NCERT Solutions for Class 12 Chemistry Chapter 5 Coordination Compounds

The NCERT Solutions Class 12 Coordination Compounds Chapter 5 are provided below by SimplyAcad to help students understand the vital concepts of the chapter thoroughly. These solutions are structurally organised with proper numbering according to the NCERT Chemistry textbook prescribed by CBSE. 

 

Coordination Compounds Chapter 5 is an extremely crucial segment of the Chemistry syllabus and students must revise them daily. The solutions are according to the recent syllabus of CBSE 2024-25 to avoid any confusions and doubts. NCERT Solutions Coordination Compounds Chapter 5 deals with atoms and several other related chemical reactions that form important questions in paper.

 

NCERT Solutions for Class 12 Coordination Compounds Chapter 5

Question1. Write the formulas for the following coordination compounds:

(i) Tetraamminediaquacobalt(III) chloride

(ii) Potassium tetracyanonickelate(II)

(iii) Tris(ethane−1,2−diamine) chromium(III) chloride

(iv) Amminebromidochloridonitrito-N-platinate(II)

(v) Dichloridobis(ethane−1,2−diamine)platinum(IV) nitrate

(vi) Iron(III) hexacyanoferrate(II)

Solution :

Question 2. Write the IUPAC names of the following coordination compounds:

(i) [Co(NH3)6]Cl3

(ii) [Co(NH3)5Cl]Cl2

(iii) K3[Fe(CN)6]

(iv) K3[Fe(C2O4)3] 

(v) K2[PdCl4] 

(vi) [Pt(NH3)2Cl(NH2CH3)]Cl

Solution :
(i) Hexaamminecobalt(III) chloride

(ii) Pentaamminechloridocobalt(III) chloride

(iii) Potassium hexacyanoferrate(III)

(iv) Potassium trioxalatoferrate(III)

(v) Potassium tetrachloridopalladate(II)

(vi) Diamminechlorido(methylamine)platinum(II) chloride

Question3. Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers: \begin{enumerate}\item (\text{K}[\text{Cr}(\text{H}_2\text{O})_2(\text{C}_2\text{O}_4)_2])\item([\text{Co}(\text{en})_3]\text{Cl}_3)\item(\text{Co}(\text{NH}_3)_5(\text{NO}_2)_2) \item ([\text{Pt}(\text{NH}_3)(\text{H}_2\text{O})\text{Cl}_2]) \end{enumerate} \textb

{Solution}\begin{enumerate}\item(\text{K}[\text{Cr}(\text{H}_2\text{O})_2(\text{C}_2\text{O}_4)_2]) This complex exhibits \textbf{geometrical isomerism} (cis and trans forms). \item ([\text{Co}(\text{en})_3]\text{Cl}_3) This complex exhibits \textbf{optical isomerism} due to the presence of three bidentate ethylenediamine (en) ligands. \item (\text{Co}(\text{NH}_3)_5(\text{NO}_2)_2) This complex exhibits \textbf{linkage isomerism} due to the ambidentate nature of the (\text{NO}_2) ligand, which can bind through either nitrogen (nitro) or oxygen (nitrito). \item ([\text{Pt}(\text{NH}_3)(\text{H}_2\text{O})\text{Cl}_2]) This complex exhibits \textbf{geometrical isomerism} (cis and trans forms).

Question4. Give evidence that ([\text{Co}(\text{NH}_3)_5\text{Cl}]\text{SO}_4) and ([\text{Co}(\text{NH}_3)_5(\text{SO}_4)]\text{Cl}) are ionization isomers. \textbf

{Solution} \textbf{Ionization isomers} Ionization isomers are isomers which, upon dissolving, yield different ions and these ions react differently with various reagents. This type of isomerism occurs when different ions are present outside the coordination sphere for two compounds with the same formula. \textbf{Evidence} For ([\text{Co}(\text{NH}_3)_5\text{Cl}]\text{SO}_4): – Upon dissolution, it produces (\text{SO}_4^{2-}) ions. When (\text{SO}_4^{2-}) ions react with (\text{Ba}^{2+}), a white precipitate of (\text{BaSO}_4) is formed, while no reaction occurs with (\text{Ag}^+). [ [\text{Co}(\text{NH}_3)_5\text{Cl}]\text{SO}_4 + \text{Ba}^{2+} \rightarrow \text{BaSO}_4 \downarrow \quad (\text{White precipitate}) ] [ [\text{Co}(\text{NH}_3)_5\text{Cl}]\text{SO}_4 + \text{Ag}^+ \rightarrow \text{No reaction} ] For ([\text{Co}(\text{NH}_3)_5(\text{SO}_4)]\text{Cl}): – Upon dissolution, it produces (\text{Cl}^-) ions. When (\text{Cl}^-) ions react with (\text{Ag}^+), a white precipitate of (\text{AgCl}) is formed, while no reaction occurs with (\text{Ba}^{2+}) because (\text{BaCl}_2) is soluble. [ [\text{Co}(\text{NH}_3)_5(\text{SO}_4)]\text{Cl} + \text{Ba}^{2+} \rightarrow \text{No reaction} ] [ [\text{Co}(\text{NH}_3)_5(\text{SO}_4)]\text{Cl} + \text{Ag}^+ \rightarrow \text{AgCl} \downarrow \quad (\text{White precipitate}) ] These reactions show that the two compounds yield different ions upon dissolution, providing evidence that they are ionization isomers.

