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Coordination Compounds Test - 7

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Coordination Compounds Test - 7
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  • Question 1
    1 / -0

    Which among the following will not show geometrical isomerism?

    Solution

    Geometrical isomerism arises in heterolyptic complexes due to different possible geometric arrangements of the ligands. In [Co(en)3]3+ all ligands are same i.e. ethane-1,2-diamine. It is symmetrical so it does not show geometrical isomerism.

  • Question 2
    1 / -0

    What is used as the complexing agent in volumetric analysis of ions like Mg2+and Ca2+?

    Solution

    EDTA (ethylenediaminetetraacetato) is a hexadentate ligand and it forms stable chelate complex with given metal ions. The selective estimation of these ions can be done due to difference in stability constants of complexes of these ions with EDTA.

  • Question 3
    1 / -0

    Which of the following is a homolyptic complex?

    Solution

    Complexes in which the central metal is bound to only one kind of donor groups are called homolyptic complexes. [Cu(NH​​​​​3)6]+2 is a homolyptic complex because in this only ammonia group is the donor group bound to Cu+2.

  • Question 4
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    Blue solution of CuSO4 becomes darker when treated with ammonia because

    Solution

    Ammonia forms stable complex with Cu2+ which is dark blue in colour. So the blue colour of CuSO 4 solution becomes darker when ammonia is added to it.
    Cu+2 + 4NH3 →→ [Cu(NH3)4]+2

  • Question 5
    1 / -0

    According to Werner’s theory, the secondary valences of the central atom correspond to its

    Solution

    According to Werner’s theory, secondary valences are non-ionisable and are satisfied by neutral molecules or negative ions. Secondary valence is equal to the coordination number and is fixed for a metal.

  • Question 6
    1 / -0

    K3CoF6 is high spin complex. What is the hybrid state of Co atom in this complex?

    Solution

    Given complex can be written as K3[CoF6]. There are 3 Potassium ions K+ means an overall +3 charge on cations and so the charge on the complex anion is -3. Each Fligand has -1 charge so there is a total of -6 charge on ligands. Let oxidation state of Co (Z=27) be x

    x + ( -6) = -3
    x = -3 + 6 = +3

    So oxidation state of Co=+3 and its electronic configuration is 1s22s22p63s23p63d6 . Since its a high spin complex means there is no pairing of electrons in 3d subshell. Coordination number of Co is 6 as there are 6 ligands bound to it, so this octahedral complex has hybridization sp3d2.

  • Question 7
    1 / -0

    Tetraaminecopper(II) ion is a square planar complex with one unpaired electron. According to valence bond theory the hybrid state of copper should be

    Solution

    Tetraaminecopper(II) ion is square planar. Square planar complexes have dsp​​​​2 hybridisation. So hybridization is dsp2.

  • Question 8
    1 / -0

    The IUPAC name of KAl(SO4)2.12H2O is

    Solution

    This is a double salt so its name is written in alphabetical order and the number of molecules of water of crystallisation are mentioned in the end. So the name is aluminium potassium sulphate dodecahydrate.

  • Question 9
    1 / -0

    The IUPAC name of (NH​​​​​4)2Fe(SO​​​​​4)2.6H2O is

    Solution

    This is a double salt so its IUPAC name is written in alphabetical order and number of the molecules of water of crystallisation is mentioned in the end. So the name is ammonium ferrous sulphate hexahydrate.

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