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NCERT Textual Exercise (Solved)

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Coordination Compounds

1. Explain the bonding in coordination compounds in terms of Werner’s postulates.

Sol. The main postulates of Werner’s theory of coordination compounds are as follows:

(a) Metals possess two types of valencies called (i) primary valency which are ionisable; (ii) secondary valency which are non-ionisable. (b) Primaryvalencyissatisfiedbythenegativeionsanditisthatwhicha

metal exhibits in the formation of its simple salts (e.g. in CoCl3⋅6NH3, valency between Co and Cl is primary valency and valency between Co and NH3 is secondary).

(c) Secondaryvalenciesaresatisfiedbyneutralligandornegativeligandand are those which metal exercises in the formation of its complex ions.Everycationhasafixednumberofsecondaryvalenciesandaredirectedinspaceaboutcentralmetalionincertainfixeddirections.For example in CoCl3⋅6NH3, six ammonia molecules linked to Co by secondary valencies are directed to six corners of a regular octahedron and thus account for structure of CoCl3⋅6NH3 as follows:

In modern theory, it is now referred as coordination number of central

metal atom or ion. 2. FeSO4 solution mixed with (NH4)2SO4 solution in 1 : 1 molar ratio given the

test of Fe2+ but CuSO4 solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu2+ ion. Explain why.

Sol. FeSO4 solution mixed with (NH4)2SO4 solution in 1 : 1 molar ratio forms a double salt, FeSO4⋅ (NH)2SO4⋅6H2O (Mohr’s salt) which ionises in the solution to give Fe2+ ions. Hence it gives the tests of Fe2+ ions. CuSO4 solution mixed with aqueous ammonia in 1 : 4 molar ratio forms a complex salt, with the formula [Cu (NH3)4] SO4. The complex ion [Cu (NH3)4]

2+ does not ionise to give Cu2+ ions. Hence, it does not give the tests of Cu2+ ion.

4. What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.

Sol. A molecule or an ion which has only one donor atoms to form one coordinate


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bond with the central metal atom is called unidentate ligand (e.g. Cl– and NH3).

A molecule or ion which contains two donor atoms and hence forms two coordinate bonds with the central metal atom is called a didentate ligand. For example,

CH NH and COO

CH NH COO

2 2

2 2

−

−| |

A molecule or an ion which contains two donor atoms but only one of them forms a coordinate bond at a time with the central metal atom is called ambidentate ligand (e.g. :CN– or NC : and NO ONO2

− or : ). 5. Specify the oxidation numbers of the metals in the following coordination

entities: (a) [Co(H2O) (CN) (en)2]

2+ (b) [Pt Cl4]2–

(c) [Cr (NH3)3 Cl3] (d) [Co Br2 (en)2]+

(e) K3 [Fe (CN)6] Sol. (a) [Co (H2O) (CN) (en)2]

2+ (b) [PtCl4]2–

x + 0 + (–1) + 0 = + 2 x – 4 = – 2 x = + 3 x = + 2 (c) [Cr (NH3)3 Cl3] (d) [Co Br2 (en2)]

+

x + 0 + 3 (–1) = 0 x + 2 (–1) + 0= + 1 x = + 3 x = + 3 (e) K3 [Fe (CN)6] ≡ [Fe (CN)6]

3–

x + 6 (–1)= – 3 x = + 3. 6. Using IUPAC norms, write the formulae of the following: (a) Tetrahydroxozincate (II) (b) Hexaamminecobalt (III) sulphate (c) Potassium tetrachloridopalladate (II) (d) Potassium trioxalatochromate (III) (e) Diaminedichlorido platinum (II) (f) Hexammine platinum (IV) (g) Potassium tetracyanonickelate (II) (h) Tetrabromidocuprate (II) (i) Pentaamminenitrito-O-cobalt (III)


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(j) Pentaamminenitrito-N-cobalt (III) Sol. (a) [Zn (OH)4]

2– (b) [Co (NH3)6]2(SO4)3

(c) K2 [PdCl4] (d) K3 [Cr (C2O4)3] (e) [Pt (NH3)2Cl2] (f) [Pt (NH3)6]

