Coordination Compounds Class 12 Chemistry NCERT Solutions

Q1Exercises

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

Solution

Alfred Werner's theory explains the bonding in coordination compounds through the following main postulates:

Two Types of Valences: In coordination compounds, metals exhibit two types of valences: a primary valence and a secondary valence.
Primary Valence:
- This corresponds to the oxidation state of the central metal ion.
- It is ionisable.
- It is satisfied by negative ions.
- In modern terms, this represents the ionic bond between the central metal ion and the counter ions outside the coordination sphere.
- For example, in $[Co(NH_3)_6]Cl_3$ , the primary valence of Cobalt is 3, satisfied by the three chloride ions.
Secondary Valence:
- This corresponds to the coordination number of the central metal ion.
- It is non-ionisable.
- It is satisfied by neutral molecules or negative ions (ligands).
- The secondary valence is fixed for a particular metal ion.
- In modern terms, this represents the coordinate covalent bonds between the central metal ion and the ligands within the coordination sphere.
- For example, in $[Co(NH_3)_6]Cl_3$ , the secondary valence of Cobalt is 6, satisfied by the six ammonia molecules.
Spatial Arrangement:
- The groups bound by the secondary valences (ligands) are directed towards fixed positions in space around the central metal ion.
- This gives the complex a definite geometry or coordination polyhedron. For example, a secondary valence of 6 corresponds to an octahedral geometry, while a secondary valence of 4 can correspond to either a tetrahedral or a square planar geometry.

In summary, Werner proposed that a central metal ion is surrounded by ligands held by secondary valences, forming a coordination sphere. This sphere as a whole can be an ion, which is then balanced by counter ions held by primary valences.

Q2Exercises

$FeSO_4$ solution mixed with $(NH_4)_2SO_4$ solution in 1:1 molar ratio gives the test of $Fe^{2+}$ ion but $CuSO_4$ solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of $Cu^{2+}$ ion. Explain why?

Solution

The difference in behavior is due to the formation of a double salt in the first case and a coordination compound (complex) in the second case.

$FeSO_4$ and $(NH_4)_2SO_4$ mixture: When ferrous sulphate ( $FeSO_4$ ) solution is mixed with ammonium sulphate ( $(NH_4)_2SO_4$ ) solution in a 1:1 molar ratio, a double salt called Mohr's salt, $FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$ , is formed. Double salts are stable only in the solid state. When dissolved in water, they dissociate completely into their constituent simple ions. $FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O (s) \xrightarrow{Water} Fe^{2+} (aq) + 2NH_4^+ (aq) + 2SO_4^{2−} (aq) + 6H_2O (l)$ Since free $Fe^{2+}$ ions are present in the solution, it gives a positive test for the $Fe^{2+}$ ion (e.g., forming a Turnbull's blue precipitate with potassium ferricyanide solution).
$CuSO_4$ and aqueous ammonia mixture: When copper sulphate ( $CuSO_4$ ) solution is mixed with aqueous ammonia ( $NH_3$ ) in a 1:4 molar ratio, a coordination compound, tetraamminecopper(II) sulphate, $[Cu(NH_3)_4]SO_4$ , is formed. In this compound, the copper ion and ammonia molecules form a stable complex ion, $[Cu(NH_3)_4]^{2+}$ , called the coordination entity. When dissolved in water, this compound dissociates, but the complex ion remains intact. $[Cu(NH_3)_4]SO_4 (s) \xrightarrow{Water} [Cu(NH_3)_4]^{2+} (aq) + SO_4^{2−} (aq)$ The $Cu^{2+}$ ion is strongly bonded to the four ammonia ligands within the coordination sphere and does not dissociate into free $Cu^{2+}$ ions. Because the concentration of free $Cu^{2+}$ ions is extremely low, the solution does not give the characteristic tests for the $Cu^{2+}$ ion (e.g., it will not form a precipitate of CuS with $H_2S$ gas).

Q3Exercises

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

Solution

Coordination Entity: A coordination entity constitutes a central metal atom or ion bonded to a fixed number of ions or molecules (ligands).
- Examples: $[CoCl_3(NH_3)_3]$ , $[Ni(CO)_4]$ .
Ligand: The ions or molecules bound to the central atom/ion in the coordination entity are called ligands. They are electron-pair donors (Lewis bases).
- Examples: $H_2O$ (aqua), $CN^−$ (cyanido).
Coordination Number: The coordination number of a central metal ion in a complex is the number of ligand donor atoms to which the metal is directly bonded. It is the number of sigma bonds between the ligands and the central atom.
- Examples: In $[Co(NH_3)_6]^{3+}$ , the coordination number of Co is 6. In $[Ni(CN)_4]^{2−}$ , the coordination number of Ni is 4.
Coordination Polyhedron: The spatial arrangement of the ligand atoms which are directly attached to the central atom/ion defines a coordination polyhedron.
- Examples: $[Co(NH_3)_6]^{3+}$ has an octahedral geometry. $[Ni(CO)_4]$ has a tetrahedral geometry.
Homoleptic Complex: A complex in which the central metal atom/ion is bonded to only one kind of donor group or ligand.
- Examples: $[Co(NH_3)_6]^{3+}$ (only $NH_3$ ligands), $[Fe(CN)_6]^{4−}$ (only $CN^−$ ligands).
Heteroleptic Complex: A complex in which the central metal atom/ion is bonded to more than one kind of donor group or ligand.
- Examples: $[Co(NH_3)_4Cl_2]^+$ (both $NH_3$ and $Cl^−$ ligands), $[Pt(NH_3)_2Cl_2]$ (both $NH_3$ and $Cl^−$ ligands).

Q4Exercises

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

Solution

Unidentate Ligand: A ligand that can bind to a central metal atom/ion through a single donor atom. It forms one coordinate bond.
- Examples: $Cl^−$ (chloride), $H_2O$ (aqua), $NH_3$ (ammine).
Didentate Ligand: A ligand that can bind to a central metal atom/ion through two donor atoms. It forms two coordinate bonds, typically forming a chelate ring.
- Examples: Ethane-1,2-diamine, $H_2N-CH_2-CH_2-NH_2$ (abbreviated as 'en'), which donates through its two nitrogen atoms. Oxalate ion, $(C_2O_4)^{2−}$ , which donates through two oxygen atoms.
Ambidentate Ligand: A unidentate ligand that can coordinate to the central metal atom/ion through two different donor atoms. However, at any given time, it uses only one of the donor atoms to form a coordinate bond.
- Examples: Nitrite ion ( $NO_2^−$ ), which can coordinate through the Nitrogen atom (nitro, $-NO_2$ ) or through an Oxygen atom (nitrito, $-ONO$ ). Thiocyanate ion ( $SCN^−$ ), which can coordinate through the Sulphur atom (thiocyanato, $-SCN$ ) or through the Nitrogen atom (isothiocyanato, $-NCS$ ).

Q5Exercises

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

(i)

$[Co(H_2O)(CN)(en)_2]^{2+}$

(ii)

$[CoBr_2(en)_2]^−$

(iii)

$[PtCl_4]^{2−}$

(iv)

$K_3[Fe(CN)_6]$

(v)

$[Cr(NH_3)_3Cl_3]$

Solution

The oxidation number is calculated by setting the sum of the charges of the metal ion and the ligands equal to the overall charge of the coordination entity.

(i)

$[Co(H_2O)(CN)(en)_2]^{2+}$ Let the oxidation number of Co be x. Charge of

H_2O

= 0 Charge of

CN^−

= -1 Charge of en (ethane-1,2-diamine) = 0

x + (0) + (−1) + 2(0) = +2

x − 1 = +2

x = +3

Oxidation number of Co is +3.

(ii)

$[CoBr_2(en)_2]^+$ (Note: The charge in the question is likely a typo, it should be + for Co(III). Assuming +1 charge) Let the oxidation number of Co be x. Charge of

Br^−

= -1 Charge of en = 0

x + 2(−1) + 2(0) = +1

x − 2 = +1

x = +3

Oxidation number of Co is +3.

(iii)

$[PtCl_4]^{2−}$ Let the oxidation number of Pt be x. Charge of

Cl^−

= -1

x + 4(−1) = −2

x − 4 = −2

x = +2

Oxidation number of Pt is +2.

(iv)

$K_3[Fe(CN)_6]$ Let the oxidation number of Fe be x. Charge of

K^+

= +1 Charge of

CN^−

= -1

3(+1) + x + 6(−1) = 0

3 + x − 6 = 0

x = +3

Oxidation number of Fe is +3.

(v)

$[Cr(NH_3)_3Cl_3]$ Let the oxidation number of Cr be x. Charge of

NH_3

= 0 Charge of

Cl^−

= -1

x + 3(0) + 3(−1) = 0

x − 3 = 0

x = +3

Oxidation number of Cr is +3.

Q6Exercises

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)

Solution

(i)

Tetrahydroxidozincate(II):

K_2[Zn(OH)_4]

or

[Zn(OH)_4]^{2−}

(if counter-ion is not specified, showing the complex ion is sufficient. Assuming Potassium as counter-ion for a neutral compound:

K_2[Zn(OH)_4]

).

(ii)

Potassium tetrachloridopalladate(II):

K_2[PdCl_4]

(iii)

Diamminedichloridoplatinum(II):

[Pt(NH_3)_2Cl_2]

(iv)

Potassium tetracyanidonickelate(II):

K_2[Ni(CN)_4]

(v)

Pentaamminenitrito-O-cobalt(III):

[Co(NH_3)_5(ONO)]^{2+}

(The full compound name would likely specify a counter-ion, e.g., chloride, making it

[Co(NH_3)_5(ONO)]Cl_2

).

(vi)

Hexaamminecobalt(III) sulphate:

[Co(NH_3)_6]_2(SO_4)_3

(vii)

Potassium tri(oxalato)chromate(III):

K_3[Cr(C_2O_4)_3]

(viii)

Hexaammineplatinum(IV):

[Pt(NH_3)_6]^{4+}

(The full compound name would specify a counter-ion, e.g., chloride, making it

[Pt(NH_3)_6]Cl_4

).

(ix)

Tetrabromidocuprate(II):

[CuBr_4]^{2−}

(x)

Pentaamminenitrito-N-cobalt(III):

[Co(NH_3)_5(NO_2)]^{2+}

(The full compound name would specify a counter-ion, e.g., chloride, making it

[Co(NH_3)_5(NO_2)]Cl_2

).

Q7Exercises

Using IUPAC norms write the systematic names of the following:

(i)

$[Co(NH_3)_6]Cl_3$

(ii)

$[Pt(NH_3)_2Cl(NH_2CH_3)]Cl$

(iii)

$[Ti(H_2O)_6]^{3+}$

(iv)

$[Co(NH_3)_4Cl(NO_2)]Cl$

(v)

$[Mn(H_2O)_6]^{2+}$

(vi)

$[NiCl_4]^{2−}$

(vii)

$[Ni(NH_3)_6]Cl_2$

(viii)

$[Co(en)_3]^{3+}$

(ix)

$[Ni(CO)_4]$

Solution

(i)

$[Co(NH_3)_6]Cl_3$ : Hexaamminecobalt(III) chloride

(ii)

$[Pt(NH_3)_2Cl(NH_2CH_3)]Cl$ : Diamminechlorido(methanamine)platinum(II) chloride

(iii)

$[Ti(H_2O)_6]^{3+}$ : Hexaaquatitanium(III) ion

(iv)

$[Co(NH_3)_4Cl(NO_2)]Cl$ : Tetraamminechloridonitrito-N-cobalt(III) chloride (Assuming standard nitro linkage. If it were nitrito, it would be specified as nitrito-O).

