CBSE Class 12 Chemistry Coordination Compounds Worksheet Set 01

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

Concept 1: One Mark Questions

Question 1. Write the formula for Tetraamineaquachloridocobalt(III) chloride
Answer: The chemical formula of the coordination compound is \( [\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})\text{Cl}]\text{Cl}_2 \).
In simple words: This formula represents a cobalt complex with four ammonia molecules, one water molecule, and one chlorine atom inside the bracket, and two chlorine atoms outside.

Exam Tip: Be sure to write the ligands in alphabetical order (ammine before aqua, then chlorido) within the coordination sphere.

Question 2. Write the IUPAC name of [Co (NH3)4 Br2]2 [Zn Cl4]
Answer: The IUPAC name is tetraamminedibromocobalt(III) tetrachlorozincate(II).
In simple words: The compound consists of a positive cobalt-based complex ion and a negative zinc-based complex ion.

Exam Tip: Since the zinc complex is anionic, its name must end with the suffix "-ate" (zincate), and both metal oxidation states must be specified in Roman numerals.

Question 3. Which of these cannot act as ligand and why: NH3, H2O, CO, CH4. Give reason?
Answer: Out of these, methane (\( \text{CH}_4 \)) cannot function as a ligand because its carbon atom is fully saturated with single covalent bonds and lacks any unshared lone pairs of electrons to donate to a metal.
In simple words: Methane does not have any extra electron pairs that it can share with a metal atom to form a bond.

Exam Tip: To act as a ligand, a species must possess at least one lone pair of electrons on its donor atom.

Question 4. How many EDTA (ethylenediamine tetraacetic acid) molecules are required to make an octahedral complex with a Ca2+ ion.
Answer: Since EDTA is a hexadentate ligand, it can form six coordinate bonds with a single metal ion. Therefore, only one EDTA molecule is needed to complete an octahedral complex with \( \text{Ca}^{2+} \).
In simple words: One molecule of EDTA is enough to completely wrap around the calcium ion because it has six grabbing hands.

Exam Tip: State clearly that EDTA is a "hexadentate ligand" to secure full marks on this question.

Question 5. Why tetrahedral complexes do not exhibit geometrical isomerism ?
Answer: In a tetrahedral complex, all four coordination sites are adjacent to one another. Since the spatial positions of all ligands remain identical relative to one another, no separate geometrical isomers can be formed.
In simple words: No matter how you rearrange the groups on a tetrahedron, they always end up right next to each other.

Exam Tip: Mention that all four positions in a tetrahedral geometry are "cis" to each other to explain the absence of geometrical isomerism.

Question 6. What is the hybridisation of central metal ion and shape of Wilkinson’s catalyst ?
Answer: Wilkinson's catalyst is \( [\text{Rh}(\text{PPh}_3)_3\text{Cl}] \). The central rhodium ion undergoes \( dsp^2 \) hybridization, resulting in a square planar shape.
In simple words: Wilkinson's catalyst has a flat, square shape and its central rhodium atom uses square planar hybridization.

Exam Tip: Make sure to write the formula \( [\text{Rh}(\text{PPh}_3)_3\text{Cl}] \) and state both hybridization and geometry clearly.

Question 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]+
Answer: The oxidation number of the metal in both complexes is \( +3 \):
(i) In \( [\text{Co}(\text{H}_2\text{O})(\text{CN})(\text{en})_2]^{2+} \): \( x + 0 - 1 + 0 = +2 \implies x = +3 \).
(ii) In \( [\text{CoBr}_2(\text{en})_2]^+ \): \( x - 2 + 0 = +1 \implies x = +3 \).
In simple words: The central cobalt ion has a charge of +3 in both of these compounds.

Exam Tip: Neutral ligands like water (\( \text{H}_2\text{O} \)) and ethylenediamine (\( \text{en} \)) do not contribute to the charge calculation.

Question 8. Draw the structure of optical isomers of [Co(en)3]3+.
Answer: The complex \( [\text{Co}(\text{en})_3]^{3+} \) exists as a pair of non-superimposable mirror images representing dextro (\( d \)) and laevo (\( l \)) optical isomers.
In simple words: Optical isomers are molecules that are mirror images of each other and cannot be perfectly superimposed, like our left and right hands.

Exam Tip: When drawing these structures, draw a vertical dotted line in between to represent the mirror and ensure that the bidentate ethylenediamine chains are clearly mirrored on both sides.

Question 9. Name the types of isomerism exhibited by [Co(NH3)5(NO2)](NO3)2
Answer: This compound exhibits both **linkage isomerism** and **ionization isomerism**:
- Linkage isomers: \( [\text{Co}(\text{NH}_3)_5(\text{NO}_2)](\text{NO}_3)_2 \) and \( [\text{Co}(\text{NH}_3)_5(\text{ONO})](\text{NO}_3)_2 \)
- Ionization isomers: \( [\text{Co}(\text{NH}_3)_5(\text{NO}_2)](\text{NO}_3)_2 \) and \( [\text{Co}(\text{NH}_3)_5(\text{NO}_3)](\text{NO}_3)(\text{NO}_2) \)
In simple words: This complex can show linkage isomerism because the nitro group can attach through nitrogen or oxygen, and ionization isomerism because a nitrate group outside can swap with a ligand inside.

Exam Tip: Clearly write out the chemical formulas for both isomers of each type to demonstrate your knowledge to the examiner.

Question 10. Write the formula of Amminebromidochloridonitrito-N-platinate(II) ion
Answer: The chemical formula of the complex ion is \( [\text{Pt}(\text{NH}_3)\text{BrCl}(\text{NO}_2)]^- \).
In simple words: This formula shows a negative platinum complex containing ammonia, bromine, chlorine, and nitrogen-bonded nitro ligands.

Exam Tip: Be sure to include the negative charge superscript on the bracket as the question asks for an "ion" and not a neutral compound.

Question 11. which one is more stable complex among(i) [Fe(H2O)6]3+ (ii) [Fe(NH3)6]3+(iii) [Fe(C2O4)3]3−
Answer: The most stable complex among these is \( [\text{Fe}(\text{C}_2\text{O}_4)_3]^{3-} \). This high stability is due to the **chelate effect**, as the oxalate ion is a bidentate ligand that binds to the iron atom to form multiple stable five-membered rings.
In simple words: The oxalate complex is the most stable because the oxalate ligands form closed ring loops around the central iron atom, holding onto it much more tightly.

Exam Tip: Always cite the "chelate effect" as the primary reason for increased stability when bidentate or polydentate ligands are compared with monodentate ligands.

Question 12. How many ions are produced from the complex Co(NH3)6Cl2 in solution?
Answer: In an aqueous solution, the coordination compound \( [\text{Co}(\text{NH}_3)_6]\text{Cl}_2 \) dissociates into a total of three ions: one \( [\text{Co}(\text{NH}_3)_6]^{2+} \) cation and two \( \text{Cl}^- \) anions.
In simple words: When dissolved in water, the compound splits into one large positive complex ion and two negative chloride ions, making three ions in total.

Exam Tip: Remember that ligands inside the square brackets stay bonded together as a single ion and do not dissociate in solution.

Question 13. What do you meant by degenerate d-orbitals?
Answer: Degenerate d-orbitals refer to a set of d-orbitals that share the exact same energy level. In a free, isolated gaseous transition metal atom or ion, all five d-orbitals are degenerate.
In simple words: Degenerate d-orbitals are d-orbitals that are identical in their energy level.

Exam Tip: Note that this degeneracy is broken when ligands approach the metal ion in a coordination environment, causing splitting into different energy levels.

Question 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-
Answer: Only \( \text{cis-}[\text{CrCl}_2(\text{ox})_2]^{3-} \) is chiral and optically active. The trans-isomer possesses a plane of symmetry, making its mirror image superimposable and optically inactive.
In simple words: The cis-isomer is optically active because it lacks symmetry, meaning its mirror image cannot be turned to match itself.

Exam Tip: Geometrical isomers containing bidentate ligands in a *trans* configuration are almost always symmetric and optically inactive.

Question 15. The spin only magnetic moment of [MnBr4]2- is 5.9 BM. Predict the geometry of the complex ion?
Answer: A magnetic moment of \( 5.9\text{ BM} \) indicates the presence of five unpaired electrons, corresponding to \( d^5 \) high-spin \( \text{Mn}^{2+} \). Since bromide is a weak ligand, \( sp^3 \) hybridization occurs, resulting in a **tetrahedral** geometry.
In simple words: The five unpaired electrons confirm that the d-orbitals are left unhybridized for bonding, which points to a tetrahedral layout using sp3 orbitals.

Exam Tip: Link the spin-only formula \( \mu = \sqrt{n(n+2)} \) to show how \( n=5 \) is deduced from \( 5.9\text{ BM} \).

Question 16. What is the oxidation state of Ni in [Ni(CO)4].
Answer: In this complex, carbon monoxide is a neutral ligand, which means the oxidation state of the central nickel atom is zero.
In simple words: The nickel atom carries no charge in this neutral carbonyl compound.

Exam Tip: Be sure to write the answer clearly as "Zero (0)" to avoid losing marks on simple definitions.

Question 17. What is the magnetic behavior of [Ni(CN)4]2- .
Answer: The complex \( [\text{Ni}(\text{CN})_4]^{2-} \) is **diamagnetic** in nature.
In simple words: All the electrons in this complex are fully paired up, so it is not attracted by magnets.

Exam Tip: Since cyanide is a strong-field ligand, it forces the pairing of d-electrons in the \( \text{Ni}^{2+} \) ion, removing all unpaired spins.

Question 18. Name the metal ion present in vitamin B12.
Answer: The central metal ion present in Vitamin B12 is cobalt in its \( +3 \) oxidation state (\( \text{Co}^{3+} \)).
In simple words: Vitamin B12 contains a cobalt ion at the center of its complex structure.

