CBSE Class 12 Chemistry Coordination Compounds Worksheet

ONE MARK QUESTIONS:

Q.1Write the formula forTetraamineaquachloridocobalt(III) chloride

Q.2 Write the IUPAC name of [Co (NH₃)₄ Br₂]₂ [Zn Cl₄]

Q.3 Which of these cannot act as ligand and why: NH₃, H₂O, CO, CH₄. Give reason?

Q.4 How many EDTA (lethylendiamine tetra acetic acid) molecules are required to make an octahedral complex with a Ca²⁺ ion.

Q.5 Why tetrahedral complexes do not exhibit geometrical isomerism ?

Q.6 What is the hybridisation of central metal ion and shape of Wilkinson’s catalyst ?

Q.7 Specify the oxidation numbers of the metals in the following coordination entities:

(i) [Co(H₂O)(CN)(en)₂]²⁺ and (ii) [CoBr₂(en)₂]⁺

Q.8 Draw the structure of optical isomers of [Co(en)₃]³⁺.

Q.9 Name the types of isomerism exhibited by [Co(NH₃)₅(NO₂)](NO₃)₂

Q.10 Write the formula of Amminebromidochloridonitrito-N-platinate(II) ion

Q.11 which one is more stable complex among(i) [Fe(H₂O)₆]³⁺ (ii) [Fe(NH₃)₆]³⁺(iii) [Fe(C₂O₄)₃]³⁻

Q.12 How many ions are produced from the complex Co(NH₃)₆Cl₂ in solution?

Q.13 What do you meant by degenerate d-orbitals?

Q.14 Out of the following two coordination entities which is chiral (optically active)? (a) cis-[CrCl2 (ox)₂]^3- 

(b) trans-[CrCl2 (ox)₂]^3-

Q.15 The spin only magnetic moment of [MnBr4]^2- is 5.9 BM. Predict the geometry of the complex ion?

1 mark questions

Question. Explain coordination entity with example.
Answer : it constitute a central metal atom or ions bonded to a fixed number of molecules or ions (ligands) .eg. [Co(NH₃)₃Cl₃].

Question. What do you understand by coordination compounds?
Answer : coordination compounds are the compounds which contains complex ions. These compounds contain a central metal atom or cation which is attached with a fixed number of anions or molecules called ligands through coordinate bonds. eg. [Co(NH₃)₃Cl₃]

Question. What is coordination number?
Answer : the coordination number of a metal ion in a complex may be defined as the total number of ligand donor atoms to which the metal ion is directly bonded. Eg. In the complex ion [Co(NH₃)₆]³⁺ has 6 coordination number.

Question. Name the different types of isomerisms in coordination compounds.
Answer : structural isomerism and stereoisomerism.

Question. Draw the structure of xenon difluoride.
Answer : structure :trigonalbipyramidal
Shape: linear

6. What is spectrochemical series?
Answer : the series in which ligands are arranged in the order of increasing field strength is called spectrochemical series. The order is :
I^- < Br^- < SCN^- < Cl^- < S^2 - < F^- < OH^- < C₂O₄^2- < H₂O < NCS^- < EDTA₄^-< NH³ < en < CN^- < CO

Question. What do you understand by denticity of a ligand?
Answer : the number of coordinating groups present in ligand is called denticity of ligand.
Eg.Bidentateligand ethane-1,2-diamine has 2 donor nitrogen atoms which can link to central metal atom.

Question. Why is CO a stronger ligand than Cl^-?
Answer : because CO has π bonds.

Question. Why are low spin tetrahedral complexes not formed?
Answer : because for tetrahedral complexes, the crystal field stabilisation energy is lower than pairing energy.

Question. Square planar complexes with coordination number 4 exhibit geometrical isomerism whereas tetrahedral complexes do not. Why?
Answer : tetrahedral complexes do not show geometrical isomerism because the relative positions of the ligands attached to the central metal atom are same with respect to each other.

Question. What are crystal fields?
Answer : the ligands has around them negatively charged field because of which they are called crystal fields.

Question. What is meant by chelate effect? Give an example .
Answer : when a didentate or polydentate ligand contains donor atoms positioned in such a way that when they coordinate with the central metal atom, a 5 or 6 membered ring is formed , the effect is called chelate effect. Eg. [PtCl₂ ( en)]

Question. What do you understand by ambidentate ligand?
Answer : a ligand which contains two donor atoms but only one of them forms a coordinate bond at a time with central metal atom or ion is called an ambidentate ligand. Eg.nitrito-N and nitrito-O.

Question. What is the difference between homoleptic and heteroleptic complexes?
Answer : in homoleptic complexes the central metal atom is bound to only one kind of donor groups whereas in heteroleptic complexes the central metal atom is bound to more than one type of donor atoms.

Question. Give one limitation for crystal field theory.
Answer : i) as the ligands are considered as point charges, the anionic ligands should exert greater splitting effect. However the anionic ligands are found at the low end of the spectrochemical series.
ii) it does not take into account the covalent character of metal ligand bond. 

Question. How many ions are produced from the complex: [Co(NH₃)₆]Cl₂
Answer : 3 ions

Question. The oxidation number of cobalt in K[Co(CO)₄]
Answer : -1

Question. Which compound is used to estimate the hardness of water volumetrically?
Answer : EDTA

Question. Magnetic moment of [MnCl₄]^2- is 5.92B.M explain with reason.
Answer : the magnetic moment of 5.9 B.M. corresponds to the presence of 5 unpaired electrons in the dorbitals of Mn²⁺ ion. As a result the hybridisation involved is sp3 rather than dsp2. Thus tetrahedral structure of [MnCl₄]^2- complex will show 5.92 B.M magnetic moment value.

