Class Xii Coordination Compounds Test

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| By Dr. Mukesh
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Dr. Mukesh
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Quizzes Created: 2 | Total Attempts: 577
Questions: 30 | Attempts: 292

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Class Xii Coordination Compounds Test - Quiz

TIME ALLOTTED: 60 MINUTES
NEGATIVE MARKING OF -1 MARKS FOR EVERY INCORRECT ANSWER
ALL THE BEST


Questions and Answers
  • 1. 

    Which one of the following shows maximum paramagnetic character?

    • A.

      [Fe(CN)6]3-

    • B.

      [Fe(CN)6]4-

    • C.

      [Cr(H2O)6]3+

    • D.

      [Cu(H2O)6]2+

    Correct Answer
    C. [Cr(H2O)6]3+
    Explanation
    The compound [Cr(H2O)6]3+ shows maximum paramagnetic character because it has an unpaired electron in its d orbital. Paramagnetic substances are attracted to a magnetic field due to the presence of unpaired electrons. In this compound, the chromium ion (Cr3+) has one unpaired electron in its d orbital, making it paramagnetic. The other compounds listed do not have any unpaired electrons and therefore exhibit less paramagnetic character.

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  • 2. 

    Which of the following will give maximum number of isomers?

    • A.

      [Co(NH3)4Cl2]+

    • B.

      [Ni(en)(NH3)4]2+

    • C.

      [Ni(C2O4)(en)2]-2

    • D.

      [Cr(SCN)2(NH3)4]+

    Correct Answer
    D. [Cr(SCN)2(NH3)4]+
    Explanation
    The complex [Cr(SCN)2(NH3)4]+ has the potential to form the maximum number of isomers because it contains both a coordination number of 6 and a chelating ligand (SCN). The coordination number of 6 allows for a greater number of possible arrangements of ligands around the central chromium ion. Additionally, the chelating ligand SCN can form multiple isomers due to its ability to bind to the metal ion through different atoms (sulfur or nitrogen) and in different orientations. Therefore, [Cr(SCN)2(NH3)4]+ has the highest potential for isomerism among the given options.

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  • 3. 

    Which statement is incorrect

    • A.

      [Ni(CO)4] - Tetrahedral, paramagnetic

    • B.

      [Ni(CN)4]2-  Square planar, diamagnetic

    • C.

      [Ni(CO)4]  Tetrahedral, diamagnetic

    • D.

      [NiCl4]2-  Tetrhaedral, paramagnetic

    Correct Answer
    A. [Ni(CO)4] - Tetrahedral, paramagnetic
    Explanation
    The given answer is incorrect because [Ni(CO)4] is not tetrahedral, it is actually square planar. Additionally, it is paramagnetic, not diamagnetic.

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  • 4. 

    Which of the following will exhibit maximum ionic conductivity

    • A.

      K4[Fe(CN)6]

    • B.

      [Co(NH3)6]Cl3

    • C.

      [Cu(NH3)4]Cl2

    • D.

      [Ni(CO)4]

    Correct Answer
    A. K4[Fe(CN)6]
    Explanation
    K4[Fe(CN)6] will exhibit maximum ionic conductivity because it is a salt composed of potassium ions (K+) and hexacyanoferrate(II) ions ([Fe(CN)6]4-). The presence of ions allows for the movement of charged particles, enabling the conduction of electricity. In comparison, the other compounds listed do not contain ions or have fewer ions present, resulting in lower conductivity.

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  • 5. 

    CN- is a strong field ligand. This is due to the fact that

    • A.

      It forms high spin complexes with metal species

    • B.

      It give negative charge

    • C.

      It is psedohalide

    • D.

      It can easily accept electrons from metal species

    Correct Answer
    D. It can easily accept electrons from metal species
    Explanation
    CN- is a strong field ligand because it can easily accept electrons from metal species. This means that CN- has a high affinity for metal ions and can form stable complexes with them. The ability to accept electrons from metal species is a characteristic of strong field ligands, as it results in a large energy difference between the d orbitals of the metal ion. This large energy difference leads to the formation of high spin complexes, where the electrons are distributed across the d orbitals in a way that maximizes the number of unpaired electrons.

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  • 6. 

    Considering H2O as a weak field ligand, the number of unpaired electrons in [Mn(H2O)6]2+ will be:  (atomic number of Mn is 25)

    • A.

      4

    • B.

      3

    • C.

      5

    • D.