Exam Tip: Keep this biochemical fact in mind as it is a common general chemistry question in board exams.

Question 19. Name the isomerism shown by complex K[Cr(H2O)2(C2O4)2]
Answer: This complex exhibits both **geometrical isomerism** (existing in *cis-* and *trans-* configurations) and **optical isomerism** (specifically for the *cis-* isomer which lacks symmetry).
In simple words: This compound can exist in cis and trans shapes, and its cis shape has left- and right-handed mirror images.

Exam Tip: If asked, detail that only the cis-isomer is chiral, while the trans-isomer is symmetric and optically inactive.

Question 20. Name the compound used for inhibiting the growth of tumours. (cancer treatment)
Answer: The coordination compound used for tumor inhibition is **cisplatin**, which has the formula \( \text{cis-}[\text{Pt}(\text{NH}_3)_2\text{Cl}_2] \).
In simple words: Cisplatin is a platinum-based coordination compound used in chemotherapy to treat cancer.

Exam Tip: Specify the *cis*-isomer prefix, as the *trans*-isomer (transplatin) is biologically inactive against tumors.

Concept 2: Two Mark Questions

Question 1. Write the IUPAC names of the following coordination compounds: (i) [Pt(NH3)2Cl(NO2 )] (ii) K3 [Cr(C2O4)3 ]
Answer: The systematic IUPAC names are:
(i) Diamminechloridonitrito-N-platinum(II)
(ii) Potassium trioxalatochromate(III)
In simple words: The first is a neutral platinum complex with ammonia, chlorine, and nitro ligands. The second is a potassium salt containing an anionic chromium complex with three oxalate groups.

Exam Tip: Be sure to write coordinated ligands alphabetically and identify anionic complex metals with the suffix "-ate".

Question 2. A cationic complex has two isomers A & B. Each has one Co3+, five NH3, one Br and one SO42-. A gives a white precipitate with BaCl2 solution while B gives a yellow precipitate with AgNO3 solution. (a) What are the possible structures of the complexes A and B? (b) Write the name of structural isomerism shown by A and B.
Answer:
(a) The chemical structures of the isomers are:
- **Isomer A:** \( [\text{Co}(\text{NH}_3)_5\text{Br}]\text{SO}_4 \) (gives a white precipitate of \( \text{BaSO}_4 \) with \( \text{BaCl}_2 \)).
- **Isomer B:** \( [\text{Co}(\text{NH}_3)_5(\text{SO}_4)]\text{Br} \) (gives a pale yellow precipitate of \( \text{AgBr} \) with \( \text{AgNO}_3 \)).
(b) The relationship between the two complexes is **ionization isomerism**.
In simple words: Isomer A has sulfate on the outside which precipitates with barium, while Isomer B has bromine on the outside which precipitates with silver. They show ionization isomerism because different ions are released in water.

Exam Tip: Always write the chemical reactions of precipitation to back up your structures in descriptive answers.

Question 3. FeSO4 solution mixed with (NH4)2SO4 solution in 1 : 1 molar ratio gives the test of Fe2+ ion but CuSO4 solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu2+ ion. Explain why ?
Answer: When \( \text{FeSO}_4 \) and \( (\text{NH}_4)_2\text{SO}_4 \) are mixed, they form Mohr's salt, \( \text{FeSO}_4\cdot(\text{NH}_4)_2\text{SO}_4\cdot6\text{H}_2\text{O} \), which is a double salt and completely dissociates into simple ions in water. However, mixing \( \text{CuSO}_4 \) and ammonia in a 1:4 ratio produces a coordination complex \( [\text{Cu}(\text{NH}_3)_4]\text{SO}_4 \), which does not release free \( \text{Cu}^{2+} \) ions in solution.
In simple words: Mohr's salt is a double salt that falls apart completely in water, releasing simple iron ions. The copper-ammonia mixture forms a coordinate complex that locks copper inside a complex ion, so it cannot react as free copper ions.

Exam Tip: Differentiate clearly between double salts (which dissociate completely) and coordination complexes (which retain their identity in solution).

Question 4. What is meant by ambidentate ligands? Give two examples.
Answer: Ambidentate ligands are monodentate ligands that contain more than one potential donor atom but coordinate to the central metal ion through only one donor site at a time.
Examples:
1. **Nitro/Nitrito ligand:** Can bond via nitrogen (\( \text{-NO}_2 \)) or via oxygen (\( \text{-ONO} \)).
2. **Thiocyanate/Isothiocyanate ligand:** Can bond via sulfur (\( \text{-SCN} \)) or via nitrogen (\( \text{-NCS} \)).
In simple words: These are ligands that have two different atoms they can use to bind to a metal, but they only use one of them at a time.

Exam Tip: Draw the simple binding symbols showing the donor atoms clearly to make your explanation robust.

Question 5. [Co(NH3)6]3+ is diamagnetic whereas [Co(F6)]3- is paramagnetic. Give reasons.
Answer: In both complexes, the cobalt ion is in the \( +3 \) state (\( 3d^6 \)):
- In \( [\text{Co}(\text{NH}_3)_6]^{3+} \), \( \text{NH}_3 \) is a strong-field ligand that causes the pairing of the \( 3d \) electrons, leaving no unpaired spins. Thus, it undergoes \( d^2sp^3 \) hybridization and is **diamagnetic**.
- In \( [\text{CoF}_6]^{3-} \box \), the fluoride ion is a weak-field ligand and cannot force pairing. There remain four unpaired electrons in its outer \( sp^3d^2 \) hybridization, making the complex **paramagnetic**.
In simple words: Ammonia forces all of cobalt's d-electrons to pair up, leaving no magnetic spin. Fluoride is a weak ligand and lets the electrons remain unpaired, which creates a magnetic attraction.

Exam Tip: Mention the hybridization of the metal ion (\( d^2sp^3 \) vs \( sp^3d^2 \)) to support your crystal field theory arguments.

Question 6. [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2− is diamagnetic. Explain why?
Answer: In \( [\text{Cr}(\text{NH}_3)_6]^{3+} \), the chromium ion has a \( d^3 \) configuration, leaving three unpaired electrons in the \( t_{2g} \) level, resulting in paramagnetism. In \( [\text{Ni}(\text{CN})_4]^{2-} \), nickel has a \( d^8 \) configuration. The strong-field cyanide ligand forces the two unpaired electrons of \( \text{Ni}^{2+} \) to pair up within a single \( 3d \) orbital, resulting in diamagnetism.
In simple words: Chromium has three electrons that must remain unpaired in its orbitals, creating magnetic behavior. Nickel's electrons are forced to fully pair up by the strong cyanide ligands, making it non-magnetic.

Exam Tip: Detail the electron configurations of \( \text{Cr}^{3+} \) (\( 3d^3 \)) and \( \text{Ni}^{2+} \) (\( 3d^8 \)) to make your explanation mathematically sound.

Question 7. Draw figure to show the splitting of d orbitals in an octahedral crystal field. How is the magnitude of ∆0 affected by (i) Nature of ligand. (ii) Oxidation State of metal ion.
Answer: In an octahedral field, the electrostatic repulsion splits the five degenerate \( d \)-orbitals into two higher-energy \( e_g \) orbitals (\( d_{x^2-y^2} \) and \( d_{z^2} \)) and three lower-energy \( t_{2g} \) orbitals (\( d_{xy} \), \( d_{yz} \), and \( d_{zx} \)). (i) **Nature of Ligand:** Strong-field ligands increase electrostatic repulsion, leading to a much larger splitting (\( \Delta_0 \)) compared to weak-field ligands.
(ii) **Oxidation State:** Higher oxidation states of the metal draw the ligands closer, increasing electrostatic repulsion and thereby increasing the magnitude of \( \Delta_0 \).
In simple words: The five d-orbitals split into two upper and three lower energy groups. Strong ligands and high metal charges push these energy groups further apart.

Exam Tip: Be sure to label the Barycentre and specify the energy shifts of \( +\frac{3}{5}\Delta_0 \) for \( e_g \) and \( -\frac{2}{5}\Delta_0 \) for \( t_{2g} \).

Question 8. Metal carbonyl are much more stable than normal complexes, why?
Answer: Metal carbonyl stability is driven by **synergic bonding**: the carbon atom of \( \text{CO} \) donates its lone pair of electrons to a vacant metal orbital forming a \( \sigma \)-bond, while the filled metal \( d \)-orbitals donate electron density back into the empty \( \pi^* \) antibonding orbitals of \( \text{CO} \), forming a \( \pi \)-bond. This mutual reinforcement strengthens the overall bond.
In simple words: Carbon monoxide donates electrons to the metal, and the metal donates electrons back to the carbon monoxide. This dual back-and-forth bonding makes the bond exceptionally strong and stable.

Exam Tip: Use the term "synergic effect" and write about both \( \sigma \)-donation and \( \pi \)-backbonding to score full marks.

Question 9. Give evidence that [Co(NH3)5Cl]SO4 and [Co(NH3)5SO4]Cl are ionization isomers.
Answer: These isomers dissolve in water to yield distinct free ions, which can be identified by specific precipitation tests:
- \( [\text{Co}(\text{NH}_3)_5\text{Cl}]\text{SO}_4 \) reacts with \( \text{Ba}^{2+} \) to form a white precipitate of \( \text{BaSO}_4 \), but does not react with \( \text{Ag}^+ \).
- \( [\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Cl} \) reacts with \( \text{Ag}^+ \) to form a white precipitate of \( \text{AgCl} \), but does not react with \( \text{Ba}^{2+} \).
In simple words: The first compound releases sulfate ions in water which turn white when mixed with barium, while the second releases chloride ions which turn white when mixed with silver.

Exam Tip: Providing the two distinct ionic reactions and their resulting precipitate colors is required to support your answer.

