Question

NCERT Solutions Class 12 Chemistry Chapter 9 Coordination Compounds is discussed in a very detailed manner. NCERT Solutions is based on the topics of chemistry particularly the coordination compounds. NCERT Solutions Class 12 is made in a pointwise method to make it easy for a student to go through all difficult concepts.

• Coordination compounds – coordination compounds are such chemical compounds in which the central metal atom is chemically bonded by the atoms of a non-metal group of non-metals and these non-metal groups are called the ligands.
• Werner’s Theory of Coordination Compounds – Werner the inorganic chemist proposed a theory in 1893 that explained the structure, formation, and the nature of bonds formed in coordination compounds. He was the first-ever chemist who won the Nobel Prize in 1913 for his extraordinary works in the field of chemistry. He worked out the study of different chemical compounds which are obtained upon reaction between cobalt chloride and ammonia.
• Definitions of Some Important Terms About Coordination Compounds – there are different important terms related to the coordination compounds such as:
  • Coordination entity – in the coordination entity the central ion or atom is bound to set several atoms, molecules, or ions.
  • Central Atoms and Central Ions – the metallic atom around which the non-metallic atoms are bound or held are called the central atoms or the central ions.
  • Ligands – the atoms or the ions which are chemically bound to the central metallic atom are called the ligands.
• Nomenclature of Coordination Compounds – while naming the coordination compounds the neutral ligands are named first then the central atom is named as the suffix.
• Isomerism in Coordination Compounds – isomers are those different compounds that have the same chemical formula but the arrangement of atoms is different. In the development of coordination chemistry, it is very important to consider the arrangement of ligands.

Students must refer to our NCERT Solutions Class 12 Chemistry for understanding the tough concepts of chemistry. This solution is prepared by experts.  

Answer 1

NCERT Solutions Class 12 Chemistry Chapter 9 Coordination Compounds

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

Answer:

(a) A metal shows two kinds of valencies viz primary valency and secondary valency. Negative ions satisfy primary valencies and secondary valencies are filled by both neutral ions and negative ions.

(b) A metal ion has a fixed amount of secondary valencies about the central atom. These valencies also orient themselves in a particular direction in the space provided to the definite geometry of the coordination compound.

(c) Secondary valencies cannot be ionized, while primary valencies can usually be ionized.

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:

FeSO₄ solution when mixed with (NH₄)₂SO₄in 1: 1 molar ratio produces a double salt FeSO₄(NH₄)₂SO₄-6H₂O. This salt is responsible for giving the Fe²⁺.

CuSO₄ mixed with aqueous ammonia in the ratio of  1:4 gives a complex salt.  The complex salt does not ionize to give Cu²⁺, hence failing the test.

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

Answer:

(a) Ligands – they are neutral molecules or negative ions bound to a metal atom in the coordination entity. Example- Cl^–,  ^–OH

(b) Coordination entity –  they are electrically charged radicals or species. They constitute a central ion or atom surrounded by neutral molecules or ions. Example –  [ Ni(CO)₄] , [COCL₃(NH₃)₃]

(c) Coordination number– it is the number of bonds formed between ligands and central atom/ion.

Example:

(i) In K₂[PtCl₆], 6 chloride ions are attached to Pt in the coordinate sphere. Thus, 6 is the coordination number of Pt.

(ii) In [Ni(NH₃)₄]Cl₂, the coordination number of the central metal ion (Ni) is 4.

(d) Coordination polyhedron –  it is the spatial positioning of ligands that are directly connected to the central atom in the coordination sphere. 

Example –

(v) Heteroleptic: they are complexes with their metal ion being bounded to more than one kind of donor group. 

Example – [Co(NH₃)₄Cl₂]⁺ , [ Ni(CO)₄]

(vi) Homoleptic: they are complexes with their metal ion being bounded to only one type of donor. 

Example – [PtCl₄]^2-, [Co(NH₃)₆]³⁺

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

Answer:

(i) Unidentate ligands: these are ligands with one donor site. 

Example Cl^–, NH₃

(ii)  Ambidentate ligands: these are ligands that fasten themselves to the central metal ion/ atom via two different atoms.

Example NO^–₂or ONO^–, CN^– or NC^–

(iii) Bidentate: these are ligands with two donor sites.