Question 5. Explain on the basis of valence bond theory that [Ni(CN)4]2− ion with square planar structure is diamagnetic and the [NiCl4]2− ion with tetrahedral geometry is paramagnetic. : Coordination Compounds Chapter 5

Solution :
Ni is in the +2 oxidation state i.e., in d8 configuration.

There are 4 CN− ions. Thus, it can either have a tetrahedral geometry or square planar geometry. Since CN− ion is a strong field ligand, it causes the pairing of unpaired 3d electrons.

It now undergoes dsp2 hybridization. Since all electrons are paired, it is diamagnetic.

In case of [NiCl4]2−, Cl− ion is a weak field ligand. Therefore, it does not lead to the pairing of unpaired 3d electrons. Therefore, it undergoes sp3 hybridization.

Since there are 2 unpaired electrons in this case, it is paramagnetic in nature.

Question6. [NiCl4]2− is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why?  : Coordination Compounds Chapter 5

Solution :
Though both [NiCl4]2− and [Ni(CO)4] are tetrahedral, their magnetic characters are different. This is due to a difference in the nature of ligands. Cl− is a weak field ligand and it does not cause the pairing of unpaired 3d electrons. Hence, [NiCl4]2− is paramagnetic.

In Ni(CO)4, Ni is in the zero oxidation state i.e., it has a configuration of 3d8 4s2.

But CO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons. Also, it causes the 4s electrons to shift to the 3d orbital, thereby giving rise to sp3 hybridization. Since no unpaired electrons are present in this case, [Ni(CO)4] is diamagnetic.

Question7. ([ \text{Fe}(\text{H}_2\text{O})_6 ]^{3+}) is strongly paramagnetic, whereas ([ \text{Fe}(\text{CN})_6 ]^{3-}) is weakly paramagnetic. Explain. \textbf

{Solution} Paramagnetic strength is directly proportional to the number of unpaired electrons. In both complexes, iron (Fe) has a (+3) oxidation state, which corresponds to a (d^5) electron configuration. \textbf{Case 1:} ([ \text{Fe}(\text{H}_2\text{O})_6 ]^{3+}) – The (\text{H}_2\text{O}) ligand is a weak field ligand, which does not cause pairing of the (d)-electrons. – Therefore, all five (d)-electrons remain unpaired. \textbf{Case 2:} ([ \text{Fe}(\text{CN})_6 ]^{3-}) – The (\text{CN}^-) ligand is a strong field ligand, which causes pairing of the (d)-electrons. – As a result, four electrons pair up, leaving only one unpaired electron. Since the paramagnetic strength is directly proportional to the number of unpaired electrons, ([ \text{Fe}(\text{H}_2\text{O})_6 ]^{3+}) with five unpaired electrons is strongly paramagnetic, while ([ \text{Fe}(\text{CN})_6 ]^{3-}) with only one unpaired electron is weakly paramagnetic. 

Question8. Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex. : Coordination Compounds Chapter 5

Solution :

Oxidation state of cobalt = +3	Oxidation state of Ni = +2
Electronic configuration of cobalt = d6	Electronic configuration of nickel = d8
NH3 being a strong field ligand causes the pairing. Therefore, Cobalt can undergo d2sp3 hybridization. Hence, it is an inner orbital complex.	If NH3 causes the pairing, then only one 3d orbital is empty. Thus, it cannot undergo d2sp3 hybridization. Therefore, it undergoes sp3d2 hybridization. Hence, it forms an outer orbital complex.

Question9. Predict the number of unpaired electrons in the square planar [Pt(CN)4]2− ion. : Coordination Compounds Chapter 5

Solution :

In this complex, Pt is in the +2 state. It forms a square planar structure. This means that it undergoes dsp2 hybridization. Now, the electronic configuration of Pd(+2) is 5d8.

CN− being a strong field ligand causes the pairing of unpaired electrons. Hence, there are no unpaired electrons in

Question 10. The hexaquo manganese(II) ion contains five unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory. : Coordination Compounds Chapter 5

Solution :

Mn is in the +2 oxidation state.	Mn is in the +2 oxidation state.
The electronic configuration is d5.	The electronic configuration is d5.
The crystal field is octahedral. Water is a weak field ligand. Therefore, the arrangement of the electrons in is t2g3eg2.	The crystal field is octahedral. Cyanide is a strong field ligand. Therefore, the arrangement of the electrons in is T2g5eg0.

Hence, hexaaquo manganese (II) ion has five unpaired electrons, while hexacyano ion has only one unpaired electron.

Question11. Calculate the overall complex dissociation equilibrium constant for the Cu(NH3)42+ ion, given that β4 for this complex is 2.1 × 1013. : Coordination Compounds Chapter 5

Solution :
β4 = 2.1 × 1013

The overall complex dissociation equilibrium constant is the reciprocal of the overall stability constant, β4.

1/β4 = 1/2.1 × 1013 = 4.7 x 10 -14

Question12. Explain the bonding in coordination compounds in terms of Werner’s postulates. : Coordination Compounds Chapter 5

Solution :
Werner’s postulates explain the bonding in coordination compounds as follows:

(i) A metal exhibits two types of valencies namely, primary and secondary valencies. Primary valencies are satisfied by negative ions while secondary valencies are satisfied by both negative and neutral ions.