Question 10. Write the formulas for the following coordination compounds and name the isomerism shown by them: i. potassium tetracyanonickelate(II) ii. tris(ethane−1,2−diamine) chromium(III) chloride
Answer: The chemical formulas and their respective isomerisms are:
i. \( \text{K}_2[\text{Ni}(\text{CN})_4] \) - This symmetric square planar complex does not display any isomerism.
ii. \( [\text{Cr}(\text{en})_3]\text{Cl}_3 \) - This complex lacks a plane of symmetry and therefore exhibits **optical isomerism**.
In simple words: The first is a symmetric nickel compound with no isomers. The second is an octahedral chromium complex that exists as left- and right-handed optical isomers.

Exam Tip: Be sure to write the correct coordination bracket formulas alongside your isomeric descriptions.

Concept 3: Three Mark Questions

Question 1. How many geometrical isomers are possible in the following coordination entities? (i) [Cr(C2O4)3]3− (ii) [Co(NH3)3Cl3]
Answer:
(i) **For \( [\text{Cr}(\text{C}_2\text{O}_4)_3]^{3-} \):** Zero. Geometrical isomerism is impossible because all three bidentate ligands occupy identical relative positions around the metal.
(ii) **For \( [\text{Co}(\text{NH}_3)_3\text{Cl}_3] \):** Two geometrical isomers are possible:
- **Facial (*fac*):** - **Meridional (*mer*):**
In simple words: The first complex has identical loops so it cannot show different shapes. The second complex can form facial (grouped on one face) and meridional (grouped in a ring) isomers.

Exam Tip: Draw clear stereochemical diagrams for both *fac-* and *mer-* isomers, labeling the positions of the chlorine and ammonia groups explicitly.

Question 2. (a) Write the IUPAC name of [Ti(H2O)6]+3. (b) [Ti(H2O)6]+3 is coloured why? (c) A Complex having scandium in +3 oxidation-state was found colorless why?
Answer:
(a) The IUPAC name is hexaaquatitanium(III) ion.
(b) The \( \text{Ti}^{3+} \) ion has a \( 3d^1 \) configuration. The absorption of visible light excites this single electron from the lower-energy \( t_{2g} \) set to the higher-energy \( e_g \) set via a \( d\text{-}d \) transition, yielding a purple color.
(c) Scandium in its \( +3 \) oxidation state has a \( 3d^0 \) electronic configuration. Due to the absolute absence of \( d \)-electrons, no \( d\text{-}d \) transition can take place, rendering the complex colorless.
In simple words: Titanium(III) has a single electron in its d-orbitals that can absorb light and jump to a higher level. Scandium(III) has no d-electrons at all, so it cannot absorb light in this way and remains colorless.

Exam Tip: Always relate the presence or absence of color in transition metal complexes directly to the availability of d-electrons for \( d\text{-}d \) transitions.

Question 3. (a) Write IUPAC name of [Co(en)3]3+ (b) [NiCl4]2- is paramagnetic while [Ni(CO)]4 is diamagnetic though both are tetrahedral. Why? (c) Explain why K3[Fe(CN)6] is more stable than K4[Fe(CN)6].
Answer:
(a) The IUPAC name is tris(ethane-1,2-diamine)cobalt(III) ion.
(b) In \( [\text{NiCl}_4]^{2-} \), nickel is in the \( +2 \) state (\( 3d^8 \)). Since chloride is a weak-field ligand, it does not force pairing, leaving two unpaired electrons and making the complex paramagnetic. In \( [\text{Ni}(\text{CO})_4] \), nickel is in the zero state (\( 3d^8 4s^2 \)). Carbon monoxide is a strong-field ligand that forces all \( 4s \) electrons to pair into the \( 3d \) subshell, leaving no unpaired electrons, resulting in diamagnetism.
(c) The iron in \( \text{K}_3[\text{Fe}(\text{CN})_6] \) is \( \text{Fe}^{3+} \), while in \( \text{K}_4[\text{Fe}(\text{CN})_6] \) it is \( \text{Fe}^{2+} \). Due to its higher positive charge, \( \text{Fe}^{3+} \) has a higher charge density, which allows it to attract and bind ligands much more strongly, making the complex more stable.
In simple words: Weak chloride ligands leave nickel's electrons unpaired and magnetic, whereas strong carbon monoxide ligands force all electrons to pair up. For stability, iron with a higher +3 charge binds cyanide more tightly than iron with a lower +2 charge.

Exam Tip: Be sure to write out the electronic configurations of the metal ions to fully justify differences in their magnetic properties.

Question 4. (a) What is the hybridization state of nickel in [Ni(CN)4]2− (b) Draw the structure of [Ni(CN)4]2− (c) A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2− is colourless. Explain.
Answer:
(a) The hybridization state of nickel is \( dsp^2 \).
(b) The structure is square planar: (c) In \( [\text{Ni}(\text{H}_2\text{O})_6]^{2+} \), water is a weak-field ligand, preserving two unpaired electrons in the \( 3d \) orbitals, allowing visible-range \( d\text{-}d \) transitions which produce a green color. In \( [\text{Ni}(\text{CN})_4]^{2-} \), the strong-field cyanide ligand pairs all d-electrons, preventing \( d\text{-}d \) transition in the visible range, making the solution colorless.
In simple words: The nickel-cyanide complex has a flat, square planar shape. It is colorless because its electrons are all locked in pairs, while the nickel-water complex has unpaired electrons that can absorb light to appear green.

Exam Tip: Always contrast weak-field and strong-field ligand behavior to explain why one complex is colored while the other is colorless.

Question 5. Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex.
Answer: In \( [\text{Co}(\text{NH}_3)_6]^{3+} \), the cobalt ion has a \( 3d^6 \) configuration. The strong-field ligand \( \text{NH}_3 \) forces pairing, leaving two vacant inner \( 3d \) orbitals for \( d^2sp^3 \) hybridization, creating an **inner orbital complex**. In \( [\text{Ni}(\text{NH}_3)_6]^{2+} \), the nickel ion has a \( 3d^8 \) configuration. Pairing the electrons can at most leave only one \( 3d \) orbital vacant, which cannot accommodate the two vacant \( d \)-orbitals required for octahedral hybridization. Consequently, nickel must use its outer \( 4d \) orbitals for \( sp^3d^2 \) hybridization, forming an **outer orbital complex**.
In simple words: Cobalt can easily free up its inner d-orbitals for bonding by pairing its electrons. Nickel has too many d-electrons, so even after pairing it must use its outer d-orbitals to connect with the six ammonia ligands.

Exam Tip: Use the orbital box diagrams representing the d-electron pairing in both complexes to clearly demonstrate inner vs outer orbital hybridization.

Question 6. For the complexes (i) [Ni(CN)4]2− (ii) [Ni(Cl)4]2− (iii) [Ni(CO)4] Identify: (a) The oxidation No. of nickel (b) The hybrid orbitals and the shape of the complexes
Answer:
(a) **Oxidation Numbers of Nickel:**
(i) In \( [\text{Ni}(\text{CN})_4]^{2-} \): \( x + 4(-1) = -2 \implies x = +2 \).
(ii) In \( [\text{Ni}(\text{Cl})_4]^{2-} \): \( x + 4(-1) = -2 \implies x = +2 \).
(iii) In \( [\text{Ni}(\text{CO})_4] \): Since \( \text{CO} \) is a neutral ligand, \( x = 0 \).

(b) **Hybridization and Shapes:**
(i) \( [\text{Ni}(\text{CN})_4]^{2-} \) undergoes \( dsp^2 \) hybridization, resulting in a **square planar** geometry.
(ii) \( [\text{Ni}(\text{Cl})_4]^{2-} \) undergoes \( sp^3 \) hybridization, resulting in a **tetrahedral** geometry.
(iii) \( [\text{Ni}(\text{CO})_4] \) undergoes \( sp^3 \) hybridization, resulting in a **tetrahedral** geometry.
In simple words: The nickel has a +2 charge in the first two complexes and no charge (0) in the third. The cyanide complex is flat and square-shaped, while the chlorine and carbonyl complexes have a pyramid-like tetrahedral shape.

Exam Tip: Be sure to write the answer in a structured tabular format if asked to compare their properties directly in the exam.

Question 7. Specify the (i) oxidation numbers (ii) coordination numbers and (iii) IUPAC name of the following coordination entities: (a) [Co(H2O)(CN)(en)2]2+ (b) [PtCl4]2−
Answer:
(a) **For \( [\text{Co}(\text{H}_2\text{O})(\text{CN})(\text{en})_2]^{2+} \):**
(i) Oxidation number = +3
(ii) Coordination number = 6 (as ethylenediamine is bidentate: \( 1 + 1 + 2(2) = 6 \))
(iii) IUPAC name = aquacyanidobis(ethane-1,2-diamine)cobalt(III) ion.

(b) **For \( [\text{PtCl}_4]^{2-} \):**
(i) Oxidation number = +2
(ii) Coordination number = 4
(iii) IUPAC name = tetrachloridoplatinate(II) ion.
In simple words: The cobalt complex has a +3 charge and binds to 6 donor sites, while the platinum complex has a +2 charge and binds to 4 donor sites.

Exam Tip: When writing IUPAC names, keep the ligands in alphabetical order (aqua before cyanide) and make sure the anionic metal ends in "-ate" (platinate).

Question 8. Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory: (a) [Fe(CN)6]4− (b) [FeF6]3−
Answer:
(a) **Bonding in \( [\text{Fe}(\text{CN})_6]^{4-} \):** The central iron is in the \( +2 \) state (\( 3d^6 \)). Since cyanide (\( \text{CN}^- \)) is a strong-field ligand, it forces the pairing of the \( 3d \) electrons, vacating two inner \( 3d \) orbitals. This allows \( d^2sp^3 \) hybridization, resulting in an **inner orbital octahedral complex** that is **diamagnetic** (no unpaired electrons).