Question. How many donor atoms are present in EDTA ligand?
Answer : 6

2 marks questions

Question. Give the electronic configuration of the following complexes on the basis of crystal field splitting theory.
i) [CoF₆]^3-
ii) [Fe(CN)₆]^4-
Answer : i) Co³⁺ (d⁶) t_2g⁴e_g²
iii) Fe²⁺ d⁶t₂g⁶e_g⁰

Question. Explain the following with examples:
i) Linkage isomerism
ii) Outer orbital complex
Answer : i) this type of isomerism arises due to the presence of ambidentate ligand in a coordination compound. Eg. [Co(NH₃)₅NO₂]Cl₂ and [Co(NH₃)₅ONO]Cl₂
iii) When ns, np and nd orbitals are involved in hybridisation , outer orbital complex is formed. Eg. [CoF₆]^2- in which cobalt is sp3d2 hybridised.

Question. i) Low spin octahedral complexes of nickel are not found . Explain why?
ii) theπ complexes are known for transition elements only.explain.
Answer : i) nickel in its atomic or ionic state cannot afford 2 vacant 3d orbitals and hence d²sp³ hybridisation is not possible.
ii) transition metals have vacant d orbitals in their atoms or ions into which the electron pairs can be donated by ligands containing πelectrons.eg. benzene, ethylene etc. thus dπ-pπ bonding is possible.

Question. How would you account for the following:
i) [Ti(H₂O)₆]³⁺ is coloured while [Sc(H₂O)₆]³⁺ is colourless.
ii) [Ni(CO)₄] possess tetrahedral geometry whereas [Ni(CN)₄]^2- is square planar.
Answer : i) due to the presence of 1 electron in 3d subshell in [Ti(H₂O)₆]³⁺ complex d-d transition takes place by the absorption of visible light. Hence the complex appears coloured. On the other hand, [Sc(H₂O)₆]³⁺ does not possess any unpaired electron .Hence d-d transition is not possible (which is responsible for colour) in this complex is not possible, therefore it is colourless.
ii) Ni in [Ni(CO)₄] is sp³ hybridised. Hence it is tetrahedral. Whereas for [Ni(CN)₄]^2- is dsp²
hybridised hence it has square planar geometry.

Question. State reasons for each of the following:
i) All the P—Cl bonds in PCl₅ molecule are not equivalent.
ii) S has greater tendency for catenation than O.
Answer : i) in P Cl₅ the 2 axial bonds are longer than 3 equatorial bonds. This is due to the fact that the axial bond pairs suffers more repulsion as compared to equatorial bond pairs.
ii) The property of catenation depends upon the bond strength of the element. As S—S bond is much stronger (213kJ / mole) than O—O bond (138 kJ/mole), S has greater tendency for
catenation than O.

Question. Give the stereochemistry and the magnetic behaviour of the following complexes:
i) [Co(NH₃)₅Cl]Cl₂
ii) K₂[Ni(CN)₄]
Answer : i) d²sp³ hybridisation, structure and shape = octahedral Magnetic behaviour- diamagnetic
ii) dsp² hybridisation, structure and shape = square planar magnetic behaviour- diamagnetic

Question. Draw the structures of isomers if any and write the names of the following complexes:
i) [Cr(NH₃)₄Cl₂]⁺
ii) [Co(en)₃]³⁺
Answer : i) tetraamminedichloridochromium(III) ion
ii) tris(ethane-1,2-diammine)cobalt(III)ion

Question. State reasons for each of the following:
i) The N—O bond in NO₂^- is shorter than the N—O bond in NO^3-
ii) SF6 is kinetically an inert substance.
Answer : i) this is because the N—O bone in NO₂^- is an average of a single bond and a double bond whereas N—O bond in NO₃^- is an average of 2 single bonds and a double bond.
iii) In SF₆ the S atom is sterically protected by 6 fluorine atoms and does not allow water molecules to attack the S atom. Further F atoms does not contain d orbitals to accept the electrons denoted by water molecules. Due to these reasons , SF₆ is kinetically an inert substance.

Question. Hydrated copper sulphate is blue in colour whereas anhydrous copper sulphate is colourless. Why?
Answer : because water molecules act as ligands which splits the d orbital of the Cu²⁺ metal ion. This result in d-d transition in which t_2g⁶e_g³ excited to t_2g⁵e_g⁴ and this impart blue colour to the crystal.
Whereas when we talk about anhydrous copper sulphate it does not contain any ligand which
could split the d orbital to have CFSE effect.

Question. Calculate the magnetic moment of the metal ions present in the following complexes:
i) [Cu(NH₃)₄]SO₄
ii) [Ni(CN)₄]^2-
Answer : i) electronicconfig. t_2g⁶e_g³, n=1, μs= √n(n+2) = 1.732 B.M
ii) electronicconfig. t₂g⁶e_g² , n = 2, μ_s= √n(n+2) = 2.828 B.M

3 marks questions

Question. (a) What is a ligand? Give an example of a bidentate ligand.
(b) explain as to how the 2 complexes of nickel,[Ni(CN)₄]^2- and Ni(CO)₄ have different structures but donot differ in their magnetic behaviour.( Ni = 28)
Answer : (a) the ion , atom or molecule bound to the central atom or ion in the coordination entity is called ligand. A ligand should have lone pair of electrons in their valence orbital which can be donated to central metal atom or ion.
Eg.Bidentate ligand-  H₂N-H-CH₂-NH₂    ethylenediammine
(b) dsp², square planar, diamagnetic (n=0)
Sp³ hybridisation , tetrahedral geometry, diamagnetic (n=0)

