Please refer to Coordination Compounds HOTs Class 12 Chemistry provided below with Coordination Compounds. All HOTs for Class 12 Chemistry with answers provided below have been designed as per the latest syllabus and examination petter issued by CBSE, NCERT, KVS. Students of Standard 12 Chemistry should learn the solved HOTS for Class 12 Chemistry provided below to gain better marks in examinations.
Coordination Compounds Class 12 Chemistry HOTs
Question. Which of the following is a tridentate ligand?
(a) EDTA4–
(b) (COO)22–
(c) dien
(d) NO2 –
Answer
C
Question. The magnitude of magnetic moment (spin only) of [NiCl4]2– will be
(a) 2.82 B.M.
(b) 0
(c) 1.23 B.M.
(d) 5.64 B.M.
Answer
A
Question. Identify the statement which is not correct?
(a) Coordinate compounds are mainly known for transition metals.
(b) Coordination number and oxidation state of a metal are same.
(c) Tetrahedral complexes form low spin complex.
(d) A ligand donates at least one electron pair to the metal atom to form a bond.
Answer
B
Question. Hexaamminenickel(II) hexanitrocobaltate(III) can be written as
(a) [Ni(NH3)6][Co(NO2)6]
(b) [Ni(NH3)6]3[Co(NO2)6]2
(c) [Ni(NH3)6] [Co(NO2)6]
(d) [Ni(NH3)6(NO2)6]Co
Answer
B
Question. The increasing order of crystal field splitting strength of the given ligands is
(a) NH3 < Cl– < CN– < F– < CO < H2O
(b) F– < Cl– < NH3 < CN– < H2O < CO
(c) Cl– < F– < H2O < NH3 < CN– < CO
(d) CO < CN– < NH3 < H2O < F– < Cl–
Answer
D
Question. A substance appears coloured because
(a) it absorbs light at specific wavelength in the visible part and reflects rest of the wavelengths
(b) ligands absorb different wavelengths of light which give colour to the complex
(c) it absorbs white light and shows different colours at different wavelength
(d) it is diamagnetic in nature.
Answer
A
Question. When one mole of each of the following complexes is treated with excess of AgNO3 which will give maximum amount of AgCl?
(a) [Co(NH3)6]Cl3
(b) [Co(NH3)5Cl]Cl2
(c) [Co(NH3)4Cl2]Cl
(d) [Co(NH3)3Cl3]
Answer
A
Question. What will be the correct order of absorption of wavelength of light in the visible region for the complexes, [Co(NH3)6]3+, [Co(CN)6]3–,[Co(H2O)6]3+ ?
(a) [Co(CN)6]3– > [Co(NH3)6]3+ > [Co(H2O)6]3+
(b) [Co(NH3)6]3+ > [Co(H2O)6]3+ > [Co(CN)6]3–
(c) [Co(H2O)6]3+ > [Co(NH3)6]3+ > [Co(CN)6]3–
(d) [Co(CN)6]3– > [Co(H2O)6]3+ > [Co(NH3)6]3+
Answer
C
Question. Some details of few Nickel complexes are given below:
Complex I : Diamagnetic and square planar
Complex II : Paramagnetic and tetrahedral
Complex III : Diamagnetic and tetrahedral
Complex IV : Paramagnetic and Octahedral
Which is not correct option for the given complexes?
(a) The ligand in complex I is CN– and it has dsp2 hybridisation.
(b) The ligand in complex II is Cl– and it has sp3 hybridisation.
(c) The ligand in complex IV is H2O and it has d2sp3 hybridisation.
(d) The ligand in complex III is CO and it has sp3 hybridiation.
Answer
C
Question. Arrange the following complexes in increasing order of conductivity of their solutions.
(i) [Co(NH3)3Cl3]
(ii) [Co(NH3)4Cl2]Cl
(iii) [Co(NH3)6]Cl3
(iv) [Co(NH3)5Cl]Cl2
(a) (i) < (ii) < (iv) < (iii)
(b) (ii) < (i) < (iii) < (iv)
(c) (i) < (iii) < (ii) < (iv)
(d) (iv) < (i) < (ii) < (iii)
Answer
A
Question. Which of the following complexes will show maximum paramagnetism?
(a) 3d4
(b) 3d5
(c) 3d6
(d) 3d7
Answer
B
Question. Among the following, which are ambidentate ligands?
(i) SCN–
(ii) NO3–
(iii) NO2–
(iv) C2O42–
(a) (i) and (iii)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv)
Answer
A
Question. The name of [Co(NH3)5NO2]Cl2 will be
(a) pentaamminonitrocobalt(II) chloride
(b) pentaamminenitrochloridecobaltate(III)
(c) pentaamminenitrito-N-cobalt(III) chloride
(d) pentanitrosoamminechlorocobaltate(III).