(b) **Bonding in \( [\text{FeF}_6]^{3-} \):** The central iron is in the \( +3 \) state (\( 3d^5 \)). Since fluoride (\( \text{F}^- \)) is a weak-field ligand, it cannot force the pairing of electrons. It utilizes outer \( 4s \), \( 4p \), and \( 4d \) orbitals for \( sp^3d^2 \) hybridization, forming an **outer orbital octahedral complex** that is highly **paramagnetic** (5 unpaired electrons).
In simple words: Cyanide forces iron's electrons to pair up, making the complex non-magnetic and compact. Fluoride is a weak ligand and leaves the electrons unpaired, making the complex magnetic.

Exam Tip: Illustrating the orbital box diagrams for the metal ion before and after hybridization is highly recommended for securing full marks on bonding questions.

Question 9. Dimethyl glyoxime is added to alcoholic solution of NiCl2. When ammonium hydroxide is slowly added to it, a rosy red precipitate of a complex appears. (a) Give the str. of the complex showing hydrogen bond. (b) Give oxidation state and hybridization of central metal ion. (c) Identify whether it is paramagnetic or diamagnetic.
Answer:
(a) **Structure of \( \text{Ni(dmg)}_2 \):** The complex is stabilized by symmetric intramolecular hydrogen bonds. (b) **Oxidation State and Hybridization:** The oxidation state of nickel is \( +2 \), and the central metal ion is \( dsp^2 \) hybridized.
(c) **Magnetic Behavior:** The complex is **diamagnetic** because the strong-field dimethylglyoximate ligands force all d-electrons to pair up.
In simple words: This nickel complex has a square planar shape and is held together tightly by hydrogen bonds, leaving no unpaired electrons to be magnetic.

Exam Tip: Remember that the rosy-red precipitate is a specific analytical test for Ni(II) ions in a basic ammonia solution.

Question 10. Draw all the isomers (geometrical and optical) of: (a) [CoCl2(en)2]+ (b) [Co(NH3)Cl(en)2]2+
Answer:
(a) **Isomers of \( [\text{CoCl}_2(\text{en})_2]^+ \):** A total of three isomers exist: a symmetric *trans*-isomer (optically inactive) and a chiral *cis*-isomer which exists as a pair of non-superimposable mirror images (dextro and laevo optical isomers).

(b) **Isomers of \( [\text{Co}(\text{NH}_3)\text{Cl}(\text{en})_2]^{2+} \):** It also forms *cis-* and *trans-* geometrical isomers. The *trans*-isomer is symmetric and optically inactive, whereas the *cis*-isomer is asymmetric and exhibits optical activity.
In simple words: Both complexes form trans shapes which are symmetric, and cis shapes which are asymmetric and have left- and right-handed optical configurations.

Exam Tip: Be sure to draw and label both *cis-* and *trans-* isomers clearly, showing the mirror plane for the optically active *cis-* isomer.

Five Mark Questions

Question 1. For the complex [Fe(en)2Cl2]Cl identify : (2009) (a) the oxidation No. of Iron. (b) the hybrid orbitals and the shape of the complex. (c) the magnetic behavior of the complex. (d) No. of geometrical isomers. (e) whether there is an optical isomer also (f) name of the complex.
Answer:
(a) **Oxidation Number:** Iron is in the \( +3 \) oxidation state (\( x + 0 - 2 = +1 \implies x = +3 \)).
(b) **Hybridization and Shape:** It undergoes \( d^2sp^3 \) hybridization, resulting in an **octahedral** shape.
(c) **Magnetic Behavior:** It is **paramagnetic** (contains one unpaired electron because the strong-field ethylenediamine ligands pair up the \( d^5 \) electrons of \( \text{Fe}^{3+} \) as much as possible).
(d) **Geometrical Isomers:** Two isomers are possible, *cis* and *trans*.
(e) **Optical Isomerism:** Yes, the *cis*-isomer lacks symmetry and possesses a non-superimposable optical isomer.
(f) **IUPAC Name:** dichloridobis(ethane-1,2-diamine)iron(III) chloride.
In simple words: The iron has a +3 charge and forms a standard octahedral structure. The cis shape can form left- and right-handed mirror images, and the compound is weakly magnetic.

Exam Tip: In multi-part questions, number each sub-part clearly (a to f) and give direct, concise answers to ensure you receive full credit.

Question 2. State a reason for each of the following situations (a) Co2+ ion is easily oxidized to Co3+ in the presence of strong ligand. (2010,12) (b) CO is a stronger complexing reagent than NH3 . (2009,12) (c) The molecular shape of [Ni(CO)4 ] is not the same as that of [Ni(CN)4]2−. (2012) (d) Ni does not form low spin octahedral complexes. (2009,10) (e) The \(\pi\) complexes are known for the transition metals only. (2009,10)
Answer:
(a) In the presence of a strong ligand, \( \text{Co}^{3+} \) has a highly stable \( t_{2g}^6 e_g^0 \) configuration, whereas \( \text{Co}^{2+} \) has an unstable \( t_{2g}^6 e_g^1 \) configuration, making the oxidation process thermodynamically highly favorable.
(b) \( \text{CO} \) acts as a \( \pi \)-acceptor ligand that can donate electrons to form a \( \sigma \)-bond and simultaneously accept electrons back from filled metal d-orbitals into its empty \( \pi^* \) antibonding orbitals. This **synergic effect** greatly strengthens the metal-ligand bond.
(c) In \( [\text{Ni}(\text{CO})_4] \), nickel is in a zero oxidation state and undergoes \( sp^3 \) hybridization (**tetrahedral**). In \( [\text{Ni}(\text{CN})_4]^{2-} \), nickel is in a \( +2 \) state and undergoes \( dsp^2 \) hybridization (**square planar**) because the strong-field \( \text{CN}^- \) ligand forces electron pairing.
(d) Nickel(II) is a \( d^8 \) system. Inner-orbital octahedral hybridization requires \( d^2sp^3 \), needing two empty \( 3d \) orbitals. Even with strong ligands, pairing the eight electrons can leave at most one \( 3d \) orbital vacant, making \( d^2sp^3 \) impossible.
(e) Transition metals are unique because they have both empty d-orbitals to accept \( \sigma \)-donation and filled d-orbitals capable of back-donating electrons to empty \( \pi^* \) molecular orbitals of ligands (like ethylene, benzene, or CO).
In simple words: Strong ligands stabilize the +3 charge of cobalt. Carbon monoxide back-bonds to metals, making a much stronger connection than ammonia. Carbonyl nickel is tetrahedral, but nickel cyanide is flat. Nickel has too many d-electrons to vacate two d-orbitals for inner-orbital bonding. Only transition metals can do the dual back-and-forth bonding required for pi-complexes.

Exam Tip: Be sure to write the term "synergic bonding" when discussing carbon monoxide as a ligand to secure maximum marks.

Question 3. Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex: (i) K[Cr(H2O)2(C2O4)2].3H2O (ii) [Co(NH3)5Cl]Cl2 (iii) CrCl3(py)3
Answer:
(i) **For \( \text{K}[\text{Cr}(\text{H}_2\text{O})_2(\text{C}_2\text{O}_4)_2] \cdot 3\text{H}_2\text{O} \):**
- IUPAC Name: Potassium diaquadioxalatochromate(III) trihydrate.
- Oxidation state of Cr = +3.
- Coordination number = 6.
- Electronic configuration: \( 3d^3 \) (\( t_{2g}^3 \)).
- Stereochemistry: Exists as *cis-* and *trans-* geometrical isomers. The *cis*-isomer is chiral.
- Magnetic moment: \( \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\text{ BM} \).

(ii) **For \( [\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2 \):**
- IUPAC Name: Pentaamminechloridocobalt(III) chloride.
- Oxidation state of Co = +3.
- Coordination number = 6.
- Electronic configuration: \( 3d^6 \) (\( t_{2g}^6 \)).
- Stereochemistry: Octahedral shape, symmetric, and optically inactive.
- Magnetic moment: 0 (diamagnetic).

(iii) **For \( \text{CrCl}_3(\text{py})_3 \):**
- IUPAC Name: Trichloridotripyridinechromium(III).
- Oxidation state of Cr = +3.
- Coordination number = 6.
- Electronic configuration: \( 3d^3 \) (\( t_{2g}^3 \)).
- Stereochemistry: Exists as *facial* (*fac-*) and *meridional* (*mer-*) geometrical isomers.
- Magnetic moment: \( \mu = \sqrt{15} \approx 3.87\text{ BM} \).
In simple words: All three complexes have a central metal with a +3 charge and coordinate to 6 sites. The first and third complexes are magnetic with three unpaired electrons, while the second cobalt complex is non-magnetic.

Exam Tip: Remember to calculate the spin-only magnetic moment using the formula \( \mu = \sqrt{n(n+2)} \), showing the steps to secure full marks.

Question 4. (i) Write the IUPAC name and type of isomerism shown by following complex compounds a) [Co(NH3)5(NO2)](NO3)2 b) [Pt(NH3)(H2O)Cl2] (ii) Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?
Answer:
(i)
a) **IUPAC Name:** Pentaamminenitrito-N-cobalt(III) nitrate. **Isomerism:** Linkage isomerism and Ionization isomerism.
b) **IUPAC Name:** Ammineaquachloridoplatinum(II). **Isomerism:** Geometrical isomerism (*cis-* and *trans-*).

(ii) For the square planar complex \( [\text{Pt}(\text{NH}_3)(\text{Br})(\text{Cl})(\text{py})] \), three geometrical isomers exist. None of these will show optical isomerism because square planar complexes have a plane of symmetry, making them achiral and optically inactive.
In simple words: The cobalt complex shows linkage and ionization isomers, and the platinum complex shows cis and trans shapes. The four-ligand platinum complex has three different flat shapes, but none of them are chiral.

