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Coordination Compounds Test - 6

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Coordination Compounds Test - 6
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  • Question 1
    4 / -1

     

    Among the following complexes the one which shows zero crystal field stabilisation energy is

     

    Solution

     

    In [Fe(H2O)6]3+, iron has 3d5 configuration. Since, H2O is weak ligand the distribution is .
    Hence, CFSE = - 3 x 0.4 + 2 x 0.6 = 0.0

     

  • Question 2
    4 / -1

    The wavelength of light absorbed is highest in

    Solution

    Among all the coordination aforesaid in complex [Co(NH3)5Cl]2+ unpaired electrons are present with the central element cobalt so it required less energy for excitation. Thus, the wavlength of light absorbed will be the highest.

  • Question 3
    4 / -1

    The CFSE for octahedral [CoCI6]4- is 18000 cm-1. Then the CFSE for tetrahedral [CoCI4]2- will be

    Solution

    CFSE for [CoCI4]2- will be 
    i.e.  = 8000cm-1

  • Question 4
    4 / -1

    The crystal field splitting energy (Δo) of

    I. [CoBr6]3- II. [CoF6]3- 
    III. [Co(NCS)6]3- IV. [Co(CN)6]3- is in the order of

    Solution

    Crystal field splitting energy depends on strength of ligand. Strong ligands have more value of CFSE (Δo).
    Hence, Br- < F- < NCS < CN-

  • Question 5
    4 / -1

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

    Solution

    The correct answer is Option A.
    Electronic configuration of Cu2+  ion in [Cu(NH3)4]2+.
    Cu2+ ion =[Ar]3d94s0.
    ∴Cu2+ ion has one unpaired electron.
    Magnetic moment of [Cu(NH3)4]2+ (μ) = BM
    where, n = no. of unpaired electrons

    Whereas Ni2+ in [Ni(CN)4]2− , Ti4+ in TiCl4 and Co2+ ion [COCl6]4− has 2,0 and 3 unpaired electrons respectively.

  • Question 6
    4 / -1

     

    The correct order of ligand field strength is

     

  • Question 7
    4 / -1

    The crystal field splitting energies (CFSE) of high spin and low spin d6 metal complexes in octahedral complex in terms of Δo respectively are

    Solution

    For high spin d6, CFSE = - 4 x 0.4 + 2 x 0.6 = -0.4
    For low spin d6, CFSE= - 6 x 0.4 = -2.4

  • Question 8
    4 / -1

    The Δt of the following complexes follows the order

    I. [CoCI4]2-

    II. [CoBr4]2-

    III. [Co(NCS)4]2-

    Solution

    Correct Answer :- d

    Explanation : Bonding Energy :

    [CoCI4]2      -2084.67

    [CoBr4]2      -3139.39

    [Co(NCS)4]2-  - 8559.48

    Correct order is : III > II > I

  • Question 9
    4 / -1

    In which of the following coordination entities, the magnitude Δo (CFSE in octahedral field) will be maximum ?

    Solution

    Increasing order of ligand field strength
    l- < Br- < SCN- < CI- < S2- < F- < OH-< H2O < NCS- < edta4- < NH3 < en < CN- < CO

  • Question 10
    4 / -1

    The increasing order of wavelength of absorption for the complex ions

    I. [Cr(NH3)6]3+

    II. [CrCI6]3-

    III. [Cr(H2O)6]3+
    IV. [Cr(CN)6]3-

    is :

    Solution

    Stronger ligand causes greater splitting between two energy levels.

    The energy of light absorbed is more and the wavelength of light absorbed is less for the stronger ligand.

    Hence order of wavelength absorption will be reverse of the order of the strength of ligands.

    Hence A is correct.

  • Question 11
    4 / -1

     

    The formula of the complex hexamminecobalt (III) chloride sulphate is

     

    Solution

     

    The formula of complex hexamminecobalt (III) chloride sulphate is [Co(NH3)6]CISO4.

     

  • Question 12
    4 / -1

     

    The donor sites of EDTA ligand are

     

    Solution

     

    Ethylenediaminetetraacetate ion (EDTA)4- is an important hexadentate ligand. It can bind through two nitrogen and four oxygen atoms to a central metal ion.

     

  • Question 13
    4 / -1

     

    What is the IUPAC name of compound NaBH4?

     

    Solution

     

    The IUPAC name of NaBH4 is sodium tetrahydridoborate 

     

  • Question 14
    4 / -1

     

    Ferrocyanide ion is a type of​

     

    Solution

    Ferricyanide is the anion [Fe(CN)6]3−. It is also called hexacyanoferrate(III) and in rare, but systematic nomenclature, hexacyanidoferrate(III). The most common salt of this anion is potassium ferricyanide, a red crystalline material that is used as an oxidant in organic chemistry.

