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The d- and f- Block Elements Test - 34

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The d- and f- Block Elements Test - 34
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  • Question 1
    1 / -0
    Which of the following ions has a magnetic moment of 5.93 BM?

    [At . No. : V = 23 , Cr = 24, Mn = 25, Fe = 26]
    Solution
    Magnetic moment $$ = \sqrt{n(n+2)} BM $$

    where n = number of unpaired electrons 

    $$ 5.93 = \sqrt{n(n+2)} $$

    $$n = 5 $$

    $$ Mn^{2+} $$ ion $$ (3d^5) $$ has $$5$$ unpaired electrons and thus, magnetic moment is $$ 5.93 BM $$

    Hence, option $$A$$ is correct.
  • Question 2
    1 / -0
    Spin only magnetic moment value of $${ Cr }^{ 3+ }$$ ion is:
    Solution
    $${ Cr }^{ 3+ }=3{ d }^{ 3 }$$;

    Hence, $$\mu =\sqrt { n\left( n+2 \right)  } $$

    BM $$=\sqrt { 3\left( 3+2 \right)  } =\sqrt { 15 } =3.87$$ BM

    Hence, option $$B$$ is correct.
  • Question 3
    1 / -0
    Select the coloured and paramagnetic ions.
    Solution
    Ion is coloured if there are electrons in d-suborbit. Paramagnetic nature is also due to unpaired electrons. Thus, every coloured ion is also paramagnetic.
    $$Cu^{2+}=[Ar]3d^9$$, one unpaired electron in $$3$$d.
    $$Cr^+=[Ar]3d^5$$, five unpaired electrons in $$3$$d.
    $$Mn^{2+}=[Ar]3d^5$$ five unpaired electrons in $$3d$$.
  • Question 4
    1 / -0
    The type of magnetism exhibited by $$[Mn(H_2O)_6]^{2+}$$ ion is:
    Solution
    The given complex is  [Mn (H$$_2O)_6]^{2+}$$
    The electronic configuration of Mn is [Ar] $$3d^5 4s^2$$ 
       for Mn$$^{2+}$$ $$\rightarrow$$  {Ar]  $$3d^5 4s^0$$ 
                      [Ar]     $$\upharpoonleft $$    $$\upharpoonleft $$    $$\upharpoonleft $$    $$\upharpoonleft $$    $$\upharpoonleft $$ 
                                    3d                          4s 

                                                            3d                      4s    4p               4d
    So,  [Mn (H$$_2O)_6]^{2+}$$ $$\rightarrow$$   [Ar]     $$\upharpoonleft $$    $$\upharpoonleft $$    $$\upharpoonleft $$    $$\upharpoonleft $$    $$\upharpoonleft $$     xx    xx  xx  xx    xx xx    
                                                                                               $$H_2O$$
                                                                                      sp$$^3 d^2$$ hybridised
    So, due to presence of 5 unpaired d$$e^-$$ s, it is paramagnetic.
  • Question 5
    1 / -0
    The spin only magnetic moment of $$Fe^{3+}$$ ion (in BM) is approximately:
    Solution
    $$_{26}Fe=[Ar]3d^64s^2$$

    $$Fe^{3+}=[Ar]3d^5$$

    Number of unpaired electrons $$= 5$$ 

    $$\mu=\sqrt{n(n+2)}$$

    $$\mu =\sqrt{5\times (5+2)}$$

    $$\mu=\sqrt{35}= 5.91\, BM$$

    $$\mu= 6\, BM$$
  • Question 6
    1 / -0
    Magnetic moment of $$[Cu(NH_3)_4]^{2+}$$ ion is?
    Solution
    The given complex is $$ [Cu(NH_3)_4]^{2+}$$
    The electronic configuration of Cu is [Ar] $$3d^{10} 4s^1$$ 
     for  Cu$$^{2+}$$  $$\rightarrow$$ [Ar]$$3d^{9} 4s^0$$
    [Ar] $$\upharpoonleft \downharpoonright$$   $$\upharpoonleft \downharpoonright$$   $$\upharpoonleft \downharpoonright$$  $$\upharpoonleft \downharpoonright$$  $$\upharpoonleft$$
                       3d                                        4s

                                                             3d                 4s    4p
    So, [Cu(NH$$_3)_4]^{2+}$$  $$\rightarrow$$  [Ar] $$\upharpoonleft \downharpoonright$$   $$\upharpoonleft \downharpoonright$$   $$\upharpoonleft \downharpoonright$$  $$\upharpoonleft \downharpoonright$$   xx    xx         xx      xx      $$\upharpoonleft$$
                                                                        $$NH_3 $$ $$NH_2$$  $$NH_3 $$ $$NH_3 $$
                                                                      dsp$$^2$$ hybridisation
    No. of unpaired $$e^-$$ s =1
    Magnetic moment= $$\sqrt{n(n+2)}$$ BM
    Where, n= no.of unpaired $$e^-$$ s
    So, magnetic moment = $$\sqrt{1(1+2)}  =\sqrt{3}$$ = 1.73 BM

  • Question 7
    1 / -0
    Ammonia will not form complex with:
    Solution
    Generally d-block elements form complex and lead is p-block element.
  • Question 8
    1 / -0
    There are $$14$$ elements in actinoid series. Which of the following elements does not belong to this series?
    Solution

    Explanation:

    • Actinoid series is a series of $$14$$ elements.
    • The name of the series is given by the name of the first element actinium.
    • Actinoid series has elements from atomic number $$90$$ to $$103$$.
    • Uranium $$U$$ has atomic number $$92$$.
    • Neptunium $$Np$$ has atomic number $$93$$.
    • Fermium $$Fm$$ has atomic number $$100$$.
    • Thulium $$Tm$$ has atomic number $$69$$.
    • Hence, $$Tm$$ does not belong to the actinoid series.

    Hence, the correct answer is option $$C$$.

  • Question 9
    1 / -0
    Metallic radii of some transition elements are given below. Which of these elements will have the highest density?
    Element                    $$Fe$$     $$CO$$       $$Ni$$     $$Cu$$ 
    Metallic radii/pm     126     125     125    128
    Solution
    As we all know that on moving left to right along period, metallic radius decreases while mass increases. Decrease in metallic radius coupled with increase in atomic mass results in increase in density of metal.
    Hence, among the given four choices $$Cu$$ belongs to right side of Periodic Table in transition metal, and it must have highest density. So the correct option is D
  • Question 10
    1 / -0
    Which of the following arrangement does not represent the correct order of property stated against it ? 
    Solution
    Paramagnetic property depends upon the number of unpaired electrons.
    As the number of unpaired electrons unpaired electron increases, the paramagnetic property increases.
    $${ Fe }^{ 2+ }\Rightarrow n=4\quad \quad { Mn }^{ 2+ }\Rightarrow n=5$$
    $$\Rightarrow \quad { Mn }^{ 2+ }>{ Fe }^{ 2+ }$$ paramagnetic behaviour.
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