Exam Tip: Tetrahedral and square planar complexes containing four monodentate ligands rarely show optical isomerism due to structural symmetry.

Question 5. (i) If to an aqueous solution of CuSO4 in two tubes, we add ammonia solution in one tube and HCl(aq) to the other tube, how the colour of the solutions will change? Explain with the help of reaction. (ii) Write the IUPAC name of complex compounds formed during the process
Answer:
(i) **Color Changes and Reactions:**
- **Tube 1 (addition of ammonia):** The light blue color of \( [\text{Cu}(\text{H}_2\text{O})_4]^{2+} \) shifts to deep blue because ammonia acts as a stronger ligand, replacing water to form a tetraammine complex: \[ [\text{Cu}(\text{H}_2\text{O})_4]^{2+} + 4\text{NH}_3 \rightarrow [\text{Cu}(\text{NH}_3)_4]^{2+} + 4\text{H}_2\text{O} \]
- **Tube 2 (addition of HCl):** The light blue color changes to yellow due to the formation of a tetrachlorido complex: \[ [\text{Cu}(\text{H}_2\text{O})_4]^{2+} + 4\text{Cl}^- \rightarrow [\text{CuCl}_4]^{2-} + 4\text{H}_2\text{O} \]
(ii) **IUPAC Names:**
- (a) \( [\text{Cu}(\text{NH}_3)_4]^{2+} \): Tetraamminecopper(II) ion.
- (b) \( [\text{CuCl}_4]^{2-} \): Tetrachloridocuprate(II) ion.
In simple words: Adding ammonia turns copper sulfate deep blue because a copper-ammonia complex is made. Adding hydrochloric acid turns it yellow because a copper-chlorine complex is made.

Exam Tip: Be sure to write the charge of the copper complex correctly (\( 2+ \) for the cationic ammine complex, and \( 2- \) for the anionic chlorido complex).

Coordination Compounds (Worksheet 2)

1 Mark Questions

Question 21. Explain coordination entity with example.
Answer: A coordination entity consists of a central metal atom or ion bonded to a fixed number of ions or molecules (known as ligands). For example, \( [\text{Co}(\text{NH}_3)_3\text{Cl}_3] \) is a neutral coordination entity.
In simple words: A coordination entity is a metal atom packaged together inside brackets with a specific group of surrounding molecules.

Exam Tip: Always provide a simple neutral or ionic complex as an example to illustrate your definition.

Question 22. What do you understand by coordination compounds?
Answer: Coordination compounds are chemical compounds that contain complex ions, where a central metal atom or ion is bonded to a fixed number of ions or neutral molecules (ligands) via coordinate covalent bonds. For example, \( [\text{Co}(\text{NH}_3)_3\text{Cl}_3] \).
In simple words: These are chemical compounds that contain complex molecular units held together by coordinate bonds.

Exam Tip: Use the term "coordinate covalent bond" or "dative bond" to define the connection between the metal and ligands.

Question 23. What is coordination number?
Answer: The coordination number of a metal ion in a complex is the total number of ligand donor atoms to which the metal ion is directly bonded. For example, the coordination number of cobalt in \( [\text{Co}(\text{NH}_3)_6]^{3+} \) is 6.
In simple words: Coordination number is the total number of chemical bonds the central metal forms with surrounding ligands.

Exam Tip: Remember that bidentate ligands (like ethylenediamine or oxalate) contribute two donor bonds each to the coordination number.

Question 24. Name the different types of isomerisms in coordination compounds.
Answer: Coordination compounds exhibit structural isomerism (including linkage, ionization, hydrate, and coordination isomerism) and stereoisomerism (geometrical and optical isomerism).
In simple words: Isomerism in complexes is divided into structural forms (different bonding connections) and stereo forms (different spatial layouts).

Exam Tip: Always draw a simple flowchart of the different types of isomerisms to make your descriptive answers highly structured.

Question 25. Draw the structure of xenon difluoride.
Answer: Xenon difluoride has a trigonal bipyramidal geometry with three equatorial lone pairs, resulting in a **linear** molecular shape. In simple words: Xenon difluoride has a straight, linear structure with xenon in the center and fluorine on both ends.

Exam Tip: Specify that although the electron pair geometry is trigonal bipyramidal, the physical molecular shape is linear.

Question 26. What is spectrochemical series?
Answer: The spectrochemical series is a list of ligands arranged in the order of their increasing crystal field splitting strength (\( \Delta \)): \[ \text{I}^- < \text{Br}^- < \text{SCN}^- < \text{Cl}^- < \text{S}^{2-} < \text{F}^- < \text{OH}^- < \text{C}_2\text{O}_4^{2-} < \text{H}_2\text{O} < \text{NCS}^- < \text{EDTA}^{4-} < \text{NH}_3 < \text{en} < \text{CN}^- < \text{CO} \]
In simple words: It is a scale that ranks ligands from weak to strong based on how much they split the metal's d-orbitals.

Exam Tip: Keep in mind that \( \text{I}^- \) is the weakest ligand on this scale, while \( \text{CO} \) is the strongest.

Question 27. What do you understand by denticity of a ligand?
Answer: Denticity refers to the number of donor atoms through which a single ligand binds to the central metal atom or ion in a coordination complex. For example, ethane-1,2-diamine is a bidentate ligand (denticity = 2).
In simple words: Denticity is the number of bonding connections a single ligand can make with a metal atom at the same time.

Exam Tip: Use terms like monodentate, bidentate, and polydentate to categorize ligands based on their denticity.

Question 28. Why is CO a stronger ligand than Cl-?
Answer: Carbon monoxide (\( \text{CO} \)) is a strong-field ligand because of its ability to participate in \( \pi \)-backbonding (synergic effect) with the central metal, whereas chloride (\( \text{Cl}^- \)) is a weak-field ligand capable only of basic \( \sigma \)-donation.
In simple words: Carbon monoxide can accept electrons back from the metal, forming a very strong double-sided bond. Chloride can only donate electrons in one direction.

Exam Tip: Explain that \( \pi \)-backbonding involves filled metal d-orbitals overlapping with empty ligand \( \pi^* \) antibonding orbitals.

Question 29. Why are low spin tetrahedral complexes not formed?
Answer: Low-spin tetrahedral complexes are not formed because the crystal field splitting energy in a tetrahedral field (\( \Delta_t \)) is very small (about \( \frac{4}{9}\Delta_o \)) and always less than the electron pairing energy (\( P \)), meaning electrons prefer to occupy the upper orbitals rather than pair up.
In simple words: The energy gap in tetrahedral shapes is too small, so electrons would rather jump to the higher level than pair up in the lower level.

Exam Tip: Always state the mathematical relation \( \Delta_t \approx \frac{4}{9}\Delta_o \) to justify why pairing does not occur.

Question 30. Square planar complexes with coordination number 4 exhibit geometrical isomerism whereas tetrahedral complexes do not. Why?
Answer: In a tetrahedral complex, all four positions are adjacent and equidistant from each other, meaning no distinct *cis* and *trans* configurations exist. In square planar complexes, adjacent ligands are *cis* (\( 90^\circ \)) and opposite ligands are *trans* (\( 180^\circ \)), allowing for distinct geometrical isomers.
In simple words: All four positions in a tetrahedral shape are identical relative to each other, but a square planar shape has distinct side-by-side (cis) and opposite (trans) positions.

Exam Tip: Draw a simple 2D sketch of both geometries to make this structural difference clear.

Question 31. What are crystal fields?
Answer: The crystal field is the negative electrostatic field generated around the central metal ion by incoming anionic ligands (point charges) or neutral dipolar ligands (point dipoles). This field splits the degenerate d-orbitals of the metal.
In simple words: The crystal field is the electrostatic charge cloud created by ligands around a metal, which splits the metal's d-orbitals into different energy levels.

Exam Tip: Focus on the ionic model assumptions of crystal field theory to define this field correctly.

Question 32. What is meant by chelate effect? Give an example .
Answer: The chelate effect refers to the enhanced thermodynamic stability of a complex formed by bi- or polydentate ligands that coordinate through multiple donor atoms to form stable 5- or 6-membered heterocyclic rings. For example, \( [\text{PtCl}_2(\text{en})] \).
In simple words: The chelate effect is the extra stability gained when ligands form ring loops around the metal atom.

Exam Tip: Mention that chelating ligands are much more stable than comparable monodentate complexes.

Question 33. What do you understand by ambidentate ligand?
Answer: An ambidentate ligand is a monodentate ligand that has two different donor atoms and can coordinate through either of them, but only one at a time. Examples include nitrito-N (\( \text{-NO}_2 \)) and nitrito-O (\( \text{-ONO} \)).
In simple words: These are ligands that have two different atoms they can use to bind to a metal, but they only use one of them at a time.

Exam Tip: Draw the simple binding symbols showing the donor atoms clearly to make your explanation robust.

Question 34. What is the difference between homoleptic and heteroleptic complexes?
Answer: In homoleptic complexes, the central metal ion is bound to only one kind of donor groups (e.g., \( [\text{Co}(\text{NH}_3)_6]^{3+} \)). In heteroleptic complexes, the metal is bound to more than one type of donor groups (e.g., \( [\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+ \)).
In simple words: Homoleptic complexes have only one type of ligand attached, while heteroleptic complexes have a mixture of different ligands.

Exam Tip: Provide one simple example for each class to show a thorough understanding.

Question 35. Give one limitation for crystal field theory.
Answer: Since CFT treats ligands as point charges, anionic ligands should exert a larger splitting effect than neutral ones, yet they are found at the lower end of the spectrochemical series. It also completely ignores the covalent character of metal-ligand bonds.
In simple words: The theory treats bonds as purely ionic, which fails to explain why some neutral ligands split d-orbitals more strongly than negative ions.

