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Class- XII-CBSE-Chemistry, , Coordination Compounds, , CBSE NCERT Solutions for Class 12 Chemistry Chapter 9, Back of Chapter Questions, , 1., , Write the formulas for the following coordination compounds:, (i), , Tetraamminediaquacobalt (III) chloride, , (ii), , Potassium tetracyanidonickelate (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) hexacyanidoferrate(II), , Solution:, Rule to write the name of a coordination compound, 1., , The name of the cation is named first in both positively and negatively, charged entities., , 2., , The ligands name should be arranged in alphabetical order before the name, of the central atom., , 3., , Names of the anionic ligand end in -o, those of the neutral are the same, and, the cation ends in -ium., , 4., , Prefixes mono, di, tri are used to indicate the number of the individual, ligands., , 5., , When the name of the ligands include numerical prefix then the terms bis,, tris, tetrakis are used, , 6., , If the complex ion is a cation, the metal is named the same as the element, , 7., , If the complex is anion the name of the metal ends with the suffix -ate., Therefore, the formula of coordination compounds mentioned are:, (i), , Tetraamminediaquacobalt (III) chloride, Ligands: Ammine(NH3 ) – neutral, Aqua(H2 O) −neutral, Charge on coordination sphere:, 3+2×0+4×0=𝑥, 𝑥=3, , Practice more on Coordination Compounds, , Page - 1, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , Solution:, Rules to write the IUPAC name of coordination compound :, (i), , find the oxidation number of the central atom involved., , (ii), , The cation is named first as is the case with other. This rule applies to both, positively and negatively charged coordination entities., , (iii), , Usually, the number of free cation or anions is not given as can easily be, calculated by charge, , (iv), , For writing the names of complex cation use the following rule., Names of the ligand in alphabetical order with their quantity as prefix (but, do not use mono for one and do not consider first letters of these number, denoting prefixes in alphabetisation). This is followed by the name of metal, followed by its oxidation number (in Roman numeral) in a bracket like, (0) (I), (II), (III), (-I), (-II) etc., , (v), , For writing the names of complex anion use the following rule., Names of the ligand in alphabetical order with their quantity as prefix (but, do not use mono for one and do not consider first letters of these number, denoting prefixes in alphabetisation). This is followed by the name of metal, (modified as per rule already specified) followed by its oxidation number, (in Roman numeral) in a bracket like (I), (II), (III), (0), (-I), (-II), etc., , (vi), , The neutral complex molecule is named similar to that of the complex, cation., By following the above rules, we get the IUPAC name of the compounds, as follows:, (i), , The IUPAC name of the coordination compound [Co(NH3 )6 ]Cl3 :, Central atom = cobalt, Ligand = NH3 (ammine), Oxidation state = x + 6(0) + 3(−1) = 0, x=3, Complex type = cationic, IUPAC name = Hexamminecobalt(III) chloride, , (ii), , The IUPAC name of the coordination compound [Co(NH3 )5 Cl]Cl2, Central atom = cobalt, Ligand = NH3 (ammine) is neutral ligand and Cl (chloride) has -1, charge, , Practice more on Coordination Compounds, , Page - 4, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , x=2, Complex type = cationic, IUPAC name =, Diamminechlorido(methylamine)platinum(II)chloride, 3., , Indicate the types of isomerism exhibited by the following complexes and draw the, structures for these isomers, (i), , K[Cr(H2 O)2 (C2 O4 )2 ], , (ii), , [Co(en)3 ]Cl3, , (iii), , [Co(NH3 )5 (NO2 )](NO3 )2, , (iv), , [Pt(NH3 )(H2 O)Cl2 ], , Solution:, (i), , Both cis or trans geometrical isomers for K[Cr(H2 O)2 (C2 O4 )2 ] can exist., Also, optical isomers for cis-isomer exist., , Trans-isomer of the compound is optically inactive. While cis-isomer is, optically active., , (ii), , Two optical isomers for [CO(en)3 ]Cl3 exist., , This structure has two optical isomers., , Practice more on Coordination Compounds, , Page - 6, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , 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 the 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., 6., , [NiCl4 ]2− is paramagnetic while [Ni(CO)4 ] is diamagnetic though both are, tetrahedral. Why?, Solution:, Although [NiCl4 ]2− and [Ni(CO)4 ] are both tetrahedral, their magnetic characters, are different. This is because of the 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., Practice more on Coordination Compounds, , Page - 8, , www.embibe.com