Answer
C
Question. The formula of the complex diamminechlorido (ethylenediamine)nitroplatinum(IV) chloride is
(a) [Pt(NH3)2Cl(en)NO2]Cl2
(b) Pt[Pt(NH3)2(en)Cl2NO2]
(c) Pt[(NH3)2(en)NO2]Cl2
(d) Pt[(NH3)2(en)NO2Cl2]
Answer
A
Question. Correct formula of tetraamminechloridonitroplatinum( IV) sulphate can be written as
(a) [Pt(NH3)4(ONO)Cl]SO4
(b) [Pt(NH3)4Cl2NO2]2SO4
(c) [Pt(NH3)4(NO2)Cl]SO4
(d) [PtCl(ONO)NH3(SO4)]
Answer
C
Question. Copper sulphate dissolves in ammonia due to the formation of
(a) Cu2O
(b) [Cu(NH3)4]SO4
(c) [Cu(NH3)4]OH
(d) [Cu(H2O)4]SO4
Answer
B
Question. Which of the following is correct?
(a) Valence bond theory explains the colour of the coordination compounds.
(b) [NiCl4]2– is diamagnetic in nature.
(c) EDTA is a chelating ligand.
(d) A bidentate ligand can have four coordination sites.
Answer
C
Question. According to Werner’s theory of coordination compounds,
(a) primary valency is ionisable
(b) secondary valency is ionisable
(c) primary and secondary valencies are ionisable
(d) neither primary nor secondary valency is ionisable.
Answer
A
Question. Which of the following shall form an octahedral complex?
(a) d 4(low spin)
(b) d 8(high spin)
(c) d 6(low spin)
(d) None of these
Answer
D
Question. The number of unpaired electrons in [Ni(CO)4] is
(a) one
(b) two
(c) three
(d) zero
Answer
D
Question. Which of the following is not a neutral ligand?
(a) H2O
(b) NH3
(c) ONO
(d) CO
Answer
C
Question. In which of the following compounds, the transition metal is in oxidation state of zero?
(a) [Fe(H2O)3(OH)3]
(b) [Ni(CO)4]
(c) [Fe(H2O)6]SO4
(d) [Co(NH3)6]Cl3
Answer
B
Case Based MCQs
Case I : Read the passage given below and answer the following questions.
The extent to which the set of d-orbitals is split in the electrostatic field produced by the ligands depends upon several factors. Two of the most important factors are the nature of the ligands and the nature of the metal ion. In order to see this effect, consider the complex ion [Ti(H2O)6]3+. The Ti3+ ion has a single electron in the 3d-orbital, and we refer to it as d1 ion. In the octahedral field generated by six H2O molecules, the single electron will reside in one of the three degerate t2g orbitals. Under spectral excitation, the electron is promoted to an e.g., orbital giving rise to on absorption spectrum consisting of a single peak that can be represented as shown :
The maximum absorption in the spectrum for [Ti(H2O)6]3+ occurs at 20,300 cm–1 which is equal to 243 kJ mol–1.This gives the value of Δo directly, but only in case of simple d1 ions. Other complexes containing the Ti3+ ion (e.g., [Ti(NH3)6]3+, [TiF6]3–, etc.) could also be prepared and spectra obtained for these complexes. If this was done, it would be observe that the absorption maximum occurs at a different energy for each complex. Because the maximum corresponds to the splitting of d-orbitals, the ligands could be ranked in terms of their ability to cause the splitting of orbital energies. Such a ranking is known as the spectrochemical series and for several common ligands the following order of decreasing energy is observed, CO > CN– >NO2– > en > NH3 > H2O > OH– > F–, Cl– > Br–. In general, the splitting in tetrahedral fields is only about half as large as that in octahedral fields.
Question. The visible spectra of salts of the following complexes are measured in aqueous solution for which complex would the spectrum contain absorption with highest Emax values?
(a) [Co(H2O)6]2+
(b) [Co(H2O)6]3+
(c) [Co(NH3)6]3+
(d) [Co(CN)6]3–
Answer
D
Question. Which of the following ligands has lowest Do value?
(a) CN–
(b) CO
(c) F–
(d) NH3
Answer
C
Question. Which of the following statements is incorrect for complex [Ti(H2O)6]3+ ?
(a) [Ti(H2O)6]3+ is violet in colour.
(b) [Ti(H2O)6]3+ is an octahedral complex.
(c) Exitation of electron in [Ti(H2O)6]3+ occurs
(d) The colour of the complex [Ti(H2O)6]3+ arises due to d-d and f-f transition of the electron.