(In modern terminology, the primary valency corresponds to the oxidation number of the metal ion, whereas the secondary valency refers to the coordination number of the metal ion.

(ii) A metal ion has a definite number of secondary valencies around the central atom. Also, these valencies project in a specific direction in the space assigned to the definite geometry of the coordination compound.

(iii) Primary valencies are usually ionizable, while secondary valencies are non-ionizable.

Question13. (\text{FeSO}_4) solution mixed with ((\text{NH}_4)_2\text{SO}_4) solution in a 1:1 molar ratio gives the test of (\text{Fe}^{2+}) ion but (\text{CuSO}_4) solution mixed with aqueous ammonia in a 1:4 molar ratio does not give the test of (\text{Cu}^{2+}) ion. Explain why? \textbf

{Solution} Ferrous sulphate and ammonium sulphate form Mohr’s salt (a double salt), which gives free ferrous ions in solution. Copper forms a complex with ammonia, so free cupric ions are not present in solution. [ \text{FeSO}_4 + (\text{NH}_4)_2\text{SO}_4 + 6\text{H}_2\text{O} \rightarrow \text{FeSO}_4(\text{NH}_4)_2\text{SO}_4\cdot6\text{H}_2\text{O} \, (\text{Mohr’s salt}) ] [ \text{CuSO}_4 + 4\text{NH}_3 + 5\text{H}_2\text{O} \rightarrow [\text{Cu}(\text{NH}_3)_4]\text{SO}_4 \cdot 5\text{H}_2\text{O} ] Mohr’s salt dissociates in solution to release free (\text{Fe}^{2+}) ions, which can be detected in a test. On the other hand, in the reaction involving (\text{CuSO}_4) and ammonia, a complex ion ([\text{Cu}(\text{NH}_3)_4]^{2+}) forms, and the (\text{Cu}^{2+}) ions are not free in solution, hence they do not give a positive test for (\text{Cu}^{2+}) ions. 

Question14. Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic. \textbf

{Solution} \textbf{Coordination Entity:} A coordination entity consists of a central metal atom or ion bonded to a fixed number of oppositely charged ions or neutral molecules. Examples include: \begin{itemize} \item ([ \text{Ni}(\text{CO})_4 ]) \item ([ \text{CoCl}_3(\text{NH}_3)_3 ]) \end{itemize} \textbf{Ligand:} A ligand is an atom, molecule, or ion capable of donating a pair of electrons to the central metal atom or ion, forming a coordinate bond. Examples include: \begin{itemize} \item Chloride ion (( \text{Cl}^- )) \item Ammonia (( \text{NH}_3 )) \end{itemize} \textbf{Coordination Number:} The coordination number is the total number of ligand atoms, ions, or molecules that are directly bonded to the central metal atom or ion in a coordination compound. Examples include: \begin{itemize} \item In ([ \text{Ni}(\text{CO})_4 ]), the coordination number of Ni is 4. \item In ([ \text{CoCl}_3(\text{NH}_3)_3 ]), the coordination number of Co is 6. \end{itemize} \textbf{Coordination Polyhedron:} The coordination polyhedron refers to the spatial arrangement of the ligand atoms around the central atom. Examples include: \begin{itemize} \item ([ \text{Co}(\text{NH}_3)_6 ]^{3+}) has an octahedral geometry. \item ([ \text{Ni}(\text{CO})_4 ]) has a tetrahedral geometry. \end{itemize} \textbf{Homoleptic Complex:} In a homoleptic complex, the central metal atom is bonded to only one type of donor group. Examples include: \begin{itemize} \item ([ \text{Ni}(\text{NH}_3)_6 ]^{2+}) \item ([ \text{Co}(\text{NH}_3)_6 ]^{3+}) \end{itemize} \textbf{Heteroleptic Complex:} In a heteroleptic complex, the central metal atom is bonded to more than one type of donor group. Examples include: \begin{itemize} \item ([ \text{CoCl}_3(\text{NH}_3)_3 ]) \item ([ \text{NiCl}_2(\text{H}_2\text{O})_4 ]) \end{itemize} 

Question15. What is meant by unidentate, bidentate, and ambidentate ligands? Give two examples for each. \textbf

{Solution} \textbf{Unidentate Ligand:} Unidentate ligands have only one donor atom through which they can coordinate to the central metal ion. Examples include: \begin{itemize} \item Ammonia (( \text{NH}_3 )) \item Chloride ion (( \text{Cl}^- )) \end{itemize} \textbf{Bidentate Ligand:} Bidentate ligands have two donor atoms, allowing them to form two bonds with the central metal ion. Examples include: \begin{itemize} \item Ethylenediamine (en, ( \text{NH}_2\text{CH}_2\text{CH}_2\text{NH}_2 )) \item Oxalate ion (( \text{C}_2\text{O}_4^{2-} )) \end{itemize} \textbf{Ambidentate Ligand:} Ambidentate ligands can coordinate to a central metal ion through two different atoms, but not simultaneously. Examples include: \begin{itemize} \item Nitro group (nitro, ( \text{-NO}_2 ), with nitrogen as the donor atom) and nitrito group (nitrito, ( \text{-ONO} ), with oxygen as the donor atom) \item Thiocyanate (( \text{SCN}^- ), with sulfur as the donor atom) and isothiocyanate (( \text{NCS}^- ), with nitrogen as the donor atom) \end{itemize} \end{document}