Question. Nomenclate the following complexes:
i) [Co(NH₃)₅(CO₃)]Cl
ii) [COCl₂(en)₂]Cl
iii) Fe₄[Fe(CN) ₆]
Answer : i) pentaam mine carbon a to cobalt (III)chloride
ii) dichloridobis(ethane-1,2-diamine)cobalt(III)chloride
iii) iron(III) hexacyanidoferrate(II)

Question. (a) why do compounds with similar geometry have different magnetic moment?
(b) what is the relationship between the observed colour and wavelength of light absorbed by the complex?
Answer : (a) it is due to the presence of weak and strong ligands in complexes, if CFSE is high the complex will show low value of magnetic moment and if it is low the value of magnetic moment is high. Eg. [CoF₆]^3- and [Co(NH₃)₆]³⁺ , the former is paramagnetic and the latter is diamagnetic.
(b) higher the CFS lower will be the wavelength of absorbed light. Colour of the complex is obtained from the wavelength of the leftover light.

Question. Explain the following terms giving a suitable example.
(a) ambident ligand
(b) denticity of a ligand
(c) crystal field splitting in an octahedral field
Answer : (a) Aligand which contains two donor atoms but only one of them forms a coordinate bond at a time with central metal atom or ion is called an ambidentate ligand. Eg.nitrito-N and nitrito-O.
(b) The number of coordinating groups present in ligand is called denticity of ligand. Eg.Bidentateligand ethane-1,2-diamine has 2 donor nitrogen atoms which can link to central metal atom.
(c) the splitting of the degenerated d orbital into 3 orbitals of lower energy t₂g and 2 orbitals of higher energy eg due to presence of a ligand in a octahedral crystal field is known as crystal field splitting in an octahedral complex.

Question. (a) Copper sulphate pentahydrate is blue in colour while anhydrous copper sulphate is colourless. Why?
(b) Sulphur has greater tendency for catenation than oxygen.Why?
Answer : (a) because water molecules act as ligands which splits the d orbital of the Cu²⁺ metal ion. This result in d-d transition in which t_2g⁶e_g³ excited to t_2g⁵e_g⁴ and this impart blue colour to the crystal. Whereas when we talk about anhydrous copper sulphate it does not contain any ligand which could split the d orbital to have CFSE effect.
(b) The property of catenation depends upon the bond strength of the element. As S—S bond is much stronger (213kJ / mole) than O—O bond (138 kJ/mole), S has greater tendency for catenation than O.

Question. how would you account for the following:
(i) [Ti(H₂O)₆]³⁺is coloured while [Sc(H₂O)₆]³⁺ is colourless .
(II) [ Fe(CN)₆]^3- is weakly paramagnetic while [ Fe(CN)₆]^4- is diamagnetic.
(III) Ni(CO)₄ possess tetrahedral geometry while [Ni (CN)₄]^2- is square planar.
Answer : i) due to the presence of 1 electron in 3d subshell in [Ti(H₂O)₆]³⁺ complex d-d transition takes place by the absorption of visible light. Hence the complex appears coloured. On the other hand, [Sc(H₂O)₆]³⁺ does not possess any unpaired electron .Hence d-d transition is not possible (which is responsible for colour) in this complex is not possible, therefore it is colourless.
(ii) paramagnetism is attributed to the presence o f unpaired electrons. Greater the number of unpaired electron greater is the paramagnetism. Due to the presence of one electron in the 3d subshell in [ Fe(CN)₆]^3- it is weakly paramagnetic. On the other hand [ Fe(CN)₆]^4- is diamagnetic because all electrons are paired.
iii) Ni in [Ni(CO)₄] is sp³ hybridised. Hence it is tetrahedral. Whereas for [Ni(CN)₄]^2- is dsp² hybridised hence it has square planar geometry.

Question. Explain the following ::
(i) low spin octahedral complexes of Ni are not known.
(ii) The pi – complexes are known for the transition elements only.
(iii) CO is a stronger ligand than NH3 for many metals
Answer : i) nickel in its atomic or ionic state cannot afford 2 vacant 3d orbitals and hence d²sp³ hybridisation is not possible.
(ii) transition metals have vacant d orbitals in their atoms or ions into which the electron pairs can be donated by ligands containing πelectrons.eg. benzene, ethylene etc. thus dπ-pπ bonding is possible.
(iii) because in case of CO back bonding takes place in which the central atom uses its filled d orbitals with empty anti bonding π*molecular orbital of CO.

Question. What is meant by stability of a coordination compounds in solutions? State the factors which govern the stability of complexes.
Answer : the stability of a complex in solution refers to the degree of association between the two species involved in the state of equilibrium. The magnitude of the equilibrium constant for the association expresses the stability .
M + 4 L → ML₄
K = [ML4]/[M][L]₄
Factors on which stability of complex depends (i) charge on central metal ion (ii) nature of the metal ion (iii) basic nature of the ligand (iv) presence of the chelate ring (v) effect of multidentate cyclic ligand .

