Coordination Compounds Class 12th Chemistry Part I CBSE Solution

Class 12th Chemistry Part I CBSE Solution
Intext Questions Pg-244
  1. Write the formulas for the following coordination compounds: (i)
  2. Write the IUPAC names of the following coordination compounds: (i) [Co(NH3)6]Cl3 (ii)…
Intext Questions Pg-247
  1. Indicate the types of isomerism exhibited by the following complexes and draw the…
  2. Give evidence that [Co(NH3)5Cl]SO4 and [Co(NH3)5(SO4)]Cl are ionisation isomers.…
Intext Questions Pg-254
  1. Explain on the basis of valence bond theory that [Ni(CN)4]2- ion with square planar…
  2. [Ni(Cl)4]2- is paramagnetic while [Ni(Co)4]is diamagnetic though both are tetrahedral.…
  3. [Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3-is weakly paramagnetic. Explain.…
  4. Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital…
  5. Predict the number of unpaired electrons in the square planar Pt[(CN)4]2- ion.…
  6. The hexaquo manganese (II) ion contains five unpaired electrons, while the hexacyano ion…
Intext Questions Pg-256
  1. Calculate the overall complex dissociation equilibrium constant for the ion [Cu(NH3)4]2+ ,…
Exercises
  1. Explain the bonding in coordination compounds in terms of Werner’s postulates.…
  2. FeSO4 solution mixed with (NH4)2SO4 solution in 1:1 molar ratio gives the test of Fe2+ ion…
  3. Explain with two examples each of the following: coordination entity, ligand, coordination…
  4. What is meant by unidentate, didentate and ambidentate ligands? Give two examples for…
  5. Specify the oxidation numbers of the metals in the following coordination entities: (i)…
  6. Using IUPAC norms write the formulas for the following: (i) Tetrahydroxidozincate (II)…
  7. Using IUPAC norms write the systematic names of the following: (i) [Co(NH3)6]Cl3 (ii)…
  8. List various types of isomerism possible for coordination compounds, giving an example of…
  9. How many geometrical isomers are possible in the following coordination entities? (i)…
  10. Draw the structures of optical isomers of: (i) [Cr(C2O4)3]3- (ii) [PtCl2(en)2]2+ (iii)…
  11. Draw all the isomers (geometrical and optical) of: (i) [CoCl2(en)2] + (ii)…
  12. Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will…
  13. Aqueous copper sulphate solution (blue in colour) gives: (i) a green precipitate with…
  14. What is the coordination entity formed when excess of aqueous KCN is added to an aqueous…
  15. Discuss the nature of bonding in the following coordination entities on the basis of…
  16. Draw figure to show the splitting of d orbitals in an octahedral crystal field.…
  17. What is spectrochemical series? Explain the difference between a weak field ligand and a…
  18. What is crystal field splitting energy? How does the magnitude of Δ0 decide the actual…
  19. [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2- is diamagnetic. Explain why?…
  20. A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2- is colourless. Explain.…
  21. [Fe(CN)6]4- and [Fe(H2O)6]2 + are of different colours in dilute solutions. Why?…
  22. Discuss the nature of bonding in metal carbonyls.
  23. Give the oxidation state, d orbital occupation and coordination number of the central…
  24. Write down the IUPAC name for each of the following complexes and indicate the oxidation…
  25. What is meant by stability of a coordination compound in solution? State the factors which…
  26. What is meant by the chelate effect? Give an example.
  27. Discuss briefly giving an example in each case the role of coordination compounds in: (i)…
  28. How many ions are produced from the complex Co(NH3)6Cl2 in solution? (i) 6 (ii) 4 (iii) 3…
  29. Amongst the following ions which one has the highest magnetic moment value? (i)…
  30. The oxidation number of cobalt in K[Co(CO)4] is (i) + 1 (ii) + 3 (iii) -1 (iv) -3…
  31. Amongst the following the most stable complex is (i) [Fe(H2O)6]3+ (ii) [Fe(NH3)6]3+ (iii)…
  32. What will be the correct order for the wavelength of absorption in the visible region for…

Intext Questions Pg-244
Question 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)


Answer:

(i) [Co(NH3)4(H2O)2]Cl3


(ii) K2[Ni(CN)4]


(iii) [Cr(en)3]NO3


(iv) [Pt(NH3)BrCl(NO2)]


(v) [Pt(Cl)2(en)2](NO3)2


(vi) Fe4[Fe(CN)6]3


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


Answer:

(i) Hexaminecobalt(III)chloride


(ii) Pentaaminechloridecobalt(III)chloride


(iii) Potassium hexacyanoferrate(III)


(iv) Potassium trioxalatoferrate(III)


(v) Potassium tetrachloridopalladate(II)


(vi) Diaminechloride(methylamine)platinum(II)chloride




Intext Questions Pg-247
Question 1.

Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers:

(i) K[Cr(H2O)2(C2O4)2

(ii) [Co(en)3]Cl3

(iii) [Co(NH3)5(NO2)](NO3)2

(iv) [Pt(NH3)(H2O)Cl2]


Answer:

(i) K[Cr(H2O)2(C2O4)2 exhibits geometrical isomerism (Cis and trans) and optical isomerism of cis and trans type.


(ii) Optical isomerism exhibiting mirror images.


(iii) Ionisation isomerism- [Co(NH3)5(NO3)](NO3)(NO2) and


Linkage isomerism-[Co(NH3)5(ONO)](NO3)2


(iv) Geometrical isomers are seen in [Pt(NH3)(H2O)Cl2]



Question 2.

Give evidence that [Co(NH3)5Cl]SO4 and [Co(NH3)5(SO4)]Cl are ionisation isomers.


Answer:

These compounds give different ions in aqueous solution. This can be tested by using AgNO3 solution and BaCl2

[Co(NH3)5Cl]SO4(aq) + BaCl2(aq) → BaSO4(ppt)


[Co(NH3)5Cl]SO4(aq) + AgNO3(aq) → no reaction


[Co(NH3)5(SO4)]Cl(aq) + BaCl2(aq) → no reaction


[Co(NH3)5(SO4)]Cl(aq) + AgNO3(aq) → AgCl(ppt)


Hence they give different precipitates with different solutions. Thus they are ionisation isomers.




Intext Questions Pg-254
Question 1.

Explain on the basis of valence bond theory that [Ni(CN)4]2- ion with square planar structure is diamagnetic and [Ni(Cl)4]2- the ion with tetrahedral geometry is paramagnetic.


Answer:

According to the valence band theory , the central metal atom or ion under the influence of ligands can use its (n-1)d , ns, np (inner orbital complex) or ns, np, nd (outer orbital complex)orbitals for hybridisation to form equivalent set of orbitals of definite geometry.

In [Ni(CN)4]2- , oxidation state of Ni can be calculated as :


Using overall charge balance as the whole ion has overall -2 charge:


x + 4(-1) = -2 (∵ CN- has -1 negative charge)


x = + 2


Ni is in + 2 oxidation state.


Electronic configuration of Ni is: [Ar]3d84s2


Where, [Ar] = 1s22s22p63s23p6


Electronic configuration of Ni+2 = [Ar]3d8


Outer electronic configuration of Ni+2 = 3d8


Since there are 4 CN ions so they can either form tetrahedral or square planar geometry. And CN- is a strong field ligand (according to experimental data of spectro-chemical series) it causes pairing of the 2 unpaired electrons.



It undergoes dsp2 (one d orbital, one s and two p orbitals used by the ligands) hybridization and forms square planar structure.


Since all the electrons are paired so it is diamagnetic.


Paramagnetic compounds- those compounds which have one or more no. of unpaired electrons in their atomic orbitals.


Diamagnetic compounds-those compounds in which all the electrons in their atomic orbitals are paired.



In case of [Ni(Cl)4]2- ion, Cl- is a weak field ligand so it will not pair the unpaired electrons of Ni+2 ion. Therefore it undergoes sp3 hybridization.


Overall charge balance:


X + 4(-1) = -2


X = + 2.



Since there are 2 unpaired electrons in the d orbital so it is a paramagnetic compound.



Question 2.

[Ni(Cl)4]2- is paramagnetic while [Ni(Co)4]is diamagnetic though both are tetrahedral. Why?


Answer:

In [Ni(Cl)4]2- ion, Cl- is a weak field ligand so it will not pair the unpaired electrons of Ni+2 ion.

Electronic configuration of Ni is: [Ar]3d84s2 where [Ar] = 1s22s22p63s23p6


Electronic configuration of Ni+2 = [Ar]3d8


Outer electronic configuration of Ni+2 = 3d8


Overall charge balance:


X + 4(-1) = -2


X = + 2.