Example – Ethane-1,2-diamine , Oxalate ion ( C₂O₄^2- )

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) [Co (H₂O) (CN) (en)₂]²⁺

⇒  x + 0  + ( -1)  + 2 (0) = +2

x -1 = +2

x = +3

(ii)  [Co Br₂ (en)₂]¹⁺

⇒ x + 2 ( -1 ) + 2 ( 0 ) = +1

x  – 2 = +1

x = +3

(iii) [PtCl₄] ^2-

⇒ x + 4 ( -1 ) = -2

x = +2

(iv) K₃ [Fe (CN)₆]

⇒  [Fe ( CN )₆]^3-

⇒ x + 6 ( -1 ) = -3

x = +3

(v) [Cr(NH₃)₃Cl₃]

⇒ x + 3( 0 ) + 3 ( -1 ) = 0

x – 3 = 0

x = 3

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) [Zn(OH)₄]^{2 –}

(ii) K₂[ Pd Cl₄]

(iii) [Pt (NH₃)₂Cl₂]

(iv) K₂[ Ni(CN )₄]

(v) [Co(NO₂) ( 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(ONO)(NH₃)₅] ²⁺

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) 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)

Answer 2

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

Answer:

The various types of isomerism present in coordination compounds are :

(i) Geometrical isomerism:

(ii) Optical isomerism:

(iii) Linkage isomerism: This is found in complexes that have ambidentate ligands.

For e.g. :[Co( NH₃ )₅( NO₂)]Cl₂ and [Co( NH₃)₅( ONO)]Cl₂

(iv) Coordination isomerism:

This kind of isomerism comes up when ligands are interchanged between anionic and cationic entities of different metal ions present in the complex.

For e.g. : [Cr( NH₃)₆] [Co( CN)₆]

(v) Ionisation isomerism:

This is the kind of isomerism where a counter ion takes the place of a ligand inside the coordination sphere. 

For e.g.: [Co( NH₃)₅Br]SO₄and [Co( NH₃)₅SO₄]Br

(vi) Solvate isomerism:

[Cr( H₂O)₅ Cl]Cl.H₂O

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

(i) [Cr(C₂O₄)₃]^3–

(ii) [Co(NH₃)₃Cl₃]

Answer:

(i) In  [ Cr(C₂O₄)₃] ³⁻ no geometric isomers are present because it is a bidentate ligand.

(ii)  In [Co( NH₃)₃ Cl₃]two isomers are possible.

 

10. Draw the structures of optical isomers of:

(i) [Cr(C₂O₄)₃]^3–

(ii) [PtCl₂(en)₂]²⁺

(iii) [Cr(NH₃)₂Cl₂(en)]⁺

Answer:

(i) [Cr( C₂O₄ )₃] ³⁻

(ii) [PtCl₂(en)₂]²⁺

(iii) [Cr(NH₃)₂Cl₂(en)]⁺

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

(i) [CoCl₂(en)₂]⁺

(ii) [Co(NH₃)Cl(en)₂]²⁺

(iii) [Co(NH₃)₂Cl₂(en)]⁺

Answer:

(i) [CoCl₂(en)₂]⁺

(ii) [Co(NH₃)Cl(en)₂]²⁺

(iii) [Co(NH₃)₂Cl₂(en)]⁺

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

Answer:

[Pt(NH₃)( Br)(Cl)(py)]

None of the above isomers will exhibit optical isomerism.

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

Answer:

The blue colour of aqueous CuSO₄ solution is because of the presence of [Cu( H₂O)₄]²⁺ ions.

(i) So when KF is added, H₂O ligands are replaced by F^– ligands which yield green coloured [CuF₄]²⁺ ions.

\(\left[Cu(H_2O)_4\right]^{2+} + 4F^- \longrightarrow \left[CuF_4\right]^{2-} + 4H_2O\)

(ii) So when KCL is added, H₂O ligands are replaced by Cl^– ligands which yield bright green coloured [CuCl₄]²⁺ ions.

\([Cu(H_2O)_4]^{2+} + 4Cl^- \longrightarrow [CuCl_4]^{2-} + 4H_2 O\)

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 H₂S(g) is passed through this solution?

Answer:

\(2\left[CuSO_4\right](aq) + 10KCN(aq) \longrightarrow 2K_2[Cu(CN)_4](aq) + 2K_2 SO_4 (aq)+ (CN)_2\)

Therefore, the coordination entity obtained in the above process is  K₂[ Cu(CN)₄ ] .

As the above coordination entity is highly stable it does not ionize to yield Cu²⁺ ions. Thus, no precipitate is obtained when

hydrogen Sulphide gas is bubbled through it.