      2

    Correct Answer
    C. 5
    Explanation
    In the given complex, [Mn(H2O)6]2+, Mn is in the +2 oxidation state. As Mn is a transition metal, it can have a maximum of 5 unpaired electrons in its d orbitals. Since H2O is a weak field ligand, it does not cause strong crystal field splitting, resulting in a higher number of unpaired electrons. Therefore, the complex [Mn(H2O)6]2+ will have 5 unpaired electrons.

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  • 7. 

    Which of the following does not have a metal-carbon bond?

    • A.

      [Ni(CO)4]

    • B.

      Al(OC2H5)3

    • C.

      C2H5MgBr

    • D.

      K[Pt(C2H4)Cl3]

    Correct Answer
    B. Al(OC2H5)3
    Explanation
    Al(OC2H5)3 does not have a metal-carbon bond because it contains aluminum (Al) which is not a metal-carbon bond. The compound consists of aluminum (Al) bonded to three ethoxide (OC2H5) groups.

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  • 8. 

    Which one of the following is an inner orbital complex as well as diamagnetic in behavior:

    • A.

      [Zn(NH3)6]2+

    • B.

      [Ni(NH3)6]2+

    • C.

      [Cr(NH3)6]3+

    • D.

      [Co(NH3)6]3+

    Correct Answer
    D. [Co(NH3)6]3+
    Explanation
    The complex [Co(NH3)6]3+ is an inner orbital complex because all the ligands (NH3) are bonded to the central metal ion (Co) through their lone pairs of electrons, occupying the inner d orbitals. It is diamagnetic because all the electrons in the d orbitals are paired, resulting in no unpaired electrons and therefore no magnetic moment.

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  • 9. 

    The aqueous solution containing which of the following ions will be colourless?

    • A.

      Fe2+

    • B.

      Mn2+

    • C.

      Ti3+

    • D.

      Sc3+

    Correct Answer
    D. Sc3+
    Explanation
    The aqueous solution containing Sc3+ ions will be colorless because Scandium (Sc) is a transition metal that does not exhibit any strong color in its aqueous form. Transition metals often exhibit colorful compounds due to the presence of partially filled d-orbitals, which can absorb certain wavelengths of light. However, Scandium does not have any partially filled d-orbitals in its ground state, resulting in a colorless solution.

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  • 10. 

    If Δo < P, the correct electronic configuration for d4 system will be-

    • A.

      T42g e0g

    • B.

      T32g e1g

    • C.

      T02g e4g

    • D.

      T22g e2g

    Correct Answer
    B. T32g e1g
    Explanation
    The correct electronic configuration for a d4 system when Δo < P is t32g e1g. In a d4 system, there are four d electrons. The t represents the number of electrons in the t2g orbitals, and the e represents the number of electrons in the eg orbitals. In this configuration, there are three electrons in the t2g orbitals and one electron in the eg orbitals. This configuration is consistent with the given condition that Δo < P.

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  • 11. 

    Which one of the following pairs represents stereoisomerism?

    • A.

      Linkage isomerism and Geometric isomerism

    • B.

      Chain isomerism and rotational isomerism

    • C.

      Optical isomerism and Geometric isomerism

    • D.

      Structural isomerism and Geometric isomerism

    Correct Answer
    C. Optical isomerism and Geometric isomerism
    Explanation
    Optical isomerism and geometric isomerism represent stereoisomerism. Stereoisomerism refers to the phenomenon where two or more compounds have the same molecular formula and connectivity but differ in their spatial arrangement. Optical isomerism occurs when a compound has a chiral center and exists in two non-superimposable mirror image forms (enantiomers). Geometric isomerism, on the other hand, occurs when compounds have restricted rotation around a double bond or a ring and exist as cis and trans isomers. Both optical isomerism and geometric isomerism are examples of stereoisomerism because they involve different spatial arrangements of atoms.

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  • 12. 

    Select the diamagnetic species among the following

    • A.

      [Ni(CN)4]2-

    • B.

      [NiCl4]2-

    • C.

      [CoCl4]2-

    • D.

      [CoF6]2-

    Correct Answer
    A. [Ni(CN)4]2-
    Explanation
    The correct answer is [Ni(CN)4]2-. Diamagnetic species are those that have all of their electrons paired up in their atomic or molecular orbitals, resulting in no net magnetic moment. In [Ni(CN)4]2-, the nickel ion (Ni2+) has a d8 electron configuration with all of its electrons paired up, making it diamagnetic. The other species, [NiCl4]2-, [CoCl4]2-, and [CoF6]2-, have unpaired electrons in their d-orbitals, making them paramagnetic.

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  • 13. 

    The IUPAC name of the coordination compound K3[Fe(CN)6] is

    • A.

      Potassium hexacyanoferrate (II)

    • B.