Question16. Specify the oxidation numbers of the metals in the following coordination entities: \begin{enumerate} \item ([Co(H_2O)(CN)(en)_2]^{2+}) \item ([CoBr_2(en)_2]^+) \item ([PtCl_4]^{2-}) \item (K_3[Fe(CN)_6]) \item ([Cr(NH_3)_3Cl_3]) \end{enumerate} \section*

{Solution} Let ( X ) be the oxidation number of the metal in the complex. \begin{enumerate} \item ([Co(H_2O)(CN)(en)_2]^{2+}) \begin{align*} & X + 0 + (-1) + 2(0) = +2 \ & X – 1 = 2 \ & X = +3 \end{align*} \textbf{Oxidation number of Co: +3} \item ([CoBr_2(en)_2]^+) \begin{align*} & X + 2(-1) + 2(0) = +1 \ & X – 2 = 1 \ & X = +3 \end{align*} \textbf{Oxidation number of Co: +3} \item ([PtCl_4]^{2-}) \begin{align*} & X + 4(-1) = -2 \ & X – 4 = -2 \ & X = +2 \end{align*} \textbf{Oxidation number of Pt: +2} \item (K_3[Fe(CN)_6]) \begin{align*} & 3(+1) + (X + 6(-1)) = 0 \ & 3 + (X – 6) = 0 \ & X – 3 = 0 \ & X = +3 \end{align*} \textbf{Oxidation number of Fe: +3} \item ([Cr(NH_3)_3Cl_3]) \begin{align*} & X + 3(0) + 3(-1) = 0 \ & X – 3 = 0 \ & X = +3 \end{align*} \textbf{Oxidation number of Cr: +3} \end{enumerate}

Question17. Using IUPAC norms write the formulas for the following:

(i) Tetrahydroxozincate(II)

(ii) Potassium tetrachloridopalladate(II)

(iii) Diamminedichloridoplatinum(II)

(iv) Potassium tetracyanonickelate(II)

(v) Pentaamminenitrito-O-cobalt(III)

(vi) Hexaamminecobalt(III) sulphate

(vii) Potassium tri(oxalato)chromate(III)

(viii) Hexaammineplatinum(IV)

(ix) Tetrabromidocuprate(II)

(x) Pentaamminenitrito-N-cobalt(III)

Solution :
(i) [Zn(OH)4]2−

(ii) K2[PdCl4]

(iii) [Pt(NH3)2Cl2]

(iv) K2[Ni(CN)4]

(v) [Co(ONO) (NH3)5]2+

(vi) [Co(NH3)6]2 (SO4)3

(vii) K3[Cr(C2O4)3]

(viii) [Pt(NH3)6]4+

(ix) [Cu(Br)4]2−

(x) [Co[NO2)(NH3)5]2+

Question18. Using IUPAC norms write the systematic names of the following:

(i) [Co(NH3)6]Cl3

(ii) [Pt(NH3)2Cl(NH2CH3)]Cl 

(iii) [Ti(H2O)6]3+

(iv) [Co(NH3)4Cl(NO2)]Cl 

(v) [Mn(H2O)6]2+

(vi) [NiCl4]2−

(vii) [Ni(NH3)6]Cl2

(viii) [Co(en)3]3+

(ix) [Ni(CO)4]

Solution :
(i) Hexaamminecobalt(III) chloride

(ii) Diamminechlorido(methylamine) platinum(II) chloride

(iii) Hexaquatitanium(III) ion

(iv) Tetraamminichloridonitrito-N-Cobalt(III) chloride

(v) Hexaquamanganese(II) ion

(vi) Tetrachloridonickelate(II) ion

(vii) Hexaamminenickel(II) chloride

(viii) Tris(ethane-1, 2-diammine) cobalt(III) ion

(ix) Tetracarbonylnickel(0)

Question19. List various types of isomerism possible for coordination compounds, giving an example of each. \section*

{Solution} Isomerism in coordination compounds is mainly of two types: 1. \textbf{Structural isomerism} \begin{itemize} \item \textbf{Linkage isomerism} \item \textbf{Coordination isomerism} \item \textbf{Ionization isomerism} \item \textbf{Solvate isomerism} \end{itemize} 2. \textbf{Stereoisomerism} \begin{itemize} \item \textbf{Geometrical isomerism} \item \textbf{Optical isomerism} \end{itemize} \textbf{Examples:} 1. \textbf{Structural Isomerism} \begin{itemize} \item \textbf{Linkage isomerism}: \begin{itemize} \item ([Co(NH_3)_5NO_2]Cl_2) and ([Co(NH_3)_5ONO]Cl_2) \item ([Mn(CO)_5SCN]) and ([Mn(CO)_5NCS]) \end{itemize} \item \textbf{Coordination isomerism}: \begin{itemize} \item ([Co(NH_3)_6][Cr(CN)_6]) and ([Cr(NH_3)_6][Co(CN)_6]) \end{itemize} \item \textbf{Ionization isomerism}: \begin{itemize} \item ([Co(NH_3)_5SO_4]Br) and ([Co(NH_3)_5Br]SO_4) \end{itemize} \item \textbf{Solvate isomerism}: \begin{itemize} \item ([Cr(H_2O)_6]Cl_3) (violet) and ([Cr(H_2O)_5Cl]Cl_2 \cdot H_2O) (grey-green) \end{itemize} \end{itemize} 2. \textbf{Stereoisomerism} \begin{itemize} \item \textbf{Geometrical isomerism}: \begin{itemize} \item Cis- and trans-isomers of ([CoCl_2(en)_2]) \end{itemize} \item \textbf{Optical isomerism}: \begin{itemize} \item Dextro- and laevo-rotatory forms of ([Co(en)_3]^{3+})