Question. draw structures of geometrical isomers of the following complexes:
(a) [Fe(NH₃)₂(CN)₄]^-
(b) [CrCl₂(ox)₂]^3-
(c) [Co(en)₃]Cl₃

Question. write the state of hybridisation, the shape and the magnetic behaviour of the following complexes:
(i) [Co(en)₃]Cl₃
(II) K₂[Ni(CN)₄]
(III) [Fe(CN)₆]^3-

5 marks questions

1. Draw the structures of the following molecules:
(a) [Fe(NH₃)₂(CN)₄]^-
(b) [CrCl₂(ox)₂]^3-
(c) [Co(en)₃]Cl₃
(d) [Co(en)³]Cl³(e)[Fe(CN)⁶]^3-

2. What is crystal field theory for octahedral complexes? Also write the limitations of this theory.

3. Write the state of hybridisation the shape and the magnetic behaviour of the following complex entities:
(i) [Cr(NH₃)₄Cl₂]Cl
(ii) [Co(en)₃]Cl₃
(iii) K₂[NiCl₄]
(iv) [Fe(H₂O)₆]²⁺
(v) [NiCl₄]^2-

4. Using valence bond theory explain the following questions in relation to [Co(NH₃)₆]^3+.
(i) Nomenclature
(ii) Type of hybridisation
(iii) Inner or outer orbital complex
(iv) Magnetic behaviour
(v) Spin only magnetic moment

5. Compare the following complexes with respect to structural shape of units, magnetic behaviour and hybrid orbitals involved in units: [Co(NH₃)₆]₃₊, [Cr(NH₃)₆]³⁺,[ Ni(CO)₄]

ONE MARK QUESTIONS

1.What are ambident ligands? Explain giving example.

2.Write the IUPAC name of the ionization isomer of [Pt(NH₃)₃Br] Cl

3.Write the formula of CrCl₃.5H₂O that furnishes 2 moles of Chloride ions per mole of salt.

TWO MARK QUESTIONS

1.i) Write down the IUPAC name of the following complex : [Pt(NH₃)(H₂O)Cl₂]

(ii) Write the formula for the following complex : tris(ethane-1,2-diamine)chromium(III) chloride (2015)

2.Write IUPAC names of the following:

a) [Co (NH₃)₅ Cl ] Cl₂ b) [Cr(NH₃)₆]³⁺

THREE MARKQUESTIONS

1.a) What type of isomerism is shown by [Co (NH₃)₅ONO]Cl2 ? 

b) On the basis of crystal field theory, write the electronic configuration for d4 ion if Δo < P. 

c) Write the hybridization and shape of [Fe (CN)₆]^3-. 

(Atomic number of Fe = 26) (2015)

2.Give the formula of the compound

a) Nitrito – N-pentaamminecobalt(III)nitrate

b) Potassium hexacyanocobaltate(III)

c) Hexaammineplatinum(IV)chloride

3. Account for the following
a) [Fe (CN)₆]^3- is weakly paramagnetic while [Fe(CN)₆]^4- is diamagnetic.
b) [Ni (CO)₄] is tetrahedral while [Ni(CN)₄]^2- is square planar.
c) [Ti(H₂O)₆]³⁺ is coloured while [Sc(H₂O)₆³⁺ is colourless.

4. a) For the complex [Fe(CO)₅], write the hybridization, magnetic character and spin of the complex. (At. Number : Fe = 26 ) b) Define crystal field splitting energy.

5. Describe the state of hybridization, the shape and magnetic behavior of the following complexes:
a) [Cr(H₂O)₂(C₂O₄)₂]^─
b) [Co(NH₃)₂(en)₂]³⁺
(At no’s: Cr = 24 , Co = 27)

FIVE MARKQUESTIONS

1. a) What is a ligand? Give an example of a bidentate ligand.
b) Explain as to how the two complexes of nickel, [Ni(CN)₄]^2─ and [Ni(CO)₄], have different structures but do not differ in their magnetic behavior. (At no: of Ni = 28)
c) Discuss the nature of bonding in metal carbonyls.

VALUE BASED QUESTION

1. Swetha’s father is working in a battery factory. These people are engaged in recycling lead acid batteries. Since few days Swetha’s father is feeling sick. Swetha has taken his father to a doctor. Doctor found him suffering from lead poisoning. Swetha then goes to the factory and asks the seniors to take necessary steps for the health of the workers of the factory.
a) Which coordination compound is used for the treatment of lead poisoning?
b) How does it work in our body?
c) Write the value shown by Swetha in the above paragraph.
d) Write down two examples of coordination compounds which are of great importance to biological systems.

Question. Cobalt(III)chloride forms several octahedral complexes with ammonia .Which of the following will not give test for chloride ions with silver nitrate at 2500C ?
(a) CoCl₃.4NH₃
(b) CoCl₃.5NH₃
(c) CoCl₃.6NH₃
(d) CoCl₃.3NH₃
Answer. D

Question. Which one of the following is an outer orbital complex and exhibits paramagnetic behaviour :
(a) [Cr (NH₃)₆]³⁺
(b) [Co (NH₃)₆]³⁺
(c) [Ni (NH₃)₆]²⁺
(d) [Zn((NH₃)₆]²⁺
Answer. C

Question. The oxidation number of Cobalt in K[Co (CO)₄] is
(a) +1
(b) +3
(c) -1
(d) -3
Answer. C

Question. Fac-mer isomerism is associated with which one of the following complexes?
(a) [M(AA)₂]
(b) [MA₃B₃]
(c) [M(AA)₃]
(d) [MA₄B₂]
Answer. B

Question. Which type of isomerism is shown by the complex compounds [Co (NH₃)5Br]SO₄ and [Co (NH₃)₅SO₄]Br
(a) Ionisation
(b) Optical
(c) Linkage
(d) Coordination
Answer. A

Question. Amongst the following ions, which one is highly paramagnetic?
(a) [Cr (H₂O)₆]³⁺
(b) [Fe (H₂O)₆]²⁺
(c) [Cu (H₂O)₆]²⁺
(d) [Zn (H₂O)₆]²⁺
Answer. B

Question. The geometry of [Ni(CN)₄]^2- and [NiCl₄]^2- are
(a) Both square planar
(b) Both tetrahedral
(c) Tetrahedral and square planar respectively
(d) Square planar and tetrahedral respectively
Answer. D

ASSERTION – REASON TYPE QUESTIONS
In the following questions, a statement of Assertion followed by a statement of Reason is given.
Choose the correct option out of the following choices.
(a) Assertion and Reason both are true, Reason is the correct explanation of Assertion .
(b) Assertion and Reason both are true but Reason is not the correct explanation of Assertion .(c) Assertion is true, Reason is false.
(d) Assertion is false, Reason is true.