(v)

$[Mn(H_2O)_6]^{2+}$ : Hexaaquamanganese(II) ion

(vi)

$[NiCl_4]^{2−}$ : Tetrachloridonickelate(II) ion

(vii)

$[Ni(NH_3)_6]Cl_2$ : Hexaamminenickel(II) chloride

(viii)

$[Co(en)_3]^{3+}$ : Tris(ethane-1,2-diamine)cobalt(III) ion

(ix)

$[Ni(CO)_4]$ : Tetracarbonylnickel(0)

Q8Exercises

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

Solution

Isomerism in coordination compounds can be broadly classified into two main types: Structural Isomerism and Stereoisomerism.

A. Structural Isomerism (Compounds with the same molecular formula but different bonding arrangements)

Ionisation Isomerism: Arises when the counter ion in a complex salt is itself a potential ligand and can exchange places with a ligand within the coordination sphere.
- Example: $[Co(NH_3)_5Br]SO_4$ and $[Co(NH_3)_5(SO_4)]Br$
Solvate (or Hydrate) Isomerism: Arises when a solvent molecule (often water) can be present either as a ligand in the coordination sphere or as a free molecule in the crystal lattice.
- Example: $[Cr(H_2O)_6]Cl_3$ (violet) and $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$ (grey-green)
Linkage Isomerism: Arises in complexes containing an ambidentate ligand, which can coordinate to the metal through different donor atoms.
- Example: $[Co(NH_3)_5(NO_2)]Cl_2$ (nitro, yellow) and $[Co(NH_3)_5(ONO)]Cl_2$ (nitrito, red)
Coordination Isomerism: Arises from the interchange of ligands between cationic and anionic coordination entities of different metal ions in a complex salt.
- Example: $[Co(NH_3)_6][Cr(CN)_6]$ and $[Cr(NH_3)_6][Co(CN)_6]$

B. Stereoisomerism (Compounds with the same chemical formula and bonds but different spatial arrangements)

Geometrical (cis-trans) Isomerism: Arises in heteroleptic complexes due to different possible geometric arrangements of the ligands.
- Example: The square planar complex $[Pt(NH_3)_2Cl_2]$ exists as cis- and trans-isomers.
Optical Isomerism: Arises when a complex and its mirror image are non-superimposable. These isomers are called enantiomers and are optically active.
- Example: The octahedral complex $[Co(en)_3]^{3+}$ exists as a pair of enantiomers (d- and l- forms).

Q9Exercises

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

(i)

$[Cr(C_2O_4)_3]^{3−}$

(ii)

$[Co(NH_3)_3Cl_3]$

Solution

(i)

$[Cr(C_2O_4)_3]^{3−}$ This coordination entity is of the type

[M(AA)_3]

, where 'AA' represents a symmetrical bidentate ligand (oxalate,

C_2O_4^{2−}

). Complexes of this type do not exhibit geometrical isomerism. All six coordination positions are equivalent with respect to each other. Therefore, zero geometrical isomers are possible. (Note: This complex is chiral and does exhibit optical isomerism.)

(ii)

$[Co(NH_3)_3Cl_3]$ This is an octahedral coordination entity of the type

[Ma_3b_3]

. Such complexes can have two different arrangements of ligands, leading to two geometrical isomers.

Facial (fac) isomer: The three identical ligands (e.g., the three $NH_3$ groups) occupy the corners of one face of the octahedron.
Meridional (mer) isomer: The three identical ligands occupy positions around the meridian of the octahedron, with two ligands trans to each other and the third cis to both. Therefore, two geometrical isomers (fac and mer) are possible.

Q10Exercises

Draw the structures of optical isomers of:

(i)

$[Cr(C_2O_4)_3]^{3−}$

(ii)

$[PtCl_2(en)_2]^{2+}$

(iii)

$[Cr(NH_3)_2Cl_2(en)]^+$

Solution

Optical isomers are non-superimposable mirror images of each other, called enantiomers.

(i)

$[Cr(C_2O_4)_3]^{3−}$ This complex is of the type

[M(AA)_3]

and is chiral. It exists as a pair of enantiomers. (A diagram showing a central Cr atom with three bidentate oxalate ligands forming a propeller-like structure, and its non-superimposable mirror image would be drawn here.)

(ii)

$[PtCl_2(en)_2]^{2+}$ This octahedral complex has two geometrical isomers: cis and trans. Only the cis-isomer is chiral and exhibits optical isomerism. The trans-isomer has a plane of symmetry and is optically inactive. The optical isomers are the d- and l- forms of cis-

[PtCl_2(en)_2]^{2+}

. (A diagram showing the cis-isomer with two Cl atoms adjacent, and its non-superimposable mirror image would be drawn here.)

(iii)

$[Cr(NH_3)_2Cl_2(en)]^+$ This is an octahedral complex of the type

[M(AA)a_2b_2]

. Several geometrical isomers are possible. The one that is most commonly discussed for optical activity is the isomer where the two Cl atoms are cis to each other and the two

NH_3

molecules are also cis to each other. This cis-dichloro, cis-diammine isomer is chiral and exists as a pair of enantiomers. (A diagram showing this specific cis-cis isomer and its non-superimposable mirror image would be drawn here.)

Q11Exercises

Draw all the isomers (geometrical and optical) of:

(i)

$[CoCl_2(en)_2]^−$

(ii)

$[Co(NH_3)Cl(en)_2]^{2+}$

(iii)

$[Co(NH_3)_2Cl_2(en)]^+$

Solution

(i)

$[CoCl_2(en)_2]^+$ (Note: charge should be +1 for Co(III)) This complex is of the type

[M(AA)_2b_2]

.

Geometrical Isomers: It has two geometrical isomers: cis and trans.
- trans-isomer: The two chloride ligands are opposite to each other. This isomer is achiral (has a plane of symmetry) and optically inactive.
- cis-isomer: The two chloride ligands are adjacent to each other. This isomer is chiral (lacks a plane of symmetry) and optically active.
Optical Isomers: The cis-isomer exists as a pair of non-superimposable mirror images (enantiomers). In total, there are three distinct isomers: trans, d-cis, and l-cis.

(ii)

$[Co(NH_3)Cl(en)_2]^{2+}$ This complex is of the type

[M(AA)_2ab]

.

Geometrical Isomers: It has two geometrical isomers, based on the positions of the unidentate ligands $NH_3$ $N H_{3}$ and $Cl$ $Cl$ .
- cis-isomer: The $NH_3$ and $Cl$ ligands are adjacent to each other.
- trans-isomer: The $NH_3$ and $Cl$ ligands are opposite to each other.
Optical Isomers: In this case, both the cis and trans isomers are chiral and optically active. Neither has a plane of symmetry.
- The cis-isomer exists as a pair of enantiomers.
- The trans-isomer also exists as a pair of enantiomers. In total, there are four distinct isomers (two pairs of enantiomers).

(iii)

$[Co(NH_3)_2Cl_2(en)]^+$ This complex is of the type

[M(AA)a_2b_2]

.

Geometrical Isomers: There are several possibilities based on the relative positions of the two $NH_3$ $N H_{3}$ and two $Cl$ $Cl$ ligands.
- Isomer 1: $NH_3$ ligands are trans, $Cl$ ligands are trans. (Achiral)
- Isomer 2: $NH_3$ ligands are cis, $Cl$ ligands are trans. (Achiral)
- Isomer 3: $NH_3$ ligands are trans, $Cl$ ligands are cis. (Achiral)
- Isomer 4: $NH_3$ ligands are cis, $Cl$ ligands are cis. This arrangement is chiral.
Optical Isomers: Only the geometrical isomer where both pairs of unidentate ligands are cis to each other is chiral and exists as a pair of enantiomers. In total, there are five distinct isomers (three achiral geometrical isomers and one pair of enantiomers).

Q12Exercises

Write all the geometrical isomers of $[Pt(NH_3)(Br)(Cl)(py)]$ and how many of these will exhibit optical isomers?

Solution

The complex

[Pt(NH_3)(Br)(Cl)(py)]

is a square planar complex of the type

[Mabcd]

, where a, b, c, and d are four different unidentate ligands (

NH_3

,

Br^−

,

Cl^−

, and pyridine).

Geometrical Isomers: For a square planar complex of this type, there are three possible geometrical isomers. We can find them by keeping one ligand in a fixed position and arranging the other three ligands trans to it. Let's fix the position of

NH_3

. The three isomers are distinguished by which ligand is trans to

NH_3

:

Isomer 1: Bromido ( $Br^−$ ) is trans to ammine ( $NH_3$ ).
Isomer 2: Chlorido ( $Cl^−$ ) is trans to ammine ( $NH_3$ ).
Isomer 3: Pyridine (py) is trans to ammine ( $NH_3$ ).

Optical Isomers: Square planar complexes are generally considered not to exhibit optical isomerism. This is because they possess a plane of symmetry which is the plane of the molecule itself. Therefore, the molecule and its mirror image are always superimposable.

Thus, none of these geometrical isomers will exhibit optical isomerism.

Q13Exercises

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

The blue colour of aqueous copper sulphate solution is due to the presence of the complex ion

[Cu(H_2O)_4]^{2+}

. The observed changes are due to ligand substitution reactions, where the water ligands are replaced by fluoride or chloride ions. The colour of a complex depends on the ligands attached to the central metal ion, as different ligands cause different degrees of crystal field splitting.

(i)

Reaction with aqueous potassium fluoride (KF): When KF is added, the fluoride ions (

F^−

) replace the weaker field water ligands to form the tetrafluoridocuprate(II) complex,

[CuF_4]^{2−}

.

[Cu(H_2O)_4]^{2+} (aq, blue) + 4F^− (aq) \rightarrow [CuF_4]^{2−} (aq, green) + 4H_2O (l)

This complex,

[CuF_4]^{2−}

, is green. The question states a precipitate is formed, which suggests that the resulting potassium salt,

K_2[CuF_4]

, is sparingly soluble or insoluble in water, thus forming a green precipitate.

(ii)

Reaction with aqueous potassium chloride (KCl): When KCl is added, the chloride ions (

Cl^−

) replace the water ligands to form the tetrachloridocuprate(II) complex,

[CuCl_4]^{2−}

.

[Cu(H_2O)_4]^{2+} (aq, blue) + 4Cl^− (aq) \rightarrow [CuCl_4]^{2−} (aq, yellow-green) + 4H_2O (l)

The

[CuCl_4]^{2−}

complex ion is yellow-green. In solution, there is often an equilibrium between the blue

[Cu(H_2O)_4]^{2+}

and the yellow-green

[CuCl_4]^{2−}

. The mixture of these two colours results in the solution appearing bright green.