Answer 3

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

(i) [Fe(CN)₆]^4–

(ii) [FeF₆]^3–

(iii) [Co(C₂O₄)₃]^3–

(iv) [CoF₆]^3-

Answer:

(i) Fe(CN)₆]⁴⁻

In this coordination complex, the oxidation state of Fe is +3.

Fe ²⁺ : Electronic configuration is 3d⁶

Orbitals of Fe²⁺ ion :

Since CN⁻ is a strong field ligand, it causes the unpaired 3d electrons to pair up:

As there are six ligands around the central metal ion, the most practical hybridization is d²sp³, d²sp³ hybridized orbitals of Fe²⁺ are:

6 electron pairs from CN ⁻ ions take the place of the six hybrid d²sp³ orbitals.

Then,

Thus, the geometry of the complex is octahedral and it is a diamagnetic complex (since all the electrons are paired).

(ii) [FeF₆] ³⁻

In this coordinate entity, the oxidation state of iron is  +3.

Orbitals of Fe ⁺³ ion:

There are 6 F⁻ ions. Hence, it will go through d²sp³ or sp³d² hybridization.

Since F⁻ is a weak field ligand, it does not cause the pairing of the electrons in the 3d orbital. Thus, the most practical hybridization is sp³d². sp³d ² hybridized orbitals of Fe are :

Thus, the geometry of this coordinate entity is octahedral.

(iii) [Co( C₂O₄)₃]³⁻

In this complex, the oxidation state of cobalt is +3.

Orbitals of Co ³⁺ ion :

Oxalate is a weak field ligand. Thus, it will not cause the pairing of the 3d orbital electrons.

As there are 6 ligands, hybridization has to be either sp³d ² or d ²sp³ hybridization.

sp³d² hybridization of Co ³⁺ :

The 6 electron pairs from the 3 oxalate ions (oxalate anion is a bidentate ligand) occupy these sp³d ² orbitals :

Thus, the complex shows octahedral geometry.

(iv) [CoF₆]³⁻

In this complex, Cobalt has an oxidation state of +3.

Orbitals of Co³⁺ ion:

As fluoride ion is a weak field ligand it will not cause the 3d electrons to pair. Hence, the Co³⁺  ion will go through sp³d² hybridization.

sp³d² hybridized orbitals of Co³⁺ ion are :

Thus, the complex has a geometric configuration of an octahedral and it is paramagnetic.

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

Answer:

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

Answer:

A series of common ligands in ascending order of their crystal-field splitting energy (CFSE) is termed as the Spectrochemical series.

Strong field ligands have larger values of CFSE. Whereas, weak field ligands have smaller values of CFSE.

18. What is crystal field splitting energy? How does the magnitude of ∆o decide the actual configuration of d orbitals in a coordination entity?

Answer:

Crystal-field splitting energy is the difference in the energy between the two levels ( i.e., t_2g and e_g ) that have split from a degenerated d orbital because of the presence of a ligand. It is symbolized as  ∆o.

Once the orbitals split up, electrons start filling the vacant spaces. An electron each goes into the three t_2g orbitals, the fourth electron, however, can enter either of the two orbitals:

(1) It can go to the e_g orbital ( producing t_2g³e_g ¹ like electronic configuration), or

(2) it can go to the t_2g orbitals ( producing t_2g⁴e_g ⁰ like electronic configuration).

This filling of the fourth electron is based on the energy level of ∆o. If a ligand has an ∆o value smaller than the pairing energy, then the fourth electron enters the e_g orbital. However, if the value of ∆o is greater than the value of pairing energy, the electron enters t_2gorbital.

19. [Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]^2– is diamagnetic. Explain why.

Answer:

In [ Ni ( CN)₄ ] ²⁻, Ni has an oxidation state of +2. Thus, it has d ⁸configuration.

Ni ²⁺ :

CN ⁻ being a strong field ligand causes the electrons in 3d orbitals to pair. This causes, Ni ²⁺ to undergo dsp² hybridization.

Since all the electrons are paired, it is diamagnetic in nature.

Cr has an oxidation state of +3. Thus, it has a  d ³ configuration. As NH₃ is not a strong field ligand it does not cause the electrons in the 3d orbital to pair.

Cr³⁺ :

It undergoes d²sp³ hybridization and the 3d orbital electrons remain unpaired. Thus, [Ni(CN)₄]²⁻ is paramagnetic in nature.

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

Answer:

[ Ni ( H₂O)₆ ] ²⁺ consists of Ni⁺² ion with 3d⁸ electronic configuration. In this configuration, there are two unpaired electrons which cannot pair up because H₂O is a weak ligand. Thus, the d – d transition absorbs the incoming light and it emits a green light. Thereby, giving a  green colour to the solution.