Exam Tip: Always list "ignoring covalent character" or "failure to explain anionic ligand trends" as primary limitations.

Question 36. How many ions are produced from the complex: [Co(NH3)6]Cl2
Answer: In aqueous solution, \( [\text{Co}(\text{NH}_3)_6]\text{Cl}_2 \) dissociates to produce a total of three ions: one \( [\text{Co}(\text{NH}_3)_6]^{2+} \) complex cation and two \( \text{Cl}^- \) anions.
In simple words: When dissolved in water, the compound splits into one large complex ion and two simple chlorine ions.

Exam Tip: Remember that ligands inside the square brackets stay bonded together as a single ion and do not dissociate in solution.

Question 37. The oxidation number of cobalt in K[Co(CO)4]
Answer: The oxidation number of cobalt in \( \text{K}[\text{Co}(\text{CO})_4] \) is \( -1 \) (since \( \text{K} = +1 \) and \( \text{CO} \) is neutral).
In simple words: The cobalt in this complex carries a rare negative charge of -1.

Exam Tip: Carbonyl complexes of transition metals can exhibit zero or negative oxidation states.

Question 38. Which compound is used to estimate the hardness of water volumetrically?
Answer: Ethylenediaminetetraacetic acid (EDTA) is used as a complexing agent to estimate water hardness volumetrically.
In simple words: EDTA is used to measure the amount of calcium and magnesium ions in hard water.

Exam Tip: EDTA forms highly stable complexes with \( \text{Ca}^{2+} \) and \( \text{Mg}^{2+} \) ions, which is the basis of this volumetric titration.

Question 39. Magnetic moment of [MnCl4]2- is 5.92B.M explain with reason.
Answer: A magnetic moment of \( 5.92\text{ BM} \) indicates the presence of 5 unpaired electrons in the d-orbitals of \( \text{Mn}^{2+} \) (\( 3d^5 \)). Since chloride is a weak-field ligand, it does not force electron pairing, resulting in high-spin \( sp^3 \) tetrahedral hybridization with 5 unpaired spins.
In simple words: The magnetic moment confirms that manganese has five unpaired electrons, which occurs because the weak chloride ligands cannot force them to pair up.

Exam Tip: Use the spin-only formula \( \mu = \sqrt{n(n+2)} \) to show how \( n=5 \) is deduced from \( 5.92\text{ BM} \).

Question 40. How many donor atoms are present in EDTA ligand?
Answer: The EDTA ligand has a total of six (6) donor atoms: two nitrogen atoms and four oxygen atoms from the carboxylate groups.
In simple words: EDTA has six grabbing hands (two nitrogens and four oxygens) to bind with a metal ion.

Exam Tip: Identify EDTA as a "hexadentate" ligand to demonstrate your knowledge to the examiner.

2 Mark Questions

Question 11. Give the electronic configuration of the following complexes on the basis of crystal field splitting theory.
iv) [CoF6]3-
v) [Fe(CN)6]4-
Answer:
iv) For \( [\text{CoF}_6]^{3-} \): The metal ion is \( \text{Co}^{3+} \) (\( d^6 \)). Since fluoride (\( \text{F}^- default \)) is a weak-field ligand, the crystal field splitting energy is small (\( \Delta_o < P \)), resulting in a high-spin state with configuration: \( t_{2g}^4 e_g^2 \).
v) For \( [\text{Fe}(\text{CN})_6]^{4-} \): The metal ion is \( \text{Fe}^{2+} \) (\( d^6 \)). Since cyanide (\( \text{CN}^- \)) is a strong-field ligand, the crystal field splitting energy is large (\( \Delta_o > P \)), forcing electron pairing in a low-spin configuration: \( t_{2g}^6 e_g^0 \).
In simple words: The cobalt complex has weak ligands and spreads its six d-electrons across both energy levels. The iron complex has strong ligands and forces all six d-electrons into the lower level.

Exam Tip: Use the subscripts \( t_{2g} \) and \( e_g \) properly in your electronic configuration answers to reflect CFT principles.

Question 12. Explain the following with examples:
iv) Linkage isomerism
v) Outer orbital complex
Answer:
iv) **Linkage Isomerism:** This isomerism arises when a coordination complex contains an ambidentate ligand that can bind through different donor atoms. For example, \( [\text{Co}(\text{NH}_3)_5(\text{NO}_2)]\text{Cl}_2 \) and \( [\text{Co}(\text{NH}_3)_5(\text{ONO})]\text{Cl}_2 \).
v) **Outer Orbital Complex:** An outer orbital complex is formed when the outer \( nd \) orbitals (along with \( ns \) and \( np \) orbitals) participate in hybridization because the weak-field ligands cannot force the pairing of electrons in the inner \( (n-1)d \) subshell. For example, \( [\text{CoF}_6]^{3-} \) undergoes \( sp^3d^2 \) hybridization.
In simple words: Linkage isomers have the same formula but the ligand connects through different atoms. Outer orbital complexes use higher-energy outer d-orbitals for bonding because the ligands are too weak to pair up inner electrons.

Exam Tip: Differentiate clearly between inner orbital complexes (\( d^2sp^3 \)) and outer orbital complexes (\( sp^3d^2 \)) using their ligand field strengths.

Question 13. i)Low spin octahedral complexes of nickel are not found . Explain why?
ii)the\(\pi\) complexes are known for transition elements only.explain.
Answer:
i) Nickel(II) is a \( d^8 \) system. octahedral low-spin hybridization requires \( d^2sp^3 \), which needs two vacant inner \( 3d \) orbitals. Even under the strongest field ligands, the pairing of eight electrons can vacate at most only one \( 3d \) orbital, making \( d^2sp^3 \) impossible.
ii) Transition elements are unique because they have both empty d-orbitals to accept \( \sigma \)-donation from the ligand, and filled d-orbitals capable of back-donating electrons into the empty \( \pi^* \) antibonding molecular orbitals of ligands (synergic effect).
In simple words: Nickel has too many d-electrons, so it cannot free up two inner orbitals even with strong ligands. Only transition metals can perform the back-and-forth bonding required for pi-complexes.

Exam Tip: Highlight that \( \text{Ni}^{2+} \) always forms outer orbital octahedral complexes with \( sp^3d^2 \) hybridization.

Question 14. How would you account for the following:
iii) [Ti(H2O)6]3+ is coloured while [Sc(H2O)6]3+ is colourless.
iv) [Ni(CO)4] possess tetrahedral geometry whereas [Ni(CN)4]2- is square planar.
Answer:
iii) \( \text{Ti}^{3+} \) is a \( d^1 \) system; the single unpaired d-electron can undergo visible \( d\text{-}d \) transition, causing color. \( \text{Sc}^{3+} \) is a \( d^0 \) system with no d-electrons, so \( d\text{-}d \) transition is impossible, making it colorless.
iv) In \( [\text{Ni}(\text{CO})_4] \), nickel is in zero oxidation state and undergoes \( sp^3 \) hybridization (tetrahedral). In \( [\text{Ni}(\text{CN})_4]^{2-} \), nickel is in the \( +2 \) state and the strong ligand \( \text{CN}^- \) forces pairing, allowing \( dsp^2 \) hybridization (square planar).
In simple words: Titanium(III) has a d-electron that can absorb light and jump levels, making it colored. Scandium(III) has no d-electrons and is colorless. Carbonyl nickel has no charge and is tetrahedral, while nickel-cyanide has a +2 charge and is flat.

Exam Tip: Connect the geometry of \( [\text{Ni}(\text{CN})_4]^{2-} \) to \( dsp^2 \) hybridization and that of \( [\text{Ni}(\text{CO})_4] \) to \( sp^3 \) hybridization.

Question 15. State reasons for each of the following:
iii) All the P-Cl bonds in PCl5 molecule are not equivalent.
iv) S has greater tendency for catenation than O.
Answer:
iii) In \( \text{PCl}_5 \) (trigonal bipyramidal), the two axial bonds suffer greater electrostatic repulsion from the equatorial bond pairs, making them longer and weaker than the three equatorial bonds.
iv) Catenation depends on the element-element bond strength. The S-S bond is much stronger (\( 213\text{ kJ/mol} \)) than the O-O bond (\( 138\text{ kJ/mol} \)) due to the absence of severe lone pair repulsions present in smaller oxygen atoms.
In simple words: The axial bonds in phosphorus pentachloride are pushed harder by other bonds, making them longer. Sulfur can chain with itself better because its bonds are stronger, as it doesn't suffer the electronic crowding that oxygen does.

Exam Tip: Be sure to write the bond energy values of S-S (\( 213\text{ kJ/mol} \)) and O-O (\( 138\text{ kJ/mol} \)) to strengthen your answer.

Question 16. Give the stereochemistry and the magnetic behaviour of the following complexes:
iii) [Co(NH3)5Cl]Cl2
iv) K2[Ni(CN)4]
Answer:
iii) **\( [\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2 \):** Stereochemistry is **octahedral** (optically inactive). Magnetic behavior is **diamagnetic** (all electrons are paired due to strong-field ammine ligands).
iv) **\( \text{K}_2[\text{Ni}(\text{CN})_4] \):** Stereochemistry is **square planar** (optically inactive). Magnetic behavior is **diamagnetic** (all electrons are paired due to strong-field cyanide ligands).
In simple words: The first is a symmetric octahedral shape and is non-magnetic. The second is a symmetric flat square shape and is also non-magnetic.

Exam Tip: Always list both the structural shape and whether the complex contains unpaired spins when asked for magnetic behavior.

