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Coordination Compounds Test - 4

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Coordination Compounds Test - 4
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Weekly Quiz Competition
  • Question 1
    4 / -1

     

    Which one of the following is wrongly matched?

    Solution

     

    Due to strong ligand, this is inner orbital complex and the hybridisation of Fe is d2sp3.

  • Question 2
    4 / -1

    In which of the following the central atom has sp3d2-hybridisation?

    Solution

    Since, F- ion is a weak ligand hence, pairing of electrons does not occur and outer orbital complex is formed in [CoF6]3-.

  • Question 3
    4 / -1

    Which is the diamagnetic?

    Solution

    [Ni(CN)4]2- in this, Ni has dsp2 hybridisation with square planar geometry and it is diamagnetic.

  • Question 4
    4 / -1

    A magnetic moment of 1.73 BM will be shown by one among following.

    Solution

    A magnetic moment of 1.73 BM corresponds to one unpaired electron. Among the given, (a) and (b) are diamagnetic and (c) is paramagnetic with 3 unpaired electrons (3d7 configuration).

  • Question 5
    4 / -1

    Which one of the following complex species does not involve inner orbital hybridisation?

    Solution

    F- ion is weak ligand hence, pairing of electrons does not occur and outer orbital complex is formed in [CoF6]3-.

  • Question 6
    4 / -1

    Which of the following complex has zero magnetic moment (spin only)?

    Solution

    The potassium ferrocyanide, K4 [Fe(CN)6] due to strong ligand pairing of electrons occur and it has no unpaired electrons.

  • Question 7
    4 / -1

    The number of unpaired electrons calculated in [Co(NH3)6]3+ and [CoF6]3- are

    Solution

     In [Co(NH3)6]3+ due to pairing cobalt configuration and it is diamagnetic.
    In [CoF6]3- , F- being weak ligand pairing does not occur and Co3+ ion has configuration . It has 4 unpaired electrons.

  • Question 8
    4 / -1

    The tetrachloro complexes of Ni(ll) and Pd(ll) respectively are (atomic number of Ni and Pd are 28 and 46 respectively).

    Solution

    ln [NiCI4]2- , Ni has sp3 hybridisation with 2 unpaired electrons, i.e. paramagnetic.
    In [PdCI4]2- Pd has dsp2 hybridisation and it has no unpaired electrons, i.e. diamagnetic.

  • Question 9
    4 / -1

    Which one of the following is an inner orbital complex as well as diamagnetic in behaviour? (Atomic number, Zn = 30, Cr= 24, Co = 27, Ni = 28)

    Solution

    Zn and Ni always give outer orbital complexes with coordination number 6.
    [Co(NH3)6]3+ is diamagnetic complex.
    [Cr(NH3)6]3+ is param agnetic with 3d3 configuration.

  • Question 10
    4 / -1

    The pair of compounds having the same hybridisation for the central atom is

    Solution

    [Co(NH3)6]3+ : For this complex action the 6 valance electrons of cobalt will occupy the 3d orbital, whereas the 6 ligands will occupy 3d,4s and 4p orbitals in d2sp3 hybridisation.

    [Co(H2O)6]3+ : This intermixing is based on quantum mechanics. Transition metals may exhibit d2sp3 hybridization where the d orbitals are from the 3d and the s and p orbitals are the 4s and 3d. it acts as strong ligand when metal ion is in +3 oxidation for metals cobalt.

     

  • Question 11
    4 / -1

     

    When 0.1 mol CoCl3(NH3)5 is combined with excess AgNO3, then 0.2 mol AgCl is obtained. The conductivity of the solution suits the

  • Question 12
    4 / -1

     

    A chelating agent has two or more than two donor atoms to bind to a single metal ion. Which of the following is not a chelating agent?

  • Question 13
    4 / -1

     

    IUPAC name of [Pt(NH3)2Cl(NO2)] is

  • Question 14
    4 / -1

     

    In the complex [E(en)2(C2O4)]NO2 (where (en) is ethylenediamine) _______are the coordination number and the oxidation state of the element ‘E’ respectively.

