Page No 259: - Chapter 9 Coordination Compunds Exercise Solutions class 12 ncert solutions Chemistry - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 9.11:

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]⁺ Geometrical isomerism

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

In total, three isomers are possible.

(ii) [Co(NH₃)Cl(en)₂]²⁺

Geometrical isomerism

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

(iii) [Co(NH₃)₂Cl₂(en)]⁺

Question 9.12:

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)

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

Question 9.13:

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:

Aqueous CuSO₄ exists as [Cu(H₂O)₄]SO₄. It is blue in colour due to the presence of

[Cu[H₂O)₄]²⁺ ions.

(i) When KF is added:

(ii) When KCl is added:

In both these cases, the weak field ligand water is replaced by the F⁻ and Cl⁻ ions.

Question 9.14:

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

Answer:

i.e., 

Thus, the coordination entity formed in the process is K₂[Cu(CN)₄].   is a very stable complex, which does not ionize to give Cu²⁺ ions when added to water. Hence, Cu²⁺ ions are not precipitated when H₂S₍g₎ is passed through the solution.

Question 9.15:

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 the above coordination complex, iron exists in the +II oxidation state.

Fe²⁺ : Electronic configuration is 3d⁶

Orbitals of Fe²⁺ ion:

As CN⁻ is a strong field ligand, it causes the pairing of the unpaired 3d electrons.

Since there are six ligands around the central metal ion, the most feasible hybridization is d²sp³.

d²sp³ hybridized orbitals of Fe²⁺ are:

6 electron pairs from CN⁻ ions occupy the six hybrid d²sp³orbitals.

Then,

Hence, the geometry of the complex is octahedral and the complex is diamagnetic (as there are no unpaired electrons).

(ii) [FeF₆]³⁻

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

Orbitals of Fe⁺³ ion:

There are 6 F⁻ ions. Thus, it will undergo d²sp³ or sp³d² hybridization. As F⁻ is a weak field ligand, it does not cause the pairing of the electrons in the 3d orbital. Hence, the most feasible hybridization is sp³d².

sp³d² hybridized orbitals of Fe are:

Hence, the geometry of the complex is found to be octahedral.

(iii) [Co(C₂O₄)₃]³⁻

Cobalt exists in the +3 oxidation state in the given complex.

Orbitals of Co³⁺ ion:

Oxalate is a weak field ligand. Therefore, it cannot cause the pairing of the 3d orbital electrons. As there are 6 ligands, hybridization has to be either sp³d² or d²sp³ hybridization.

sp³d² hybridization of Co^3+:

The 6 electron pairs from the 3 oxalate ions (oxalate anion is a bidentate ligand) occupy these sp³d² orbitals.

Hence, the geometry of the complex is found to be octahedral.

(iv) [CoF₆]³⁻

Cobalt exists in the +3 oxidation state.

Orbitals of Co³⁺ ion:

Again, fluoride ion is a weak field ligand. It cannot cause the pairing of the 3d electrons. As a result, the Co³⁺ ion will undergo sp³d² hybridization.

sp³d² hybridized orbitals of Co³⁺ ion are:

Hence, the geometry of the complex is octahedral and paramagnetic.

Question 9.16:

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

Answer:

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

Question 9.17:

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

Answer:

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

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

Question 9.18:

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

Answer:

The degenerate d-orbitals (in a spherical field environment) split into two 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 (e_g and t_2g) is called the crystal-field splitting energy. It is denoted by Δ_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 e_g orbital (giving rise to t_2g³ e_g¹ like electronic configuration) or the pairing of the electrons can take place in the t_2g orbitals (giving rise to t_2g⁴ e_g⁰ like electronic configuration). If the Δ_o value of a ligand is less than the pairing energy (P), then the electrons enter the e_g orbital. On the other hand, if the Δ_o value of a ligand is more than the pairing energy (P), then the electrons enter the t_2g orbital.

Question 9.19:

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

Answer:

Cr is in the +3 oxidation state i.e., d³ configuration. Also, NH₃ is a weak field ligand that does not cause the pairing of the electrons in the 3d orbital.

Cr³⁺

Therefore, it undergoes d²sp³ hybridization and the electrons in the 3d orbitals remain unpaired. Hence, it is paramagnetic in nature.