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Class- XII-CBSE-Chemistry, , 7., , Coordination Compounds, , [Fe(H2 O)6 ]3+ is strongly paramagnetic whereas [Fe(CN)6 ]3− is weakly, paramagnetic. Explain., Solution:, In both [Fe(H2 O)6 ]3+ and [Fe(CN)6 ]3− , Fe exists in the +3 oxidation state i.e., in, d5 configuration., , Since CN − is a strong field ligand, it causes the pairing of unpaired electrons., Therefore, there is only one unpaired electron left in the d-orbital., , Therefore,, μ = √n(n + 2), = √1(1 + 2), = √3, = 1.732 BM, On the other hand, H2 O is a weak field ligand. Therefore, it cannot cause the pairing, of electrons. This means that the number of unpaired electrons is 5., Therefore,, μ = √n(n + 2), = √5(5 + 2), = √35, ≅ 6BM, Thus, it is evident that [Fe(H2 O)6 ]3+ is strongly paramagnetic, while [Fe(CN)6 ]3−, is weakly paramagnetic., 8., , Explain [Co(NH3 )6 ]3+ is an inner orbital complex whereas [Ni(NH3 )6 ]2+ is an, outer orbital complex., Solution:, [𝐂𝐨(𝐍𝐇𝟑 )𝟔 ]𝟑+, , Oxidation state of cobalt = +3, , Practice more on Coordination Compounds, , [𝐍𝐢(𝐍𝐇𝟑 )𝟔 ]𝟐+, , Oxidation state of Ni = +2, , Page - 9, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , Electronic configuration of cobalt = d5, , Electronic configuration of nickel = d5, , NH3 being a strong field ligand causes the If NH3 causes the pairing, then only one 3d, pairing. Therefore, Ni can undergo d2 sp3 orbital is empty. Thus, it cannot undergo, hybridization., d2 sp3 hybridization. Therefore, it, undergoes sp3 d2 hybridization., , If the complex is formed by the use of inner, d-orbitals for hybridization it is called, If the complex is formed by the use of the, inner orbital complex., outer d-orbitals for hybridization it is, Hence, it is an inner orbital complex., called an outer orbital complex., Hence, it forms an outer orbital complex., , 9., , Predict the number of unpaired electrons in the square planar [Pt(CN)4 ]2− ion., Solution:, [Pt(CN)4 ]2−, In this complex, Pt is in the +2 state. It forms a square planar structure and it, undergoes dsp2 hybridization. Now, the electronic configuration of Pt(+2) is 5d8 ., , CN− being a strong field ligand causes the pairing of unpaired electrons. Since, all, the electrons get paired, there are no unpaired electrons in [Pt(CN)4 ]2− ., 10., , The hexaquo manganese (II) ion contains five unpaired electrons, while the, hexacyanoion contains only one unpaired electron. Explain using Crystal Field, Theory., Solution:, [𝐌𝐧(𝐇𝟐 𝐎)𝟔 ]𝟐+, , [𝐌𝐧(𝐂𝐍)𝟔 ]𝟒−, , 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 The crystal field is octahedral. Cyanide is, weak field ligand. Therefore, the a strong field ligand. Therefore, the, Practice more on Coordination Compounds, , Page - 10, , www.embibe.com

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Class- XII-CBSE-Chemistry, , arrangement of the, [Mn(H2 O)6 ]2+ is t 32g e2g ., , Coordination Compounds, , electrons, , in arrangement of the, [Mn(CN)6 ]4− is t 52g e0g ., , electrons, , in, , Hence, hexaaquo manganese (II) ion has five unpaired electrons, while hexacyano, ion has only one unpaired electron., In octahedral field splitting of d orbital is such that 3d orbitals are in ground state, where 2d orbitals are in higher energy thus` representing as t 2g , eg ., First the ground state is filled then the higher state., 11., , Calculate the overall complex dissociation equilibrium constant for the, 2+, Cu(NH3 )4 ion, given that β4 for this complex is 2.1 × 1013 ., Solution:, β4 = 2.1 × 1013, The overall complex dissociation equilibrium constant is the reciprocal of the, overall stability constant, β4 ., 1, 1, =, β4 2.1 × 1013, ∴ = 4.7 × 10−14, , 1., , Explain the bonding in coordination compounds in terms of Werner’s postulates., Solution:, Werner's postulates explain the bonding in coordination compounds as follows:, , 2., , (i), , A metal exhibits two types of valencies namely, primary and secondary, valencies. Primary valencies are satisfied by negative ions while secondary, valencles are satisfied by both negative and neutral ions., , (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 nonionizable., , FeSO4 solution mixed with (NH4 )2 SO4 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?, Solution:, (NH4 )2 SO4 + FeSO4 + 6H2 O ⟶ FeSO4 . (NH4 )2 SO4 . 6H2 O (Mohr Salt), CuSO4 + 4NH3 + 5H2 O ⟶ [Cu(NH3 )4 SO4 ].5H2 O, , Practice more on Coordination Compounds, , Page - 11, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , Both the compounds i.e., FeSO4 . (NH4 )2 SO4 . 6H2 O and [Cu(NH3 )4 SO4 ].5H2 O fall, under the category of addition compounds with only one major difference i.e., the, former is an example of a double salt, while the latter is a coordination compound., A double salt is an addition compound that is stable in the solid state but that which, breaks up into its constituent ions in the dissolved state. These compounds exhibit, individual properties of their constituents., For example: FeSO4 . (NH4 )2 SO4 . 6H2 O breaks into Fe2+ , NH4+ and SO2−, 4 ions., 2+, Hence, it gives a positive test for Fe ions., A coordination compound is an addition compound which retains its identity in the, solid as well as in the dissolved state. However, the individual properties of the, constituents are lost. This happens because [Cu(NH3 )4 ]SO4 . 5H2 O does not show, the test for Cu2+ . The ions present in the solution of [Cu(NH3 )4 ]SO4 . 5H2 O are, [Cu(NH3 )4 ]2+ and SO2−, 4 ., 3., , Explain with two examples each of the following: coordination entity, ligand,, coordination number, coordination polyhedron, homoleptic and heteroleptic., Solution:, (i), , Coordination entity:, A coordination entity is an electrically charged radical or species carrying a, positive or negative charge. In a coordination entity, the central atom or ion, is surrounded by a suitable number of neutral molecules or negative ions, (called ligands). For example:, [Ni(NH3 )6 ]2+ , [Fe(CN)6 ]4+ = cationic complex, [PtCl4 ]2− , [Ag(CN)2 ]− = anionic complex, [Ni(CO)4 ], [Co(NH3 )4 Cl2 ] = neutral complex, , (ii), , Ligands, The neutral molecules or negatively charged ions that surround the metal, atom in a coordination entity or a coordinal complex are known as ligands., For example, N̈H3 , H2 Ö, Cl− , −OH. Ligands are usually polar in nature and, possess at least one unshared pair of valence electrons., , (iii), , Coordination number:, The total number of lone pairs donated by ligands (either neutral molecules, or negative ions) attached to the central metal atom in the coordination, sphere is called the coordination number of the central metal atom. It is also, called as ligancy., For example:, , Practice more on Coordination Compounds, , Page - 12, , www.embibe.com