Answer
D
Case II : Read the passage given below and answer the following questions.
Werner, a Swiss chemist in 1892 prepared and characterised a large number of coordination compounds and studied their physical and chemical behaviour. He proposed that, in coordination compounds, metals possess two types of valencies, viz. primary valencies, which are normally ionisable and secondary valencies which are non-ionisable. In a series of compounds of cobalt (III) chloride with ammonia, it was found that some of the chloride ions could be precipitated as AgCl on adding excess of AgNO3 solution in cold, but some remained in solution. The number of ions furnished by a complex in a solution can be determined by precipitation reactions. The measurement of molar conductance of solutions of coordination compounds helps to estimate the number of ions furnished by the compound in solution.
In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices on the basis of the above passage.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
The following questions are multiple choice questions. Choose the most appropriate answer :
Question. Assertion : The complex [Co(NH3)4Cl2]Cl gives precipitate corresponding to 2 mol of AgCl with AgNO3 solution.
Reason : It ionises as [Co(NH3)4Cl2]+ + Cl–.
Answer
D
Question. Assertion : The complex [Co(NH3)3Cl3] does not give precipitate with silver nitrate solution.
Reason : The given complex is non-ionisable.
Answer
A
Question. Assertion : CoCl3⋅3NH3 is not conducting while CoCl3⋅5NH3 is conducting.
Reason : The complex of CoCl3⋅3NH3 is [CoCl3(NH3)3] while that of CoCl3⋅5NH3 is [CoCl(NH3)5]Cl2.
Answer
A
Question. Assertion : CoCl3⋅4NH3 gives 1 mol of AgCl on reacting with AgNO3, its secondary valency is 6.
Reason : Secondary valency corresponds to coordination number.
Answer
B
Question. Assertion : 1 mol of [CrCl2(H2O)4]Cl ⋅ 2H2O will give 1 mol of AgCl on treating with AgNO3.
Reason : Cl– ions satisfying secondary valanceis will not be precipitated.
Answer
A
Case III : Read the passage given below and answer the following questions.
Valence bond theory considers the bonding between the metal ion and the ligands as purely covalent. On the other hand, crystal field theory considers the metal-ligand bond to be ionic arising from electrostatic interaction between the metal ion and the ligands. In coordination compounds, the interaction between the ligand and the metal ion causes the five d-orbitals to split-up. This is called crystal field splitting and the energy difference between the two sets of energy level is called crystal field splitting energy. The crystal field splitting energy (Do) depends upon the nature of the ligand. The actual configuration of complexes is divided by the relative values of Do and P (pairing energy).
If Δo < P, then complex will be high spin.
If Δo > P, then complex will be low spin.
Question. Electronic configuration of d-orbitals in [Ti(H2O)6]3+ ion in an octahedral crystal field is
Answer
A
Question. On the basis of crystal field theory, the electronic configuration of d4 in two situations :
(a) Δo > P and (b) Δo < P are
Answer
A
Question. The crystal field splitting energy for octahedral (Do) and tetrahedral (Dt) complex is related as
Answer
B
Question. Using crystal field theory, calculate magnetic moment of central metal ion of [FeF6]4–.
(a) 1.79 B.M.
(b) 2.83 B.M.
(c) 3.85 B.M.
(d) 4.9 B.M.
Answer
D
Case IV : Read the passage given below and answer the following questions.
Coordination compounds are formulated and named according to the IUPAC system.
Few rules for naming coordination compounds are :
(I) In ionic complex, the cation is named first and then the anion.
(II) In the coordination entity, the ligands are named first and then the central metal ion.
(III)When more than one type of ligands are present, they are named in alphabetical order of preference without any consideration of charge.
Question. The IUPAC name of [Ni(CO)4] is
(a) tetracarbonylnickel(II)
(b) tetracarbonylnickel(0)
(c) tetracarbonylnickelate(II)
(d) tetracarbonylnickelate(0).
Answer
B
Question. Correct formula of pentaamminenitrito-Ocobalt( III) sulphate is
(a) [Co(NO2)(NH3)5]SO4
(b) [Co(ONO)(NH3)5]SO4
(c) [Co(NO2)(NH3)4](SO4)2
(d) [Co(ONO)(NH3)4](SO4)2
Answer
B
Question. As per IUPAC nomenclature, the name of the complex [Co(H2O)4(NH3)2]Cl3 is
(a) tetraaquadiamminecobalt(II) chloride
(b) tetraaquadiamminecobalt(III) chloride
(c) diamminetetraaquacobalt(II) chloride
(d) diamminetetraaquacobalt(III) chloride.