4+

(g) K2 [Ni (CN)4] (h) [Cu Br4]2–

(i) [Co (NH3)5 (ONO)]2+ (j) [Co (NH3)5 (NO2)]2+

7. Using IUPAC norms, write the systematic names of the following: (a) [Co (NH3)6]Cl3 (b) [Co (NH3)4 Cl (NO2)]Cl (c) [Ni (NH3)6] Cl2 (d) [Pt Cl (NH3)2 (NH2CH3)] Cl (e) [Mn (H2O)6]

2+ (f) [Co (en)3]3+

(g) [Ti (H2O)6]3+ (h) [Ni Cl4]

2–

(i) [Ni (CO)4] Sol. (a) Hexaamminecobalt(III) chloride. (b) Tetraamminechloridonitrito-N-cobalt(III) chloride. (c) Hexaamminenickel(II) chloride. (d) Diaminechlorido(methylamine)platinum(II) chloride. (e) Hexaaquamanganese(II) ion. (f) Tris(ethane-1, 2-diamine)cobalt(III) ion. (g) Hexaaquatitanium(III) ion. (h) Tetrachloridonickelate(II) ion. (i) Tetracarbonylnickel(0). 8. Draw the structures of optical isomers of (a) [Cr (C2O4)3]

3– (b) [Pt Cl2 (en)2]2+

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

Sol. (a) [Cr (C2O4)3]3– ⇒ [Cr (oxo)3]

3–

Practice Yourself

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(b) cis - [Pt Cl2 (en)2]2+

(c) cis - [Cr (NH3)2 Cl2 (en)]+

and trans C NH Cl

r en3 2 2

9. How many geometrical isomers are possible in the following coordination

entities? (a) [Cr (C2O4)3]

3– (b) [Co Cl3 (NH3)3] Sol. (a) [Cr (C2O4)3]

3– ⇒ No geometrical isomers are possible in this coordination entity.

(b) [Co Cl3 (NH3)3] ⇒ Two geometrical isomers are possible in this coordination entity (fac and mer)

11. Draw all the isomers (geometrical and optical) of (a) [CoCl2 (en)2] (b) [Co (NH3) Cl (en)2]

2+

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

Sol. (a) [CoCl2 (en)2]

Trans [CoCl ( )]en2 2

Co

Cl

enen

Cl

Co

Cl

enen

Cl


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(b) [Co (NH3) Cl (en)2]

2+

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

12. Write all the geometrical isomers of [Pt (NH3) (Br) (Cl) (Py)] and how many of these will exhibit optical isomerism?

Sol. Three isomers of [Pt (NH3) (Br) (Cl) (Py)] are possible.

This type of isomers do not show any optical isomerism. Optical isomerism

only rarely occurs in square planar or tetrahedral complexes and that too when they contain unsymmetrical chelating ligand.


380


13. Aqueous copper sulphate solution (blue in colour) gives (a) agreenprecipitatewithaqueouspotassiumfluoride,and (b) a bright green solution with aqueous potassium chloride. Explain these experimental results. Sol. Aqueous CuSO4 solution exists as [Cu (H2O)4]SO4 which has blue colour

due to [Cu (H2O)4]2+ ions.

(a) When KF is added, the weak H2O ligands are replaced by F– ligands forming [CuF4]

2– ions which is a green precipitate.

[Cu (H2O)4]

2+ + 4F–

→ +−[ ]Cu F 4H O

Tetrafluorocuperate (II)(Green ppt.)

42

2

(b) When KCl is added, Cl– ligands replace the weak H2O ligands forming [CuCl4]

2– ion which has bright green colour.

[Cu (H2O)4]

2+ + 4Cl– → +−[ ]( )

( )

Cu Cl H OTetrachloro cuperate II

Green solution

42

24

14. What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is that no precipitate of copper sulphide is obtained when H2S (g) is passed through this solution?

Sol. First cupric cyanide is formed which decomposes to give cuprous cyanide and cyanogen gas. Cuprous cyanide dissolves in excess of potassium cyanide to form the complex, K3 [Cu (CN)4].