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Class- XII-CBSE-Chemistry, , (iv), , Coordination Compounds, , (a), , In the complex, K 2 [PtCl6 ], there are six chloride ions attached to Pt, in the coordinate sphere. Therefore, the coordination number of Pt, is 6., , (b), , Similarly, in the complex [Ni(NH3 )4 ]Cl2 , the coordination number, of the central atom (Ni) is 4., , Coordination polyhedron:, Coordination polyhedrons about the central atom can be defined as the, spatial arrangement of the ligands that are directly attached to the central, metal ion in the coordination sphere., For example:, (a), , (b), , (v), , Tetrahedral, , Homoleptic complexes:, These are those complexes in which the metal ion is bound to only one kind, of a donor group. For e.g.: [Co(NH3 )6 ]3+ , [PtCl4 ]2− etc., , (vi), , Heteroleptic complexes:, Heteroleptic complexes are those complexes where the central metal ion is, bound to more than one type of a donor group., For e.g.: [Co(NH3 )4 Cl2 ]+ , [Co(NH3 )5 Cl]2+, , 4., , What is meant by unidentate, didentate and ambidentate ligands? Give two, examples for each., , Practice more on Coordination Compounds, , Page - 13, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , Solution:, A ligand may contain one or more unshared pairs of electrons which are called the, donor sites of ligands. Depending upon the number of these donor sites, ligands can, be classified as follows:, (a), , Unidentate ligands: Ligands with only one donor sites are called unidentate, ligands. For e.g., N̈H3 , Cl− etc., , (b), , Didentate ligands: ligands that have two donor sites are called didentate, ligands. E.g.,, , (c), , (1), , Ethane-1, 2-diamine, , (2), , Oxalate ion, , Ambidentate ligands:, Ligands that can attach themselves to the central metal atom through two, different atoms are called ambidentate ligands. For example:, , (The donor atom is N), , (The donor atom is oxygen), , (The donor atom is S), 5., , Specify the oxidation numbers of the metals in the following coordination entities:, (i), , [Co(H2 O)(CN)(en)2 ]2+, , Practice more on Coordination Compounds, , Page - 14, , www.embibe.com

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Class- XII-CBSE-Chemistry, , (ii), , [CoBr2 (en)2 ]+, , (iii), , [PtCl4 ]2−, , (iv), , K 3 [Fe(CN)6 ], , (v), , [Cr(NH3 )3 Cl3 ], , Coordination Compounds, , Solution:, (i), , [Co(H2 O)(CN)(en)2 ]2+, Let the oxidation number of Co be x., The charge on the complex is +2, H2 O is neutral ligand so its oxidation number = 0, CN is negatively charged ligand and its oxidation number= −1, (en) is neutral ligand so its oxidation number = 0, [Co (H2 O) (CN) (en)2 ]2+, x + 0 + (−1) + 2(0) = +2, x − 1 = +2, x = +3, , (ii), , [Pt(Cl)4 ]2−, Let the oxidation number of Pt be x., Cl is negatively charged ligand and its oxidation number= −1, The charge on the complex is −2., [Pt (Cl)4 ]2−, x + 4(−1) = −2, x = +2, , (iii), , [Co (Br)2 (en)2 ]2+, Br is negatively charged ligand and its oxidation number= −1, (en) is neutral ligand so its oxidation number = 0, [Co (Br)2 (en)2 ]2+, x + 2(−1) + 2(0) = +1, x − 2 = +1, x = +3, , Practice more on Coordination Compounds, , Page - 15, , www.embibe.com

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Class- XII-CBSE-Chemistry, , (iv), , Coordination Compounds, , K 3 [Fe(CN)6 ], Let the oxidation number of Fe be x., CN is negatively charged ligand and its oxidation number= −1, The oxidation number of k = +1., K 3 [Fe(CN)6 ], 3(+1) + x + 6(−1) = 0, x = +3, , (v), , [Cr(NH3 )3 Cl3 ], Let the oxidation number of Cr be x., Cl is negatively charged ligand and its oxidation number = −1, NH3 is neutral ligand and its oxidation number = 0, x + 3(0) + 3(−1) = 0, x = +3, , 6., , 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), , Tetrahydroxozincate(II) :, Central metal atom is zinc and the oxidation number of Zn = +2., Hydroxide (OH) is negatively charged ligand and its oxidation number =, −1, Charge on complex = 4 (−1) + (+2) = 𝑥, , Practice more on Coordination Compounds, , Page - 16, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , 𝑥 = −2 (complex is negatively charged ), Hence the complex is [Zn(OH)4 ]−2, (ii), , Potassium tetrachloridopalladate(II) :, Central metal atom is palladate (Pd) and the oxidation number of Pd = +2., chlorido (Cl) is negatively charged ligand and its oxidation number = −1, Charge on complex = 4 (−1) + (+2) = 𝑥, 𝑥 = −2 (complex is negatively charged ), Potassium has +1 oxidation number so to balance the charge on complex, two potassium ions present in complex., Hence the complex is K 2 [PdCl4 ], , (iii), , Diamminedichloridoplatinum(ll) :, Central metal atom is platinum (Pt) and the oxidation number of Pt = +2., chlorido (Cl) is negatively