Question20. How many geometrical isomers are possible in the following coordination entities?  : Coordination Compounds Chapter 5

(i) [Cr(C2O4)3]3− (ii) [Co(NH3)3Cl3]

Solution :
(i) For [Cr(C2O4)3]3−, no geometric isomer is possible as it is a bidentate ligand.

(ii) [Co(NH3)3Cl3]

Two geometrical isomers are possible.

Question 21. Draw the structures of optical isomers of:

(i) [Cr(C2O4)3]3− 

(ii) [PtCl2(en)2]2+

(iii) [Cr(NH3)2Cl2(en)]+

Solution :
(i) [Cr(C2O4)3]3−

(ii) [PtCl2(en)2]2+

(iii) [Cr(NH3)2Cl2(en)]+

Question 22. Draw all the isomers (geometrical and optical) of:

(i) [CoCl2(en)2]+

(ii) [Co(NH3)Cl(en)2]2+

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

Solution :
(i) [CoCl2(en)2]+ Geometrical isomerism

Optical isomerism Since only cis isomer is optically active, it shows optical isomerism.

In total, three isomers are possible.

(ii) [Co(NH3)Cl(en)2]2+

Geometrical isomerism

Optical isomerism Since only cis isomer is optically active, it shows optical isomerism.

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

Question 23. Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomers? : Coordination Compounds Chapter 5

Solution :
[Pt(NH3)(Br)(Cl)(py)

From the above isomers, none will exhibit optical isomers. Tetrahedral complexes rarely show optical isomerization. They do so only in the presence of unsymmetrical chelating agents.

Question 24. Aqueous copper sulphate solution (blue in colour) gives: \begin{enumerate} \item a green precipitate with aqueous potassium fluoride and \item a bright green solution with aqueous potassium chloride. \end{enumerate} Explain these experimental results. \section*

{Solution} Aqueous copper sulphate solution is blue in colour due to the presence of ([Cu(H_2O)_4]^{2+}) ions. \begin{enumerate} \item It reacts with KF to form green-coloured ([CuF_4]^{2-}): [ [Cu(H_2O)_4]^{2+} + 4F^- \rightarrow [CuF_4]^{2-} ] \item It reacts with KCl to form bright green-coloured ([CuCl_4]^{2-}): [ [Cu(H_2O)_4]^{2+} + 4Cl^- \rightarrow [CuCl_4]^{2-} ] \end{enumerate} In both cases, the weak field ligand water ((H_2O)) is replaced by fluoride ((F^-)) and chloride ((Cl^-)) ions, leading to the formation of different colored complexes. 

Question 25. What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H$_2$S(g) is passed through this solution? \section*

{Solution} Copper sulphate reacts with excess KCN to form ([Cu(CN)_4]^{2-}). Cyanide ion (CN$^-$) is a strong field ligand and forms a highly stable complex ion. As a result, the complex ion does not ionize to give free Cu$^{2+}$ ions. When H$_2$S gas is passed through the solution, free Cu$^{2+}$ ions are not available to react with sulfide ions (S$^{2-}$) to form CuS precipitate. The reaction can be represented as: [ \text{CuSO}_4(aq) + 4\text{KCN}(aq) \rightarrow [Cu(CN)_4]^{2-}(aq) + \text{K}_2\text{SO}_4(aq) ] Since ([Cu(CN)_4]^{2-}) is highly stable and does not dissociate to release Cu$^{2+}$ ions, no CuS precipitate forms when H$_2$S gas is introduced. \end{document}

Question 26. Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory: \begin{enumerate} \item ([Fe(CN)_6]^{4-}) \item ([FeF_6]^{3-}) \item ([Co(C_2O_4)_3]^{3-}) \item ([CoF_6]^{3-}) \end{enumerate} \section*

{Solution} \begin{enumerate} \item ([Fe(CN)_6]^{4-}) Fe in the complex has an oxidation state of +2. The electronic configuration of Fe(II) is 3d$^6$. With cyanide (CN$^-$) being a strong field ligand, the electrons in the 3d orbitals pair up, leading to (d^2sp^3) hybridization. This forms a diamagnetic octahedral complex. It is an inner orbital, low-spin (spin-paired) complex. \item ([FeF_6]^{3-}) Fe in the complex has an oxidation state of +3. The electronic configuration of Fe(III) is 3d$^5$. Fluoride (F$^-$) is a weak field ligand, so the electrons do not pair up, leading to (sp^3d^2) hybridization. This forms a paramagnetic octahedral complex. It is an outer orbital, high-spin (spin-free) complex. \item ([Co(C_2O_4)_3]^{3-}) Co in the complex has an oxidation state of +3. The electronic configuration of Co(III) is 3d$^6$. Oxalate (C$_2$O$_4^{2-}$) is a strong field ligand, causing the electrons in the 3d orbitals to pair up, leading to (d^2sp^3) hybridization. This forms a diamagnetic octahedral complex. It is an inner orbital, low-spin (spin-paired) complex. \item ([CoF_6]^{3-}) Co in the complex has an oxidation state of +3. The electronic configuration of Co(III) is 3d$^6$. Fluoride (F$^-$) is a weak field ligand, so the electrons do not pair up, leading to (sp^3d^2) hybridization. This forms a paramagnetic octahedral complex. It is an outer orbital, high-spin (spin-free) complex. \end{enumerate}

Question27. Draw figure to show the splitting of d orbitals in an octahedral crystal field. 