Q14Exercises

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 $H_2S(g)$ is passed through this solution?

Solution

When an excess of aqueous potassium cyanide (KCN) is added to an aqueous solution of copper sulphate (

CuSO_4

), a series of reactions occurs. Initially, copper(II) cyanide precipitates, but it then reacts with excess cyanide. In this process,

Cu^{2+}

is reduced to

Cu^+

by the cyanide ion, which is oxidized to cyanogen gas,

(CN)_2

. The

Cu^+

then forms a highly stable complex with the excess cyanide ions.

The overall reaction leads to the formation of the coordination entity tetracyanidocuprate(I) ion, $[Cu(CN)_4]^{3−}$ .

2CuSO_4 (aq) + 10KCN (aq) \rightarrow 2K_3[Cu(CN)_4] (aq) + 2K_2SO_4 (aq) + (CN)_2 (g)

The reason no precipitate of copper(I) sulphide (

Cu_2S

) is obtained when hydrogen sulphide (

H_2S

) gas is passed through this solution is the extremely high stability of the complex ion $[Cu(CN)_4]^{3−}$ . The stability constant for this complex is very large. This means that in the equilibrium:

[Cu(CN)_4]^{3−} (aq) \rightleftharpoons Cu^+ (aq) + 4CN^− (aq)

the position of equilibrium lies far to the left. As a result, the concentration of free

Cu^+

ions in the solution is extremely low.

For a precipitate of copper(I) sulphide to form, the ionic product of its ions,

[Cu^+]^2[S^{2−}]

, must exceed its solubility product,

K_{sp}

. Although

H_2S

provides sulphide ions (

S^{2−}

), the concentration of free

Cu^+

ions is so low that the ionic product remains less than the

K_{sp}

of

Cu_2S

. Therefore, no precipitation occurs.

Q15Exercises

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

(i)

$[Fe(CN)_6]^{4−}$

(ii)

$[FeF_6]^{3−}$

(iii)

$[Co(C_2O_4)_3]^{3−}$

(iv)

$[CoF_6]^{3−}$

Solution

(i)

$[Fe(CN)_6]^{4−}$

Oxidation state of Fe: $x + 6(−1) = −4 \Rightarrow x = +2$ . So, $Fe^{2+}$ .
Electronic configuration of $Fe^{2+}$ : $[Ar] 3d^6$ .
Ligand: $CN^−$ is a strong field ligand. It causes pairing of the 6 electrons in the 3d orbitals.
The six 3d electrons pair up in three d-orbitals, leaving two 3d orbitals empty.
Hybridization: For an octahedral complex, Fe(II) uses two inner 3d, one 4s, and three 4p orbitals, leading to $d^2sp^3$ hybridization.
Magnetic property: Since all electrons are paired, the complex is diamagnetic.
Geometry: Octahedral. It is an inner orbital (low spin) complex.

(ii)

$[FeF_6]^{3−}$

Oxidation state of Fe: $x + 6(−1) = −3 \Rightarrow x = +3$ . So, $Fe^{3+}$ .
Electronic configuration of $Fe^{3+}$ : $[Ar] 3d^5$ .
Ligand: $F^−$ is a weak field ligand. It does not cause electron pairing.
The five 3d electrons remain unpaired in the five d-orbitals.
Hybridization: For an octahedral complex, Fe(III) uses one 4s, three 4p, and two outer 4d orbitals, leading to $sp^3d^2$ hybridization.
Magnetic property: There are five unpaired electrons, so the complex is highly paramagnetic.
Geometry: Octahedral. It is an outer orbital (high spin) complex.

(iii)

$[Co(C_2O_4)_3]^{3−}$

Oxidation state of Co: $x + 3(−2) = −3 \Rightarrow x = +3$ . So, $Co^{3+}$ .
Electronic configuration of $Co^{3+}$ : $[Ar] 3d^6$ .
Ligand: Oxalate ( $C_2O_4^{2−}$ ) is a strong field ligand for $Co^{3+}$ . It causes pairing of the 6 electrons in the 3d orbitals.
The six 3d electrons pair up, leaving two 3d orbitals empty.
Hybridization: For an octahedral complex, Co(III) undergoes $d^2sp^3$ hybridization.
Magnetic property: Since all electrons are paired, the complex is diamagnetic.
Geometry: Octahedral. It is an inner orbital (low spin) complex.

(iv)

$[CoF_6]^{3−}$

Oxidation state of Co: $x + 6(−1) = −3 \Rightarrow x = +3$ . So, $Co^{3+}$ .
Electronic configuration of $Co^{3+}$ : $[Ar] 3d^6$ .
Ligand: $F^−$ is a weak field ligand. It does not cause electron pairing.
The six 3d electrons are arranged with four unpaired electrons.
Hybridization: For an octahedral complex, Co(III) uses outer orbitals, leading to $sp^3d^2$ hybridization.
Magnetic property: There are four unpaired electrons, so the complex is paramagnetic.
Geometry: Octahedral. It is an outer orbital (high spin) complex.

Q16Exercises

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

Solution

In an isolated gaseous metal atom/ion, the five d-orbitals (

d_{xy}, d_{yz}, d_{xz}, d_{x^2-y^2}, d_{z^2}

) are degenerate (have the same energy). When this ion is surrounded by six ligands in an octahedral arrangement, the ligands approach along the x, y, and z axes. This causes the degeneracy of the d-orbitals to be lifted.

The two d-orbitals whose lobes point directly towards the ligands along the axes ( $d_{x^2-y^2}$ and $d_{z^2}$ ) experience greater repulsion from the ligand electrons. Their energy is raised. This set of orbitals is called the $e_g$ set.
The three d-orbitals whose lobes lie between the axes ( $d_{xy}, d_{yz}, d_{xz}$ ) experience less repulsion. Their energy is lowered relative to the average energy (barycentre). This set of orbitals is called the $t_{2g}$ set.

This splitting of d-orbitals is called crystal field splitting. The energy difference between the

t_{2g}

and

e_g

sets is denoted by

Δ_o

(o for octahedral).

The energy of the

e_g

orbitals increases by

(3/5)Δ_o

or

+0.6Δ_o

. The energy of the

t_{2g}

orbitals decreases by

(2/5)Δ_o

or

-0.4Δ_o

.

(A diagram should be drawn showing five degenerate d-orbitals on the left, an arrow pointing to a higher average energy level for a spherical field, and then another arrow showing this level splitting into a lower triplet

t_{2g}

level and a higher doublet

e_g

level. The energy gap should be labeled

Δ_o

.)

Q17Exercises

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

Solution

Spectrochemical Series: The spectrochemical series is an experimentally determined series in which ligands are arranged in order of their increasing ability to cause crystal field splitting. A larger splitting corresponds to a stronger field ligand. A simplified version of the series is:

I^− < Br^− < Cl^− < F^− < OH^− < H_2O < NH_3 < en < CN^− < CO

Difference between a Weak Field Ligand and a Strong Field Ligand: The difference lies in the magnitude of the crystal field splitting energy (

Δ_o

in octahedral fields) they produce.

Weak Field Ligand:
- These ligands cause a small degree of crystal field splitting.
- The crystal field splitting energy ( $Δ_o$ ) is less than the mean electron pairing energy (P), i.e., $Δ_o < P$ .
- In filling the d-orbitals (for $d^4$ to $d^7$ configurations), electrons will occupy the higher energy $e_g$ orbitals before pairing up in the lower energy $t_{2g}$ orbitals.
- They tend to form high-spin complexes, which have the maximum number of unpaired electrons.
- Examples: Halide ions ( $I^−, Br^−, Cl^−, F^−$ ), $H_2O$ .
Strong Field Ligand:
- These ligands cause a large degree of crystal field splitting.
- The crystal field splitting energy ( $Δ_o$ ) is greater than the mean electron pairing energy (P), i.e., $Δ_o > P$ .
- In filling the d-orbitals, it is energetically more favorable for electrons to pair up in the lower energy $t_{2g}$ orbitals before occupying the higher energy $e_g$ orbitals.
- They tend to form low-spin complexes, which have a reduced number of unpaired electrons.
- Examples: Cyanide ( $CN^−$ ), Carbonyl (CO), Ethylenediamine (en).

Q18Exercises

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

Solution

Crystal Field Splitting Energy ( $Δ_o$ ) Crystal field splitting energy is the energy difference between the two sets of d-orbitals (

t_{2g}

and

e_g

) that are generated due to the splitting of degenerate d-orbitals by the electrostatic field of the ligands in a coordination complex. For an octahedral complex, this energy separation is denoted by

Δ_o

.

How $Δ_o$ decides the d-orbital configuration: The magnitude of

Δ_o

relative to another energy parameter, the mean pairing energy (P), determines how the d-electrons will be distributed in the

t_{2g}

and

e_g

orbitals. The pairing energy (P) is the energy required to place two electrons in the same orbital, overcoming the electrostatic repulsion between them.

This decision is critical for metal ions with d-electron configurations from

d^4

to

d^7

.

Case 1: Weak Field Ligand ( $Δ_o < P$ )
- When the crystal field splitting energy ( $Δ_o$ ) is small (caused by a weak field ligand), it costs less energy for an electron to occupy a higher energy $e_g$ orbital than to pair up in a lower energy $t_{2g}$ orbital.
- Therefore, after the $t_{2g}$ orbitals are half-filled (for $d^4$ ), the fourth electron will enter an $e_g$ orbital rather than pairing up.
- This leads to a high-spin configuration with the maximum number of unpaired electrons (e.g., $d^4$ is $t_{2g}^3 e_g^1$ ; $d^5$ is $t_{2g}^3 e_g^2$ ).
Case 2: Strong Field Ligand ( $Δ_o > P$ )
- When the crystal field splitting energy ( $Δ_o$ ) is large (caused by a strong field ligand), it costs more energy for an electron to jump to the $e_g$ level than to pair up in the $t_{2g}$ level.
- Therefore, after the $t_{2g}$ orbitals are half-filled, the fourth, fifth, and sixth electrons will pair up in the $t_{2g}$ orbitals before any occupy the $e_g$ orbitals.
- This leads to a low-spin configuration with a minimum number of unpaired electrons (e.g., $d^4$ is $t_{2g}^4 e_g^0$ ; $d^5$ is $t_{2g}^5 e_g^0$ ).

For

d^1, d^2, d^3, d^8, d^9,

and

d^{10}

configurations, there is only one possible arrangement of electrons, regardless of the magnitude of

Δ_o

.

Q19Exercises

$[Cr(NH_3)_6]^{3+}$ is paramagnetic while $[Ni(CN)_4]^{2−}$ is diamagnetic. Explain why?

Solution

1. For $[Cr(NH_3)_6]^{3+}$ (Paramagnetic):

Central Ion: The oxidation state of Chromium (Cr) is +3. So, we have $Cr^{3+}$ .
Electronic Configuration: The configuration of $Cr^{3+}$ is $[Ar] 3d^3$ .
Geometry and Splitting: This is an octahedral complex. According to Crystal Field Theory, the three 3d electrons will occupy the lower energy $t_{2g}$ orbitals singly, following Hund's rule.
Electron Distribution: The configuration is $t_{2g}^3 e_g^0$ .
Magnetic Property: There are three unpaired electrons. The presence of unpaired electrons makes the complex paramagnetic.