CuSO KCN Cu CN K SOCu CN Cu CN CN

Cu

4 2 2 4

2 2 2 2

2

2 22+ → + ×

→ +( ) ]

( ) ( ) ( )(CCN KCN K Cu CN

CuSO KCN K Cu CN K SO) [ ( ) ]

[ ( ) ]2 3 4

4 3 4 2

6 22 10 2 2

+ →

+ → + 44 2+ ( )CN

Thus, coordination entity formed is [Cu(CN)4]3–. As CN– is a strong ligand,

the complex ion is highly stable and does dissociate/ionise to give Cu2+ ions. Hence, no precipitate with H2S is formed.

15. Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:

(a) [Fe (CN)6]4– (b) [Fe F6]

3–

(c) [Co (C2O4)3]3– (d) [Ni Cl4]

2–

Sol. (a) [Fe (CN)6]4– is d2sp3 hybridised, octahedral in shape and diamagnetic

in nature.


381


(b) [Fe F6]

3– is sp3d2 hybridised, octahedral in shape and paramagnetic in nature.

3d 4s 4p 4d

Fe (Z = 26) ground state

Fe3+

sp d hybrid orbitals2 3

×× ×× ×× ×× ×× ××

sp d32

[Fe F ]6

3–

six pair of electronsfrom F

–

(c) [Co(C2O4)3]3– is d2sp3 hybridised, octahedral in shape and diamagnetic

in nature.

Co (Z = 27) ground state

Co3+

d sp hybrid orbitals2 3

[Co (C O ) ]2 4 3

3–

3d

×

d sp2 3

Six pair of electrons from C O2 42–

4s 4p

× ×× ×× ×× ×× ××

(d) [Ni Cl4]2– is sp3 hybridised, tetrahedral in shape and paramagnetic

in nature.


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16. Drawfiguretoshowsplittingofdegeneratedorbitalsinanoctahedralcrystal

field. Sol.

d dxy xzdyz

AverageEnergy

Level(Bari centre)

0.6�0

0.4�0

eg

t2g

d-orbitals infree ion

Average energyof -orbitals inda spherical crystalfield

State I

State II

�0

2 2 2x y zd ,d

�

17. Whatisspectrochemicalseries?Explainthedifferencebetweenaweakfieldligandandastrongfieldligand.

Sol. The arrangement of ligands in order of their increasingfield strengths,that is increasing crystal field splitting energy (CFSE) values, is calledspectrochemical series.

The ligands with small value of CFSE (∆0)arecalledweakfieldligands,whereasthosewithlargevalueofCFSEarecalledstrongfieldligands.

18. Whatiscrystalfieldsplittingenergy?Howdoesthemagnitudeof∆0 decide theactualconfigurationofd-orbitals is a coordination entity?

Sol. When the ligands approach a transition metal ion, the d-orbitals split into two sets, one with lower energy and the other with higher energy. The difference ofenergybetweenthetwosetsoforbitals iscalledcrystalfieldsplittingenergy (∆0foroctahedralfield).If∆0 < P (pairing energy), the fourth electron


383


enters one of the eg,orbitalsgivingtheconfiguration t eg g23 1 thus forming

high spin complexes. Such ligands for which ∆0<Parecalledweakfieldligands. If ∆0 > P, the fourth electron pairs up in one of the t2g orbitals giving theconfiguration t eg g2

4 thereby forming low spin complexes. Such ligands for which ∆0>Parecalledstrongfieldligands.

19. [Cr (NH3)6]3+ is paramagnetic while [Ni (CN)4]

2– is diamagnetic. Explain why?

Sol. The presence of three unpaired electrons in [Cr (NH3)6]3+ explains its

paramagnetic character.

Cr (Z = 24) inground state

Cr3+

[Cr (NH ) ]3 6

3+

3d��

�

�

�

�

�

�

�

� ×× × × ×

4s 4p

× ×× × × × ×

[Ni (CN)4]2– is diamagnetic, since there is no unpaired electrons. dsp2

Ni (Z = 28) in ground state

Ni2+

[Ni (CN) ]4

2–

3d

× ×××

4s 4p

× × × ×

[Ni (CN) ]4

2–

dsp2

20. A solution of [Ni (H2O)6]2+ is green but a solution of [Ni (CN)]4

2– is colourless. Explain.