charged ligand and its oxidation number= −1, ammine (NH3 ) is neutral ligand and its oxidation number= 0, Charge on complex = 2 (0) + 2 (−1) + (+2) = 𝑥, 𝑥 = 0 (complex is neutral ), Hence the complex is [Pt(NH3 )2 Cl2 ], , (iv), , Potassium tetracyanonickelate(II), The central metal atom is nickel (Ni) and the oxidation number of Ni = +2., cyano (CN) is negatively charged ligand and its oxidation number = −1, Charge on complex = 4 (−1) + (+2) = 𝑥, 𝑥 = −2, Potassium has +1 oxidation number so to balance the charge on complex, two potassium ions present in complex., Hence the complex is K 2 [Ni(CN)4 ], , (v), , Pentaamminenitrito-o-cobalt(III) :, Central metal atom is cobalt (Co) and the oxidation number of Co = +3., nitrito-o- (ONO) is negatively charged ligand and its oxidation number =, −1, ammine (NH3 ) is neutral ligand and its oxidation number= 0, , Practice more on Coordination Compounds, , Page - 17, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , Charge on complex = 5 (0) + 1 (-1) + (+3)= x, 𝑋 = 2 (complex is positively ), Hence the complex is [Co(ONO) (NH3 )5 ]2+, (vi), , Hexaamminecobalt(III) sulphate :, The central metal atom is cobalt (Co) and the oxidation number of Co =+3., ammine (NH3 ) is neutral ligand and its oxidation number= 0, [Co(NH3 )6 ]2 (SO4 )3, , (vii), , Potassium tri(oxalato)chromate(III), Central metal atom is chromium (Cr) and the oxidation number of Cr = +3., Oxalate (C2 O4 ) is negatively charged ligand and its oxidation number= −2, Charge on complex = 3 (−2) + (+3) = 𝑥, 𝑥 = −3, Potassium has +1 oxidation number so to balance the charge on complex, three potassium ions present in complex, Hence the complex is K 3 [Cr(C2 O4 )3 ], , (viii) Hexaammineplatinum(IV) :, The central metal atom is platinum (Pt) and the oxidation number of Pt =, +4., ammine (NH3 ) is neutral ligand and its oxidation number= 0, Charge on complex = 6 (0) + (+4) = 𝑥, 𝑥 = +4, Hence the complex is [Pt(NH3 )6 ]4+, (ix), , Tetrabromidocuprate(II) :, Central metal atom is copper (Cu) and the oxidation number of Cu = +2., bromido (Br ) is negatively charged ligand and its oxidation number= −1, Charge on complex = 4(−1) + (+2) = 𝑥, 𝑥 = −2, Hence the complex is [Cu(Br)4 ]2−, , (x), , Pentaamminenitrito-N-cobalt(III) :, , Practice more on Coordination Compounds, , Page - 18, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , The central metal atom is cobalt (Co) and the oxidation number of Co =, +3., nitro (NO2 ) is negatively ligand and its oxidation number= −1, ammine (NH3 ) is neutral ligand and its oxidation number= 0, Charge on complex = +3 + (−1) + 5(0) = 𝑥, 𝑥 = +2, Hence the complex is [Co(NO2 )(NH3 )5 ]+2, 7., , Using IUPAC norms write the systematic names of the following:, (i), , [Co(NH3 )6 ]Cl3, , (ii), , [Pt(NH3 )2 Cl(NH2 CH3 )]Cl, , (iii), , [Ti(H2 O)6 ]3+, , (iv), , [Co(NH3 )4 Cl(NO2 )]Cl, , (v), , [Mn(H2 O)6 ]2+, , (vi), , [NiCl4 ]2−, , (vii), , [Ni(NH3 )6 ]Cl2, , (viii) [Co(en)3 ]3+, (ix), , [Ni(CO)4 ], , Solution:, (i), , [Co(NH3 )6 ]Cl3, Let the oxidation number of Co be x., NH3 is neutral ligand its name is ammine and its oxidation number = 0, the oxidation number of Cl = −1., The oxidation number of central metal atom =, x + 6(0) + 3(−1) = 0, x = +3, The complex is positively charged, So the name of the compound is Hexaamminecobalt(III) chloride, , (ii), , [Pt(NH3 )2 Cl(NH2 CH3 )]Cl, Let the oxidation number of Pt be x., NH3 is neutral ligand its name is amine and its oxidation number = 0, , Practice more on Coordination Compounds, , Page - 19, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , NH2 CH3 is a neutral ligand its name is methylamine and its oxidation, number = 0, The oxidation number of Cl = −1., The oxidation number of central metal atom =, x + 2(0) + 1(−1) + 1(0) = +1, x=2, The complex is positively charged, So the name of the compound is Diamminechlorido(methylamine), platinum(II) chloride, (iii), , [Ti(H2 O)6 ]3+, Let the oxidation number of Ti be x., H2 O is neutral ligand its name is aqua and its oxidation number= 0, The oxidation number of central metal atom =, x + 6(0) = +3, x = +3, The complex is positively charged, So the name of the compound is Hexaquatitanium(III) ion, , (iv), , [Co(NH3 )4 Cl(NO2 )]Cl :, Let the oxidation number of Co be x., NH3 is a neutral ligand its name is amine and its oxidation number= 0, Cl and NO2 is negatively charged ligand. Their name is chlorido and nitritoN respectively and their oxidation number is = 0, The oxidation number of central metal atom =, x + 4(0) + 1(−1) + 1(−1) = +1, x = +3, The complex is positively charged, So the name of the compound is Tetraamminichloridonitrito-N-Cobalt (III), chloride, , (v), , [Mn(H2 O)6 ]2+, Let the oxidation number of Mn be x., H2 O is neutral ligand its name is aqua and its oxidation number= 0, , Practice more on Coordination Compounds, , Page - 20, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , The oxidation number of central metal atom =, x + 6(0) = +2, x = +2, the complex is positively charged, So the name of the compound is Hexaquamanganese(II) ion, (vi), , [NiCl4 ]2−, Let the oxidation number of Ni be x., Cl is negatively charged ligand its name is chlorido and its oxidation, number= −1, The oxidation number of central metal atom =, x + 4(−1) = −2, x = +2, The complex is negatively charged, So the name of the compound is Tetrachloridonickeltate(II) ion, , (vii), , [Ni(NH3 )6 ]Cl2, Let the oxidation number of Ni be x., NH3 is neutral ligand its name is ammine and its oxidation number= 0, The oxidation number of central metal atom =, x + 6(0) + 2(−1) = 0, x = +2, The complex is positively charged, So the name of the compound is Hexaamminenickel(II) chloride, , (viii) [Co(en)3 ]3+, Let the oxidation number of Co be x., en is neutral ligand its name is ethane-1, 2-diamine and its oxidation, number = 0, The oxidation number of central metal atom =, x + 3(0) = +3, x = +3, the complex is positively charged, , Practice more on Coordination Compounds, , Page - 21, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , So name of compound is Tris(ethane-1, 2-diamine) cobalt(III) ion, (ix), , [Ni(CO)4 ], Let the oxidation number of Ni be x., CO is neutral ligand its name is carbonyl and its oxidation number= 0, The oxidation number of central metal atom =, x + 4(0) = 0, x=0, Complex is neutral, So the name of the compound is Tetracarbonylnickel(0), , 8., , List various types of isomerism possible for coordination compounds, giving an, example of each., Solution:, , (a), , Geometric isomerism:, This type of isomerism is common in heteroleptic complexes. It arises due, to the different possible geometric arrangements of the ligands. For, example:, When the coordination number is 4. The tetrahedral compounds do not, show geometrical isomerism as we all know that all four positions are, equivalent in a tetrahedral geometry. So moving on to square planar, complexes there are different possibilities depending upon the formula of, the compound, , Practice more on Coordination Compounds, , Page - 22, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , Diagram 1 is facial or (fac), and Diagram 2 is meridional., (b), , Optical isomerism: A Coordination compound which can rotate the plane, of polarised light is said to be optically active. When the coordination, compounds have the same formula but differ in their ability to rotate, directions of the plane of polarised light are said to exhibit optical, isomerism, and the molecules are optical isomers. The optical isomers are, pair of molecules which are non-super impossible mirror images of each, other. This is due to the absence of elements of symmetry in the complex., This type of isomerism arises in chiral molecules; Isomers are mirror images, of each other and are non-superimposable., , (c), , Linkage isomerism: This type of isomerism is found in complexes that, contain ambidentate ligands., For example, [Co(NO2 )(NH3 )5 ]2+ has two isomeric forms having, red and yellow colours. The red coloured isomer is [Co(ONO)(NH3 )5 ]2+, named as pentaamminenitrito-O-cobalt (III) cation while the yellow isomer, is formulated as [Co(NO2 )(NH3 )5 ]2+ +. It is named as pentaamminenitritoN-cobalt (III) cation. In the first cation, we have Co − (ONO) link, while in, the second case, NO2 coordinates to the metal ion through the nitrogen, atom, Co − (NO2 )., , (d), , Coordination isomerism:, This type of isomerism arises when the ligands are interchanged between, cationic and anionic entities of different metal ions present in the complex., , Practice more on Coordination Compounds, , Page - 23, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , [Co(NH3 )6 ][Cr(CN)6 ] and [Cr(NH3 )6 ][Co(CN)6 ], These isomers are expected to have significant differences in their physical, and chemical properties., (e), , Ionisation isomerism:, This type of isomerism arises when a counter ion replaces a ligand within, the coordinator sphere. Thus, complexes that have the same composition,, but furnish different ions when dissolved in water are called ionisation, isomers. For e.g.,(i) Co(NH3 )5 SO4 )Br and Co(NH3 )5 Br]SO4 ., , (f), , Solvate isomerism:, Solvate isomers differ by whether or not the solvent molecule is directly, bonded to the metal ion or merely present as a free solvent molecule in the, crystal lattice., [Cr[H2 O)6 ]Cl3 , Violet, [Cr(H2 O)5 Cl]Cl2 ∙ H2 O , Blue − green, [Cr(H2 O)5 Cl2 ]Cl ∙ 2H2 O, Dark green, , 9., , How many geometrical isomers are possible in the following coordination entities?, (i), , [Cr(C2 O4 )3 ]3−, , (ii), , [Co(NH3 )3 Cl3 ], , Solution:, (i), , For [Cr(C2 O4 )3 ]3− , no geometric isomer is possible as it is a bidentate, ligand., , (ii), , [Co(NH3 )3 Cl3 ], Two geometrical isomers are possible., Facial or (fac), two triangular faces such that each corner is occupied by, similar ligands. or meridional(If the three ligands and the metal ion are in, one plane, the isomer is said to be meridional), , Practice more on Coordination Compounds, , Page - 24, , www.embibe.com

