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The d- and f- Block Elements Test - 41

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The d- and f- Block Elements Test - 41
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  • Question 1
    1 / -0
    Which of the following is diamagnetic ion?
    Solution
    $${ Zn }^{ 2+ }\equiv \left[ Ne \right] { 3s }^{ 2 }{ 3p }^{ 6 }{ 4s }^{ 0 }{ 3d }^{ 10 }$$        
    It is diamagnetic, so no unpaired $${ e }^{ - }s$$.
  • Question 2
    1 / -0
    $$\mu = \sqrt{15}$$ is true for the pair:
    Solution
    The electronic configuration of $$Co^{2+}$$ is;$$[Ar]4s^0 3d^7$$
    Hence there are three unpaired electrons in $$Co^{2+}$$
    Now the elctronic configuration of $$Cr^{3+}$$ is;$$[Ar]3d^3$$
    Hence there are 3 unpaired electrons in $$Cr^{3+}$$
    We know magnetic moment $$\mu=\sqrt{(n(n+2)}$$$$
    where n is number of unpaired electrons.
    So when n=3,
    $$\mu=\sqrt{(3(3+2)}=\sqrt{15}$$
    Hence option A is correct answer.
  • Question 3
    1 / -0
    Atomic number of three elements A, B, and C are respectively, 

         $$Pm\ (Z=61),\ Sm\ (Z=62),\ Eu\ (Z=63).$$
        
    Which one has a maximum atomic radius?
    Solution

  • Question 4
    1 / -0
    Which of the following paramagnetic ions would exhibit magnetic moment (Spin only) of the order 5 BM? (Atomic number Mn = 25, Cr = 24, V = 23, Ti = 22)
    Solution
    Configuration of ,
    $$V^{2+}[Ar]4s^03d^3$$    $$n=3$$
    $$Ti^{2+}[Ar]4s^03d^2$$    $$n=2$$
    $$Mn^{2+}[Ar]4s^03d^5$$    $$n=5$$
    $$Cr^{2+}[Ar]4s^03d^4$$    $$n=4$$
    Hence spin moment of $$Cr^{2+}=\sqrt{4(4+2)}\simeq4.92BN\simeq5BM$$ 
  • Question 5
    1 / -0
    The ratio of number of exchange pairs present in $$d^{9}$$ and magnetic moment of Cr in B.M. is:-
    Solution

    Magnetic moment of $$Cr$$ in $$d^9$$:
    $$\mu=\sqrt {n(n+2)}$$
        $$=\sqrt {1(1+2)}=\sqrt {3}$$    (Refer to Image)
    Two exchange pairs present in $$d^9$$ of $$Cr$$:
    $$\therefore \cfrac {2}{\sqrt {3}}$$ .

  • Question 6
    1 / -0
    Amongst the following ions, which is considered as most stable in $${ M }^{ 2+ }$$ state?
    Solution
    Half reactions with lowest standard reduction have stable reactants (ions).
    Here, $$Ti^{2+}+2e^-\longrightarrow Ti\ [Reduction]$$
    The standard reduction potential of this reaction is least. So, the reactant $$Ti^{2+}$$ is more stable.
  • Question 7
    1 / -0
    Which of the following pairs of atom have the most similar radii ?
  • Question 8
    1 / -0
    Which of the following will be paramagnetic?
    Solution
    • Configuration of $$Cr^{3+}$$ is:$$[Ar]3d^3$$
    • Since there are five-3orbitals, they are not all at least singly filled yet, and thus, all three electrons in the lowest-energy configuration are unpaired. So, this is paramagnetic.
  • Question 9
    1 / -0
    Which of the following ions in its aqueous solution possesses the value of magnetic moment as $$3.87$$?
    Solution

  • Question 10
    1 / -0
    Which of the following complex compounds will exibit highest paramagnetic behavior?
    Solution
    $${ T }i^{ 3+ }-\left[ Ne \right] { 3s }^{ 2 }{ 3p }^{ 6 }{ 4s }^{ 0 }{ 3d }^{ 1 }$$
    $${ Zn }^{ 2+ }-\left[ Ne \right] { 3s }^{ 2 }{ 3p }^{ 6 }{ 4s }^{ 0 }{ 3d }^{ 10 }\quad \} -$$ not much paramagnetic
    $${ Cr }^{ 3+ }\rightarrow { d }_{ 3 }$$ configuration $$\rightarrow { t }_{ 2 g}$$ means paramagnetic
    $${ Co }^{ 3+ }\rightarrow { d }_{ 6 }$$ configuration $$\rightarrow { NH }_{ 3 }$$ strong field ligand
    So, it would be $${ d }_{ 6 }$$ (low spin) diamagnetic.
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