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Coordination Chemistry Test - 2

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Coordination Chemistry Test - 2
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Weekly Quiz Competition
  • Question 1
    1 / -0
    Which of the following complexes are diamagnetic?

    (1) [Co(NH3)6]3+
    (2) [Ni(CO)6]
    (3) [Cr(NH3)6]3+
    (4) [Fe(CN)6]3-
    Solution
    [Co(NH3)6]3+ - t2g6 eg0; no unpaired electrons; diamagnetic
    [Ni(CO)6] - t2g6 eg0; no unpaired electrons; diamagnetic
  • Question 2
    1 / -0
    Which of the following sets represents square planar complexes only?
    Solution
    PtCl42-, AuCl4-, PdCl42-, Ni(CN)42- : All are square planar complexes.
  • Question 3
    1 / -0
    Which of the following complexes cannot exhibit geometrical isomerism?
    i. [Ni(CN)4]2-
    ii. [Zn(NH3)2Cl2]
    iii. [Pt(NH3)2Cl2]
    iv. [Cr(NH3)6]3+
    Solution

    [Ni(CN)4]2- homoleptic complex with tetrahedral geometry. Homoleptic complex does not show geometrical isomerism.
    [Zn(NH3)2Cl2] is a heteroleptic complex with tetrahedral geometry. Tetrahedral complex does not exhibit geometrical isomerism, as all the four ligand positions are equivalent.
    [Pt(NH3)2Cl2] is a heteroleptic square planar complex of Ma2b2 type. In a square planar complex of formula [Ma2b2], the two similar ligands may be arranged adjacent to each other in a cis isomer or opposite to each other in a trans isomer.
    [Co(NH3)6]3+homoleptic complex ion with octahedral geometry. Homoleptic complex does not show geometrical isomerism.

  • Question 4
    1 / -0
    Which of the following complexes have square planar geometry?

    i. [PtCl4]2-
    ii. [NiCl4]2-
    iii. [Pt(NH3)2Cl2]
    iv. [Ni(CO)4]
    v. [Ni(CN)4]2-
    Solution
    [PtCl4]2-: Inner orbital complex; square planar geometry; dsp2 hybridisation
    [Pt(NH3)2Cl2] -: Inner orbital complex; square planar geometry; dsp2 hybridisation
    [Ni(CN)4]2-: Inner orbital complex; square planar geometry; dsp2 hybridisation
  • Question 5
    1 / -0
    Which of the following ligands do not show linkage isomerism?

    (1) SO32-
    (2) NO2-
    (3) EDTA4-
    (4) C2O42-
    (5) SCN-
    Solution

    Linkage isomerism is only shown by ambidentate ligands.
    EDTA4- (hexadentate ligand) is not an ambidentate ligand. Hence, it will not show linkage isomerism.
    C2O42- (bidentate ligand) is not an ambidentate ligand. Hence, it will not show linkage isomerism.

  • Question 6
    1 / -0
    Which of the following statements is false?
    Solution
    The secondary valency represents the coordination number of the central metal ion.
  • Question 7
    1 / -0
    Which of the following complexes is correctly named?
    Solution
    It is correctly named.
    [Cr(NCS)4(NH3)2]- : Diamminetetrathiocyanatochromate(III) ion
  • Question 8
    1 / -0
    Which of the following is/are inner orbital complex(es)?

    1. [Cr(NH3)6]3+
    2. [Fe(CN)6]4-
    3. [Fe(H2O)6]2+
    Solution
    As both NH3 and CN- are strong field ligands, so energy splitting is high. Therefore, in [Cr(NH3)6]3+ and [Fe(CN)6]4-, inner d-orbitals take part in hybridisation (d2sp3). Thus, both form inner orbital complexes.
  • Question 9
    1 / -0
    Which of the following sets represents bidentate ligands only?
    Solution

    Ethylenediammine (en): Bidentate ligand
    Oxalate ion (C2O42-): Bidentate ligand
    Acetylacetonate ion (acac): Bidentate ligand
    Dimethylphosphinomethane (dmpe): Bidentate ligand

  • Question 10
    1 / -0
    Which of the following is/are paramagnetic inner orbital complex/complexes?
    i. [Co(C2O4)3]3–
    ii. [Fe(CN)6]3–
    iii. [Mn(CN)6]3–
    Solution
    [Fe(CN)6]3– : Inner orbital complex; d2sp3 hybridisation; paramagnetic
    [Mn(CN)6]3– : Inner orbital complex; d2sp3 hybridisation; paramagnetic
  • Question 11
    1 / -0
    Which of the following 'Formula : Name' combinations is incorrect?
    Solution
    Correct IUPAC name of [Cr(NH3)4(OH)2]Br is tetraamminedihydroxochromium(III) bromide.
  • Question 12
    1 / -0
    Which of the following pairs represents inner orbital complexes only?
    Solution
    [Co(NH3)6]3+ is an inner orbital complex with octahedral geometry (d2sp3 hybridisation).
    [Cr(H2O)6]3+ is an inner orbital complex with octahedral geometry (d2sp3 hybridisation).
  • Question 13
    1 / -0
    Which of the following complexes are diamagnetic in nature?

    (1) [CoF6]3-
    (2) [Fe(CN)6]4-
    (3) [Ni(CN)4]2-
    (4) [Zn(H2O)6]2+
    (5) [Cr(H2O)6]3+
    Solution
    [Fe(CN)6]4- is an inner orbital complex with octahedral geometry (d2sp3 hybridisation); no unpaired electrons; diamagnetic.
    [Ni(CN)4]2- is an inner orbital complex with square planar geometry (dsp2 hybridisation); no unpaired electrons; diamagnetic.
    [Zn(H2O)6]2+     is an outer orbital complex with square planar geometry (sp3d2 hybridisation); no unpaired electrons; diamagnetic.
  • Question 14
    1 / -0
    Which of the following coordination complexes is not correctly named?
    Solution
    The correct IUPAC name of [Pt(en)(ONO)Cl] is chloro(ethylenediamine)nitrito-O-platinum(II).
  • Question 15
    1 / -0
    The coordination number and the oxidation state of cobalt in [Cr(H2O)(en)2OH]+2 respectively are
    Solution
    The coordination number of Co in [Cr(H2O)(en)2OH]+2 is 6 as (en)-ethylene diamine is bidentate ligand, it will attach to central metal atom through two active sites, while H2O and OH are monodentate ligands (attached to central metal atom with one active site).
    Now, let the oxidation number of Co in [Cr(H2O)(en)2OH]+2 be x.
    x + 1(0) + 2(0) + (-1) = +2
    x = +3
    Therefore, the coordination number and the oxidation state of cobalt in [Cr(H2O)(en)2OH]+2 are 6 and +3, respectively.
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