Therefore it undergoes sp3 hybridization. So it will have tetrahedral geometry.



Since there are 2 unpaired electrons in the d orbital so it is a paramagnetic compound.


In [Ni(Co)4]:


Overall charge is neutral and oxidation state of Ni can be calculated as:


X + 4(0) = 0


X = 0


Ni is in zero oxidation state.



Co is a strong field ligand it causes pairing of the 4 unpaired electrons in d orbital. Also, it causes the 4s electrons to shift to the 3d orbital, thereby undergoes sp3 hybridisation forming tetrahedral geometry. Since it has no unpaired electron, therefore it is diamagnetic compound.



Question 3.

[Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3-is weakly paramagnetic. Explain.


Answer:

In [Fe(H2O)6]3+

Electronic configuration of Fe is: [Ar]3d64s2


[Ar] = 1s22s22p63s23p6


Electronic configuration of Fe+3 = [Ar]3d5


Outer electronic configuration of Fe+3 = 3d5


Overall charge balance:


X + 6(0) = 3


X = + 3


In [Fe(CN)6]3-


Overall charge balance:


X + 6(-1) = -3


X = + 3


In both the compounds Fe is in + 3 oxidation state.



In case of [Fe(H2O)6]3+


H2O is weak field ligand so it does not pair the unpaired electron. Total no. of the unpaired electron, n = 5.


Spin only magnetic moment is given by:


μ = [n(n + 2)]1/2


μ = [5×7]1/2


μ = 5.916BM


In case of [Fe(CN)6]3-


CN- is a strong field ligand so it pairs up the electron.


Total no. of unpaired electrons = 1


Spin only magnetic moment is given by:


μ = [n(n + 2)]1/2


μ = [1×3]1/2


μ = 1.732BM


as we can see spin only magnetic moment of [Fe(H2O)6]3+ is more than [Fe(CN)6]3- .


so, [Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3 weakly paramagnetic.



Fe in ground state


Fe in + 3 oxidation state


Electrons from 5 CN- ligands



Question 4.

Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex.


Answer:




Question 5.

Predict the number of unpaired electrons in the square planar Pt[(CN)4]2- ion.


Answer:

In Pt[(CN)4]2- ion:

Overall charge balance:


X + 4(-1) = -2


X = + 2.


The oxidation state of Pt is + 2.


Since CN- is a strong field ligand, it causes pairing of the unpaired electrons.



Therefore, now the 2 unpaired electrons from 5d orbital get paired and it undergoes dsp2 hybridisation. It forms square planar geometry. Since all the electrons are paired,


No. of unpaired electrons = 0.



Question 6.

The hexaquo manganese (II) ion contains five unpaired electrons, while the hexacyano ion contains only one unpaired electron. Explain using Crystal Field Theory.


Answer:


Mn in + 2 oxidation state:


In presence of CN- Ligands:



d2sp3 hybridisation:



Electronic configuration of [Mn(CN)6]4-:





Intext Questions Pg-256
Question 1.

Calculate the overall complex dissociation equilibrium constant for the ion [Cu(NH3)4]2+ , given that β4 for this complex is 2.1×1013.


Answer:

given: β4 = 2.1×1013

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



⇒ The overall complex dissociation equilibrium = 4.7×10-14




Exercises
Question 1.

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


Answer:

Bonding in coordination compounds in terms of Werner’s postulates is explained as:

a) Metals can show two types of valencies which are Primary valency and Secondary valency.


1. Primary Valency: Primary Valency shows Oxidation state. Primary valencies are ionizable.


2. Secondary Valency: Secondary Valency shows coordination number. These are non-ionizable.


b) Both Primary and secondary valency of the metal are to be satisfied which is done by negative ions in case of primary valency and negative or neutral species in case of secondary valency.


c) Metals have a fixed number of secondary valencies/ Coordination number around the central atom, these secondary valencies are placed in such a way which leads to a specific geometry of the coordination compound.




Question 2.

FeSO4 solution mixed with (NH4)2SO4 solution in 1:1 molar ratio gives the test of Fe2+ ion but CuSO4 solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu2+ ion. Explain why?