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Class- XII-CBSE-Chemistry, , 11., , Coordination Compounds, , Draw all the isomers (geometrical and optical) of:, (i), , [CoCl2 (en)2 ]+, , (ii), , [Co(NH3 )Cl(en)2 ]2+, , (iii), , [Co(NH3 )2 Cl2 (en)]+, , Solution:, (i), , [CoCl2 (en)2 ]+, , (ii), , [Co(NH3 )Cl(en)2 ]2+, , Practice more on Coordination Compounds, , Page - 26, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , Trans-isomers are optically inactive., Cis - isomers are optically active, (iii), , 12., , [Co(NH3 )2 Cl2 (en)]+, , Write all the geometrical isomers of [Pt(NH3 )(Br)(Cl)(Py)] and how many of, these will exhibit optical isomers?, Solution:, [Pt(NH3 )(Br)(Cl)(py), , Practice more on Coordination Compounds, , Page - 27, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , From the above isomers, none will exhibit optical isomerism. Tetrahedral, complexes rarely show optical isomerisation. They do so only in the presence of, unsymmetrical chelating agents., 13., , Aqueous copper sulphate solution (blue in colour) gives:, (i), , a green precipitate with aqueous potassium fluoride, and, , (ii), , a bright green solution with aqueous potassium chloride, , Explain these experimental results., Solution:, Aqueous CuSO4 exists as [Cu(H2 O)4 ]SO4 . It is blue in colour due to the presence, of [Cu[H2 O)4 ]2+ ions., (i), , When KF is added:, [Cu(H2 O)4 ]2+ + 4F − ⟶ +4H2 O, , (ii), , When KCl is added:, [Cu(H2 O)4 ]2+ + 4Cl− ⟶ 4H2 O, In both these cases, the weak field ligand water is replaced by the F − and, Cl− ions., , 14., , What is the coordination entity formed when an 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 H2 S(g) is passed through this solution?, Solution:, CuSO4 (s) + 4KCN(aq) ⟶ K 2 [Cu(CN)4 ] (aq) + K 2 SO4 (aq), i.e., [Cu(H2 O)4 ]2+ + 4CN − ⟶ [Cu(CN)4 ]2− + 4H2 O, Thus, the coordination entity formed in the process is K 2 [Cu(CN)4 ]. It is a very, stable complex, which does not ionise to give Cu2+ ions when added to water., Hence, Cu2+ ions do not precipitate when H2 S(g) is passed through the solution., , 15., , Discuss the nature of bonding in the following coordination entities based on, valence bond theory:, (i), , [Fe(CN)6 ]4−, , Practice more on Coordination Compounds, , Page - 28, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , The splitting of the d orbitals in an octahedral field takes place in such a way that, dx2 −y2 , dz2 experience a rise in energy and form the eg level, while dxy , dyz and dzx, experience a fall in energy and form the t 2g level., 17., , What is the 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 right-hand side of the series are strong field ligands while that on the light hand, side are weak field ligands. Also, strong-field ligands cause higher splitting in the, d orbitals than weak field ligands., −, I − < Br − < S 2− < SCN − < Cl− < N3− < F − < OH − < C2 O2−, 4 < H2 O < NCS, −, < CN− < NH3 < en < SO2−, 3 < NO2 < phen < CO, , 18., , What is crystal field splitting energy? How does the magnitude of ∆0 decide the, actual configuration of d-orbitals in a coordination entity?, Solution:, The degenerate d-orbitals (having same energy) split into two energy levels i.e., eg, and t 2g 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 t 2g ) is called the crystal-field splitting energy. It is, denoted by ∆o ., After the orbitals have split, the filling of the electrons takes place. After 1 electron, (each) has been filled in the three t 2g orbitals, the filling of the fourth electron takes, place in two ways. It can enter the eg orbital (giving rise to t 32g e1g like electronic, configuration) or the pairing of the electrons can take place in the t 2g orbitals, , Practice more on Coordination Compounds, , Page - 32, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , (giving rise to t 42g e0g like electronic configuration). If the ∆o value of a ligand is, less than the pairing energy (P), then the electrons enter the eg orbital. On the other, hand, if the ∆0 value of a ligand is more than the pairing energy (P), then the, electrons enter the t 2g orbital., 19., , [Cr(NH3 )6 ]3+ is paramagnetic while [Ni(CN)4 ]2− is diamagnetic. Explain why?, Solution:, Cr is in the +3 oxidation state i.e., 1s2 , 2s 2 2p6 , 3s2 3p6 3d3 , 4s0, electronic configuration. NH3 act as a weak field ligand that does not cause the, pairing of the electrons in the 3d orbital., , Therefore, it undergoes d2 sp3 hybridization and the three electrons in the 3d, orbitals remain unpaired. Hence, it is paramagnetic in nature., [Ni(CN)4 ]2− , Ni, In, exists, in, the, 2, 2, 6, 2, 6, 8, 0, 1s , 2s 2p , 3s 3p 3d , 4s configuration., , +2, , oxidation, , state, , i.e.,, , CN− is a strong field ligand. It causes the pairing of the 3d orbital electrons. Then,, Ni2+ undergoes dsp2 hybridization., , As there are no unpaired electrons, it is diamagnetic., 20., , A solution of [Ni(H2 O)6 ]2+ is green but a solution of [Ni(CN)4 ]2− is colourless., Explain., Solution:, In [Ni(H2 O)6 ]2+ , H2 Ö acts as 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(H2 O)6 ]2+ is coloured., In [Ni(CN)4 ]2− , all the electrons are paired as CN − is a strong field ligand., Therefore, a d-d transition is not possible in [Ni(CN)4 ]2− . Hence, it is colourless., , Practice more on Coordination Compounds, , Page - 33, , www.embibe.com

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Class- XII-CBSE-Chemistry, , 21., , Coordination Compounds, , [Fe(CN)6 ]4− and [Fe(H2 O)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( ∆o ) . This CFSE, in turn, depends on the nature of, the ligand. In the case of [Fe(CN)6 ]4− and [Fe(H2 O)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., , 22., , 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 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., , 23., , Give the oxidation state, d-orbital occupation and coordination number of the, central metal ion in the following complexes:, (i), , K 3 [Co(C2 O4 )3 ], , (ii), , cis − [Cr(en)2 Cl2 ]Cl, , (iii), , (NH4 )2 [CoF4 ], , (iv), , [Mn(H2 O)6 ]SO4, , Solution:, (i), , K 3 [Co(C2 O4 )3 ], The central metal ion is Co. Let its oxidation