2. For $[Ni(CN)_4]^{2−}$ (Diamagnetic):

Central Ion: The oxidation state of Nickel (Ni) is +2. So, we have $Ni^{2+}$ .
Electronic Configuration: The configuration of $Ni^{2+}$ is $[Ar] 3d^8$ .
Geometry and Ligand: The complex has a square planar geometry, and the cyanide ligand ( $CN^−$ ) is a strong field ligand.
Electron Distribution: In a square planar field with a strong ligand, the eight 3d electrons are forced to pair up in the four lowest energy d-orbitals.
Hybridization (VBT view): To achieve $dsp^2$ hybridization for the square planar shape, one of the 3d orbitals must be empty. The strong $CN^−$ ligand forces the pairing of the 8 electrons in four 3d orbitals, making one 3d orbital available for hybridization.
Magnetic Property: As all eight electrons are paired up, there are no unpaired electrons. Therefore, the complex is diamagnetic.

Q20Exercises

A solution of $[Ni(H_2O)_6]^{2+}$ is green but a solution of $[Ni(CN)_4]^{2−}$ is colourless. Explain.

Solution

The colour of transition metal complexes is generally attributed to the absorption of light in the visible region, which causes the promotion of an electron from a lower energy d-orbital to a higher energy d-orbital (a d-d transition).

1. For $[Ni(H_2O)_6]^{2+}$ (Green):

Ion and Ligand: The complex contains the $Ni^{2+}$ ion ( $3d^8$ ) and six water ( $H_2O$ ) ligands. $H_2O$ is a relatively weak field ligand.
d-d Transition: The water ligands cause a crystal field splitting ( $Δ_o$ ) of a certain magnitude. This energy gap corresponds to the energy of light in the red region of the visible spectrum.
Colour: The complex absorbs red light to promote an electron from the $t_{2g}$ level to the $e_g$ level. The transmitted light is the complementary colour of red, which is green. Hence, the solution appears green.

2. For $[Ni(CN)_4]^{2−}$ (Colourless):

Ion and Ligand: The complex contains the $Ni^{2+}$ ion ( $3d^8$ ) and four cyanide ( $CN^−$ ) ligands. $CN^−$ is a very strong field ligand.
d-d Transition: The strong field cyanide ligands cause a very large crystal field splitting ( $Δ_{sp}$ ). The energy gap between the d-orbitals is much larger than in the aqua complex.
Colour: This large energy gap corresponds to the energy of light in the ultraviolet (UV) region, which is outside the visible spectrum. Since the complex does not absorb any light from the visible region, it transmits all visible light and thus appears colourless to the human eye.

Q21Exercises

$[Fe(CN)_6]^{4−}$ and $[Fe(H_2O)_6]^{2+}$ are of different colours in dilute solutions. Why?

Solution

The colour of coordination compounds arises from d-d electronic transitions, where an electron absorbs energy from visible light to move from a lower energy d-orbital to a higher energy d-orbital. The energy absorbed, and thus the colour observed, depends on the magnitude of the crystal field splitting energy (

Δ_o

).

In both complexes, the central metal ion is Iron in the +2 oxidation state (

Fe^{2+}

), which has a

d^6

electronic configuration.

$[Fe(H_2O)_6]^{2+}$ :
- The ligand is water ( $H_2O$ ), which is a weak field ligand.
- It causes a small crystal field splitting ( $Δ_o$ ).
- To excite an electron, the complex will absorb light of lower energy (longer wavelength) from the visible spectrum. It typically absorbs in the red-orange region and appears pale green.
$[Fe(CN)_6]^{4−}$ :
- The ligand is cyanide ( $CN^−$ ), which is a strong field ligand.
- It causes a large crystal field splitting ( $Δ_o$ ).
- To excite an electron, the complex needs to absorb light of higher energy (shorter wavelength) from the visible spectrum. It absorbs in the violet-blue region and appears yellow.

Conclusion: Although both complexes contain the same central metal ion (

Fe^{2+}

) with the same oxidation state, they have different ligands (

H_2O

and

CN^−

). According to the spectrochemical series, these ligands have different field strengths, leading to different magnitudes of crystal field splitting (

Δ_o

). Because the

Δ_o

values are different, the complexes absorb different frequencies (and therefore different colours) of light, resulting in them having different observed colours.

Q22Exercises

Discuss the nature of bonding in metal carbonyls.

Solution

The bonding in metal carbonyls involves a unique mechanism called synergic bonding, which has components of both a sigma (

σ

) bond and a pi (

π

) bond. The central metal atom is typically in a zero or low oxidation state.

Let's consider the bond between a metal (M) and a carbon monoxide (CO) ligand:

M-C Sigma ( $σ$ ) Bond Formation:
- The carbon monoxide molecule acts as a Lewis base. The lone pair of electrons on the carbon atom of CO is donated into a vacant d-orbital of the metal atom.
- This forms a coordinate covalent sigma bond: M ← CO. This is a ligand-to-metal donation.
M-C Pi ( $π$ ) Bond Formation (Back-bonding):
- The metal atom, which now has increased electron density from the sigma donation, acts as a Lewis base.
- It donates a pair of electrons from one of its filled d-orbitals into the vacant antibonding pi-star ( $π^*$ ) molecular orbital of the carbon monoxide ligand.
- This forms a pi bond: M → CO. This process is called pi back-bonding or back-donation.

Synergic Effect: These two bonding processes are mutually reinforcing. The sigma donation from CO to the metal increases the electron density on the metal, which in turn enhances the metal's ability to back-donate into the

π^*

orbital of CO. This back-donation removes electron density from the metal, making it a better acceptor for the initial sigma donation from CO.

This synergic (working together) effect strengthens the bond between the metal and the carbon monoxide ligand significantly, leading to the formation of stable metal carbonyl compounds even with the metal in a zero oxidation state.

Q23Exercises

Give the oxidation state, d orbital occupation and coordination number of the central metal ion in the following complexes:

(i)

$K_3[Co(C_2O_4)_3]$

(ii)

cis- $[CrCl_2(en)_2]Cl$

(iii)

$(NH_4)_2[CoF_4]$

(iv)

$[Mn(H_2O)_6]SO_4$

Solution

(i)

$K_3[Co(C_2O_4)_3]$

Oxidation state: Let Co be x. $3(+1) + x + 3(−2) = 0 \Rightarrow x = +3$ . Oxidation state is +3.
d orbital occupation: Co(III) is $d^6$ . Oxalate is a strong field ligand for Co(III), so it's a low spin complex. Configuration is $t_{2g}^6 e_g^0$ .
Coordination number: Oxalate ( $C_2O_4^{2−}$ ) is a bidentate ligand. There are three such ligands. C.N. = $3 \times 2 = 6$ . Coordination number is 6.

(ii)

cis- $[CrCl_2(en)_2]Cl$

Oxidation state: Let Cr be x. $x + 2(−1) + 2(0) = +1 \Rightarrow x = +3$ . Oxidation state is +3.
d orbital occupation: Cr(III) is $d^3$ . Configuration is $t_{2g}^3 e_g^0$ .
Coordination number: Chloride is unidentate, en is bidentate. C.N. = $(2 \times 1) + (2 \times 2) = 2 + 4 = 6$ . Coordination number is 6.

(iii)

$(NH_4)_2[CoF_4]$

Oxidation state: Let Co be x. $2(+1) + x + 4(−1) = 0 \Rightarrow x = +2$ . Oxidation state is +2.
d orbital occupation: Co(II) is $d^7$ . Fluoride is a weak field ligand, so it's a high spin tetrahedral complex. Configuration is $e^4 t_2^3$ .
Coordination number: Fluoride ( $F^−$ ) is a unidentate ligand. There are four. Coordination number is 4.

(iv)

$[Mn(H_2O)_6]SO_4$

Oxidation state: Let Mn be x. $x + 6(0) = +2 \Rightarrow x = +2$ . Oxidation state is +2.
d orbital occupation: Mn(II) is $d^5$ . Water is a weak field ligand, so it's a high spin complex. Configuration is $t_{2g}^3 e_g^2$ .
Coordination number: Water ( $H_2O$ ) is a unidentate ligand. There are six. Coordination number is 6.

Q24Exercises

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)

$K[Cr(H_2O)_2(C_2O_4)_2] \cdot 3H_2O$

(ii)

$[Co(NH_3)_5Cl]Cl_2$

(iii)

$[CrCl_3(py)_3]$

(iv)

$Cs[FeCl_4]$

(v)

$K_4[Mn(CN)_6]$

Solution

(i)

$K[Cr(H_2O)_2(C_2O_4)_2] \cdot 3H_2O$

IUPAC Name: Potassium diaquabis(oxalato)chromate(III) trihydrate
Oxidation State: Let Cr be x. $+1 + x + 2(0) + 2(−2) = 0 \Rightarrow x = +3$ . Cr(III).
Electronic Configuration: $Cr^{3+}$ is $d^3$ ( $t_{2g}^3 e_g^0$ ).
Coordination Number: 2(aqua) + 2(oxalato, bidentate) = $2 + 2(2) = 6$ . C.N. = 6.
Stereochemistry: Octahedral geometry. Exhibits geometrical (cis/trans) and optical isomerism (cis-isomer is chiral).
Magnetic Moment: Number of unpaired electrons (n) = 3. $\mu = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ BM.

(ii)

$[Co(NH_3)_5Cl]Cl_2$

IUPAC Name: Pentaamminechloridocobalt(III) chloride
Oxidation State: Let Co be x. $x + 5(0) + (−1) = +2 \Rightarrow x = +3$ . Co(III).
Electronic Configuration: $Co^{3+}$ is $d^6$ . $NH_3$ is a strong ligand for Co(III), so low spin: $t_{2g}^6 e_g^0$ .
Coordination Number: 5(ammine) + 1(chlorido) = 6. C.N. = 6.
Stereochemistry: Octahedral geometry. No geometrical or optical isomerism.
Magnetic Moment: Number of unpaired electrons (n) = 0. Diamagnetic, $\mu = 0$ BM.

(iii)

$[CrCl_3(py)_3]$

IUPAC Name: Trichloridotris(pyridine)chromium(III)
Oxidation State: Cr is in +3 state.
Electronic Configuration: $Cr^{3+}$ is $d^3$ ( $t_{2g}^3 e_g^0$ ).
Coordination Number: 3(chlorido) + 3(pyridine) = 6. C.N. = 6.
Stereochemistry: Octahedral geometry. Exhibits geometrical isomerism (facial and meridional).
Magnetic Moment: Number of unpaired electrons (n) = 3. $\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87$ BM.

(iv)

$Cs[FeCl_4]$

IUPAC Name: Caesium tetrachloridoferrate(III)
Oxidation State: Let Fe be x. $+1 + x + 4(−1) = 0 \Rightarrow x = +3$ . Fe(III).
Electronic Configuration: $Fe^{3+}$ is $d^5$ . Weak field ligand, high spin.
Coordination Number: 4(chlorido) = 4. C.N. = 4.
Stereochemistry: Tetrahedral geometry.
Magnetic Moment: Number of unpaired electrons (n) = 5. $\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92$ BM.