Solution :

The splitting of the d orbitals in an octahedral field takes palce in such a way that ,experience a rise in energy and form the eg level, while dxy, dyzand dzx experience a fall in energy and form the t2g level.

Question28. What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.

Solution :
A spectrochemical series is the arrangement of common ligands in the increasing order of their crystal-field splitting energy (CFSE) values. The ligands present on the R.H.S of the series are strong field ligands while that on the L.H.S are weak field ligands. Also, strong field ligands cause higher splitting in the d orbitals than weak field ligands.

I− < Br− < S2− < SCN− < Cl−< N3 < F− < OH− < C2O42− ∼ H2O < NCS− ∼ H− < CN− < NH3 < en ∼ SO32− < NO2− < phen < CO

Question29. What is crystal field splitting energy? How does the magnitude of Δo decide the actual configuration of d-orbitals in a coordination entity?

Solution :
The degenerate d-orbitals (in a spherical field environment) split into two levels i.e., eg and t2g in the presence of ligands. The splitting of the degenerate levels due to the presence of ligands is called the crystal-field splitting while the energy difference between the two levels (eg and t2g) is called the crystal-field splitting energy. It is denoted by Δo.

After the orbitals have split, the filling of the electrons takes place. After 1 electron (each) has been filled in the three t2g orbitals, the filling of the fourth electron takes place in two ways. It can enter the eg orbital (giving rise to t2g3 eg1 like electronic configuration) or the pairing of the electrons can take place in the t2g orbitals (giving rise to t2g4 eg0 like electronic configuration). If the Δo value of a ligand is less than the pairing energy (P), then the electrons enter the eg orbital. On the other hand, if the Δo value of a ligand is more than the pairing energy (P), then the electrons enter the t2g orbital.

Question30. ([Cr(NH_3)_6]^{3+}) is paramagnetic while ([Ni(CN)_4]^{2-}) is diamagnetic. Explain why. \section*

{Solution} ([Cr(NH_3)_6]^{3+}) is paramagnetic due to the presence of three unpaired electrons. Chromium (Cr) has an atomic number of 24, with the electronic configuration ([Ar] 3d^5 4s^1). The Cr$^{3+}$ ion, formed by the loss of three electrons, has the electronic configuration ([Ar] 3d^3). In this complex, the Cr$^{3+}$ ion undergoes (d^2sp^3) hybridization, forming six hybrid orbitals that are occupied by electron pairs donated by six ammonia (NH$_3$) ligands. However, the 3d orbitals still contain three unpaired electrons, making the complex paramagnetic. ([Ni(CN)_4]^{2-}) is diamagnetic as all the electrons are paired. Nickel (Ni) has an atomic number of 28, with the electronic configuration ([Ar] 3d^8 4s^2). The Ni$^{2+}$ ion, formed by the loss of two electrons, has the electronic configuration ([Ar] 3d^8). In this complex, the Ni$^{2+}$ ion undergoes (dsp^2) hybridization, forming four hybrid orbitals that are occupied by electron pairs donated by four cyanide (CN$^-$) ligands. The strong field cyanide ligands cause pairing of electrons in the 3d orbitals, resulting in no unpaired electrons, which makes the complex diamagnetic. 

Question31. A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2− is colourless. Explain.

Solution :
In [Ni(H2O)6]2+, is a weak field ligand. Therefore, there are unpaired electrons in Ni2+. In this complex, the d electrons from the lower energy level can be excited to the higher energy level i.e., the possibility of d−d transition is present. Hence, Ni(H2O)6]2+ is coloured.

In [Ni(CN)4]2−, the electrons are all paired as CN- is a strong field ligand. Therefore, d-d transition is not possible in [Ni(CN)4]2−. Hence, it is colourless.

Question32. [Fe(CN)6]4− and [Fe(H2O)6]2+ are of different colours in dilute solutions. Why?

Solution :
The colour of a particular coordination compound depends on the magnitude of the crystal-field splitting energy, Δ. This CFSE in turn depends on the nature of the ligand. In case of [Fe(CN)6]4− and [Fe(H2O)6]2+, the colour differs because there is a difference in the CFSE. Now, CN− is a strong field ligand having a higher CFSE value as compared to the CFSE value of water. This means that the absorption of energy for the intra d-d transition also differs. Hence, the transmitted colour also differs.

Question33. Discuss the nature of bonding in metal carbonyls.

Solution :
The metal-carbon bonds in metal carbonyls have both σ and π characters. A σ bond is formed when the carbonyl carbon donates a lone pair of electrons to the vacant orbital of the metal. A π bond is formed by the donation of a pair of electrons from the filled metal d orbital into the vacant anti-bonding π* orbital (also known as back bonding of the carbonyl group). The σ bond strengthens the π bond and vice-versa. Thus, a synergic effect is created due to this metal-ligand bonding. This synergic effect strengthens the bond between CO and the metal.