[Ni ( CN)₄]^2- consists of Ni⁺² ion with 3d⁸ electronic configuration. But, CN ^– is present here, which is a strong ligand and in its presence, the unpaired electrons pair up. Thus, there is no d –d  transition so no colour.

Answer 4

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

Answer:

[Fe ( H₂O)₆]²⁺ and [Fe( CN)₆]⁴⁻ have two different ligands H₂O and CN^{­­ –}. CN^{­ –} being a strong field ligand has a higher value of CFSE (crystal field splitting energy ) than water. As a result, the d-d transitions absorb and give back different wavelengths of light. Thus, they have different colours in a solution.

22. Discuss the nature of bonding in metal carbonyls.

Answer:

In metal carbonyls, the metal-carbon bond contains both the σ and π bond characters. A σ bond forms when a lone pair of electrons are donated to the empty orbital of the metal by the carbonyl carbon. A π bond forms when a pair of electrons are donated to the empty antibonding π* orbital by the filled d orbital of the metal. This in entirety stabilizes and strengthens the metal-ligand bonding.

The above two types of  bonding are represented as :

23. 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-[CrCl₂(en)₂]Cl

(iii) (NH₄)₂[CoF₄]

(iv) [Mn(H₂O)₆]SO₄

Answer:

(i)K₃[Co(C₂O₄)₃]

Central metal ion: Co.

Coordination number = 6.

We know,

Oxidation state is :

x − 6 = −3

x  = + 3

The d orbital occupation  : t_2g⁶e_g ⁰.

(ii) cis – [ Cr( en)₂ Cl₂ ]Cl

Central metal ion : Cr.

Coordination number = 6.

We know,

Oxidation state is :

x +2(0) + 2 ( -1) = +1

x -2 = -1

x  = + 3

The d orbital occupation  : t_2g³.

(iii) ( NH₄)₂[ CoF₄ ]

Central metal ion : Co.

Coordination number = 4.

We know,

Oxidation state is :

x  – 4 = -2

x  = + 2

The d orbital occupation : e_g⁴t_2g³.

(iv) [ Mn (H₂O)₆ ]SO₄

Central metal ion : Mn.

Coordination number = 6.

We know, 

Oxidation state is :

x  + 0 = 2

x  = + 2

The d orbital occupation  : t_2g³ e_g².

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(H₂O)₂(C₂O₄)₂].3H₂O

(ii) [Co(NH₃)₅Cl]Cl₂

(iii) [CrCl₃(py)₃]

(iv) Cs[FeCl₄]

(v) K₄[Mn(CN)₆]

Answer:

(i) IUPAC name = Potassium diaquadioxalatochromate (III) trihydrate.

Coordination number = 6

Oxidation state of chromium :

x + 0 +2( -2) = -1

x = 3

Electronic configuration: 3d ³ =  t_2g³

Shape : Octahedral

Stereochemistry :

Magnetic moment, \(\mu = \sqrt{n(n+ 2)}\)        [ n = unpaired electrons ]

\( = \sqrt{3(3+ 2)}\)

\(= \sqrt{15}\) ≈ 4BM

(ii) IUPAC name = Pentaamminechloridocobalt(III) chloride

Coordination number = 6

Oxidation state of Co :

x + 0 – 1 = + 2

x = 3

Electronic configuration: 3d ⁶ = t_2g⁶

Shape : Octahedral

Stereochemistry :

n = 0.

Thus,  Magnetic moment  = 0

(iii) IUPAC name = Trichloridotripyridinechromium (III)

Coordination number = 6

Oxidation state of Cr :

x – 3 + 0 = 0

x = 3

Electronic configuration: 3d³ = t_2g³

Shape : Octahedral

Stereochemistry :

n = 3 

Magnetic moment, \(\mu = \sqrt{n(n+ 2)}\)

\( = \sqrt{3(3+ 2)}\)

 \(= \sqrt{15}\) ≈ 4BM

(iv) IUPAC name = Caesiumtetrachloroferrate (III)

Coordination number = 4

Oxidation state of Fe :

x – 4 = -1

x = 3

Electronic configuration: d⁶ = e_g² t_2g³

Shape : Tetrahedral

Stereochemistry :-  optically inactive

n = 5

Magnetic moment, \(\mu = \sqrt{n(n+ 2)}\)

\( = \sqrt{5(5+ 2)}\)

 \(= \sqrt{35}\) ≈ 6BM

(v) IUPAC name = Potassium hexacyanomanganate(II)

Coordination number = 6

Oxidation state of Mn :

x – 6 = -4

x = + 2

Electronic configuration: 3d⁵ =  t_2g⁵

Shape : Octahedral

Stereochemistry :-  optically inactive

n = 1

Magnetic moment, \(\mu = \sqrt{n(n+ 2)}\)

\( = \sqrt{1(1+ 2)}\)

 \(= \sqrt{3}\) ≈ 1.732 BM

Answer 5

25. What is meant by stability of a coordination compound in solution? State the factors which govern stability of complexes.