Question 17. Draw the structures of isomers if any and write the names of the following complexes:
iii) [Cr(NH3)4Cl2]+
iv) [Co(en)3]3+
Answer:
iii) **\( [\text{Cr}(\text{NH}_3)_4\text{Cl}_2]^+ \):** Name is **tetraamminedichloridochromium(III) ion**. It exists as *cis-* and *trans-* geometrical isomers.
iv) **\( [\text{Co}(\text{en})_3]^{3+} \):** Name is **tris(ethane-1,2-diamine)cobalt(III) ion**. It lacks a plane of symmetry and exists as a pair of optically active dextro (\( d \)) and laevo (\( l \)) isomers.
In simple words: The first complex forms cis and trans shapes. The second complex forms non-superimposable left- and right-handed mirror images.

Exam Tip: Make sure to include the ionic charge on the brackets when writing the structures of these complex ions.

Question 18. State reasons for each of the following:
iv) The N-O bond in NO2- is shorter than the N-O bond in NO3-
v) SF6 is kinetically an inert substance.
Answer:
iv) The N-O bond in \( \text{NO}_2^- \) is a resonance hybrid of a single and a double bond, resulting in an average bond order of 1.5. The N-O bond in \( \text{NO}_3^- \) is a resonance hybrid of two single bonds and one double bond, resulting in a lower average bond order of 1.33. Since a higher bond order yields a shorter bond, \( \text{NO}_2^- \) has shorter bonds.
v) In \( \text{SF}_6 \), the central sulfur atom is sterically protected by six fluorine atoms, preventing attacking reagents (like water) from accessing the sulfur atom. Moreover, fluorine has no vacant d-orbitals to accept extra electrons.
In simple words: The nitrogen-oxygen bonds in nitrite have more double-bond character than those in nitrate, making them shorter. Sulfur hexafluoride is unreactive because the central sulfur atom is completely shielded by six fluorine atoms.

Exam Tip: State clearly that steric protection by six fluorine atoms is the primary reason for the kinetic inertness of \( \text{SF}_6 \).

Question 19. Hydrated copper sulphate is blue in colour whereas anhydrous copper sulphate is colourless. Why?
Answer: In hydrated copper sulphate, water molecules act as ligands that split the d-orbitals of the \( \text{Cu}^{2+} \) ion. This allows for a visible-range \( d\text{-}d \) transition. Anhydrous copper sulphate lacks ligands, so no d-orbital splitting occurs, rendering it colorless.
In simple words: The water molecules in hydrated copper sulfate act as ligands that split the d-orbitals, allowing electrons to absorb light and create a blue color. Without water, there is no splitting, so it remains colorless.

Exam Tip: Use the term "absence of crystal field splitting" to explain why anhydrous salts are colorless.

Question 20. Calculate the magnetic moment of the metal ions present in the following complexes:
iii) [Cu(NH3)4]SO4
iv) [Ni(CN)4]2-
Answer:
iii) In \( [\text{Cu}(\text{NH}_3)_4]\text{SO}_4 \), \( \text{Cu}^{2+} \) has a \( 3d^9 \) configuration with \( n = 1 \) unpaired electron: \[ \mu = \sqrt{1(1+2)} = 1.732\text{ BM} \]
iv) In \( [\text{Ni}(\text{CN})_4]^{2-} \), \( \text{Ni}^{2+} \) has a \( 3d^8 \) configuration. Because \( \text{CN}^- \) is a strong-field ligand, all electrons are paired up (\( n = 0 \)): \[ \mu = 0\text{ BM} \]
In simple words: The copper complex has one unpaired electron and is magnetic, while the nickel complex has all its electrons paired up and is non-magnetic.

Exam Tip: Always show the formula \( \mu = \sqrt{n(n+2)} \) and state the number of unpaired electrons (\( n \)) before calculating.

Concept 4: Three Mark Questions

Question 2. (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) A ligand is an ion, atom, or molecule that is directly coordinated to the central metal atom or ion by donating one or more pairs of electrons. Example of a bidentate ligand: ethylenediamine (\( \text{H}_2\text{N-CH}_2\text{-CH}_2\text{-NH}_2 \)).

(b) Both complexes are **diamagnetic** (no unpaired electrons), but they have different structures:
- In \( [\text{Ni}(\text{CN})_4]^{2-} \), nickel is in the \( +2 \) state (\( 3d^8 \)). The strong-field \( \text{CN}^- \) ligand pairs up the \( 3d \) electrons, leaving one empty \( 3d \) orbital for \( dsp^2 \) hybridization, resulting in a **square planar** geometry.
- In \( [\text{Ni}(\text{CO})_4] \), nickel is in the zero state (\( 3d^8 4s^2 \)). The strong-field \( \text{CO} \) ligand forces the \( 4s \) electrons to pair into the \( 3d \) subshell, giving a completely filled \( 3d^{10} \) configuration. It undergoes \( sp^3 \) hybridization, resulting in a **tetrahedral** geometry.
In simple words: A ligand is an electron donor that binds to a metal. Both nickel complexes are non-magnetic, but the first is flat and square because of its charge and orbital arrangement, while the second is a pyramid-like tetrahedron.

Exam Tip: Highlight that both complexes are diamagnetic (\( \mu = 0 \)) despite their completely different geometries and nickel oxidation states.

Question 2. Nomenclate the following complexes:
i) [Co(NH3)5(CO3)]Cl
ii)[COCl2(en)2]Cl
iii) Fe4[Fe(CN) 6]3
Answer: The systematic IUPAC names are:
i) pentaamminecarbonatocobalt(III) chloride
ii) dichloridobis(ethane-1,2-diamine)cobalt(III) chloride
iii) iron(III) hexacyanidoferrate(II)
In simple words: These are the correct systematic names for the given cobalt and iron complexes based on IUPAC rules.

Exam Tip: For the third complex, make sure to use Roman numerals to specify both the outer iron oxidation state (III) and the inner iron oxidation state (II).

Question 3. (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) This is due to the presence of weak-field or strong-field ligands. Strong-field ligands cause electron pairing (yielding low-spin, diamagnetic, or weakly paramagnetic states), whereas weak-field ligands do not force pairing (yielding high-spin, strongly paramagnetic states). For example, \( [\text{CoF}_6]^{3-} \) is paramagnetic, while \( [\text{Co}(\text{NH}_3)_6]^{3+} \) is diamagnetic.
(b) The observed color of a complex is complementary to the wavelength of light absorbed. A larger crystal field splitting (\( \Delta \)) means shorter wavelengths of light are absorbed, altering the complementary color.
In simple words: Even with the same shape, strong ligands force electrons to pair up (reducing magnetism), while weak ligands leave them unpaired. The color we see is the leftover light that was not absorbed by the electrons.

Exam Tip: Use the term "complementary color" when describing the relationship between absorption and observation of light in coordination chemistry.

Question 4. 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) **Ambident Ligand:** A monodentate ligand that has more than one donor site but binds through only one at a time. Example: \( \text{NO}_2^- \) (nitrito-N or nitrito-O).
(b) **Denticity:** The number of donor atoms through which a ligand binds to the central metal atom. Example: bidentate oxalate ion (\( \text{C}_2\text{O}_4^{2-} \)) has a denticity of 2.
(c) **Crystal Field Splitting in Octahedral Field:** The separation of the five degenerate d-orbitals into three lower-energy \( t_{2g} \) orbitals and two higher-energy \( e_g \) orbitals due to the electrostatic field of incoming ligands.
In simple words: An ambident ligand can attach to a metal in two different ways. Denticity is the number of coordinate bonds a ligand makes. Crystal field splitting is the dividing of d-orbitals into different energy groups.

Exam Tip: Be ready to draw a simple crystal field splitting diagram for part (c) to support your written explanation.

Question 5. (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) Water molecules act as ligands in the pentahydrate, splitting the d-orbitals of \( \text{Cu}^{2+} \) to allow blue-color \( d\text{-}d \) transitions. Anhydrous copper sulphate lacks ligands, so no d-orbital splitting can occur.
(b) The S-S bond strength (\( 213\text{ kJ/mol} \)) is much greater than the O-O bond strength (\( 138\text{ kJ/mol} \)), because the small size of oxygen atoms causes severe repulsion between their non-bonding lone pairs.
In simple words: Water molecules act as ligands that split the d-orbitals, letting electrons absorb light and make the crystals blue. Sulfur chains with itself better than oxygen because its larger size reduces lone pair repulsions, making S-S bonds much stronger.

Exam Tip: Focus on the difference in bond energies (\( 213\text{ kJ/mol} \) vs \( 138\text{ kJ/mol} \)) to explain why sulfur catenates more than oxygen.

Question 6. 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
Answer:
(a) \( [\text{Fe}(\text{NH}_3)_2(\text{CN})_4]^- \) exhibits **geometrical isomerism** with distinct *cis-* and *trans-* configurations.
(b) \( [\text{CrCl}_2(\text{ox})_2]^{3-} \) also exhibits **geometrical isomerism** with *cis-* and *trans-* forms. The *cis*-isomer is optically active.
(c) \( [\text{Co}(\text{en})_3]\text{Cl}_3 \) does not show geometrical isomerism, but it exists as a pair of optically active **dextro** and **laevo** mirror images.
In simple words: The first two complexes form cis and trans shapes. The third complex has only one geometric shape, but it can exist as left- and right-handed mirror images.

Exam Tip: When drawing structures for bidentate complexes, draw thick loops to represent the bidentate ligands (like oxalate or ethylenediamine) clearly.

Question 7. 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-
Answer:
(i) **\( [\text{Co}(\text{en})_3]\text{Cl}_3 \):** Hybridization is \( d^2sp^3 \), Shape is **octahedral**, Magnetic behavior is **diamagnetic**.
(ii) **\( \text{K}_2[\text{Ni}(\text{CN})_4] \):** Hybridization is \( dsp^2 \), Shape is **square planar**, Magnetic behavior is **diamagnetic**.
(iii) **\( [\text{Fe}(\text{CN})_6]^{3-} \):** Hybridization is \( d^2sp^3 \), Shape is **octahedral**, Magnetic behavior is **paramagnetic** (contains 1 unpaired electron).
In simple words: The first and third are octahedral complexes, but only the third is magnetic. The second nickel complex is flat and non-magnetic.