(v)

$K_4[Mn(CN)_6]$

IUPAC Name: Potassium hexacyanidomanganate(II)
Oxidation State: Let Mn be x. $4(+1) + x + 6(−1) = 0 \Rightarrow x = +2$ . Mn(II).
Electronic Configuration: $Mn^{2+}$ is $d^5$ . $CN^−$ is a strong ligand, low spin: $t_{2g}^5 e_g^0$ .
Coordination Number: 6(cyanido) = 6. C.N. = 6.
Stereochemistry: Octahedral geometry.
Magnetic Moment: Number of unpaired electrons (n) = 1. $\mu = \sqrt{1(3)} = \sqrt{3} \approx 1.73$ BM.

Q25Exercises

Explain the violet colour of the complex $[Ti(H_2O)_6]^{3+}$ on the basis of crystal field theory.

Solution

The colour of the complex

[Ti(H_2O)_6]^{3+}

can be explained by d-d electronic transitions according to the Crystal Field Theory.

Electronic Configuration: The central metal ion is Titanium in the +3 oxidation state ( $Ti^{3+}$ ). The electronic configuration of $Ti^{3+}$ is $[Ar] 3d^1$ . It has one electron in its d-orbital.
Crystal Field Splitting: In the octahedral field created by the six water ligands, the five degenerate d-orbitals of the titanium ion split into two energy levels: a lower energy triplet set ( $t_{2g}$ ) and a higher energy doublet set ( $e_g$ ).
Ground State: In the ground state of the complex, the single 3d electron occupies one of the lower energy $t_{2g}$ orbitals. The electronic configuration is $t_{2g}^1 e_g^0$ .
Excitation (d-d transition): When the complex is exposed to visible light, it absorbs light of a specific energy (and wavelength) that corresponds to the crystal field splitting energy, $Δ_o$ . This absorbed energy promotes the electron from the $t_{2g}$ level to the higher energy $e_g$ level. $t_{2g}^1 e_g^0 \xrightarrow{h\nu} t_{2g}^0 e_g^1$
Observed Colour: The complex $[Ti(H_2O)_6]^{3+}$ absorbs light in the blue-green region of the visible spectrum (around 500 nm). The light that is not absorbed but transmitted passes through the solution. The transmitted light is the complementary colour of blue-green, which is violet. Therefore, the solution of the complex appears violet to our eyes.

Q26Exercises

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

Solution

Chelate Effect: The chelate effect refers to the enhanced stability of coordination compounds containing chelate rings compared to similar complexes with an equivalent number of analogous unidentate ligands. A chelate ring is formed when a polydentate (didentate or higher) ligand binds to a central metal ion at two or more points, forming a cyclic structure.

This increased stability is primarily due to a favorable entropy change. When a polydentate ligand replaces several unidentate ligands, the total number of free molecules in the system increases, leading to an increase in randomness or entropy (

ΔS

is positive). This makes the Gibbs free energy change (

ΔG = ΔH - TΔS

) more negative, indicating a more spontaneous and favorable reaction, and thus a more stable complex.

Example: Consider the formation of nickel(II) complexes with ammonia (unidentate) and ethane-1,2-diamine (en, a didentate ligand).

Reaction 1 (unidentate):

[Ni(H_2O)_6]^{2+} + 6NH_3 \rightleftharpoons [Ni(NH_3)_6]^{2+} + 6H_2O

Here, 7 particles on the left produce 7 particles on the right. The entropy change is small.

Reaction 2 (didentate/chelating):

[Ni(H_2O)_6]^{2+} + 3en \rightleftharpoons [Ni(en)_3]^{2+} + 6H_2O

Here, 4 particles on the left produce 7 particles on the right. There is a significant increase in the number of particles, leading to a large positive entropy change.

Because of this favorable entropy change, the formation constant for

[Ni(en)_3]^{2+}

is much larger than for

[Ni(NH_3)_6]^{2+}

, meaning the chelate complex is significantly more stable. This enhanced stability is the chelate effect.

Q27Exercises

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

(i)

biological systems

(ii)

medicinal chemistry and

(iii)

analytical chemistry

(iv)

extraction/metallurgy of metals.

Solution

(i)

Biological Systems: Coordination compounds are vital for life. Many biological processes depend on metal complexes. * Example: Haemoglobin, the red pigment in blood that transports oxygen, is a coordination compound of iron(II). The iron is coordinated to the nitrogen atoms of a large porphyrin ring. Another example is Chlorophyll, the pigment responsible for photosynthesis in plants, which is a coordination compound of magnesium.

(ii)

Medicinal Chemistry: Coordination compounds are increasingly used in the diagnosis and treatment of diseases. * Example: Cis-platin,

[Pt(NH_3)_2Cl_2]

, is a platinum-based coordination compound used as an effective anticancer drug for treating various types of tumors. Another example is the use of chelating agents like EDTA in the treatment of heavy metal poisoning (e.g., lead poisoning) by forming stable, soluble complexes that can be excreted from the body.

(iii)

Analytical Chemistry: The formation of coloured coordination compounds is the basis for many qualitative and quantitative analytical tests for metal ions. * Example: The hardness of water (due to

Ca^{2+}

and

Mg^{2+}

ions) is determined by titration with a standard solution of EDTA (ethylenediaminetetraacetic acid). EDTA forms stable complexes with these ions, and an indicator is used to detect the endpoint. Another example is the use of dimethylglyoxime (DMG) for the detection and estimation of

Ni^{2+}

ions, with which it forms a characteristic bright red precipitate.

(iv)

Extraction/Metallurgy of Metals: Coordination compounds play a crucial role in the extraction and purification of several metals. * Example: In the MacArthur-Forrest cyanide process for the extraction of gold and silver, the crushed ore is treated with an aqueous solution of NaCN. The metal dissolves by forming a soluble cyano complex, e.g.,

[Au(CN)_2]^−

. Gold is then recovered from this solution by displacement with a more electropositive metal like zinc. Similarly, nickel is purified by the Mond process, which involves forming the volatile complex tetracarbonylnickel(0),

[Ni(CO)_4]

, which is then decomposed to yield pure nickel.

Q28Exercises

How many ions are produced from the complex $Co(NH_3)_6Cl_2$ in solution?

(i)

6

(ii)

4

(iii)

3

(iv)

2 Solution

First, we must write the correct formula for the coordination compound. Cobalt commonly forms octahedral complexes with a coordination number of 6. The six ammonia (

NH_3

) molecules will act as ligands inside the coordination sphere. The chloride ions will act as counter ions.

The formula of the complex is $[Co(NH_3)_6]Cl_2$ .

When this complex salt is dissolved in an aqueous solution, it dissociates into its constituent ions: the complex cation and the simple anions.

[Co(NH_3)_6]Cl_2 (s) \xrightarrow{Water} [Co(NH_3)_6]^{2+} (aq) + 2Cl^− (aq)

From the dissociation, we get:

One complex cation: $[Co(NH_3)_6]^{2+}$
Two simple anions: $2Cl^−$

Total number of ions produced = 1 + 2 = 3.

Therefore, the correct option is (iii) 3.

Q29Exercises

Amongst the following ions which one has the highest magnetic moment value?

(i)

$[Cr(H_2O)_6]^{3+}$

(ii)

$[Fe(H_2O)_6]^{2+}$

(iii)

$[Zn(H_2O)_6]^{2+}$

Solution

The magnetic moment (

\mu

) of a coordination compound is primarily determined by the number of unpaired electrons (n), according to the spin-only formula:

\mu = \sqrt{n(n+2)}

Bohr Magnetons (BM). A higher number of unpaired electrons results in a higher magnetic moment.

Let's determine the number of unpaired electrons for each ion:

(i)

$[Cr(H_2O)_6]^{3+}$

Central ion: $Cr^{3+}$
Electronic configuration of $Cr^{3+}$ : $[Ar] 3d^3$ .
In an octahedral field, the three electrons will occupy the $t_{2g}$ orbitals singly.
Number of unpaired electrons (n) = 3.

(ii)

$[Fe(H_2O)_6]^{2+}$

Central ion: $Fe^{2+}$
Electronic configuration of $Fe^{2+}$ : $[Ar] 3d^6$ .
Ligand: $H_2O$ is a weak field ligand, so it will form a high-spin complex.
The six electrons are arranged as $t_{2g}^4 e_g^2$ . This configuration has 4 unpaired electrons.

(iii)

$[Zn(H_2O)_6]^{2+}$

Central ion: $Zn^{2+}$
Electronic configuration of $Zn^{2+}$ : $[Ar] 3d^{10}$ .
All d-orbitals are completely filled.
Number of unpaired electrons (n) = 0.

Comparing the number of unpaired electrons: 3 for Cr complex, 4 for Fe complex, and 0 for Zn complex. Since

[Fe(H_2O)_6]^{2+}

has the highest number of unpaired electrons (n=4), it will have the highest magnetic moment value.

Therefore, the correct option is (ii) $[Fe(H_2O)_6]^{2+}$ .

Q30Exercises

Amongst the following, the most stable complex is

(i)

$[Fe(H_2O)_6]^{3+}$

(ii)

$[Fe(NH_3)_6]^{3+}$

(iii)

$[Fe(C_2O_4)_3]^{3−}$

(iv)

$[FeCl_6]^{3−}$

Solution

The stability of a complex is significantly influenced by the chelate effect. The chelate effect states that complexes formed by polydentate ligands (chelating ligands) are more stable than complexes formed by an equivalent number of unidentate ligands.

Let's analyze the ligands in each complex:

(i)

[Fe(H_2O)_6]^{3+}

: Ligand is

H_2O

(aqua), which is a unidentate ligand.

(ii)

[Fe(NH_3)_6]^{3+}

: Ligand is

NH_3

(ammine), which is a unidentate ligand.

(iii)

[Fe(C_2O_4)_3]^{3−}

: Ligand is

C_2O_4^{2−}

(oxalate), which is a bidentate ligand. It forms a five-membered chelate ring with the central metal ion.

(iv)

[FeCl_6]^{3−}

: Ligand is

Cl^−

(chlorido), which is a unidentate ligand.

Since the oxalate ligand in

[Fe(C_2O_4)_3]^{3−}

is a chelating ligand, it forms a highly stable complex due to the chelate effect. The other complexes are formed with unidentate ligands and do not benefit from this extra stability.

Therefore, the most stable complex is (iii) $[Fe(C_2O_4)_3]^{3−}$ .

Q31Exercises

What will be the correct order for the wavelengths of absorption in the visible region for the following: $[Ni(NO_2)_6]^{4−}, [Ni(NH_3)_6]^{2+}, [Ni(H_2O)_6]^{2+}$ ?

Solution

The energy of light absorbed by a complex for a d-d transition is equal to the crystal field splitting energy (

Δ_o

). The relationship between absorbed energy (E) and wavelength (

\lambda

) is given by

E = \frac{hc}{\lambda}

. This means that a larger energy gap (

Δ_o

) corresponds to the absorption of light with a shorter wavelength (

\lambda

).