In [Ni(CN)₄]²⁻, Ni exists in the +2 oxidation state i.e., d⁸ configuration.

Ni^2+:

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.

Question 9.20:

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

Answer:

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

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

Question 9.21:

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

Answer:

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

Question 9.22:

Discuss the nature of bonding in metal carbonyls.

Answer:

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

Question 9.23:

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-[Cr(en)₂Cl₂]Cl

(iii) (NH₄)₂[CoF₄]

(iv) [Mn(H₂O)₆]SO₄

Answer:

(i) K₃[Co(C₂O₄)₃]

The central metal ion is Co.

Its coordination number is 6.

The oxidation state can be given as:

x − 6 = −3

x = + 3

The d orbital occupation for Co³⁺ is t_2g⁶eg^0.

(ii) cis-[Cr(en)₂Cl₂]Cl

The central metal ion is Cr.

The coordination number is 6.

The oxidation state can be given as:

x + 2(0) + 2(−1) = +1

x − 2 = +1

x = +3

The d orbital occupation for Cr³⁺ is t₂g³.

(iii) (NH₄)₂[CoF₄]

The central metal ion is Co.

The coordination number is 4.

The oxidation state can be given as:

x − 4 = −2

x = + 2

The d orbital occupation for Co²⁺ is eg⁴ t_2g^3.

(iv) [Mn(H₂O)₆]SO₄

The central metal ion is Mn.

The coordination number is 6.

The oxidation state can be given as:

x + 0 = +2

x = +2

The d orbital occupation for Mn is t_2g³ eg².

Question 9.24:

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) Potassium diaquadioxalatochromate (III) trihydrate.

Oxidation state of chromium = 3

Electronic configuration: 3d³: t_2g³

Coordination number = 6

Shape: octahedral

Stereochemistry:

Magnetic moment, μ 

∼ 4BM

(ii) [Co(NH₃)₅Cl]Cl₂

IUPAC name: Pentaamminechloridocobalt(III) chloride

Oxidation state of Co = +3

Coordination number = 6

Shape: octahedral.

Electronic configuration: d⁶: t_2g⁶.

Stereochemistry:

Magnetic Moment = 0

(iii) CrCl₃(py)₃

IUPAC name: Trichloridotripyridinechromium (III)

Oxidation state of chromium = +3

Electronic configuration for d³ = t_2g³

Coordination number = 6

Shape: octahedral.

Stereochemistry:

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

Magnetic moment, μ 

∼ 4BM

(iv) Cs[FeCl₄]

IUPAC name: Caesium tetrachloroferrate (III)

Oxidation state of Fe = +3

Electronic configuration of d⁶ = eg²t_2g³

Coordination number = 4

Shape: tetrahedral

Stereochemistry: optically inactive

Magnetic moment:

μ 

(v) K₄[Mn(CN)₆]

Potassium hexacyanomanganate(II)

Oxidation state of manganese = +2

Electronic configuration: d⁵⁺: t₂g⁵

Coordination number = 6

Shape: octahedral.

Streochemistry: optically inactive

Magnetic moment, μ 

Question 9.25:

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

Answer:

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.

For this reaction, the greater 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:

(a) Charge on the central metal ion: Thegreater the charge on the central metal ion, the greater is the stability of the complex.

2. Basic nature of the ligand: A more basic ligand will form a more stable complex.

2. Presence of chelate rings: Chelation increases the stability of complexes.

Question 9.26:

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

Answer:

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:

Question 9.27:

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

(i) biological system

(ii) medicinal chemistry

(iii) analytical chemistry

(iv) extraction/metallurgy of metals

Answer:

(i) Role of coordination compounds in biological systems:

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

(ii) Role of coordination compounds in medicinal chemistry:

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

(iii) Role of coordination compounds in analytical chemistry:

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

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

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

Question 9.28:

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

(i) 6

(ii) 4

(iii) 3

(iv) 2

Answer:

(iii) The given complex can be written as [Co(NH₃)₆]Cl_2.

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

Question 9.29:

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

Then, μ 

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

Then, μ 

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

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

Question 9.30:

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

(i) +1

(ii) +3

(iii) −1

(iv) −3

Answer:

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

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