  • Question 15
    4 / -1

     

    The sum of coordination number and oxidation number of the metal M in the complex [M(en)2(C2O4)]Cl (where (en) is ethylenediamine) is

  • Question 16
    4 / -1

     

    Passage

    When crystals of CuSO4 . 4NH3 are dissolved in water, there is hardly any evidence for the presence of Cu2+ ions or ammonia molecules. A new ion [Cu(NH3)4]2+ is furnished in which ammonia molecules are directly linked with the metal ion. Similarly, aqueous solution of Fe(CN)2 .4KCNdoes not give tests of Fe2+ and CN- ions but give test for new ion [Fe(CN)6]4- called ferrocyanide ion.

    Q. The hybridisation and geometry and magnetic property of [Cu(NH3)4]2+ ion are

     

    Solution

     

    Cu(Z = 29) copper configuration (3d104s1)

    Hence, [Cu(NH3)4]2+ has square planar geometry and 1 unpaired electron in Ad orbital which is transferred from 3d orbital with magnetic moment 1.73 BM.

     

  • Question 17
    4 / -1

     

    When crystals of CuSO4 . 4NH3 are dissolved in water, there is hardly any evidence for the presence of Cu2+ ions or ammonia molecules. A new ion [Cu(NH3)4]2+ is furnished in which ammonia molecules are directly linked with the metal ion. Similarly, aqueous solution of Fe(CN)2 .4KCNdoes not give tests of Fe2+ and CN- ions but give test for new ion [Fe(CN)6]4- called ferrocyanide ion.

    Q. The hybridisation, geometry and magnetic property of [Fe(CN)6]4- are

     

    Solution

     

    It is inner orbital complex, diamagnetic.

     

  • Question 18
    4 / -1

    Matching List Type

    Direction (Q. Nos. 18 and 19) Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d) out of which one is correct.

    Q. 

    Match the Column I with Column II and mark the correct option from the codes given bleow.

    Solution

    (i) → (p.s), (ii) → (p,s), (iii) → (q,t), (iv) → (q.r)

  • Question 19
    4 / -1

     

    Match the Column I with Column il and mark the correct option from the codes given bleow.

     

    Solution

     

     

    A → (iii) (b) B → (i) (c) C → (iv) (d) D → (ii)

    a) Six empty orbitals (two d, one s and three p) are available, so whether the ligand is weak (H2​O)  the six orbitals hybridize to give six equivalent d2sp3 hybrid orbitals. Hence, in  [Cr(H2​O)6​]3+  ions chromium is in a state of d2sp3 hybridization.

    b)  CN is a strong ligand it makes the unpaired electrons of cobalt to pair up and occupies the space.

    • The ligands occupy one d orbital, one s orbital and 2 p orbitals. Thus the hybridisation is “dsp2” with “square planar” geometry.
    • It has 1 unpaired electron.

    c) In [Ni(NH3)6]2+, Ni is in +2 state and has configuration 3d8. In presence of NH3, the 3d electrons do not pair up. The hybridization is sp3d2 forming an outer orbital complex.

     It has been found that the complex has two unpaired electrons.

     

     

     

  • Question 20
    4 / -1

     

    Some salts containing two different metallic elements give test for only one of them in solution, such salts are

  • Question 21
    4 / -1

     

    An example of a sigma bonded organometallic compound is

  • Question 22
    4 / -1

     

    Iron carbonyl, Fe(CO)5 is

  • Question 23
    4 / -1

     

    The type of isomerism shown by the complex [CoCl2(en)2] is

  • Question 24
    4 / -1

     

    Which of the following elements do not form a complex with EDTA?

  • Question 25
    4 / -1

     

    Statement I : Both [Ni(CN)4]2- and [NiCI4]2- have same shape and same magnetic behaviour.

    Statement II : Both are square planar and diamagnetic

    Solution

     

    Statement I [Ni(CN)4]2- is square planar and diamagnetic whereas [NiCI4]2- is tetrahedral and paramagnetic .
    Statement II [Ni(CN)4]2- is square planar while [NiCI4]2- is tetrahedral.

     

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