The magnitude of the crystal field splitting energy (

Δ_o

) depends on the strength of the ligand. Stronger field ligands produce a larger

Δ_o

.

First, we need to arrange the ligands (

NO_2^−

,

NH_3

,

H_2O

) in order of their increasing field strength using the spectrochemical series:

H_2O < NH_3 < NO_2^−

This means the order of the crystal field splitting energy (

Δ_o

) for the nickel complexes will be:

Δ_o([Ni(H_2O)_6]^{2+}) < Δ_o([Ni(NH_3)_6]^{2+}) < Δ_o([Ni(NO_2)_6]^{4−})

Since the wavelength of absorption (

\lambda

) is inversely proportional to the energy (

Δ_o

), the order for the wavelengths of absorption will be the reverse of the order of ligand strength:

\lambda([Ni(NO_2)_6]^{4−}) < \lambda([Ni(NH_3)_6]^{2+}) < \lambda([Ni(H_2O)_6]^{2+})

Therefore, the correct order for the wavelengths of absorption is: $[Ni(NO_2)_6]^{4−} < [Ni(NH_3)_6]^{2+} < [Ni(H_2O)_6]^{2+}$

Q1Intext Questions

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

(i)

tetraamminediaquacobalt(III) chloride

Central metal: Cobalt (Co)
Oxidation state: +3
Ligands: 'tetraammine' means 4 $NH_3$ molecules, 'diaqua' means 2 $H_2O$ molecules.
The coordination sphere is $[Co(NH_3)_4(H_2O)_2]$ .
Charge on coordination sphere: +3 (from Co) + 4(0) (from $NH_3$ ) + 2(0) (from $H_2O$ ) = +3.
To balance the +3 charge, 3 chloride ( $Cl^−$ ) ions are needed as counter ions.
Formula: $[Co(NH_3)_4(H_2O)_2]Cl_3$

(ii)

potassium tetracyanidonickelate(II)

Cation: Potassium ( $K^+$ )
Coordination sphere: Anionic, containing Nickel (Ni) in +2 oxidation state and 4 cyanide ( $CN^−$ ) ligands.
The coordination sphere is $[Ni(CN)_4]$ .
Charge on coordination sphere: +2 (from Ni) + 4(−1) (from $CN^−$ ) = −2.
To balance the −2 charge, 2 potassium ( $K^+$ ) ions are needed.
Formula: $K_2[Ni(CN)_4]$

(iii)

tris(ethane-1,2-diamine) chromium(III) chloride

Central metal: Chromium (Cr)
Oxidation state: +3
Ligand: 'tris(ethane-1,2-diamine)' means 3 ethane-1,2-diamine (en) molecules.
The coordination sphere is $[Cr(en)_3]$ .
Charge on coordination sphere: +3 (from Cr) + 3(0) (from en) = +3.
To balance the +3 charge, 3 chloride ( $Cl^−$ ) ions are needed.
Formula: $[Cr(en)_3]Cl_3$

(iv)

amminebromidochloridonitrito-N-platinate(II)

Central metal: Platinum (Pt) in an anionic complex ('-ate' suffix).
Oxidation state: +2
Ligands: 'ammine' ( $NH_3$ ), 'bromido' ( $Br^−$ ), 'chlorido' ( $Cl^−$ ), 'nitrito-N' ( $NO_2^−$ ).
The coordination sphere is $[Pt(NH_3)(Br)(Cl)(NO_2)]$ .
Charge on coordination sphere: +2 (from Pt) + 0 (from $NH_3$ ) + (−1) (from $Br^−$ ) + (−1) (from $Cl^−$ ) + (−1) (from $NO_2^−$ ) = −1.
Formula: $[Pt(NH_3)BrCl(NO_2)]^−$

(v)

dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate

Central metal: Platinum (Pt)
Oxidation state: +4
Ligands: 'dichlorido' (2 $Cl^−$ ), 'bis(ethane-1,2-diamine)' (2 en).
The coordination sphere is $[PtCl_2(en)_2]$ .
Charge on coordination sphere: +4 (from Pt) + 2(−1) (from $Cl^−$ ) + 2(0) (from en) = +2.
The counter ion is nitrate ( $NO_3^−$ ). To balance the +2 charge, 2 nitrate ions are needed.
Formula: $[PtCl_2(en)_2](NO_3)_2$

(vi)

iron(III) hexacyanidoferrate(II)

Cation: iron(III), which is $Fe^{3+}$ .
Anion: 'hexacyanidoferrate(II)', which is $[Fe(CN)_6]$ .
Oxidation state of Fe in the anion is +2.
Charge on the anionic sphere: +2 (from Fe) + 6(−1) (from $CN^−$ ) = −4.
To balance the charges between the cation ( $Fe^{3+}$ ) and the anion ( $[Fe(CN)_6]^{4−}$ ), we use the criss-cross method.
Formula: $Fe_4[Fe(CN)_6]_3$

Q2Intext Questions

Write the IUPAC names of the following coordination compounds:

(i)

$[Co(NH_3)_6]Cl_3$

(ii)

$[Co(NH_3)_5Cl]Cl_2$

(iii)

$K_3[Fe(CN)_6]$

(iv)

$K_3[Fe(C_2O_4)_3]$

(v)

$K_2[PdCl_4]$

(vi)

$[Pt(NH_3)_2Cl(NH_2CH_3)]Cl$

Solution

(i)

$[Co(NH_3)_6]Cl_3$

Cation is the complex ion. Ligand is $NH_3$ (ammine). There are six, so 'hexaammine'.
Central metal is Cobalt (Co). Since the complex is a cation, the name is 'cobalt'.
Let the oxidation state of Co be x. $x + 6(0) = +3$ (since there are 3 $Cl^−$ counter ions). So, $x = +3$ . The oxidation state is (III).
The anion is chloride.
Name: Hexaamminecobalt(III) chloride

(ii)

$[Co(NH_3)_5Cl]Cl_2$

Cation is the complex ion. Ligands are $NH_3$ (ammine) and $Cl^−$ (chlorido). There are five ammine and one chlorido.
Named alphabetically: 'pentaamminechlorido'.
Central metal is Cobalt (Co). 'cobalt'.
Let the oxidation state of Co be x. $x + 5(0) + (−1) = +2$ (since there are 2 $Cl^−$ counter ions). So, $x = +3$ . The oxidation state is (III).
The anion is chloride.
Name: Pentaamminechloridocobalt(III) chloride

(iii)

$K_3[Fe(CN)_6]$

Cation is potassium.
Anion is the complex ion. Ligand is $CN^−$ (cyanido). There are six, so 'hexacyanido'.
Central metal is Iron (Fe). Since the complex is an anion, the name ends in '-ate', so 'ferrate'.
Let the oxidation state of Fe be x. $3(+1) + x + 6(−1) = 0$ . So, $x = +3$ . The oxidation state is (III).
Name: Potassium hexacyanidoferrate(III)

(iv)

$K_3[Fe(C_2O_4)_3]$

Cation is potassium.
Anion is the complex ion. Ligand is $C_2O_4^{2−}$ (oxalato). It is a bidentate ligand. There are three, so 'trioxalato'.
Central metal is Iron (Fe). 'ferrate'.
Let the oxidation state of Fe be x. $3(+1) + x + 3(−2) = 0$ . So, $x = +3$ . The oxidation state is (III).
Name: Potassium trioxalatoferrate(III)

(v)

$K_2[PdCl_4]$

Cation is potassium.
Anion is the complex ion. Ligand is $Cl^−$ (chlorido). There are four, so 'tetrachlorido'.
Central metal is Palladium (Pd). 'palladate'.
Let the oxidation state of Pd be x. $2(+1) + x + 4(−1) = 0$ . So, $x = +2$ . The oxidation state is (II).
Name: Potassium tetrachloridopalladate(II)

(vi)

$[Pt(NH_3)_2Cl(NH_2CH_3)]Cl$

Cation is the complex ion. Ligands are $NH_3$ (ammine), $Cl^−$ (chlorido), and $NH_2CH_3$ (methanamine).
Named alphabetically: 'diamminechlorido(methanamine)'.
Central metal is Platinum (Pt). 'platinum'.
Let the oxidation state of Pt be x. $x + 2(0) + (−1) + 0 = +1$ (since there is 1 $Cl^−$ counter ion). So, $x = +2$ . The oxidation state is (II).
The anion is chloride.
Name: Diamminechlorido(methanamine)platinum(II) chloride

Q3Intext Questions

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

(i)

$K[Cr(H_2O)_2(C_2O_4)_2]$

(ii)

$[Co(en)_3]Cl_3$

(iii)

$[Co(NH_3)_5(NO_2)](NO_3)_2$

(iv)

$[Pt(NH_3)(H_2O)Cl_2]$

Solution

(i)

$K[Cr(H_2O)_2(C_2O_4)_2]$ The complex ion is

[Cr(H_2O)_2(C_2O_4)_2]^−

. This is an octahedral complex of the type

[M(AA)_2b_2]

, where AA is the bidentate oxalate ligand and b is the unidentate water ligand.

Geometrical Isomerism: It can exist as cis and trans isomers.
- In the cis-isomer, the two water molecules are adjacent to each other.
- In the trans-isomer, the two water molecules are opposite to each other.
Optical Isomerism: The cis-isomer is chiral (lacks a plane of symmetry) and will exist as a pair of non-superimposable mirror images (enantiomers). The trans-isomer is achiral (has a plane of symmetry) and is optically inactive.

(ii)

$[Co(en)_3]Cl_3$ The complex ion is

[Co(en)_3]^{3+}

. This is an octahedral complex of the type

[M(AA)_3]

, where AA is the bidentate ligand ethane-1,2-diamine.

Geometrical Isomerism: This type of complex does not show geometrical isomerism.
Optical Isomerism: The complex is chiral and exists as a pair of non-superimposable mirror images (dextro and laevo forms). It is optically active.

(iii)

$[Co(NH_3)_5(NO_2)](NO_3)_2$

Linkage Isomerism: The nitrite ligand ( $NO_2^−$ $N O_{2}^{-}$ ) is an ambidentate ligand. It can coordinate through the Nitrogen atom (nitro, $-NO_2$ $- N O_{2}$ ) or through the Oxygen atom (nitrito, $-ONO$ $- ONO$ ).
- $[Co(NH_3)_5(NO_2)](NO_3)_2$ (nitro complex)
- $[Co(NH_3)_5(ONO)](NO_3)_2$ (nitrito complex)
Ionisation Isomerism: The nitrate ion ( $NO_3^−$ $N O_{3}^{-}$ ) from outside the coordination sphere can exchange with the nitrite ligand ( $NO_2^−$ $N O_{2}^{-}$ ) inside the sphere.
- $[Co(NH_3)_5(NO_3)](NO_2)(NO_3)$

(iv)

$[Pt(NH_3)(H_2O)Cl_2]$ This is a square planar complex of the type

[Mab c_2]

.

Geometrical Isomerism: It can exist as cis and trans isomers, based on the relative positions of the two chloride ligands.
- In the cis-isomer, the two chloride ligands are adjacent (at 90°).
- In the trans-isomer, the two chloride ligands are opposite (at 180°).
Optical Isomerism: Square planar complexes of this type are not optically active as they possess a plane of symmetry (the molecular plane).

