octahedral in shape.
  • Question 9
    1 / -0
    Which of the following statement is not true? 
    Solution
    Colourless compounds are those which have no unpaired electrons and paramagnetic substance do have unpaired electrons. Therefore paramagnetic substances possess colour. 
  • Question 10
    1 / -0
    Which of the following is the green coloured powder is produced when ammonium dichromate is used in fireworks?
    Solution
    The green coloured powder is produced when ammonium dichromate is used in fireworks is $$ Cr_2O_3 .$$

    $$ (NH_4)_2Cr_2O_7 \quad \xrightarrow { \Delta} \quad Cr_2O_3 +N_2+4H_2O$$
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