Question. Assertion : F - ion is a weak ligand and forms outer orbital complex.
Reason : F - ion cannot force the electrons of dz² and d x²-y² orbitals of the inner shells to occupy dxy ,dyz and dxz orbitals of the same subshell
Answer. A

Question. Assertion : The crystal field theory is successful in explaining the formation, structure, colour and magnetic properties of coordination compounds.
Reason : crystal field theory considers the metal-ligand bond to be ionic.
Answer. B

Question. Assertion : [Ni (CN)₄] ^2- is a diamagnetic complex
Reason :It involves dsp² hybridisation and there is no unpaired electron
Answer. A

Question. Assertion : Out of[ Fe (H₂O)₆]³⁺and [Fe (C₂O₄)₃]^3- the most stable complex is [Fe (C₂O₄)₃]^3-
Reason : Oxalate ion is an ambidentate ligand.
Answer. C

Question. Assertion : Linkage isomerism arises in coordination compounds containing ambidentate ligands
Reason : Ambidentate ligand has two different donor atoms
Answer. A

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 : Square planar complexes with coordination number 4 exhibit geometrical isomerism but tetrahedral complexes do not show geometrical isomerism.
Reason : The relative positions of the ligands in the tetrahedral complexes are the same with respect to each other .
Answer. A

Question.Assertion : Tetrahedral complexes have high spin configuration
Reason : crystal field splitting energy so small to force pairing up of the electrons
Answer. A

Question. Assertion : CO is stronger ligand than NH₃ for many metals
Reason : NH₃ can form pi bonds by back bonding
Answer. C

Question. Assertion : Nickel form low spin complexes
Reason: d²sp³ hybridisation is not possible in Nickel to form octahedral complexes.
Answer. D

SHORT ANSWER TYPE QUESTIONS

Question.Give the electronic configuration of the following complexes on the basis of Crystal Field Splitting theory.
[CoF₆]^3-, [Fe(CN)₆]^4- and [Cu(NH₃)₆]²⁺.
Answer. [CoF₆]^3- = Co³⁺ = (d)⁶ = t2g⁴eg²
[Fe(CN)₆]^4- = Fe²⁺ = (d)⁶ = t2g⁶eg⁰
[Cu(NH₃)₆]²⁺ = Cu²⁺ = (d)⁹ = t2g⁶eg³

Question. [Ni(CN)₄]^2- is colourless where as [Ni(H₂O)₆]²⁺ is green. Why?
Answer. In [Ni(CN)₄]^2-, Ni is in +2 oxidation state with electronic configuration 3d⁸. In the presence of strong CN- ligand the two unpaired electron in 3d orbital pair up. As there is no unpaired electron, it is colourless.
In [Ni(H₂O)₆]²⁺ Ni is +2 oxidation state and electronic configuration 3d⁸. The two unpaired electrons do not pair up in the presence of weak ligand H₂O. The d-d transition absorbs red light and complementary green light is emitted.

Question. Give the formula of each of the following coordination entities:
(i) Co ³⁺ ion bound to one Cl^- , one NH₃ molecule and two ethylene diamine molecules.
(ii) Ni ²⁺ ion is bound to two water molecules and two oxalate ions.
Answer. (i) [Co (NH₃)Cl (en)₂]²⁺ (ii)Ni(H₂O)₂(ox)₂] ^2-

Question. A metal complex having composition Cr(NH₃)₄Cl₂Br has been isolated in two forms A & B. The form A reacts with AgNO₃ solution to give a white precipitate soluble in aq. Ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia. Write the formulae of A&B.
Answer. A is [Cr(NH₃)₄ClBr]Cl B is [Cr(NH₃)₄Cl₂]Br

CASE BASED QUESTIONS

I. Read the given passage and answer the questions that follow.

Complex compounds play an important role in our daily life. Werner’s theory of complex compounds says every metal atom or ion has primary valency (oxidation state) which is satisfied by –vely charged ions, ionisable where secondary valency (coordination number) is non-ionisable, satisfied by ligands (+ve, –ve, neutral) but having lone pair. Primary valency is non-directional, secondary valency is directional. Complex compounds are name according to IUPAC system. Valence bond theory helps in determining shapes of complexes based on hybridisation, magnetic properties, outer or inner orbital complex. Complex show ionisation, linkage, solvate and coordination isomerism also called structural isomerism. Some of them also show stereoisomerism i.e. geometrical and optical isomerism. Ambidentate ligand are essential to show linkage isomerism. Polydentate ligands form more stable complexes then unidentate ligands. There are called chelating agents. EDTA is used to treat lead poisoning, cis-platin as anticancer agents. Vitamin B12 is complex of cobalt. Haemoglobin, oxygen carrier is complex of Fe²⁺ and chlorophyll essential for photosynthesis is complex of Mg²⁺.