Answer
D
Question. The IUPAC name of the complex [Pt(NH3)3Br(NO2)Cl]Cl is
(a) triamminechlorobromonitroplatinum(IV) chloride
(b) triamminebromonitrochloroplatinum(IV) chloride
(c) triamminebromidochloridonitroplatinum (IV) chloride
(d) triamminenitrochlorobromoplatinum(IV) chloride.
Answer
C
Question. Which of the following represents correct formula of dichloridobis(ethane-1, 2-diamine) cobalt(III) ion?
(a) [CoCl2(en)]2+
(b) [CoCl2(en)2]2+
(c) [CoCl2(en)]+
(d) [CoCl2(en)2]+
Answer
D
Case V : Read the passage given below and answer the following questions.
To explain bonding in coordination compounds various theories were proposed. One of the important theory was valence bond theory. According to that, the central metal ion in the complex makes available a number of empty orbitals for the formation of coordination bonds with suitable ligands. The appropriate atomic orbitals of the metal hybridise to give a set of equivalent orbitals of definite geometry.
The d-orbitals involved in the hybridisation may be either inner d-orbitals i.e., (n – 1)d or outer d-orbitals i.e., nd.
For example, Co3+ forms both inner orbital and outer orbital complexes, with ammonia it forms [Co(NH3)6]3+ and with fluorine it forms [CoF6]3– complex ion.
Question. [Cr(H2O)6]Cl3 (at. no. of Cr = 24) has a magnetic moment of 3.83 B.M. The correct distribution of 3d-electrons in the central metal of the complex is
Answer
B
Question. Which of the following is an inner orbital or low spin complex?
(a) [Ni(H2O)6]3+
(b) [FeF6]3–
(c) [Co(CN)6]3–
(d) [NiCl4]2–
Answer
C
Question. Which of the following is true for [Co(NH3)6]3+ ?
(a) It is an octahedral, dimagnetic and outer orbital complex.
(b) It is an octahedral, paramagnetic and outer orbital complex.
(c) It is an octahedral, paramagnetic and inner orbital complex.
(d) It is an octahedral, dimagnetic and inner orbital complex.
Answer
D
Question. Which of the following is not true for [CoF6]3–?
(a) It is paramagnetic.
(b) It has coordination number of 6.
(c) It is outer orbital complex.
(d) It involves d2sp3 hybridisation.
Answer
D
Question. The paramagnetism of [CoF6]3– is due to
(a) 3 electrons
(b) 4 electrons
(c) 2 electrons
(d) 1 electron.
Answer
B
Assertion & Reasoning Based MCQs
For question, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
Question. 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+.
Answer
A
Question. Assertion : All the octahedral complexes of Ni2+ must be outer orbital complexes.
Reason : Outer orbital octahedral complexes are given by weak ligands.
Answer
B
Question. Assertion : Low spin complexes have less number of unpaired electrons.
Reason : [FeF6]3– is a low spin complex.
Answer
C
Question. Assertion : The second and third transition series elements have lesser tendency to form low spin complex as compared to the first transition series.
Reason : The CFSE (Do) is more for 4d and 5d.
Answer
D
Question. Assertion : The [Ni(en)3]Cl2 (en = ethylenediamine) has lower stability than [Ni(NH3)6]Cl2.
Reason : In [Ni(en)3]Cl2 the geometry of Ni is octahedral.
Answer
D
Question. Assertion : Thiocarbonyl is a neutral ligand.
Reason : Thiocarbonyl has three donor atoms but behaves as a bidentate ligand.
Answer
C
Question. Assertion : Cu(OH)2 is soluble in NH4OH but not in NaOH.
Reason : Cu(OH)2 forms a soluble complex with NH3.
Answer
A
Question. Assertion : The ligand N3– is named as nitride.
Reason : N3– is derived from HN3.
Answer
D
Question.Assertion : Ethylenediaminetetraacetate ion forms an octahedral complex with the metal ion.
Reason : It has six donor atoms which coordinate simultaneously to the metal ion.
Answer
A
Question. Assertion : [Fe(CN)6]3– has d2sp3 type hybridisation.
Reason : [Fe(CN)6]3– ion shows magnetic moment corresponding to two unpaired electrons.
Answer
C
Question. Assertion : [CrCl2(H2O)4]NO3 is dichlorotetraaquachromium(III) nitrate.
Reason : In writing the name of the complex cation is written first followed by the anion.
Answer
D
Question. Assertion : [Pt(NH3)2Cl2] is square planar.
Reason : The oxidation state of platinum is + 2.
Answer
B
Question. Assertion : [Al(NH3)6]3+ does not exist in aqueous solution.
Reason : NH3 is a neutral ligand.
Answer
B
Question. 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.
Answer
B