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Coordination Compounds Test - 11
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Coordination Compounds Test - 11
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  • Question 1/6
    1 / -0

    A complex involving dsp2 hybridisation has

    Solutions

    dsp2 type of hybridisation is seen specially in case of transition metal ions. The orbitals involved in this type of hybridisation are dx2 - y2, s and two p. The four dsp2 hybrid orbitals adopt a square planar geometry. For example, the complex ion [Ni(CN)4]2- involves dsp2 hybridisation.

     

  • Question 2/6
    1 / -0

    Which of the following is an incorrect statement about the complex ion [Co(NH3)6]3+?

    Solutions

    Oxidation state of Co in the given complex is +3.
    Co in ground state: 3d7 4s2
    So, Co+3: 3d6 4s0
    There will be no unpaired electrons in this coordination compound because NH3 provides a strong ligand field and is able to pair up electrons of the given compound, forming d2sp3 hybridisation. Hence, the given complex is diamagnetic.

     

  • Question 3/6
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    In which of the following pairs do both the species have the same magnetic moment (spin-only value)?

    Solutions

    Cr exists as Cr(II) ion and is in d4 high spin configuration (for which the number of unpaired electrons is 4). Hence, the spin only magnetic moment is √24 BM. In the other complex, Fe(II) ion is in d6 high spin configuration (for which the number of unpaired electrons is 4). Hence, the spin only magnetic moment is √24 BM.

     

  • Question 4/6
    1 / -0

    The charge on the central metal ion in the complex [Ni(CO)4] is

    Solutions

    CO is a neutral ligand.
    So, X + 4(0) = 0
    X = 0

     

  • Question 5/6
    1 / -0

    Which of the following species is not expected to be a ligand?

    Solutions

    A ligand should be negatively charged or must possess at least one lone pair of electrons so as to form dative bonds with the central metal ion. NH4+ has no lone pair of electrons and thus, cannot act as a ligand in the formation of complexes.

     

  • Question 6/6
    1 / -0

    The oxidation state and coordination number of Cr in [Cr(C2O4)3]3- ion, respectively are

    Solutions

    Oxidation state of Cr:
    x + 3(-2) = -3
    x – 6 = - 3
    x = -3 + 6 = + 3

    The coordination number is the number of ligands attached to the central ion (more specifically, the number of donor atoms). As C2O42- is a bidentate ligand, so coordination number of Cr in this compound is 6.

     

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