state = x, C2 O4−2 oxalate ion is negative ligand its oxidation state = −2, Oxidation state of potassium = +1, , Practice more on Coordination Compounds, , Page - 34, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , The C2 O−2, 4 oxalate ion is a bidentate ligand., Oxidation state can be given as:, x − 6 = −3, x = +3, C2 O4−2 oxalate ion is bidentate ligand and there are three in the complex. So,, the coordination number of the complex is 6., C2 O4−2 oxalate ion is weak field ligand so the d orbital occupation for Co3+, is t 2g6 eg0 ., (ii), , cis − [Cr(en)2 Cl2 ]Cl, The central metal ion is Cr, Let its oxidation state of Cr = x, Cl chloride ion is negatively charged ligand and its oxidation state = −1, en is neutral ligand and its oxidation state = 0, The oxidation state can be given as:, x + 2(0) + 2(−1) = +1, x − 2 = +1, x = +3, chloride ion monodentate ligand and en is a bidentate ligand, so the, coordination number is 6., The d orbital occupation for Cr 3+ is t 2g3 ., , (iii), , (NH4 )2 [CoF4 ], The central metal ion is Co., Let its oxidation state of Cr = x, F fluoride ion is negatively charged ligand and its oxidation state = −1, Ammonium (NH4+1 ) is positively charged ion and its oxidation state = 0, The oxidation state can be given as:, x − 4 = −2, x = +2, chloride ion is a mono-dentate ligand, so the coordination number is 4., The d orbital occupation for Co2+ is e4g t 2g3 ., , Practice more on Coordination Compounds, , Page - 35, , www.embibe.com

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Class- XII-CBSE-Chemistry, , (iv), , Coordination Compounds, , [Mn(H2 O)6 ]SO4, The central metal ion is Mn., Let its oxidation state of Mn = x, Water (H2 O) is neutral ligand and its oxidation state = 0, Sulphate (SO4 ) is negatively charged ion and its oxidation state = -2, Water (H2 O) is a monodentate ligand so the coordination number of, compound is 6., The oxidation state can be given as:, x + 6(0) = +2, x = +2, The d orbital occupation for Mn is t 2g3 eg2 ., , 24., , 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(H2 O)2 (C2 O4 )2 ]. 3H2 O, , (ii), , [Co(NH3 )5 Cl]Cl2, , (iii), , CrCl3 (py)3, , (iv), , Cs[FeCl4 ], , (v), , K 4 [Mn(CN)6 ], , Solution:, Rules to write the IUPAC name of coordination compound :, (i), , Calculate the oxidation number of the central atom involved., , (ii), , The cation is named first as is the case with other ionic compounds. This, rule applies to both positively and negatively charged coordination entities., , (iii), , Usually, the number of free cation or anions is not given as can easily be, calculated by charge, , (iv), , For writing the names of complex cation use the following rule., Names of the ligand in english alphabetical order with their quantity as, prefix (but do not use mono for one and do not consider first letters of these, number denoting prefixes in alphabetisation). This is followed by the name, of metal followed by its oxidation number (in Roman numeral) in a bracket, like (0) (I), (II), (III), (-I), (-II) etc., , (v), , For writing the names of complex anion use the following rule., , Practice more on Coordination Compounds, , Page - 36, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , Names of the ligand in english alphabetical order with their quantity as, prefix (but do not use mono for one and do not consider first letters of these, number denoting prefixes in alphabetisation). This is followed by the name, of metal(modified as per rule already specified) followed by its oxidation, number (in Roman numeral) in a bracket. (I), (II), (III), (0), (-I), (-II), etc., (vi), , The neutral complex molecule is named similar to that of the complex, cation., , Following the above rules we get the IUPAC names as follows:, (i), , Potassium diaquadioxalatochromate (III) trihydrate., Oxidation state of chromium = 3, Electronic configuration: 3d3 : t 2g3, Coordination number = 6, Shape: octahedral, Stereochemistry:, , Magnetic moment, μ = √n(n + 2), = √3(3 + 2), = √15, = 4 BM, (ii), , [Co(NH3 )5 Cl]Cl2, , Practice more on Coordination Compounds, , Page - 37, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , IUPAC name: Caesium tetrachloroferrate (III), Oxidation state of Fe = +3, Electronic configuration of 𝑑 5 = eg2 t 2g3, Coordination number = 4, Shape: tetrahedral, Stereochemistry: optically inactive, Magnetic moment:, μ = √n(n + 2), = √5(5 + 2), = √35~6 BM, (v), , K 4 [Mn(CN)6 ], Potassium hexacyanomanganate (II), The oxidation state of manganese = +2, Electronic configuration: d5+ ; t 2g5, Coordination number = 6, Shape: octahedral, Stereochemistry: optically inactive, Magnetic moment, μ = √n(n + 2), = √1(1 + 2), = √3, = 1.732 BM, , 25., , What is meant by the stability of a coordination compound in solution? State the, factors which govern the stability of complexes., Solution:, The stability of a complex in a solution refers to the degree of association between, the two species involved in a state of equilibrium. Stability can be expressed, quantitatively in terms of stability constant or formation constant., M + 3L ⟷ ML3, [ML ], , Stability constant (𝑘𝑠𝑡𝑎𝑏𝑖𝑙𝑖𝑡𝑦 ) = [M][L]3 3, , Practice more on Coordination Compounds, , Page - 39, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , For this reaction, the higher the value of the stability constant, the greater is the, proportion of ML3 in the solution., Stability can