Question34. Give the oxidation state, d-orbital occupation and coordination number of the central metal ion in the following complexes:

(i) K3[Co(C2O4)3] 

(ii) cis-[Cr(en)2Cl2]Cl 

(iii) (NH4)2[CoF4]

(iv) [Mn(H2O)6]SO4

Solution :
(i) K3[Co(C2O4)3]

The central metal ion is Co.

Its coordination number is 6.

The oxidation state can be given as:

x − 6 = −3

x = + 3

The d orbital occupation for Co3+ is t2g6eg0.

(ii) cis-[Cr(en)2Cl2]Cl

The central metal ion is Cr.

The coordination number is 6.

The oxidation state can be given as:

x + 2(0) + 2(−1) = +1

x − 2 = +1

x = +3

The d orbital occupation for Cr3+ is t2g3.

(iii) (NH4)2[CoF4]

The central metal ion is Co.

The coordination number is 4.

The oxidation state can be given as:

x − 4 = −2

x = + 2

The d orbital occupation for Co2+ is eg4 t2g3.

(iv) [Mn(H2O)6]SO4

The central metal ion is Mn.

The coordination number is 6.

The oxidation state can be given as:

x + 0 = +2

x = +2

The d orbital occupation for Mn is t2g3 eg2.

Question35. 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

(iv) Cs[FeCl4]

(v) K4[Mn(CN)6]

Solution :
(i) Potassium diaquadioxalatochromate (III) trihydrate.

Oxidation state of chromium = 3

Electronic configuration: 3d3: t2g3

Coordination number = 6

Shape: octahedral

Stereochemistry:

(ii) [Co(NH3)5Cl]Cl2

IUPAC name: Pentaamminechloridocobalt(III) chloride

Oxidation state of Co = +3

Coordination number = 6

Shape: octahedral.

Electronic configuration: d6: t2g6.

Stereochemistry:

Magnetic Moment = 0

(iii) CrCl3(py)3

IUPAC name: Trichloridotripyridinechromium (III)

Oxidation state of chromium = +3

Electronic configuration for d3 = t2g3

Coordination number = 6

Shape: octahedral.

Stereochemistry:

Both isomers are optically active. Therefore, a total of 4 isomers exist.

Magnetic moment, μ = √n(n + 2)

(iv) Cs[FeCl4]

IUPAC name: Caesium tetrachloroferrate (III)

Oxidation state of Fe = +3

Electronic configuration of d6 = eg2t2g3

Coordination number = 4

Shape: tetrahedral

Stereochemistry: optically inactive

Magnetic moment:

(v) K4[Mn(CN)6]

Potassium hexacyanomanganate(II)

Oxidation state of manganese = +2

Electronic configuration: d5+: t2g5

Coordination number = 6

Shape: octahedral.

Streochemistry: optically inactive

Magnetic moment, μ = √n(n + 2)

Ques 36: What is meant by the stability of a coordination compound in solution? State the factors which govern the stability of complexes. \section*

{Solution} The stability of a coordination compound in solution refers to the degree of association between the metal ion and the ligands in the equilibrium state. The magnitude of the stability (or formation) constant for the association quantitatively expresses the stability of the complex. For a reaction: [ \text{M} + 4\text{L} \rightarrow \text{ML}_4 ] The stability constant, ( K ), is given by: [ K = \frac{[\text{ML}_4]}{[\text{M}][\text{L}]^4} ] A higher value of the stability constant indicates a greater proportion of the complex (\text{ML}_4) in the solution. \textbf{Types of Stability:} \begin{enumerate} \item \textbf{Thermodynamic Stability:} This refers to the extent to which the complex will be formed or transformed into another species at equilibrium. It is determined by the Gibbs free energy change of the reaction. \item \textbf{Kinetic Stability:} This refers to the rate at which the complex forms or transforms to reach equilibrium. It depends on the activation energy barrier for the transformation process. \end{enumerate} \textbf{Factors Affecting the Stability of Complexes:} \begin{enumerate} \item \textbf{Charge on the Central Metal Ion:} A higher charge on the metal ion generally increases the stability of the complex because it attracts the ligands more strongly. \item \textbf{Nature of the Metal Ion:} For metal ions with the same charge, the stability of their complexes decreases as the size of the metal ion increases, given the same ligands. \item \textbf{Basic Nature of the Ligands:} The greater the basic strength of the ligand, the more readily it can donate electron pairs to the metal ion, enhancing the stability of the complex. \item \textbf{Presence of Chelate Rings:} Chelate rings, formed when a ligand binds to a metal ion at multiple points, increase the stability of the complex due to the chelate effect. The chelate effect is particularly strong for 5- and 6-membered rings. \item \textbf{Effect of Multidentate Cyclic Ligands:} If the ligands are multidentate and form cyclic structures without steric hindrance, the stability of the complex is further increased. \end{enumerate}

Question37. What is meant by the chelate effect? Give an example. \section*

{Solution} The chelate effect refers to the increased stability of coordination compounds containing chelating ligands compared to those with unidentate ligands. A chelating ligand can form more than one coordinate bond with the central metal ion, creating a ring-like structure known as a chelate. This multi-point attachment makes the complex more stable due to the entropy gain and the stronger interaction between the ligand and the metal ion. An example of the chelate effect is observed in the complexes ([Co(en)_3]^{3+}) and ([Co(NH_3)_6]^{3+}). Here, (en) stands for ethylenediamine, a bidentate ligand, which means it can donate two pairs of electrons to the central metal ion (Co$^{3+}$). The complex ([Co(en)_3]^{3+}) is more stable than ([Co(NH_3)_6]^{3+}) because ethylenediamine forms three five-membered chelate rings with the metal ion, providing greater stability compared to the six unidentate ammonia (NH$_3$) ligands in ([Co(NH_3)_6]^{3+}). 