Exam Tip: Specify the hybridization clearly, noting whether it uses inner (\( d^2sp^3 \)) or outer (\( sp^3d^2 \)) orbitals.

Question 8. 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) \( \text{Ti}^{3+} \) is a \( 3d^1 \) system; the single unpaired d-electron can undergo visible \( d\text{-}d \) transition, causing color. \( \text{Sc}^{3+} \) is a \( 3d^0 \) system with no d-electrons, so \( d\text{-}d \) transition is impossible, making it colorless.
(ii) In \( [\text{Fe}(\text{CN})_6]^{3-} \), \( \text{Fe}^{3+} \) has a \( 3d^5 \) configuration. The strong cyanide ligand pairs the electrons, leaving one unpaired electron (weakly paramagnetic). In \( [\text{Fe}(\text{CN})_6]^{4-} \), \( \text{Fe}^{2+} \) is \( 3d^6 \), which pairs completely under the strong ligand field (diamagnetic).
(iii) In \( \text{Ni(CO)}_4 \), nickel is in the zero oxidation state and undergoes \( sp^3 \) hybridization (tetrahedral). In \( [\text{Ni}(\text{CN})_4]^{2-} \), nickel is in the \( +2 \) state and undergoes \( dsp^2 \) hybridization (square planar).
In simple words: Titanium(III) has a d-electron that can absorb light and jump levels, making it colored. Scandium(III) has no d-electrons and is colorless. Carbonyl nickel has no charge and is tetrahedral, while nickel-cyanide has a +2 charge and is flat.

Exam Tip: Differentiate clearly between weak-field and strong-field ligand behavior to explain why one complex is colored while the other is colorless.

Question 11. Explain the following ::
(iv) low spin octahedral complexes of Ni are not known.
(v) The pi - complexes are known for the transition elements only.
(vi) CO is a stronger ligand than NH3 for many metals
Answer:
(iv) Nickel(II) is \( d^8 \). Inner-orbital \( d^2sp^3 \) hybridization requires two vacant \( 3d \) orbitals. Even with the strongest field ligands, pairing only vacates one \( 3d \) orbital, which is insufficient.
(v) Only transition elements have vacant d-orbitals to accept electron pairs from ligands, and filled d-orbitals to back-donate electrons into empty \( \pi^* \) antibonding molecular orbitals of ligands (synergic effect).
(vi) CO can form a synergic bond (metal-to-ligand backbonding), whereas \( \text{NH}_3 \) is only a \( \sigma \)-donor and cannot accept back-donated electron density.
In simple words: Nickel has too many d-electrons, so it cannot free up two inner orbitals even with strong ligands. Only transition metals can do the back-and-forth bonding required for pi-complexes.

Exam Tip: Focus on the molecular orbital and d-electron count of \( \text{Ni}^{2+} \) to explain why it cannot undergo \( d^2sp^3 \) hybridization.

Question 12. 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 metal ion and the ligands at equilibrium. It is expressed quantitatively by the stability constant (\( K \)): \[ \text{M} + 4\text{L} \rightleftharpoons \text{ML}_4 \implies K = \frac{[\text{ML}_4]}{[\text{M}][\text{L}]^4} \]
Factors governing stability:
1. **Charge on central metal ion:** Higher charge leads to stronger electrostatic attractions and greater stability.
2. **Size of metal ion:** Smaller size yields higher charge density and stronger bonding.
3. **Basic strength of ligand:** More basic ligands donate electron pairs more easily, forming more stable complexes.
4. **Chelate effect:** Complexes with bidentate or polydentate ligands that form rings are highly stable.
In simple words: Stability is how tightly a metal holds onto its ligands in water. Metals with higher positive charges and smaller sizes, as well as ligands that form closed ring loops, make the most stable complexes.

Exam Tip: State clearly that a higher value of the stability constant (\( K \)) indicates a more stable complex.

5 marks questions

Question 6. 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-
Answer:
- (a) \( [\text{Fe}(\text{NH}_3)_2(\text{CN})_4]^- \): Displays *cis-* and *trans-* geometrical isomers.
- (b) \( [\text{CrCl}_2(\text{ox})_2]^{3-} \): Geometrical isomers are *cis* and *trans*. The *cis*-isomer is chiral and optically active.
- (c) and (d) \( [\text{Co}(\text{en})_3]\text{Cl}_3 \): No geometrical isomers exist. However, it displays optical isomerism (existing as non-superimposable dextro and laevo forms).
- (e) \( [\text{Fe}(\text{CN})_6]^{3-} \): Symmetric octahedral structure with no geometrical or optical isomers.
In simple words: The first two compounds have distinct cis and trans shapes, whereas the cobalt complexes have left- and right-handed mirror images.

Exam Tip: Be sure to draw clear structural formulas, showing thick loops for chelating ligands (oxalate, ethylenediamine) to earn full marks.

Question 7. What is crystal field theory for octahedral complexes? Also write the limitations of this theory.
Answer: **Crystal Field Theory (CFT):** It is an electrostatic model that treats the metal-ligand bond as purely ionic, arising from electrostatic interactions between the metal cation and ligands (treated as point charges/dipoles). In an octahedral field, the approaching ligands split the five degenerate d-orbitals into three lower-energy \( t_{2g} \) orbitals (\( d_{xy} \), \( d_{yz} \), \( d_{zx} \)) and two higher-energy \( e_g \) orbitals (\( d_{x^2-y^2} \), \( d_{z^2} \)).

**Limitations of CFT:**
1. Since CFT treats ligands as point charges, anionic ligands should exert a larger splitting effect than neutral ones, yet they are found at the lower end of the spectrochemical series.
2. It completely ignores the covalent character of metal-ligand bonds.
In simple words: This theory assumes bonds are purely ionic and splits d-orbitals into two energy levels. However, it fails because it ignores covalent bonding and cannot explain why some neutral ligands are stronger than negative ions.

Exam Tip: Focus on explaining the d-orbital splitting and listing the two primary limitations of the theory.

Question 8. Write the state of hybridisation the shape and the magnetic behaviour of the following complex entities:
(vi) [Cr(NH3)4Cl2]Cl
(vii) [Co(en)3]Cl3
(viii) K2[NiCl4]
(ix) [Fe(H2O)6]2+
(x) [NiCl4]2-
Answer:
(vi) **\( [\text{Cr}(\text{NH}_3)_4\text{Cl}_2]\text{Cl} \):** Hybridization is \( d^2sp^3 \), Shape is **octahedral**, Magnetic behavior is **paramagnetic**.
(vii) **\( [\text{Co}(\text{en})_3]\text{Cl}_3 \):** Hybridization is \( d^2sp^3 \), Shape is **octahedral**, Magnetic behavior is **diamagnetic**.
(viii) **\( \text{K}_2[\text{NiCl}_4] \):** Hybridization is \( sp^3 \), Shape is **tetrahedral**, Magnetic behavior is **paramagnetic**.
(ix) **\( [\text{Fe}(\text{H}_2\text{O})_6]^{2+} \):** Hybridization is \( sp^3d^2 \), Shape is **octahedral**, Magnetic behavior is **paramagnetic** (4 unpaired electrons).
(x) **\( [\text{NiCl}_4]^{2-} \):** Hybridization is \( sp^3 \), Shape is **tetrahedral**, Magnetic behavior is **paramagnetic**.
In simple words: Complexes with coordination number 6 are octahedral, while those with coordination number 4 are tetrahedral. Except for the cobalt complex, most of these are magnetic due to unpaired electrons.

Exam Tip: Be sure to write the correct hybridization and shape for each complex ion to earn full marks.

Question 9. Using valence bond theory explain the following questions in relation to [Co(NH3)6]3+.
(vi) Nomenclature
(vii) Type of hybridisation
(viii) Inner or outer orbital complex
(ix) Magnetic behaviour
(x) Spin only magnetic moment
Answer:
(vi) **Nomenclature:** Hexaamminecobalt(III) ion.
(vii) **Type of Hybridization:** \( d^2sp^3 \) hybridization.
(viii) **Inner/Outer Orbital Complex:** Inner orbital complex (as it utilizes the inner \( 3d \) orbitals).
(ix) **Magnetic Behavior:** Diamagnetic (all electrons are paired).
(x) **Spin-only Magnetic Moment:** \( \mu = 0\text{ BM} \) (as \( n = 0 \)).
In simple words: This cobalt complex is an inner-orbital octahedral complex that has all its electrons paired up, making it non-magnetic.

Exam Tip: Show the step-by-step orbital diagram showing the pairing of \( 3d \) electrons to justify why it is an inner orbital, diamagnetic complex.

Question 10. 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]
Answer:
1. **\( [\text{Co}(\text{NH}_3)_6]^{3+} \):** Hybridization is \( d^2sp^3 \), Geometry is **octahedral**, Magnetic behavior is **diamagnetic** (0 unpaired electrons).
2. **\( [\text{Cr}(\text{NH}_3)_6]^{3+} \):** Hybridization is \( d^2sp^3 \), Geometry is **octahedral**, Magnetic behavior is **paramagnetic** (3 unpaired electrons).
3. **\( [\text{Ni}(\text{CO})_4] \):** Hybridization is \( sp^3 \), Geometry is **tetrahedral**, Magnetic behavior is **diamagnetic** (0 unpaired electrons).
In simple words: The first two are octahedral complexes, but only the chromium complex is magnetic. The third is a non-magnetic tetrahedral complex.

Exam Tip: Use a comparative table to list the hybridization, shape, and magnetic properties side-by-side for a high-scoring answer.