Q4Intext Questions

Give evidence that $[Co(NH_3)_5Cl]SO_4$ and $[Co(NH_3)_5(SO_4)]Cl$ are ionisation isomers.

Solution

Ionisation isomers are compounds that have the same composition but yield different ions when dissolved in a solvent. The evidence for

[Co(NH_3)_5Cl]SO_4

and

[Co(NH_3)_5(SO_4)]Cl

being ionisation isomers can be obtained by treating their aqueous solutions with reagents that can precipitate the counter ions.

When an aqueous solution of $[Co(NH_3)_5Cl]SO_4$ is treated with a solution of barium chloride ( $BaCl_2$ ), a white precipitate of barium sulphate ( $BaSO_4$ ) is formed. This confirms the presence of free sulphate ions ( $SO_4^{2−}$ ) in the solution. $[Co(NH_3)_5Cl]SO_4 (aq) \rightarrow [Co(NH_3)_5Cl]^{2+} (aq) + SO_4^{2−} (aq)$ $Ba^{2+} (aq) + SO_4^{2−} (aq) \rightarrow BaSO_4 (s)$ This solution will not give a precipitate with silver nitrate ( $AgNO_3$ ) solution, as the chloride ion is inside the coordination sphere.
When an aqueous solution of $[Co(NH_3)_5(SO_4)]Cl$ is treated with a solution of silver nitrate ( $AgNO_3$ ), a white precipitate of silver chloride ( $AgCl$ ) is formed. This confirms the presence of free chloride ions ( $Cl^−$ ) in the solution. $[Co(NH_3)_5(SO_4)]Cl (aq) \rightarrow [Co(NH_3)_5(SO_4)]^+ (aq) + Cl^− (aq)$ $Ag^+ (aq) + Cl^− (aq) \rightarrow AgCl (s)$ This solution will not give a precipitate with barium chloride ( $BaCl_2$ ) solution, as the sulphate ion is inside the coordination sphere.

Since the two compounds give different ions in solution, they are confirmed to be ionisation isomers.

Q5Intext Questions

Explain on the basis of valence bond theory that $[Ni(CN)_4]^{2−}$ ion with square planar structure is diamagnetic and the $[NiCl_4]^{2−}$ ion with tetrahedral geometry is paramagnetic.

Solution

Both complexes involve the Nickel ion in the +2 oxidation state (

Ni^{2+}

). The electronic configuration of Ni (Z=28) is

[Ar] 3d^8 4s^2

. The electronic configuration of

Ni^{2+}

is

[Ar] 3d^8

.

1. For $[Ni(CN)_4]^{2−}$ (Square planar, diamagnetic):

The ligand is cyanide ( $CN^−$ ), which is a strong field ligand.
In the presence of a strong ligand, the electrons in the 3d orbitals of $Ni^{2+}$ are forced to pair up against Hund's rule.
The 8 electrons in the 3d orbitals pair up in four orbitals, leaving one 3d orbital empty.
For square planar geometry, the hybridization required is $dsp^2$ . The central $Ni^{2+}$ ion uses its empty 3d orbital, one 4s orbital, and two 4p orbitals to form four hybrid $dsp^2$ orbitals.
These four hybrid orbitals are occupied by the four electron pairs donated by the four $CN^−$ ligands.
Since all electrons in the complex are paired, the complex is diamagnetic.

2. For $[NiCl_4]^{2−}$ (Tetrahedral, paramagnetic):

The ligand is chloride ( $Cl^−$ ), which is a weak field ligand.
In the presence of a weak ligand, the electrons in the 3d orbitals of $Ni^{2+}$ do not pair up.
The 3d orbitals contain two unpaired electrons according to Hund's rule.
For tetrahedral geometry, the hybridization required is $sp^3$ . The central $Ni^{2+}$ ion uses its one 4s orbital and three 4p orbitals to form four hybrid $sp^3$ orbitals.
These four hybrid orbitals are occupied by the four electron pairs donated by the four $Cl^−$ ligands.
Since there are two unpaired electrons in the 3d orbitals of the nickel ion, the complex is paramagnetic.

Q6Intext Questions

$[NiCl_4]^{2−}$ is paramagnetic while $[Ni(CO)_4]$ is diamagnetic though both are tetrahedral. Why?

Solution

Both complexes have a tetrahedral geometry, which implies

sp^3

hybridization.

1. For $[NiCl_4]^{2−}$ :

The oxidation state of Nickel is +2. So, we have $Ni^{2+}$ .
The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$ .
The ligand $Cl^−$ is a weak field ligand, so it does not cause pairing of electrons in the 3d orbitals.
The electronic arrangement in the 3d orbitals has two unpaired electrons.
The hybridization is $sp^3$ using the outer 4s and 4p orbitals.
Due to the presence of two unpaired electrons, $[NiCl_4]^{2−}$ is paramagnetic.

2. For $[Ni(CO)_4]$ :

The oxidation state of Nickel is 0. So, we have Ni atom.
The electronic configuration of Ni is $[Ar] 3d^8 4s^2$ .
The ligand CO (carbonyl) is a very strong field ligand.
In the presence of the strong CO ligand, the 4s electrons are pushed into the 3d orbitals to pair up with the existing 3d electrons.
This results in the electronic configuration of Ni becoming $[Ar] 3d^{10} 4s^0$ .
Now, the hybridization is $sp^3$ using the empty 4s and 4p orbitals.
Since all electrons in the 3d orbitals are paired, the complex $[Ni(CO)_4]$ is diamagnetic.

Q7Intext Questions

$[Fe(H_2O)_6]^{3+}$ is strongly paramagnetic whereas $[Fe(CN)_6]^{3−}$ is weakly paramagnetic. Explain.

Solution

In both complexes, the Iron ion is in the +3 oxidation state (

Fe^{3+}

). The electronic configuration of Fe (Z=26) is

[Ar] 3d^6 4s^2

. The electronic configuration of

Fe^{3+}

is

[Ar] 3d^5

. Both complexes are octahedral.

1. For $[Fe(H_2O)_6]^{3+}$ (Strongly paramagnetic):

The ligand is water ( $H_2O$ ), which is a weak field ligand.
A weak field ligand causes only a small crystal field splitting. The splitting energy ( $Δ_o$ ) is less than the pairing energy (P).
Therefore, the 5 electrons in the 3d orbitals will occupy the $t_{2g}$ and $e_g$ orbitals singly before any pairing occurs (following Hund's rule).
The electronic configuration is $t_{2g}^3 e_g^2$ .
There are five unpaired electrons. This large number of unpaired electrons makes the complex strongly paramagnetic.
In terms of VBT, it forms an outer orbital complex with $sp^3d^2$ hybridization.

2. For $[Fe(CN)_6]^{3−}$ (Weakly paramagnetic):

The ligand is cyanide ( $CN^−$ ), which is a strong field ligand.
A strong field ligand causes a large crystal field splitting. The splitting energy ( $Δ_o$ ) is greater than the pairing energy (P).
Therefore, the electrons will pair up in the lower energy $t_{2g}$ orbitals before occupying the higher energy $e_g$ orbitals.
The 5 electrons in the 3d orbitals will have the configuration $t_{2g}^5 e_g^0$ .
There is only one unpaired electron. This makes the complex weakly paramagnetic.
In terms of VBT, it forms an inner orbital complex with $d^2sp^3$ hybridization.

Q8Intext Questions

Explain $[Co(NH_3)_6]^{3+}$ is an inner orbital complex whereas $[Ni(NH_3)_6]^{2+}$ is an outer orbital complex.

Solution

Both complexes are octahedral and have ammonia (

NH_3

) as the ligand.

1. For $[Co(NH_3)_6]^{3+}$ (Inner orbital complex):

The central metal ion is Cobalt in the +3 oxidation state ( $Co^{3+}$ ).
The electronic configuration of Co (Z=27) is $[Ar] 3d^7 4s^2$ .
The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$ .
For $Co^{3+}$ , $NH_3$ acts as a strong field ligand. It forces the 6 electrons in the 3d orbitals to pair up in the first three 3d orbitals.
This leaves two 3d orbitals empty ( $d_{x^2-y^2}$ and $d_{z^2}$ are vacant in CFT terms, but in VBT we consider two of the five are now empty).
For octahedral geometry, the hybridization is $d^2sp^3$ , using two inner 3d orbitals, one 4s orbital, and three 4p orbitals.
Since inner (n-1)d orbitals are used in hybridization, it is called an inner orbital complex (or low spin complex).

2. For $[Ni(NH_3)_6]^{2+}$ (Outer orbital complex):

The central metal ion is Nickel in the +2 oxidation state ( $Ni^{2+}$ ).
The electronic configuration of Ni (Z=28) is $[Ar] 3d^8 4s^2$ .
The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$ .
The 8 electrons in the 3d orbitals occupy the orbitals as per Hund's rule, leaving no empty 3d orbitals. Even with $NH_3$ as a moderately strong ligand, it is not strong enough to pair up electrons against the stable $d^8$ configuration to free up an inner d-orbital.
To achieve octahedral geometry, the complex must use orbitals from the outer shell.
The hybridization is $sp^3d^2$ , using one 4s orbital, three 4p orbitals, and two outer 4d orbitals.
Since outer (nd) orbitals are used in hybridization, it is called an outer orbital complex (or high spin complex).

Q9Intext Questions

Predict the number of unpaired electrons in the square planar $[Pt(CN)_4]^{2−}$ ion.

Solution

Determine the oxidation state of Platinum (Pt): Let the oxidation state of Pt be x. The charge on the cyanide ligand ( $CN^−$ ) is -1. The overall charge on the complex ion is -2. So, $x + 4(−1) = −2$ $x − 4 = −2$ $x = +2$ The central metal ion is $Pt^{2+}$ .
Determine the electronic configuration of $Pt^{2+}$ : Platinum (Pt) is in Period 6, Group 10. Its atomic number is 78. The electronic configuration of Pt is $[Xe] 4f^{14} 5d^9 6s^1$ . The electronic configuration of $Pt^{2+}$ is $[Xe] 4f^{14} 5d^8$ .
Consider the geometry and ligand field: The complex is given as square planar. This geometry typically involves $dsp^2$ hybridization. The ligand is cyanide ( $CN^−$ ), which is a very strong field ligand. In the presence of a strong ligand and for a $d^8$ ion of a heavy transition metal like Pt, the electrons are forced to pair up in the d-orbitals to make one d-orbital available for hybridization.
Electron arrangement: The eight 5d electrons will pair up in four of the five d-orbitals. This leaves no unpaired electrons.

Conclusion: The number of unpaired electrons in the square planar

[Pt(CN)_4]^{2−}

ion is zero. The complex is diamagnetic.

Q10Intext Questions

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

Solution

In both cases, the central metal ion is Manganese in the +2 oxidation state (

Mn^{2+}

). The electronic configuration of Mn (Z=25) is

[Ar] 3d^5 4s^2

. The electronic configuration of

Mn^{2+}

is

[Ar] 3d^5

. Both complexes are octahedral.