Question38. Discuss briefly giving an example in each case the role of coordination compounds in:

(i) biological system

(ii) medicinal chemistry

(iii) analytical chemistry

(iv) extraction/metallurgy of metals

Solution :
(i) Role of coordination compounds in biological systems: 

We know that photosynthesis is made possible by the presence of the chlorophyll pigment. This pigment is a coordination compound of magnesium. In the human biological system, several coordination compounds play important roles. For example, the oxygen-carrier of blood, i.e., haemoglobin, is a coordination compound of iron.

(ii) Role of coordination compounds in medicinal chemistry:

Certain coordination compounds of platinum (for example, cis-platin) are used for inhibiting the growth of tumours.

(iii) Role of coordination compounds in analytical chemistry:

During salt analysis, a number of basic radicals are detected with the help of the colour changes they exhibit with different reagents. These colour changes are a result of the coordination compounds or complexes that the basic radicals form with different ligands.

(iii) Role of coordination compounds in extraction or metallurgy of metals:

The process of extraction of some of the metals from their ores involves the formation of complexes. For example, in aqueous solution, gold combines with cyanide ions to form [Au(CN)2]. From this solution, gold is later extracted by the addition of zinc metal.

Question39. How many ions are produced from the complex Co(NH3)6Cl2 in solution?

(i) 6 

(ii) 4 

(iii) 3 

(iv) 2

Solution :
(iii) The given complex can be written as [Co(NH3)6]Cl2.

Thus, [Co(NH3)6]+ along with two Cl− ions are produced.

Question40. Amongst the following ions which one has the highest magnetic moment value?

(i) [Cr(H2O)6]3+

(ii) [Fe(H2O)6]2+

(iii) [Zn(H2O)6]2+

Solution :
(i) No. of unpaired electrons in [Cr(H2O)6]3+ = 3

Then, μ  = √n(n + 2)

(ii) No. of unpaired electrons in [Fe(H2O)6]2+ = 4

Then, μ = √n(n + 2)

= √4(4 + 2)

= √2

= ˜ 5 BM

(iii) No. of unpaired electrons in [Zn(H2O)6]2+ = 0

Hence, [Fe(H2O)6]2+ has the highest magnetic moment value.

Question41. The oxidation number of cobalt in K[Co(CO)4] is

(i) +1

(ii) +3

(iii) −1

(iv) −3

Solution :
We know that CO is a neutral ligand and K carries a charge of +1.

Therefore, the complex can be written as K+[Co(CO)4]−. Therefore, the oxidation number of Co in the given complex is −1. Hence, option (iii) is correct.

Question42. Amongst the following, the most stable complex is: \begin{enumerate} \item [i)] ([Fe(H_2O)_6]^{3+}) \item [ii)] ([Fe(NH_3)_6]^{3+}) \item [iii)] ([Fe(C_2O_4)_3]^{3-}) \item [iv)] ([FeCl_6]^{3-}) \end{enumerate} \section*

{Solution} In all these complexes, iron (Fe) is in the +3 oxidation state. The stability of a complex is greatly influenced by the nature of the ligands. Among the options, the complex ([Fe(C_2O_4)_3]^{3-}) is the most stable. This is because the oxalate ion ((C_2O_4^{2-})) acts as a chelating ligand, forming a chelate complex. Chelate complexes are generally more stable than complexes with monodentate ligands due to the chelate effect, which involves the formation of multiple bonds between a ligand and a central metal ion, leading to increased stability. Thus, the most stable complex is ([Fe(C_2O_4)_3]^{3-}).  

43. What will be the correct order for the wavelengths of absorption in the visible region for the following complexes? \begin{enumerate} \item \([Ni(NO_2)_6]^{4-}\) \item \([Ni(NH_3)_6]^{2+}\) \item \([Ni(H_2O)_6]^{2+}\) \end{enumerate} \section*

{Solution} In all the given complexes, the metal ion is (Ni^{2+}). The increasing field strengths of the ligands present, according to the electrochemical series, are in the order: [ \text{H}_2\text{O} < \text{NH}_3 < \text{NO}_2^- ] The energies absorbed for excitation will be in the order: [ [Ni(H_2O)_6]^{2+} < [Ni(NH_3)_6]^{2+} < [Ni(NO_2)_6]^{4-} ] Since (E = \frac{hc}{\lambda}), where (E) is the energy absorbed and (\lambda) is the wavelength, the wavelength (\lambda) is inversely proportional to the energy (E). Thus, the order of wavelengths will be the opposite of the order of energy absorption. Therefore, the wavelengths of absorption will be in the following order: [ [Ni(H_2O)_6]^{2+} > [Ni(NH_3)_6]^{2+} > [Ni(NO_2)_6]^{4-} ]