      Potassium hexacyanoferrate (III)

    • C.

      Potassium hexacyanoiron (II)

    • D.

      Tripotassium hexacyanoiron (II)

    Correct Answer
    B. Potassium hexacyanoferrate (III)
    Explanation
    The IUPAC name of the coordination compound K3[Fe(CN)6] is potassium hexacyanoferrate (III) because the compound contains the Fe(CN)6 complex ion, which is a coordination compound with a central iron atom bonded to six cyanide ligands. The Roman numeral (III) indicates that the iron atom has a +3 oxidation state. The compound also contains three potassium ions (K+), hence the prefix "tripotassium".

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  • 14. 

    Which of the following compounds shows optical isomerism?

    • A.

      [Cu(NH3)4]2+

    • B.

      [ZnCl4]2-

    • C.

      [Cr(C2O4)3]3-

    • D.

      [Co(CN)6]3-

    Correct Answer
    C. [Cr(C2O4)3]3-
    Explanation
    The compound [Cr(C2O4)3]3- shows optical isomerism because it contains a central chromium atom bonded to three different oxalate ligands. These ligands are arranged in a trigonal planar geometry around the chromium atom, resulting in two possible arrangements of the ligands in space. These two arrangements are non-superimposable mirror images of each other, making [Cr(C2O4)3]3- chiral and capable of exhibiting optical isomerism.

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  • 15. 

    [Co(NH3)4(NO2)2]Cl exhibits

    • A.

      Linkage isomerism, geometrical isomerism and optical isomerism

    • B.

      Linkage isomerism, ionization isomerism and optical isomerism

    • C.

      Linkage isomerism, ionization isomerism and geometrical isomerism

    • D.

      Ionization isomerism, geometrical isomerism and optical isomerism

    Correct Answer
    C. Linkage isomerism, ionization isomerism and geometrical isomerism
    Explanation
    The correct answer is linkage isomerism, ionization isomerism, and geometrical isomerism. This is because the compound [Co(NH3)4(NO2)2]Cl can exist in different forms due to the isomerism. Linkage isomerism occurs when the ligands can coordinate to the central metal ion through different atoms. Ionization isomerism occurs when the counterion is exchanged with one of the ligands. Geometrical isomerism occurs when the ligands are arranged differently around the central metal ion, resulting in different spatial arrangements. Therefore, all three forms of isomerism are exhibited by [Co(NH3)4(NO2)2]Cl.

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  • 16. 

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

    • A.

      [Co(CN)6]3-

    • B.

      [Co(C2O4)3]3-

    • C.

      [Co(H2O)6]3+

    • D.

      [Co(NH3)6]3+

    Correct Answer
    A. [Co(CN)6]3-
    Explanation
    The magnitude of Δo (CFSE in octahedral field) is determined by the nature of the ligands surrounding the central metal ion. In general, ligands with a strong field, such as cyanide (CN-), cause a larger splitting of the d-orbitals, resulting in a larger magnitude of Δo. Therefore, in [Co(CN)6]3-, the cyanide ligands create a strong field, leading to the maximum magnitude of Δo among the given coordination entities.

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  • 17. 

    Which of the following complex does not show optical isomerism?

    • A.

      [Co(en)3]3+

    • B.

      [Co(en)2Cl2]+

    • C.

      [Co(NH3)3Cl3]

    • D.

      [Co(en)Cl2(NH3)2]+

    Correct Answer
    C. [Co(NH3)3Cl3]
    Explanation
    The complex [Co(NH3)3Cl3] does not show optical isomerism because it has a plane of symmetry. Optical isomerism occurs when a molecule or complex does not have a plane of symmetry, meaning that it cannot be superimposed on its mirror image. In this case, the three ammonia ligands and three chloride ligands are arranged in such a way that the complex can be divided into two identical halves by a plane passing through the cobalt atom. Therefore, there is no possibility of optical isomerism in [Co(NH3)3Cl3].

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  • 18. 

    Which of the following complex ion is not expected to absorb visible light?

    • A.

      [Ni(H2O)6]2+

    • B.

      [Ni(CN)4]2-

    • C.

      [Cr(NH3)6]3+

    • D.

      [Fe(H2O)6]2+

    Correct Answer
    B. [Ni(CN)4]2-
    Explanation
    [Ni(CN)4]2- is not expected to absorb visible light because it is a square planar complex with a coordination number of 4. The d-orbitals of the nickel ion in this complex are not available for d-d transitions, which are responsible for absorbing visible light. Therefore, [Ni(CN)4]2- does not have any electronic transitions in the visible light range and will not absorb visible light.

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  • 19. 