1. Hexaquo manganese(II) ion, $[Mn(H_2O)_6]^{2+}$ :

The ligand is water ( $H_2O$ ), which is a weak field ligand according to the spectrochemical series.
A weak field ligand causes a small crystal field splitting ( $Δ_o$ ). This energy gap is smaller than the mean pairing energy (P), i.e., $Δ_o < P$ .
According to Hund's rule, electrons will prefer to occupy the higher energy $e_g$ orbitals rather than pairing up in the lower energy $t_{2g}$ orbitals.
The five 3d electrons are distributed as $t_{2g}^3 e_g^2$ .
This configuration has five unpaired electrons, one in each of the five d-orbitals. The complex is a high-spin complex.

2. Hexacyano manganese(II) ion, $[Mn(CN)_6]^{4−}$ :

The ligand is cyanide ( $CN^−$ ), which is a strong field ligand.
A strong field ligand causes a large crystal field splitting ( $Δ_o$ ). This energy gap is larger than the mean pairing energy (P), i.e., $Δ_o > P$ .
It is energetically more favorable for the electrons to pair up in the lower energy $t_{2g}$ orbitals before occupying the higher energy $e_g$ orbitals.
The five 3d electrons are distributed as $t_{2g}^5 e_g^0$ .
This configuration has one unpaired electron. The complex is a low-spin complex.

Coordination CompoundsClass 12 Chemistry NCERT Solutions

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

Solution

FeSO4FeSO_4FeSO4​ solution mixed with (NH4)2SO4(NH_4)_2SO_4(NH4​)2​SO4​ solution in 1:1 molar ratio gives the test of Fe2+Fe^{2+}Fe2+ ion but CuSO4CuSO_4CuSO4​ solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu2+Cu^{2+}Cu2+ ion. Explain why?

Solution

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

Solution

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

Solution

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

[Co(H2O)(CN)(en)2]2+[Co(H_2O)(CN)(en)_2]^{2+}[Co(H2​O)(CN)(en)2​]2+

[CoBr2(en)2]−[CoBr_2(en)_2]^−[CoBr2​(en)2​]−

[PtCl4]2−[PtCl_4]^{2−}[PtCl4​]2−

K3[Fe(CN)6]K_3[Fe(CN)_6]K3​[Fe(CN)6​]

[Cr(NH3)3Cl3][Cr(NH_3)_3Cl_3][Cr(NH3​)3​Cl3​]

Solution

Using IUPAC norms write the formulas for the following:

Tetrahydroxidozincate(II)

Potassium tetrachloridopalladate(II)

Diamminedichloridoplatinum(II)

Potassium tetracyanidonickelate(II)

Pentaamminenitrito-O-cobalt(III)

Hexaamminecobalt(III) sulphate

Potassium tri(oxalato)chromate(III)

Hexaammineplatinum(IV)

Tetrabromidocuprate(II)

Pentaamminenitrito-N-cobalt(III)

Solution

Using IUPAC norms write the systematic names of the following:

[Co(NH3)6]Cl3[Co(NH_3)_6]Cl_3[Co(NH3​)6​]Cl3​

[Pt(NH3)2Cl(NH2CH3)]Cl[Pt(NH_3)_2Cl(NH_2CH_3)]Cl[Pt(NH3​)2​Cl(NH2​CH3​)]Cl

[Ti(H2O)6]3+[Ti(H_2O)_6]^{3+}[Ti(H2​O)6​]3+

[Co(NH3)4Cl(NO2)]Cl[Co(NH_3)_4Cl(NO_2)]Cl[Co(NH3​)4​Cl(NO2​)]Cl

[Mn(H2O)6]2+[Mn(H_2O)_6]^{2+}[Mn(H2​O)6​]2+

[NiCl4]2−[NiCl_4]^{2−}[NiCl4​]2−

[Ni(NH3)6]Cl2[Ni(NH_3)_6]Cl_2[Ni(NH3​)6​]Cl2​

[Co(en)3]3+[Co(en)_3]^{3+}[Co(en)3​]3+

[Ni(CO)4][Ni(CO)_4][Ni(CO)4​]

Solution

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

Solution

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

[Cr(C2O4)3]3−[Cr(C_2O_4)_3]^{3−}[Cr(C2​O4​)3​]3−

[Co(NH3)3Cl3][Co(NH_3)_3Cl_3][Co(NH3​)3​Cl3​]

Solution

Draw the structures of optical isomers of:

[Cr(C2O4)3]3−[Cr(C_2O_4)_3]^{3−}[Cr(C2​O4​)3​]3−

[PtCl2(en)2]2+[PtCl_2(en)_2]^{2+}[PtCl2​(en)2​]2+

[Cr(NH3)2Cl2(en)]+[Cr(NH_3)_2Cl_2(en)]^+[Cr(NH3​)2​Cl2​(en)]+

Solution

Draw all the isomers (geometrical and optical) of:

[CoCl2(en)2]−[CoCl_2(en)_2]^−[CoCl2​(en)2​]−

[Co(NH3)Cl(en)2]2+[Co(NH_3)Cl(en)_2]^{2+}[Co(NH3​)Cl(en)2​]2+

[Co(NH3)2Cl2(en)]+[Co(NH_3)_2Cl_2(en)]^+[Co(NH3​)2​Cl2​(en)]+

Solution

Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)][Pt(NH_3)(Br)(Cl)(py)][Pt(NH3​)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?

Solution

Aqueous copper sulphate solution (blue in colour) gives:

a green precipitate with aqueous potassium fluoride and

a bright green solution with aqueous potassium chloride. Explain these experimental results.

Solution

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)H_2S(g)H2​S(g) is passed through this solution?

Solution

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

[Fe(CN)6]4−[Fe(CN)_6]^{4−}[Fe(CN)6​]4−

[FeF6]3−[FeF_6]^{3−}[FeF6​]3−

[Co(C2O4)3]3−[Co(C_2O_4)_3]^{3−}[Co(C2​O4​)3​]3−

[CoF6]3−[CoF_6]^{3−}[CoF6​]3−

Solution

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

Solution

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

Solution

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

Solution

[Cr(NH3)6]3+[Cr(NH_3)_6]^{3+}[Cr(NH3​)6​]3+ is paramagnetic while [Ni(CN)4]2−[Ni(CN)_4]^{2−}[Ni(CN)4​]2− is diamagnetic. Explain why?

Solution

A solution of [Ni(H2O)6]2+[Ni(H_2O)_6]^{2+}[Ni(H2​O)6​]2+ is green but a solution of [Ni(CN)4]2−[Ni(CN)_4]^{2−}[Ni(CN)4​]2− is colourless. Explain.

Solution

[Fe(CN)6]4−[Fe(CN)_6]^{4−}[Fe(CN)6​]4− and [Fe(H2O)6]2+[Fe(H_2O)_6]^{2+}[Fe(H2​O)6​]2+ are of different colours in dilute solutions. Why?

$FeSO_4$ solution mixed with $(NH_4)_2SO_4$ solution in 1:1 molar ratio gives the test of $Fe^{2+}$ ion but $CuSO_4$ solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of $Cu^{2+}$ ion. Explain why?

$[Co(H_2O)(CN)(en)_2]^{2+}$

$[CoBr_2(en)_2]^−$

$[PtCl_4]^{2−}$

$K_3[Fe(CN)_6]$

$[Cr(NH_3)_3Cl_3]$

$[Co(NH_3)_6]Cl_3$

$[Pt(NH_3)_2Cl(NH_2CH_3)]Cl$

$[Ti(H_2O)_6]^{3+}$

$[Co(NH_3)_4Cl(NO_2)]Cl$

$[Mn(H_2O)_6]^{2+}$

$[NiCl_4]^{2−}$

$[Ni(NH_3)_6]Cl_2$

$[Co(en)_3]^{3+}$

$[Ni(CO)_4]$

$[Cr(C_2O_4)_3]^{3−}$

$[Co(NH_3)_3Cl_3]$

$[Cr(C_2O_4)_3]^{3−}$

$[PtCl_2(en)_2]^{2+}$

$[Cr(NH_3)_2Cl_2(en)]^+$

$[CoCl_2(en)_2]^−$

$[Co(NH_3)Cl(en)_2]^{2+}$

$[Co(NH_3)_2Cl_2(en)]^+$

Write all the geometrical isomers of $[Pt(NH_3)(Br)(Cl)(py)]$ and how many of these will exhibit optical isomers?

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 $H_2S(g)$ is passed through this solution?

$[Fe(CN)_6]^{4−}$

$[FeF_6]^{3−}$

$[Co(C_2O_4)_3]^{3−}$

$[CoF_6]^{3−}$

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

$[Cr(NH_3)_6]^{3+}$ is paramagnetic while $[Ni(CN)_4]^{2−}$ is diamagnetic. Explain why?

A solution of $[Ni(H_2O)_6]^{2+}$ is green but a solution of $[Ni(CN)_4]^{2−}$ is colourless. Explain.

$[Fe(CN)_6]^{4−}$ and $[Fe(H_2O)_6]^{2+}$ are of different colours in dilute solutions. Why?

$K_3[Co(C_2O_4)_3]$

cis- $[CrCl_2(en)_2]Cl$

$(NH_4)_2[CoF_4]$

$[Mn(H_2O)_6]SO_4$

$K[Cr(H_2O)_2(C_2O_4)_2] \cdot 3H_2O$

$[Co(NH_3)_5Cl]Cl_2$

$[CrCl_3(py)_3]$

$Cs[FeCl_4]$

$K_4[Mn(CN)_6]$

Explain the violet colour of the complex $[Ti(H_2O)_6]^{3+}$ on the basis of crystal field theory.

How many ions are produced from the complex $Co(NH_3)_6Cl_2$ in solution?

$[Cr(H_2O)_6]^{3+}$

$[Fe(H_2O)_6]^{2+}$

$[Zn(H_2O)_6]^{2+}$

$[Fe(H_2O)_6]^{3+}$

$[Fe(NH_3)_6]^{3+}$

$[Fe(C_2O_4)_3]^{3−}$

$[FeCl_6]^{3−}$

What will be the correct order for the wavelengths of absorption in the visible region for the following: $[Ni(NO_2)_6]^{4−}, [Ni(NH_3)_6]^{2+}, [Ni(H_2O)_6]^{2+}$ ?

$[Co(NH_3)_6]Cl_3$

$[Co(NH_3)_5Cl]Cl_2$

$K_3[Fe(CN)_6]$

$K_3[Fe(C_2O_4)_3]$

$K_2[PdCl_4]$

$[Pt(NH_3)_2Cl(NH_2CH_3)]Cl$

$K[Cr(H_2O)_2(C_2O_4)_2]$

$[Co(en)_3]Cl_3$

$[Co(NH_3)_5(NO_2)](NO_3)_2$

$[Pt(NH_3)(H_2O)Cl_2]$

Give evidence that $[Co(NH_3)_5Cl]SO_4$ and $[Co(NH_3)_5(SO_4)]Cl$ are ionisation isomers.

Explain on the basis of valence bond theory that $[Ni(CN)_4]^{2−}$ ion with square planar structure is diamagnetic and the $[NiCl_4]^{2−}$ ion with tetrahedral geometry is paramagnetic.