be of two types:, (a), , Thermodynamic stability:, The extent to which the complex will be formed or will be transformed into, another species at the point of equilibrium is determined by thermodynamic, stability., , (b), , Kinetic stability:, This helps in determining the speed with which the transformation will, occur to attain the state of equilibrium., Factors that affect the stability of a complex are:, , 26., , 1., , Charge on the central metal ion: The greater the charge on the, central metal ion, the higher is the stability of the complex., , 2., , Basic nature of the ligand: A more basic ligand will form a more, stable complex., , 3., , Presence of chelate rings: Chelation increases the stability of, complexes., , What is meant by the chelate effect? Give an example., Solution:, When a ligand attaches to the metal ion in a manner that forms a ring, then the, metal-ligand association is found to be more stable. In other words, we can say that, complexes containing chelate rings are more stable than complexes without rings., This is known as the chelate effect., For example:, 2+, Ni2+, (aq) + 6NH3(aq) ⟷ Ni(aq) + 3en(aq) ⟷, , 27., , Discuss briefly giving an example in each case the role of coordination compounds, in:, (i), , biological system, , Practice more on Coordination Compounds, , Page - 40, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , (ii), , medicinal chemistry, , (iii), , analytical chemistry, , (iv), , extraction/metallurgy of metals, , Solution:, (i), , Role of coordination compounds in biological systems:, Photosynthesis is made possible by the presence of the chlorophyll pigment, and 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:, Coordination compounds of platinum such as cis-platin is used for, inhibiting the growth of tumours., , (iii), , Role of coordination compounds in analytical chemistry:, During the analysis of salt, when different reagents containing ligands are, added to different basic radicals they form coordination compounds. These, coordination compounds exhibit different colours associated with each of, the basic radicals. Thus some coordination compounds can be used in the, detection of various radicals., , (iii), , Role of coordination compounds in extraction or metallurgy of metals:, In metallurgy, 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., , 28., , How many ions are produced from the complex [Co(NH3 )6 ]Cl2 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 ]+2 along with two Cl− ions are produced. So, a total of 3 ions are, produced., , Practice more on Coordination Compounds, , Page - 41, , www.embibe.com

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Class- XII-CBSE-Chemistry, , 29., , Coordination Compounds, , Amongst the following ions which one has the highest magnetic moment value?, (i), , [Cr(H2 O)6 ]3+, , (ii), , [Fe(H2 O)6 ]2+, , (iii), , [Zn(H2 O)6 ]2+, , Solution:, [Cr(H2 O)6 ]3+ ∶ H2 O is a weak field ligand., oxidation state of Cr = +3, Electronic configuration of Cr +3 =1s 2 2s2 2p6 3s 2 3p6 3d3 4s0, Number of unpaired electrons in [Cr(H2 O)6 ]3+ = 3, Then, μ = √n(n + 2), = √3(3 + 2), = √15, ~4BM, [Fe(H2 O)6 ]2+ ∶ H2 O is a weak field ligand, The oxidation state of Fe = +2, Electronic configuration of Fe+2 =1s2 2s2 2p6 3s2 3p6 3d6 4s 0, Coordination number = 6, Number of unpaired electrons in [Fe(H2 O)6 ]2+ = 4, Then, μ = √4(4 + 2), = √24, ~5 BM, [Zn(H2 O)6 ]2+ ∶ H2 O is a weak field ligand, The oxidation state of Zn = +2, Electronic configuration of Zn+2 =1s2 2s2 2p6 3s2 3p6 3d10 4s 0, Number of unpaired electrons in [Zn(H2 O)6 ]2+ = 0, Hence, [Fe(H2 O)6 ]2+ has the highest magnetic moment value., 30., , The oxidation number of cobalt in K[Co(CO)4 ] is, (i), , +1, , (ii), , +3, , Practice more on Coordination Compounds, , Page - 42, , www.embibe.com

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Class- XII-CBSE-Chemistry, , (iii), , −1, , (iv), , −3, , Coordination Compounds, , Solution: (iii), We know that CO is a neutral ligand and K carries a charge of +1., Let the oxidation number of Co = x, Oxidation number =1(+1) + 𝑥 + 4 (0) = 0, 𝑥 = −1, Hence, the option (iii) is correct., 31., , Amongst the following, the most stable complex is, (i), , [Fe(H2 O)6 ]3+, , (ii), , [Fe(NH3 )6 ]3+, , (iii), , [Fe(C2 O4 )3 ]3−, , (iv), , [FeCl6 ]3−, , Solution:, Since the stability of a complex increases by chelation, the most stable complex is, [Fe(C2 O4 )3 ]3− ., , 32., , What will be the correct order for the wavelengths of absorption in the visible, region for the following:, [Ni(NO2 )6 ]4− , [Ni(NH3 )6 ]2+ , [Ni(H2 O)6 ]2+, Solution:, , Practice more on Coordination Compounds, , Page - 43, , www.embibe.com

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Class- XII-CBSE-Chemistry, , Coordination Compounds, , Since the central metal ion in all the three complexes is the same, the absorption in, the visible region depends on the ligands. The order in which the CFSE values of, the ligands increases in the spectrochemical series is as follows:, H2 O < NH3 < NO−, 2, Hence, the amount of crystal-field splitting observed will be in the following order:, ∆o(H, , 2 O), , < ∆o(NH ) < ∆o[NO, 3, , 2−], , Therefore, the wavelengths of absorption in the visible region will be in the order:, [Ni(NO2 )6 ]4− < [Ni(NH3 )6 ]2+ < [Ni(H2 O)6 ]2+, ⧫⧫⧫, , Practice more on Coordination Compounds, , Page - 44, , www.embibe.com