    Which one of the following is an outer orbital complex and exhibits paramagnetic behavior?

    • A.

      [Cr(NH3)6]3+

    • B.

      [Co(NH3)6]3+

    • C.

      [Ni(NH3)6]2+

    • D.

      [Zn(NH3)6]2+

    Correct Answer
    C. [Ni(NH3)6]2+
    Explanation
    [Ni(NH3)6]2+ is the correct answer because it has a partially filled d-orbital in the Ni2+ ion, which allows it to exhibit paramagnetic behavior. The presence of six ammonia (NH3) ligands surrounding the nickel ion forms an outer orbital complex, where the ligands occupy the outer d-orbitals. This arrangement results in unpaired electrons in the d-orbital, causing the complex to be paramagnetic. In contrast, the other complexes listed do not have unpaired electrons in their d-orbitals and therefore do not exhibit paramagnetic behavior.

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  • 20. 

    A magnetic moment of 1.73 BM will be shown by

    • A.

      [CoCl6]4-

    • B.

      [Cu(NH3)4]2+

    • C.

      [Ni(CN)4]2-

    • D.

      TiCl4

    Correct Answer
    B. [Cu(NH3)4]2+
    Explanation
    The complex ion [Cu(NH3)4]2+ is known as a coordination complex, where the central copper ion is surrounded by four ammonia molecules. Copper has an atomic number of 29, which means it has 29 electrons. In this complex, the copper ion has lost two electrons, resulting in a +2 charge. The presence of the four ammonia molecules around the copper ion creates a magnetic field that aligns the unpaired electrons, giving rise to a magnetic moment of 1.73 BM (Bohr magnetons). This magnetic moment indicates the presence of unpaired electrons and paramagnetic behavior in the complex ion.

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  • 21. 

    What is the magnetic moment of [FeF6]3-

    • A.

      5.92

    • B.

      5.49

    • C.

      2.34

    • D.

      4

    Correct Answer
    A. 5.92
    Explanation
    The magnetic moment of [FeF6]3- is 5.92. This value indicates the strength of the magnetic field generated by the compound. The high magnetic moment suggests that the compound has unpaired electrons, which contribute to its magnetic properties. The presence of unpaired electrons in [FeF6]3- can be attributed to the d-orbital splitting caused by the ligands (F-) surrounding the central iron (Fe) atom. This splitting results in a high spin state, leading to unpaired electrons and a higher magnetic moment.

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  • 22. 

    The IUPAC name for the complex [Co(CO2)(NH3)5]Cl2 is

    • A.

      Pentaamine nitrito-N-cobalt (II) chloride

    • B.

      Pentaamine nitrito-N- cobalt(III) chloride

    • C.

      Nitrito-N-pentaaminecobalt(II) chloride

    • D.

      Nitrito-N-pentaaminecobalt(II) chloride

    Correct Answer
    B. Pentaamine nitrito-N- cobalt(III) chloride
    Explanation
    The correct answer is "pentaamine nitrito-N- cobalt(III) chloride". In the complex, there are five ammonia (NH3) ligands and one nitrito (NO2-) ligand coordinated to the cobalt (Co) ion. The ammonia ligands are named as "pentaamine" and the nitrito ligand is named as "nitrito-N". The Roman numeral III indicates that the cobalt ion has a +3 oxidation state. The chloride ions (Cl-) are outside the coordination sphere and are mentioned in the name as "chloride".

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  • 23. 

    Fac-Mer isomerism is associated with which one of the following complexes?   (M is central metal)

    • A.

      [M(AA)2]

    • B.

      [MA3B3]

    • C.

      [M(AA)3]

    • D.

      [M(ABCD)]

    Correct Answer
    B. [MA3B3]
    Explanation
    Fac-Mer isomerism is associated with the complex [MA3B3]. In this type of isomerism, the ligands can arrange themselves in two different ways around the central metal ion. In the Fac isomer, three identical ligands are arranged in a face-to-face manner, while in the Mer isomer, the three identical ligands are arranged in a meridional (around the same axis) manner. Therefore, the complex [MA3B3] can exhibit Fac-Mer isomerism.

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  • 24. 

    In solid CuSO4.5H2O , copper is coordinated to how many water molecules?

    • A.

      5

    • B.

      4

    • C.

      1

    • D.

      2

    Correct Answer
    B. 4
    Explanation
    In solid CuSO4.5H2O, copper is coordinated to four water molecules. This is because the formula CuSO4.5H2O indicates that there are five water molecules associated with one molecule of CuSO4. However, only four of these water molecules are directly coordinated to the copper ion. The fifth water molecule is not directly bonded to the copper ion, but it is associated with the crystal lattice structure of the compound. Therefore, the correct answer is 4.