Sol. In [Ni(H2O)6]2+, Ni is in +2 oxidation state and having 3d8 electronic

configuration,inwhichtherearetwounpairedelectronswhichdonotpairin the presence of the weak H2O ligand. Hence, it is coloured. The d–d transition absorbs red light and the complementary light emitted is green.

In [Ni(CN)4]2–, Ni in +2 oxidation state and having 3d8 electronic

configuration.ButinpresenceofstrongligandCN– the two unpaired electrons in the 3d orbitals pair up. Thus, there is no unpaired electron present. Hence, it is colourless.

21. [Fe(CN)6]4– and [Fe(H2O)6]

2+ are of different colours in dilute solution. Why? Sol. In both the complexes, Fe is in +2 oxidation state and have 3d6 electronic

configuration,inwhichithasfourunpairedelectrons.Inpresenceoftheweak H2O ligand, the unpaired electrons do not pair up. In presence of the


384


strong CN– ligand, they pair up leaving no unpaired electron. It is because of this difference in the number of unpaired electrons, they have different colours.

22. Discuss the nature of bonding in metal carbonyls. Sol. In metal carbonyl, the metal carbon bond (M–C) bond possess both the σ

and π-bond character. The bond are formed by overlap of atomic orbital of metal with that of C-atom of carbon monoxide in the following sequence:

(a) σ-bond isfirst formedbetweenmetalandcarbonwhenavacantd-orbital of metal atom overlaps with an orbital containing lone pair of electrons on C-atom of carbon monoxide (:C≡O:)

(b) In addition to σ-bondinmetalcarbonyl,theelectronsfromfilledd-orbitals of a transition metal atom/ion are back donated into anti-bonding π-orbitals of carbon monoxide. This stabilises the metal ligand bonding.

Theabovetwoconceptsareshowninthefollowingfigure: σ-overlap: Donation of lone pair of electrons on carbon into a vacant

orbital on the metals.

π-overlap: Donationofelectronsfromafilledmetald-orbital into

a vacant anti-bonding π-orbital of CO. 23. Give the oxidation state d-orbital occupation and coordination number of

central metal ion in the following complexes: (a) K3 [Co (C2O4)3] (b) (NH4)2 [Co F4] (c) cis [Cr (en)2 Cl2] Cl (d) [Mn (H2O)6] SO4

Sol. (a) K3 [Co (C2O4)3] ⇒ [Co (C2O4)3]3–

x – 3 (2) = – 3 Oxidation state, x = + 3 Coordination number is also 6 as C2O4

2– is didentate, Co3+ = 3d6, t2g6

eg° Diamagnetic

(b) (NH4)2 [CoF4] ( ) [ ]NH CoF4 2 42

x – 4 = –2 ∴ Oxidation state, x = +2


385


Coordination number = 4, Co2+ = 3d7, eg4 t2g

3, 3 unpaired electrons. Paramagnetic

(c) cis – [Cr (en)2 Cl2]+ Cl–

x + 0 – 2 = +1 Oxidation state, x = +3 Coordination number is 6 as en is didentate, Cr3+=3d3(t2g

3 eg°), 3 unpaired electrons, paramagnetic.

(d) [Mn (H2O)6]2+ SO4

2–

x + 0 = +2 ∴ Oxidation state, x = +2 Coordination number is 6, Mn2+ = 3d5 (t3

2g eg2), Paramagnetic

24. Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordinationnumber. Also sterochemistry and magnetic moment of the complex:

(a) K [Cr (H2O)2 (C2O4)2] ⋅ 3H2O (b) CrCl3 (Py)3

(c) K4 [Mn (CN)6] (d) [Co (NH3)5 Cl]Cl2

(e) Cs [Fe Cl4] Sol. (a) IUPAC name is potassium diaquadioxalatochromate (III) hydrate. Coordination number = 6 Oxidation state of Cr: x + 0 + 2 (–1) = –1 ∴ x = +3 Shape is octahedral ElectronicconfigurationofCo3+ = 3d3 = t eg g2

