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Coordination Compounds Test - 57

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Coordination Compounds Test - 57
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Weekly Quiz Competition
  • Question 1
    1 / -0
    The number of unpaired electron in complex $$[Cr(NO_{2})_{3}(NH_{3})_{3}]$$ is
    Solution

  • Question 2
    1 / -0
    Nickel $$(Z=28)$$ combines with a uni negative monodentate ligand $$X^{-}$$ to form a paramagnetic complex, $$[NiX_{4}]^{2-}$$. The number of unpaired electrons in the nickel and geometry of this complexion are, respectively:
  • Question 3
    1 / -0
    which has the maximum conductivity in thier 0.1M solution?
    Solution
    The maximum conductivity is shown by that compound which have maximum ions on dissociation. Hence the compound $$[Co(NH_3)_6]Cl_3$$ have the maximum number of ions $$[Co(NH_3)_6] + 3Cl^-$$ = 4 ions and hence the correct answer is D.
  • Question 4
    1 / -0
    In $$[Co(NH_3)_6]Cl_3$$, the number of covalent bonds is
    Solution

  • Question 5
    1 / -0
    Total number of possible isomers of $$\left[ Cu({ NH }_{ 3 })_{ 4 } \right] \left[ { PtCl }_{ 4 } \right]$$ is:
    Solution
    The total number of possible isomer of the $$[Cu(NH_3)_4] [PtCl_4]$$ is 4.

    The isomers can be formed by exchanging the ligands. 

    1.$$[Cu(NH_3)_3Cl] [Pt(Cl)_3NH_3]$$

    2.$$[Cu(NH_3)_2Cl_2] [PtCl_2(NH_3)_2]$$

    3.$$[Cu(NH_3)Cl_3] [PtCl(NH_3)_3]$$

    4. $$[CuCl_4] [Pt(NH_3)_4]$$

    Hence, the correct option is B
  • Question 6
    1 / -0
    Given the molecular formula of the hexa coordinated complexes (A) $${ CoCl }_{ 3 }.{ 6NH }_{ 3 }$$ (B) $${ CoCl }_{ 3 }.{ 5NH }_{ 3 }$$ (C) $${ CoCl }_{ 3 }.{ 4NH }_{ 3 }$$. If the number of coordinated $${ NH }_{ 3 }$$ molecules in $$A, B$$ and $$C$$ respectively are $$6, 5$$ and $$4$$, primary valency in $$(A), (B)$$ and $$(C)$$ are
    Solution
    The complex can be written as follows.

    $$A-[Co(NH_3)_6]Cl_3$$

    $$B- [Co(NH_3)_5Cl]Cl_2 $$

    $$C-[Co(NH_3)_4 Cl_4]Cl$$

    Hence, number of primary valency are $$3,2$$ and $$1$$ respectively.
  • Question 7
    1 / -0
    $$LiH + \underline {B}_2H_6 \rightarrow 2Li \underline{B}H_4$$

    Find the change in hybridization of the underlined atom.
  • Question 8
    1 / -0
    The complex ion having minimum wavelength of absorption in the visible region is:
  • Question 9
    1 / -0
    Which isomer of $$CrCl_{3} . 6H_{2}O$$ is dark green in colour and forms one mole of $$AgCl$$ with excess of $$AgNO_{3}$$ solution ?
  • Question 10
    1 / -0
    For the following two complexes 
    $$A:[NiCl_6]^{4-}$$ and $$[NiCl_4]^{2-}$$ the ratio of CFSE will be nearly:
    Solution
    $$[NiCl_6]^{4-}   \rightarrow$$ octahedral complex ($$t_{2g}= 6, e_{g}= 4$$)

    $$[NiCl_4]^{2-} \rightarrow $$Tetrahedral complex ($$e_g= 4, t_{2g}= 4$$)

    for octahedral CFSE =$$-0.4 \times 6+0.6\times 2=-1.2$$

    for tetradidral CFSE$$= +0.4\times 4-0.6\times 4=-0.8$$

    $$\therefore \dfrac{CFSE(1)}{CFSE(2)}=\dfrac{-1.2}{-0.8}=\dfrac{3}{2}$$

    Option A is correct.
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