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  • 25. 

    Hybridization of Ni in Ni(CO)4

    • A.

      Dsp2

    • B.

      Dsp3

    • C.

      D2sp3

    • D.

      Sp3

    Correct Answer
    D. Sp3
    Explanation
    The correct answer is sp3. In Ni(CO)4, the nickel atom is bonded to four carbon monoxide ligands. The carbon monoxide ligands are considered to be sigma donors and pi acceptors. The bonding in this complex is best described as a tetrahedral arrangement around the nickel atom, with the four ligands occupying the four sp3 hybrid orbitals of the nickel atom. Therefore, the hybridization of the nickel atom in Ni(CO)4 is sp3.

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  • 26. 

    Which among the following will be named as dibromidobis (ethylene diamine) chromium (III) bromide?

    • A.

      [Cr(en)Br2]Br

    • B.

      [Cr(en)3]Br3

    • C.

      [Cr(en)2Br2]Br

    • D.

      [Cr(en)Br4]-

    Correct Answer
    C. [Cr(en)2Br2]Br
    Explanation
    The compound [Cr(en)2Br2]Br will be named as dibromidobis(ethylene diamine) chromium (III) bromide because it contains two ethylene diamine ligands (en) and two bromide ions (Br2) coordinated to the chromium (III) ion. The additional bromide ion (Br) in the compound indicates the overall charge of the complex.

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  • 27. 

    Which one of the following species has tetrahedral shape?

    • A.

      [PdCl4]2-

    • B.

      [Ni(CN)4]2-

    • C.

      [Pd(CN)4]2

    • D.

      [NiCl4]2-

    Correct Answer
    D. [NiCl4]2-
    Explanation
    [NiCl4]2- has a tetrahedral shape because it has four ligands bonded to the central nickel atom. In a tetrahedral shape, the central atom is surrounded by four ligands, arranged in a way that the bond angles between the ligands are approximately 109.5 degrees. In [NiCl4]2-, the four chloride ligands are bonded to the nickel atom, resulting in a tetrahedral shape.

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  • 28. 

    The complex [Mn(CN)6]4- is

    • A.

      High spin complex

    • B.

      Diamagnetic ion

    • C.

      Having magnetic moment 1.73 BM

    • D.

      Outer orbital complex

    Correct Answer
    C. Having magnetic moment 1.73 BM
    Explanation
    The complex [Mn(CN)6]4- is having a magnetic moment of 1.73 BM. This indicates that it is a paramagnetic complex, meaning it has unpaired electrons. Paramagnetic complexes are attracted to a magnetic field due to the presence of unpaired electrons. The value of the magnetic moment suggests the presence of one unpaired electron in the complex.

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  • 29. 

    Select the correct answer from codes (i),(ii),(iii) and (iv)                   (i) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion. (ii) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion. (iii) Assertion is correct, but reason is wrong statement. (iv) Both Assertion and reason are wrong. Assertion(A) : In the complex K2[PtCl6], coordination number of Pt is 6. Reason(R) : In the complex, six coordinate bonds are formed between Pt and chloro compounds.

    • A.

      (i)

    • B.

      (ii)

    • C.

      (iii)

    • D.

      (iv)

    Correct Answer
    A. (i)
    Explanation
    In the complex K2[PtCl6], the coordination number of Pt is indeed 6 because there are six coordinate bonds formed between Pt and chloro compounds. Therefore, both the assertion and the reason are correct statements, and the reason is the correct explanation of the assertion.

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  • 30. 

    Select the correct answer from codes (i),(ii),(iii) and (iv)                     (i) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion. (ii) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion. (iii) Assertion is correct, but reason is wrong statement. (iv) Both Assertion and Reason are wrong. Assertion(A) : [Fe(CN)6]3- is paramagnetic in nature. Reason(R) : [Fe(CN)6]3- is a low spin complex.

    • A.

      (i)

    • B.

      (ii)

    • C.

      (iii)

    • D.

      (iv)

    Correct Answer
    B. (ii)
    Explanation
    Both the assertion and reason are correct statements. The compound [Fe(CN)6]3- is indeed paramagnetic in nature. However, the reason provided, that it is a low spin complex, is not the correct explanation for this assertion. The paramagnetic nature of [Fe(CN)6]3- is due to the presence of unpaired electrons in the d-orbitals of the iron ion, not because it is a low spin complex.

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  • Current Version
  • Mar 20, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Sep 05, 2020
    Quiz Created by
    Dr. Mukesh
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