Question. What is the oxidation state of Ni in [Ni(CO)₄]?
Answer. Zero

Question. One mole of CrCl₃ . 6H₂O reacts with excess of AgNO₃ to yield 2 mole of AgCl. Write formula of complex. Write IUPAC name also.
Answer. [Cr(H₂O)₅Cl]Cl₂ . H₂O, pentaaquachloridochromium (III) chloride.

Question. Out Cis – [Pt(en)₂ Cl₂] ²⁺ and trans [Pt(en)₂Cl₂] ²⁺ which one shows optical isomerism?
Answer. Cis – [Pt(en)₂ Cl₂] ²⁺ shows optical isomerism.

Question. Name the hexadentate ligand used for treatment of lead poisoning.
Answer. EDTA^4– (ethylenediamine tetraacetate)

Question. What is hybridisation of [CoF₆] ^3–? [Co = 27] Give its shape and magnetic properties.
Answer. Sp³d² , octahedral, paramagnetic.

Question. What type of isomerism is shown by [Cr(H₂O)₆] Cl₃ and [Cr(H₂O)₅ Cl] Cl₂ .H₂O?
Answer. Solvate isomerism.

1. Give IUPAC name of the following co-ordination compounds:
(i) [Cr(NH3)₃CL₃]
(ii) K₃[Fe(CN)₆]
(iii) [CoBr₂(en)₆]⁺

2. How would you account for the following?
(i) Transition metals exhibit variable oxidation states.
(ii) Zr(z=40) and Hf(z=72) have almost identical radii.
(iii) Transition metals and their compounds act as catalyst.

3. Complete the following equations
(i) Cr₂O₇ + 6Fe²⁺ + 14 H⁺ →
(ii) 2CrO₄^2- 2H⁺ →
(iii) 2MnO₄^- + 5C₂O₄^2- +16 H⁺ →

4. (a) Which metal in the first transition series exhibits +1 oxidation state frequently and Why?
(b) Which of the following cations are coloured in aqueous solution and why? Sc^3+, V³⁺ , Ti⁴⁺ , Mn²⁺ 

Question. The geometry and magnetic behaviour of the complex [Ni(CO)₄] are
(a) square planar geometry and diamagnetic
(b) tetrahedral geometry and diamagnetic
(c) square planar geometry and paramagnetic
(d) tetrahedral geometry and paramagnetic.
Answer. B

Question. Correct increasing order for the wavelengths of absorption in the visible region for the complexes of Co³⁺ is
(a) [Co(H₂O)₆]³⁺, [Co(en)₃]³⁺, [Co(NH₃)₆]³⁺
(b) [Co(H₂O)₆]³⁺, [Co(NH₃)₆]³⁺, [Co(en)₃]³⁺
(c) [Co(NH₃)₆]³⁺, [Co(en)₃]³⁺, [Co(H₂O)₆]³⁺
(d) [Co(en)₃]³⁺, [Co(NH₃)₆]³⁺, [Co(H₂O)₆]³⁺
Answer. D

Question. Pick out the correct statement with respect to [Mn(CN)₆]^3–.
(a) It is sp³d² hybridised and tetrahedral.
(b) It is d²sp³ hybridised and octahedral.
(c) It is dsp² hybridised and square planar.
(d) It is sp³d² hybridised and octahedral.
Answer. B

Question. Jahn–Teller effect is not observed in high spin complexes of
(a) d⁷
(b) d⁸
(c) d⁴
(d) d⁹ 
Answer. B

Question. The hybridization involved in complex [Ni(CN)₄]^2– is (At. No. Ni = 28)
(a) sp³
(b) d²sp²
(c) d²sp³
(d) dsp²
Answer. D

Question. Among the following complexes the one which shows zero crystal field stabilization energy (CFSE) is
(a) [Mn(H₂O)₆]³⁺
(b) [Fe(H₂O)₆]³⁺
(c) [Co(H₂O)₆]²⁺
(d) [Co(H₂O)₆]³⁺ 
Answer. B

Question. A magnetic moment at 1.73 BM will be shown by one among of the following
(a) TiCl₄
(b) [CoCl₆]^4–
(c) [Cu(NH₃)₄]²⁺
(d) [Ni(CN)₄]^2–
Answer. C

Question. Crystal field splitting energy for high spin d4 octahedral complex is
(a) – 1.2 Δ_o
(b) – 0.6 Δ_o
(c) – 0.8 Δ_o
(d) – 1.6 Δ_o
Answer. B

Question. Which among the following is a paramagnetic complex?
(a) [Co(NH₃)₆]³⁺
(b) [Pt(en)Cl₂]
(c) [CoBr₄]^2–
(d) Mo(CO)₆
(At. No. Mo = 42, Pt = 78)
Answer. C

Question. Which is diamagnetic?
(a) [CoF₆]^3–
(b) [Ni(CN)₄]^2–
(c) [NiCl₄]^2–
(d) [Fe(CN)₆]^3–
Answer. B

Question. Which one of the following is an outer orbital complex and exhibits paramagnetic behaviour?
(a) [Ni(NH₃)₆]²⁺
(b) [Zn(NH₃)₆]²⁺
(c) [Cr(NH₃)₆]³⁺
(d) [Co(NH₃)₆]³⁺
Answer. A

Question. Of the following complex ions, which is diamagnetic in nature?
(a) [NiCl₄]^2–
(b) [Ni(CN)₄]^2–
(c) [CuCl₄]^2–
(d) [CoF₆]^3–
Answer. B

