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Coordination Compounds Test - 26

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Coordination Compounds Test - 26
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  • Question 1
    1 / -0

    Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.

    Assertion (A): Toxic metal ions are removed by the chelating ligands.

    Reason (R): Chelate complexes tend to be more stable.

    Solution

    When a solution of chelating ligand is added to solution containing toxic metal ligands chelates the metal ions by formation of stable complex.

  • Question 2
    1 / -0

    Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.

    Assertion: A mong [CO(NH3)6]3+ and [CO(en)3]3+, coordination compound [CO(en)3]3+ is a more stable complex.

    Reason: Because (en) is a chelating ligand/bidentate ligand.

    Solution

    Since (ethylene diamine-en) is a chelating ligand/bidentate ligand, [CO(en3)]3+ is a more stable complex as compared to the other one.

  • Question 3
    1 / -0

    Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.

    Assertion (A): [Cr(H2O)6]Cl2 and [Fe(H2O)6]Cl2 are reducing in nature.

    Reason (R): Unpaired electrons are present in their d-orbitals.

    Solution

    In the complexes, Co exists as Co2+ and Fe as Fe2+. Both of the complexes become Stable by oxidation of metal ion to Co3+ and Fe3+.

  • Question 4
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    Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.

    Assertion (A): Low spin tetrahedral complexes are rarely observed.

    Reason (R): Crystal field splitting is less than pairing energy for tetrahedral complexes.

    Solution

    In tetrahedral complexes, the splitting of the d-orbitals is inverted and is smaller in comparison to octahedral complexes. The Crystal field splitting energy is not large enough to force pairing and hence, low spin complexes are rarely observed.

  • Question 5
    1 / -0

    Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.

    Assertion (A): Linkage isomerism arises in coordination compounds containing ambidentate ligand.

    Reason (R): Ambidentate ligand has two different donor atoms.

    Solution

    Linkage isomerism arises due to two different donor atoms in ambidentate ligand.

  • Question 6
    1 / -0

    Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.

    Assertion (A): [Fe(CN)6]3− ion shows magnetic moment corresponding to two unpaired electrons.

    Reason (R): Because it has d2sp3 type hybridisation.

    Solution

    [Fe(CN)6]3− ion shows magnetic moment corresponding to two unpaired electrons.

  • Question 7
    1 / -0

    Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.

    Assertion (A): Complexes of MX6 and MX5L type (X and L are unidentate) do not show geometrical isomerism.

    Reason (R): Geometrical isomerism is not shown by complexes of coordination number 6.

    Solution

    For complexes of MX6 and MX5L type, different geometric arrangements of the ligands are not possible due to presence of plane of symmetry.

  • Question 8
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    Directions: These questions consist of two statements, each printed as Assertion and Reason. While answering these questions, you are required to choose any one of the following four responses.

    Assertion : NF3 is a weaker ligand than N(CH3)3.

    Reason : NF3 ionizes to give F ions in aqueous solution.

    Solution

    It is a correct statement that NF3 ​ is a weaker ligand than N(CH3)3, the reason is that fluorine is highly electronegative therefore, it withdraws electrons from the nitrogen atom. Hence, the lone pair of nitrogen atoms cannot be ligated. While N(CH3)3​ is a strong ligand because CH3​ has an electron releasing group.

  • Question 9
    1 / -0

    Directions: These questions consist of two statements, each printed as Assertion and Reason. While answering these questions, you are required to choose any one of the following four responses.

    Assertion : [Ti(H2O)6]3+ is coloured while [Sc(H2O)6]3+ is colourless.

    Reason : d-d transition is not possible in [Sc(H2O)6]3+.

    Solution

    Ti has the electronic configuration [Ar] 3d2. In [Ti(H2O)6]3+ complex it is present as Ti3+ which has one unpaired electron in 3d orbital.Thus d-d transition is possible only in Ti(H2O)63+ and thereby it shows colour And also it undergoes d2sp3 (octahedral) hybridization to accomodate 6 pairs of water.

    So the complex is octahedral in shape, will be coloured and paramagnetic due to presence of one unpaired electron.

    Sc has the electronic configuration [Ar]3d1 . Since it is present in the complex [Sc(H2O)6]3+ as Sc3+ it has no unpaired electron. It will also undergo d2sp3 hybridization to accomodate 6 pairs of electrons from water.

    So it is also octahedral, colourless and diamagnetic due to absence of unpaired electron.

  • Question 10
    1 / -0

    Directions: These questions consist of two statements, each printed as Assertion and Reason. While answering these questions, you are required to choose any one of the following four responses.

    Assertion : [Fe(CN)6]3– is weakly paramagnetic while [Fe(CN)6]4– is diamagnetic.

    Reason : [Fe(CN)6]3– has +3 oxidation state while [Fe(CN)6]4– has +2 oxidation state.

    Solution

    Paramagnetic Character depends on the number of unpaired electrons. In the [Fe(CN)6]3– has 1 unpaired electron, therefore, it is paramagnetic whereas [Fe(CN)6]3– is diamagnetic because it does not have unpaired electrons.

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