Answer:

The reaction is given below:

FeSO4 + (NH4)2SO4 + 6H2O → FeSO4(NH4)2SO4.6H2O (Mohr Salt)


FeSO4, when reacted with (NH4)SO4, does not form any complex whereas they form a double salt, FeSO4.(NH4)2SO4.6H2O - (Mohr salt) which dissociates into ions in the solution. So, it gives the test of Fe2+ ions.


CuSO4 + 4NH3 + 5H2O→ [Cu(NH3)4SO4].5H2O


CuSO4 solution when mixed with aqueous ammonia in 1: 4 molar ratio forms a complex with formula [Cu(NH3)]SO4 in which the complex ion, [Cu(NH3)4]2+ does not dissociate to give Cu2+ ions. Therefore, it does not give the tests of the Cu2+ ion.



Question 3.

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


Answer:

1) Coordination Entity: Coordination entity is a charged entity having positive or negative charge in which the central atom is surrounded by molecules which may be neutral/negatively charged called Ligands.


Examples:


i. Cationic Complexes: [Cu(H2O)6]2+ , [Al(H2O)6]3+


ii. Anionic Complexes: [CuCl4]2- , [Al(H2O)2(OH)4]-


iii. Neutral Complexes: [Co(NH3)4 Cl2] ,[Ni(CO)4]


2) Ligands: Ligands are the neutral or negatively charged entities surrounding the central metal atom of the coordination complex which possesses at least one unshared pair of electrons


Example: F-, Cl-, Br-, I-, H20, and NH3


3) Coordination Number: Coordination Number which is also called as Ligancy is the total number of ligands that are attached to the central metal atom of the coordination complex.


Example: [Cr(NH3)2Cl2Br2] has Cr3+ as its central cation, and has a coordination number of 6


Al3+ has coordination number 4 in [AlCl4]- but 6 in [AlF6]3-.


4) Coordination polyhedron: Coordination polyhedron is defined as the spatial arrangement of the ligands around the central atom of a coordination complex.


Example:



In the figure: The pink Sphere depicts the central atom of the coordination entity.


5) Homoleptic complexes: Complexes in which the central metal atom is surrounded only by the same kind of donor groups which are the ligands.


Example: [Ag(CN)2]-,[Fe(CN6)]4+ etc


6) Heteroleptic complexes: Complexes in which the central metal atom is surrounded by more than one kind of donor groups which are the ligands.


Example: [Co(NH3)4 Cl2+ ,[Co(NH3)5 Cl]2+



Question 4.

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


Answer:

Ligands are the neutral or negatively charged entities surrounding the central metal atom of the coordination complex which possesses at least one unshared pair of electrons.


Based on the number of donor sites of these ligands, Ligands are classified as:


Unidentate ligands: These Ligands which have only one donor site are called unidentate ligands.


Example: F-,Cl  etc


Didentateligands: These Ligands which have only two donor site are called didentate ligands.


Example: Ethane-1,2-diamine, Oxalate ion etc


Ambidentate ligands: These ligands which can attach them with the central metal atom by two different atoms are called as ambidentate ligands.


Example:




Question 5.

Specify the oxidation numbers of the metals in the following coordination entities:

(i) [Co(H2O)(CN)(en)2]2+

(ii) [CoBr2(en)2]+

(iii) [PtCl4]2–

(iv) K3[Fe(CN)6]

(v) [Cr(NH3)3Cl3]


Answer:

(i) Let Oxidation no. of Co be x and charge on the complex is given as + 2


H2O has Oxidation Number: 0


CN has Oxidation Number: -1


en has Oxidation Number :0



(ii) Let Oxidation number of Co be x and charge on the complex is given as + 1


Br has Oxidation number: 1


en has oxidation number : 0



(iii) Let Oxidation number of Pt be x and charge on the complex is given as -2


Cl has oxidation number : -1



(iv) This complex can also be seen as [Fe(CN)6]3-


Let Oxidation number of Fe be x and charge given on the complex is given as -3


CN has oxidation number : -1



(v) Let Oxidation number of Cr be x and charge given on the complex is given as 0


NH3 has oxidation number : 0


Cl has oxidation number: -1




Question 6.