Question. The d-electron configurations of Cr²⁺, Mn²⁺, Fe²⁺ and Co²⁺ are d⁴, d⁵, d⁶ and d⁷ respectively.Which one of the following will exhibit minimum paramagnetic behaviour?
(a) [Mn(H₂O)₆]²⁺
(b) [Fe(H₂O)₆]²⁺
(c) [Co(H₂O)₆]²⁺
(d) [Cr(H₂O)₆]²⁺
(At. nos. Cr = 24, Mn = 25, Fe = 26, Co = 27) 
Answer. C

Question. Which of the following complex compounds will exhibit highest paramagnetic behaviour?
(a) [Ti(NH₃)₆]³⁺
(b) [Cr(NH₃)₆]³⁺
(c) [Co(NH₃)₆]³⁺
(d) [Zn(NH₃)₆]³⁺
(At. No. Ti = 22, Cr = 24, Co = 27, Zn = 30) 
Answer. B

Question. Which of the following complex ions is not expected to absorb visible light?
(a) [Ni(CN)₄]^2–
(b) [Cr(NH₃)₆]³⁺
(c) [Fe(H₂O)₆]²⁺
(d) [Ni(H₂O)₆]²⁺ 
Answer. A

Question. Crystal field stabilization energy for high spin d4 octahedral complex is
(a) – 1.8 Δ_o
(b) – 1.6 Δ_o + P
(c) – 1.2 Δ_o
(d) – 0.6 Δ_o
Answer. D

Question. Out of TiF₆^2–, CoF₆^3–, Cu₂Cl₂ and NiCl₄^2– (Z of Ti = 22, Co = 27, Cu = 29, Ni = 28) the colourless species are
(a) Cu₂Cl₂ and NiCl₄^2–
(b) TiF₆^2– and Cu₂Cl₂
(c) CoF₆^3– and NiCl₄^2–
(d) TiF₆^2– and CoF₆ ^3–.
Answer. B

Question. Which of the following complex ions is expected to absorb visible light?
(a) [Ti(en)₂(NH₃)₂]⁴⁺
(b) [Cr(NH₃)₆]³⁺
(c) [Zn(NH₃)₆]²⁺
(d) [Sc(H₂O)₃(NH₃)₃]³⁺
[At. nos. Zn = 30, Sc = 21, Ti = 22, Cr = 24] 
Answer. B

Question. Which of the following complexes exhibits the highest paramagnetic behaviour?
(a) [Co(ox)₂(OH)₂]^–
(b) [Ti(NH₃)₆]³⁺
(c) [V(gly)₂(OH)₂(NH₃)₂]⁺
(d) [Fe(en)(bpy)(NH₃)₂]²⁺
where gly = glycine, en = ethylenediamine and bpy = bipyridyl moities. (At. nos. Ti = 22, V = 23,Fe = 26, Co = 27)
Answer. A

Question. In which of the following coordination entities the magnitude of Do (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⁺
(At. No. Co = 27) 
Answer. A

Question. The d electron configurations of Cr²⁺, Mn²⁺, Fe²⁺ and Ni²⁺ are 3d⁴, 3d⁵, 3d⁶ and 3d⁸ respectively.Which one of the following aqua complexes will exhibit the minimum paramagnetic behaviour?
(a) [Fe(H₂O)₆]²⁺
(b) [Ni(H₂O)₆]²⁺
(c) [CrH₂O)₆]²⁺
(d) [Mn(H₂O)₆]²⁺.
(At. No. Cr = 24, Mn = 25, Fe = 26, Ni = 28)
Answer. B

Question. Which one of the following is an inner orbital complex as well as diamagnetic in behaviour?
(a) [Zn(NH₃)₆]²⁺
(b) [Cr(NH₃)₆]³⁺
(c) [Co(NH₃)₆]³⁺
(d) [Ni(NH₃)₆]²⁺
(Atomic number : Zn = 30, Cr = 24, Co = 27, Ni = 28) 
Answer. C

Question. Among [Ni(CO)₄], [Ni(CN)₄]^2–, [NiCl₄]^2– species, the hybridisation states at the Ni atom are, respectively
(a) sp³, dsp², dsp²
(b) sp³, dsp², sp³
(c) sp³, sp³, dsp²
(d) dsp², sp³, sp³.
[Atomic number of Ni = 28]
Answer. B

Question. CN^– is a strong field ligand. This is due to the fact that
(a) it carries negative charge
(b) it is a pseudohalide
(c) it can accept electrons from metal species
(d) it forms high spin complexes with metal species. 
Answer. B

Question. Considering H₂O as a weak field ligand, the number of unpaired electrons in [Mn(H₂O)₆]²⁺ will be (atomic number of Mn = 25)
(a) three
(b) five
(c) two
(d) four. 
Answer. B

Question. In an octahedral structure, the pair of d orbitals involved in d²sp³ hybridisation is
(a) d _x²−y² d_z² ,
(b) d_xz d_{x²− y²}
(c) d_z² , d _xz
(d) d_xy , d_yz. 
Answer. A

Question. The number of unpaired electrons in the complex ion [CoF₆]^3– is
(a) 2
(b) 3
(c) 4
(d) zero
(Atomic no. : Co = 27)
Answer. C

Question. Atomic number of Cr and Fe are respectively 24 and 26, which of the following is paramagnetic with the spin of electron?
(a) [Cr(CO)₆]
(b) [Fe(CO)₅]
(c) [Fe(CN)₆]^4–
(d) [Cr(NH₃)₆]³⁺
Answer. D

Question. 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– - tetrahedral, paramagnetic 
Answer. A

Question. Iron carbonyl, Fe(CO)₅ is
(a) tetranuclear
(b) mononuclear
(c) trinuclear
(d) dinuclear.
Answer. B