Answer:

Stability of a coordination compound in a solution is the degree/level of association among the species involved in a state of equilibrium.

Stability can also be written quantitatively in terms of formation constant or stability constant.

M + 3L   ⇔        ML₃

Stability constant , \(\beta = \frac{[ML_3]}{[M][L]^3}\)

Greater the value of β , stronger is the metal –ligand bond.

Factors responsible for the stability of a complex:

(1) Charge on the central metal ion – bigger the charge, more stable is the complex.

(2) Nature of ligand – chelating ligand produces a more stable complex.

(3) The basic strength of ligand-  more basic a ligand, more stable its complex.

26. What is meant by the chelate effect? Give an example.

Answer:

When a polydentate or a bidentate ligand fastens itself to a metal ion in such a way that it assumes the shape of a ring, the metal-ligand bond becomes more stable. These rings are called chelate rings.

From here we can infer that complexes with chelate rings are more stable than complexes without the rings. This phenomenon is termed the chelate effect.

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

(i) biological systems

(ii) medicinal chemistry 

(iii) analytical chemistry

(iv) extraction/metallurgy of metals.

Answer:

(i) Role in biological systems:

In the body of animals, there are several very important coordination compounds e.g.haemoglobin is a coordination compound of iron.
In plants, chlorophyll pigment is a coordination compound of magnesium.

(ii) Role in medicinal chemistry:

So many coordinate compounds are used for curing purposes. For e.g., a coordination compound of platinum, cis-platin is used for checking the growth ofumours.

(iii) Role in analytical chemistry:

Determination of hardness of the water.

(iv) Role in metallurgy or extraction:

During metal extraction from ores, complexes are formed. For e.g. gold combines with cyanide ions in an aqueous solution. Gold is then extracted from this complex using zinc.

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

(i) 6

(ii) 4

(iii) 3

(iv) 2

Answer:

(iii) 3

The given complex  [Co( NH₃)₆]Cl₂ ionizes to give three ions, viz one [ Co( NH₃)₆]⁺ and two Cl^– ions.

29. 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⁺

Answer:

(i) [Cr(H₂O)₆]³⁺

number of unpaired electrons, n  = 3

Magnetic moment, \(\mu = \sqrt{n(n +2)}\)

\(= \sqrt{3(3+2)}\)

\(= \sqrt{15}\) ≈ 4BM

(ii) [Fe(H₂O)₆]²⁺

number of unpaired electrons, n  = 4

Magnetic moment, \(\mu = \sqrt{n(n +2)}\)

\(= \sqrt{4(4+2)}\)

\(= \sqrt{24}\) ≈ 4BM

(iii) [Zn(H₂O)₆]²⁺

n = 0

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

30. What is the oxidation number of cobalt in K[Co(CO)₄]?

Answer:

K[Co( CO )₄] = K⁺[Co( CO )₄]^–

We know,

⇒ x + 0 =  – 1     [Where x is the oxidation number]

x = -1

31. Amongst the following, the most stable complex is

(i) [Fe(H₂O)₆]³⁺

(ii) [Fe(NH₃)₆]³⁺

(iii) [Fe(C₂O₄)₃]^3–

(iv) [FeCl₆]^3–

Answer:

In all the cases Fe has an oxidation state of +3. (C₂O₄)₃ is a bidentate chelating ligand and it forms chelating rings. Thus, (iii) is the most stable complex.

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(H₂O)₆]²⁺?

Answer:

All of the complexes have the same metal ion, so the energy absorption depends upon the CFSE values of the ligands. According to the spectro-chemical series,  the CFSE values of the ligands are in the order of H₂O <  NH₃< NO₂⁻

As

E =hc / λ

⇒ E ∝ 1/ λ

Therefore, the values of the absorbed wavelength in ascending order would be :

[Ni(H₂O)₆]²⁺ < [ Ni(NH₃)₆] ²⁺  < [ Ni(NO₂)₆] ⁴⁻