3 0 . Magnetic moment, µ = n n( ) 2 3 5 15 BM. = 3⋅87 BM

(b) CrCl3 (Py)3

IUPAC name is trichloridotripyridinechromium (III). Coordination number of Cr = 6 Oxidation state of Cr, x – 3 + 0 = 0 x = 3

ElectronicconfigurationofCr3+ = 3d3 = t eg g23 0

Unpaired electrons (n) = 3 µ = 3⋅87 BM. (c) K4 [Mn (CN)6] IUPAC name is potassium hexacyanomanganate (II) Coordination number of Mn = 6


386


Shape is octahedral Oxidation state of Mn is x – 6 = –4 x = +2

ElectronicconfigurationofMn2+ = 3d5 = t eg g25 0

n = 1 µ = 1 1 2 3 1 73( ) BM . (d) [Co (NH3)5 Cl] Cl2

IUPAC name is pentaamininechloridocobalt (III) chloride Coordination number of Co = 6 Shape is octahedral. Oxidation state of Co, x + 0 – 1 = + 2 ∴ x = + 3

ElectronicconfigurationofCo3+ = 3d6 = t eg g26

n = 3 µ = 0 (e) Cs [FeCl4] IUPAC name is caesium tetrachloroferrate (III) Coordination number of Fe is 4 Shape is tetrahedral Oxidation number of Fe, x – 4 = –1 ⇒ x = +3

ElectronicconfigurationofFe3+ = e tg g2

23

n = 5 µ = 5 (5 + 2) = 35 = 5.92 BM . 26. What is meant by chelate effect? Give an example. Sol. When a didentate or a polydentate ligand contains donor atoms positioned

insuchawaythatwhentheycoordinatewiththecentralmetalion,afiveora six membered ring is formed, the effect is called chelate effect.

For example, Cl N — CH2

N — CH2Cl

Pt

H2

H2

28. How many ions are produced from the complex [Co (NH3)6 ]Cl2 in solution? (a) 6 (b) 4 (c) 3 (d) 2 Sol. Coordination number of cobalt = 6


387


Hence, the complex is [Co (NH3)6] Cl2. It ionises in the solution as: [Co (NH3)6] Cl2

aq [Co (NH3)6]2+ + 2Cl–

Thus, three ions are produced. Hence, the correct option is (C) 29. Among the following ions, which one has the highest magnetic moment: (a) [Cr (H2O)6]

3+ (b) [Fe (H2O)6]2+

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

Sol. The oxidation states are: Cr (III), Fe (II) and Zn (II). ElectronicconfigurationofCr3+ = 3d3, unpaired electron = 3 ElectronicconfigurationofFe2+ = 3d6 , unpaired electron = 4 ElectronicconfigurationofZn2+ = 3d10, unpaired electrons = 0

µ = n n( )+ 2 Hence, (b) has highest value of magnetic moment. 30. The oxidation number of cobalt in K [Co (CO)4] is (a) +1 (b) +3 (c) –1 (d) –3 Sol. K [Co (CO)4] = K+ [Co (CO)4]

–

∴ x + 0 = –1 x = –1 Hence, (c) is the correct option. 31. Among the following, the most stable complex is: (a) [Fe (H2O)6]

3+ (b) [Fe (NH3)6]3+

(c) [Fe (C2O4)3]3– (d) [Fe Cl6]

3–

Sol. In each of the given complex, Fe is in +3 oxidation state. As C2O42– is didentate

chelating ligand, it forms chelate rings and hence is the most stable complex. 32. What will be the correct order for the wavelengths of absorption in the visible

region of the following: [Ni (NO2)6]4–, [Ni (NH3)6]

2+, [Ni (H2O)6]2+

Sol. Asmetalionisfixed,theincreasingfieldstrengths(i.e.theCFSEvaluesofthe ligands from the spectrochemical series) are in the order: H2O < NH3 < NO2

–

Thus, the energies absorbed for excitation will be in the order: [Ni (H2O)6]

2+ < [Ni (NH3)6]2+ < [Ni (NO2)6]

4–

As E = hcλ

the wavelengths absorbed will be in the opposite order.

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