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The d- and f- Block Elements Test - 40

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The d- and f- Block Elements Test - 40
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  • Question 1
    1 / -0
    The ratio of magnetic moments of $$Fe(III)$$ and $$Co(II)$$ is:
    Solution
    Electronic configuration of $$Fe^{3+}:[Ar]3d^54s^0$$ 
    So number of unpaired elctrons is $$n=5$$
    Hence, $$\mu=\sqrt{n(n+2)}=\sqrt{35}$$
    Now,
    Electronic configuration of $$Co^{2+}:[Ar]3d^74s^0$$
    So number of unpaired electros is $$n=3$$
    Hence, $$\mu=\sqrt{n(n+2)}=\sqrt{15}$$
    So, Ratio of Magnetic moments of $$Fe^{3+}$$ and $$Co^{2+}$$ is$$:\dfrac{\sqrt{7}}{\sqrt{3}}$$
  • Question 2
    1 / -0
    Aqueous solution of which of the following is green?
    Solution
    $$V^{2+}\longrightarrow$$  Mauve/Lavender.
    $$V^{3+}\longrightarrow$$  Green.
    $$VO^{2+}\longrightarrow$$  Blue.
    $$VO_2^{+}\longrightarrow$$  Yellow.
  • Question 3
    1 / -0
    Which one of the following compound is both colored and paramagnetic?
    Solution
    The species which have unpaired electron in d-subshell is both paramagnetic and coloured.
    $$Sc^{3+}\longrightarrow 4s^03d^0$$
    $$Ti^{4+}\longrightarrow 4s^03d^0$$
    $$Cr^{3+}\longrightarrow 4s^03d^3$$
    $$Cu^{+}\longrightarrow 4s^03d^{10}$$
    $$Cr^{3+}$$ is having $$3$$ unpaired $$e^-$$, thus paramagnetic and hence show color due to transition of these unpaired.
  • Question 4
    1 / -0
    In the dichromate di-anion:
    Solution
    The Dichromate di-anion consists of two tetrahedra sharing an oxygen at the common corner.The chromate ion has the tetahedral structure. The six terminal bond in the dichromate ion have same bond length due to resonance. Hence all six $$Cr-O$$ bonds are equal.

  • Question 5
    1 / -0
    Maximum ferromagnetism is found in ?
    Solution
    As Iron have highest value of saturation, Magnetization $$M_s$$ which is defined as the magnetic moment density when the material is subjected to a field strong enough to align all its moments.
    $$Fe \ 1707 \ M_s$$
    $$Co \ 1400 \ M_s$$
    $$Ni \ 485 \ M_s$$
  • Question 6
    1 / -0
    An aqueous solution containing $$V^{x+}$$ ion has magnetic moment equal to $$\sqrt {15} \ BM$$. Therefore, the x is:
    Solution
    Let $$n$$ be the number of unpaired electrons. 
    Magnetic moment $$= \sqrt {n(n+2)}= \sqrt {15}$$
    $$\Rightarrow n=3$$
    Atomic number of Vanadium is $$23$$. So, it must loose all its $$4s$$ electrons to get $$3$$ unpaired d electrons.
    $$\therefore$$ value of $$x$$ will be $$2$$
  • Question 7
    1 / -0
    Which of the following is paramagnetic?
    Solution
    $${ V\left( CO \right)  }_{ 6 }\Rightarrow V-23$$
    $${ 1s }^{ 2 }{ 2s }^{ 2 }{ 2p }^{ 6 }{ 3s }^{ 2 }{ 3p }^{ 6 }{ 4s }^{ 2 }{ 3d }^{ 3 }$$
    $$CO-$$weak ligand
    $$\Rightarrow $$ para-magnetic in nature.

  • Question 8
    1 / -0
    Smallest ionic radius is:
    Solution
    Atomic and ionic radius of lanthanides decreases from La to Lu.
    Due to the contraction of lanthanide the decrease in ionic radius of the element in the lanthanide series from lanthanum to lutetium occurs.
  • Question 9
    1 / -0
    Which of the following species do not exist?
    Solution
    $$PbF_4$$ exists but $$PbI_4$$ does not exist as $$I_2$$ reduces $$Pb(II)$$ to $$Pb$$ and oxidizes $$I$$ to $$I_2$$. Since, $$I_2$$ is not strong reducing agent to reduce $$Pb(II)$$ to $$Pb$$. Hence, $$PbI_2$$ is formed and not $$PbI_4$$.
  • Question 10
    1 / -0
    What would be magnetic moment of $$Gd^{+3}$$ ($$Z = 64$$)?
    Solution
    Electronic configuration of $$[Gd]^{3+}:[Xe]4f^{7}$$
    Hence number of unpaired electron $$n=7$$
    Hence, $$magnetic \ moment=\sqrt{n(n+2)}=\sqrt{63}=7.9BM$$
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