Using IUPAC norms write the formulas for the following:

(i) Tetrahydroxidozincate (II)

(ii) Potassium tetrachloridopalladate (II)

(iii) Diamminedichloridoplatinum (II)

(iv) Potassium tetracyanidonickelate (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)


Answer: (i) Tetrahydroxidozincate (II)

[Zn (OH)4)]2-

(ii) Potassium tetrachloridopalladate (II)

K2[PdCl4]

(iii) Diamminedichloridoplatinum (II)

[Pt(NH3)2Cl2]

(iv) Potassium tetracyanidonickelate (II)

K2[Ni(CN)4]

(v) Pentaamminenitrito-O-cobalt (III)

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

(vi) Hexaamminecobalt (III) sulphate

[Co(NH3)6]2 (SO4)3

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

K3[Cr(C2O4)3]

(viii) Hexaammineplatinum (IV)

[Pt(NH3)6]4+

(ix) Tetrabromidocuprate (II)

[Cu(Br)4]2+

(x) Pentaamminenitrito-N-cobalt (III)

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



Question 7.

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]


Answer:

(i) Starting with cation , the complex ion contains six ammonia molecules with cobalt in + 3 oxidation state. The name of compound: [Co(NH3)6]Cl3 is hexaamminecobalt(III) chloride.


(ii) The complex ion is cation, so there are 2 ammonia molecules, one chloride ion and methyammine molecule qith platinum in + 2 state. Going in alphabetical order, the name of compound: [Pt(NH3)2Cl(NH2CH3)]Cl is diamminechloridomethylammineplatinum(II) chloride.


(iii) It is a complex cation with six water molecules and titanium atom in + 3 state. The name of the compound: [Ti(H2O)6]3+ is hexaaquatitanium(III) ion


(iv) The complex ion is cation with cobalt in + 3 state and with four ammonia molecules i.e tetraammine, one chloride ion and nitrate ion. The name of the compound: [Co(NH3)4Cl(NO2)]Cl is tetraamminechloridonitritocobalt(III) chloride.


(v) The complex ion is cation with manganese in + 2 state and six water molecules. The name of compound: [Mn(H2O)6]2+ is hexaaquamanganese(II) ion.


(vi) The complex is anion with nickel in + 2 state and with four chloride ions. The name of the compound: [NiCl4]2– is tetrachloridonickeltate(II) ion.


(vii) The complex is cation with nickel in + 2 state and six ammonia molecules. The name of compound: [Ni(NH3)6]Cl2 is hexaamminenickel(II) chloride.


(viii) The complex is cation withcobalt in + 3 stateand there is a bidentate ligand called as ethylenediamine. The name of compound : [Co(en)3]3+ is tris(ethylenediamine)cobalt(III) ion.


(ix) It is neutral complex with four carbonyl molecules and cobalt in 0 state. The name of compound: [Ni(CO)4] is tetracarbonylcobalt(0).



Question 8.

List various types of isomerism possible for coordination compounds, giving an example of each.


Answer:

Isomers are the compounds which have same chemical formula but different arrangement of atoms in space. There are principle two types of isomerism:


(i) Stereo isomerism:


(a) geometrical isomerism


(b) optical isomerism


(ii) structural isomerism


(a) Ionisation isomerism


(b) Linkage isomerism


(c) Coordination isomerism


(d) Solvate isomerism


Geometrical isomerism comes into existence by the different spatial arrangement of groups around the central metal atom. Similar groups may either be arranged on the same side or on opposite sides of the central metal atom. This gives rise to two types of isomers called cis and trans isomers. When groups under consideration are arranged on the same side of the central metal atom, isomers are called cis isomers and when the groups under consideration are spatially placed on the opposite sides, isomers are called trans isomer.



Optical isomerism is exhibited by those compounds which possess chirality. The presence of an element of symmetry makes a molecule symmetric and renders it optically inactive. When molecule does not possess any element of symmetry its mirror image is non superimposable with the molecule itself. This makes the molecule optically active. Such an asymmetric molecule can show the phenomenon of optical .The two forms of the molecule which are mirror images of each other are called enantiomers.One form rotates the plane of plane polarized light in clockwise direction, while the other in anticlockwise direction. The former is called the d-form, while the latter is termed as l-form.



Ionisation isomerism: In ionisation isomerism there is an exchange of ions inside and outside the coordination sphere. Ionisation isomers have the same formula but produce different ions in solution. It is also known as ion-ion exchange isomerism.


Examples:



In the two isomers (A) and (B), there is an exchange of ions inside and outside the coordination sphere. The aqueous solution of (A) gives the precipitate of AgBr on treatment with AgNO3 as it contains Br, while that of (B) gives precipitate of BaSO4, on treatment with BaCl2.