Question. An example of a sigma bonded organometallic compound is
(a) Grignard’s reagent
(b) ferrocene
(c) cobaltocene
(d) ruthenocene.
Answer. A

Question. Which of the following has longest C—O bond length? (Free C—O bond length in CO is 1.128 Å.)
(a) [Fe(CO)₄]^2–
(b) [Mn(CO)₆]⁺
(c) Ni(CO)₄
(d) [Co(CO)₄]^–
Answer. A

Question. Which of the following carbonyls will have the strongest C – O bond?
(a) Mn(CO)₆⁺
(b) Cr(CO)₆
(c) V(CO)₆^–
(d) Fe(CO)₅ 
Answer. A

Question. Which of the following does not have a metal - carbon bond?
(a) Al(OC₂H₅)₃
(b) C₂H₅MgBr
(c) K[Pt(C₂H₄)Cl₃]
(d) Ni(CO)₄ 
Answer. A

Question. Among the following which is not the p-bonded organometallic compound?
(a) K [PtCl₃(h² – C₂H₄)]
(b) Fe (h₅ – C₅H₅)₂
(c) Cr(h₆ – C₆H₆)₂
(d) (CH₃)₄Sn 
Answer. D

Question. Which of the following organometallic compounds is s and p-bonded?
(a) [Fe(h₅ – C₅H₅)₂]
(b) K[PtCl₃(h² – C₂H₄)]
(c) [Co(CO)₅NH₃]²⁺
(d) Fe(CH₃)₃ 
Answer. C

Question. Shape of Fe(CO)₅ is
(a) octahedral
(b) square planar
(c) trigonal bipyramidal
(d) square pyramidal.
Answer. C

Question. In metal carbonyl having general formula M(CO)x where M = metal, x = 4 and the metal is bonded to
(a) carbon and oxygen
(b) C O
(c) oxygen
(d) carbon.
Answer. D

Question. Which of the following complexes is used to be as an anticancer agent?
(a) mer-[Co(NH₃)₃Cl₃]
(b) cis-[PtCl₂(NH₃)₂]
(c) cis-K₂[PtCl₂Br₂]
(d) Na₂CoCl₄
Answer. B

Question. Copper sulphate dissolves in excess of KCN to give
(a) Cu(CN)₂
(b) CuCN
(c) [Cu(CN)₄]^3–
(d) [Cu(CN)₄]^2–
Answer. C

Question. In the silver plating of copper, K[Ag(CN)₂] is used instead of AgNO₃. The reason is
(a) a thin layer of Ag is formed on Cu
(b) more voltage is required
(c) Ag+ ions are completely removed from solution
(d) less availability of Ag⁺ ions, as Cu cannot displace Ag from [Ag(CN)₂]^– ion. 
Answer. D

Question. CuSO₄ when reacts with KCN forms CuCN, which is insoluble in water. It is soluble in excess of KCN, due to formation of the following complex
(a) K₂[Cu(CN)₄]
(b) K₃[Cu(CN)₄]
(c) CuCN₂
(d) Cu[KCu(CN)₄] 
Answer. B

Question. Hypo is used in photography to
(a) reduce AgBr grains to metallic silver
(b) convert metallic silver to silver salt
(c) remove undecomposed silver bromide as a soluble complex
(d) remove reduced silver.
Answer. C

Q1. Give IUPAC names of the following:-

(a) Na₂[CrF₄O]

(b) K[Pt(NH₃)Cl₃]

(c) Na₂[SiF₆]

(d) [CoCl(en)₂(ONO)]⁺

(e) [Co(NH₃)₃(CO₃)]Cl

(f) [Pt(NH₃)₂(py)₂] [PtCl₄]

(g) [Cr(PPH₃) (CO)₅]

(h) [Mn₃(CO)₁₂]

(i) [CO(NH₃)₆]ClSO₄

(j) K₃[AL(C₂O₄)₃]

(k) Hg[Co(SCN)₄]

(l) K₂[Zn(OH)₄]

(m) [CO(en)₃] [Cr(CN)₆]

(n) Cs [Fecl₄]

Q2. Give one chemical test to distinguish between [CO (NH₃)₅Br] SO₄ and [CO (NH₃)₅SO₄] Br.

Q3. A coordination compound has formula Cocl₃ 4NH₃. It does not liberate ammonia but precipitates chloride ions as silver chloride. Give the IUPAC name of complex E-its structural formula.

Q4. The molar conductivity of CaCl₃.4NH₃.2H₂O is found to be same as that of 3:1 electrolyte. What is the structural formula of the complex?

Q5. Give reasons for the following:-

(a) [Fe(CN)₆]₃ is weakly paramagnetic while [Fe(CN)₆]^4- is diamagnetic. (VB).

(b) [Ni(CO)₄ is tetrahedral while [Ni (CN)₄] is square planar. (VB).

(c) [Co(CN)₆]^3- is low spin complex while [CoF₆]₃ is a high spin complex (VB)

(d) [Mn(H₂O)₆]²⁺ has five unpaired electrons while [Mn(CN)₆]^4- has only one unpaired. (CFST).

(e) [Ti(H₂O)₆]³⁺ is coloured while is [Sc(H₂O)₆]³⁺ colorless.

Q6. A, B and C are three complexes of chromium with empirical formula H₁₂O₆Cl₁₃Cr. All the three complexes have Cl and H₂O molecules as the ligands. Complex A does not react with conc. H₂SO₄. Complex B and C lose 6.75% and 13.5% of their original weight respectively on heating with Concn H₂SO₄. Identify A, B and C.
Hint: - Calculate molar mass E-wt of water lost, Correlate it to no. of H₂O Molecules lost.