  • Question 15
    4 / -1

     

    Valence bond theory was proposed by

     

    Solution

     

    In the valence bond (VB) theory, proposed in large part by the American scientists Linus Pauling and John C. Slater, bonding is accounted for in terms of hybridized orbitals of the… The basis of VB theory is the Lewis concept of the electron-pair bond.

     

  • Question 16
    4 / -1

     

    Comprehension Type

    Direction (Q. Nos. 16 and 17) This section contains a paragraph, describing theory, experiments, data, etc. Two questions related to the paragraph have.

    Passage

    In octahedral complexes due to repulsion between the ligands and d-orbitals, there is splitting of d-orbitals into two sets, i.e. two orbitals of higher energy called eg and three orbitals of lower energy called t2g. The difference of energy between the two sets of d-orbitals is called crystal field stabilisation energy denoted by Δo. For any given metal cation, the magnitude of Δo depends on the nature of ligands.

    Q. 

    The CFSE for d7 configuration for strong ligand field is

     

    Solution

     

    The total crystal field stabilisation energy is given by CFSE(octahedral) = are the number of electron occupying the t2g and eg orbitatls respectively.
    For d7 in strong field,


    CFSE = -0.4 x 6 + 0.6 x 1 = -2.4 + 0.6 = -1.8Δo 

     

  • Question 17
    4 / -1

    In octahedral complexes due to repulsion between the ligands and d-orbitals, there is splitting of d-orbitals into two sets, i.e. two orbitals of higher energy called eg and three orbitals of lower energy called t2g. The difference of energy between the two sets of d-orbitals is called crystal field stabilisation energy denoted by Δo. For any given metal cation, the magnitude of Δo depends on the nature of ligands.

    Q. 

    In the following complexes of manganese, the distribution of electrons in d-orbitals of manganese

    i. [Mn(H2O)6]2+

    ii. [Mn(CN)6]4-

    Solution

    In [Mn (H2O)6]2+, manganese have d5 configuration. H2O is a weak field ligand. Hence, it will form high spin complex.


    In [Mn(CN)6]4-, manganese have d5 configuration. CN- ion is a strong field ligand. It will form low spin complex by pairing of electron in t2g orbitals.

  • Question 18
    4 / -1

     

    Direction (Q. No. 18-19) 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 below.

     

    Solution

     

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

     

  • Question 19
    4 / -1

     

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

     

    Solution

     

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

     

  • Question 20
    4 / -1

     

    Direction (Q. Nos. 20-24) This section contains 5 questions. When worked out will result in an integer from 0 to 9 (both inclusive).

    Q. 

    Number of unpaired electrons in d6, low spin octahedral complex.

     

    Solution

     

    In d6 low spin octahedral complex.


    Hence, zero number of unpaired electrons.

     

  • Question 21
    4 / -1

     

    The number of ligands which have strong crystal field splitting than

    H2O among SCN-, NCS-, EDTA4- , , , Br-, PPh3, F-

     

    Solution

     

    NCS- edta4- , and PPh3 are strong field ligand than H2O.

     

  • Question 22
    4 / -1

     

    The total number of unpaired electrons in the two complexes [Cr(H2O)6]2+ and [Cr(CN)6]4- having octahedral geometry are

     

    Solution

     

    In [Cr(H2O)6]2+ chromium have d4 configuration and H2O is a weak field ligand.
    d4 (weak field)

    In [Cr(CN)6]4- chromium have d4 configuration and CN- ion is a strong field ligand. Hence, CN- ion causing pairing of electron. d4 (strong field)

    Hence, total number of unpaired electrons in both the complex are 6.

     

  • Question 23
    4 / -1

     

    Number of unpaired electrons in t2g and eg orbitals in weak octahedral ligand fields with d7 configuration.

     

    Solution

     

    d7 (weak octahedral field)

    Hence, 3 unpaired electrons.

     

  • Question 24
    4 / -1

     

    In tetrahedral complexes having weak field ligand, pairing starts with ...... electron.

     

    Solution

     

    The pairing starts with six electrons in the tetrahedral complexes which have weak field ligand.

     

  • Question 25
    4 / -1

     

    Statement Type

    Direction (Q. No. 25) This section is based on Statement I and Statement II. Select the correct answer from the codes given below.

    Q. 

    Statement I : The value of CFSE for the octahedral complex [Co(NH3)6]3+ is - 2.4Δo.

    Statement II : The ammonia molecule is strong field ligand.

     

    Solution

     

    In [Co(NH3)6]3+ Co, have d6 configuration and NH3 is strong field ligand.


    CFSE(OCt) = -0.4 x 6 + 0.6 x 0 = -2.4Δo 

     

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