NCERT Solutions for Class 12 Chemistry Chapter 5 – Coordination Compounds

Intext Questions with Solutions of Class 12 Chemistry Chapter 5 – Coordination Compounds

5.1

Ans – (i) [CO(NH₃)₄(H₂O)₂]Cl₃.
(ii) K₂[Ni(CN)₄]
(iii) [Cr(en)₃]Cl₃
(iv) [Pt (NH₃) Br Cl (N0₂)]^–
(v) [PtCl₂(en)₂](N0₃)₂
(vi) Fe₄[Fe(CN)₆]₃

5.2

Ans – (a) hexaamminecobalt (III) chloride
(b) pentaamminechloridocobalt (III) chloride
(c) potassium hexacyanoferrate (III)
(d) potassium trioxalatoferrate (III)
(e) potassium tetrachloridoplatinum (II)
(f) diamminechlorido (methylamine) platinum(II) chloride.

5.3

Ans – (i) Both geometric isomers (cis and trans) can exist for this compound.

The trans-isomers of this compound is optically inactive, while the cis-isomer is is optically active. Hence, it can demonstrate optical isomerism, as illustrated below:

(ii) It can show optical isomerism which is given below:

(iii) This compound can show three types of isomerism:
– Optical isomerism is given below,

– Linkage isomerism
[Co(NH₃)₅(NO₂)](NO₃)₂ and [Co(NH₃)₅(ONO)](NO₃)₂
– Ionization isomerism
[Co(NH₃)₅(NO₂)](NO₃)₂ and [Co(NH₃)₅(NO₃)](NO₃)(NO₂) 

(iv) Both geometric isomers (cis and trans) can exist for this compound.

5.4

Ans – When dissolved in water, they produce distinct ions in the solution that can be identified by adding AgNO₃ solution and BaCl₂ solution.

Therefore the both are are ionisation isomers.

5.5

Ans – The famous researcher’s Heitler & London developed the valence bond theory (VBT) to describe how chemical bonds, atomic orbitals & elemental electronic configurations occur. Through the procedure of hybridization, the theory reveals that, “Electrons within a molecule strive to reside in atomic orbitals rather than molecular orbital parameters.” The electrical arrangement of molecules is described by valence bond theory. A robust bond of chemical substance is created when atomic orbitals overlay one another. Because they intersect, the electrons are confined to the bonding area. The molecule’s bonding capability increases with the amount of overlapping.

The external electronic combination of nickel element (Z = 28) in the ground level zone is depicted as 3d⁸4s². Whereas, the element Nickel in this complex structure is available as + 2 oxidation state. It also attains a + 2 oxidation state by losing the 2 4s-electrons in the due course. The outcome of Ni²⁺ ion contains an outer electronic structure consisting of 3d⁸. It’s because the CN^– ion is a robust field and during its attacking nature, 2 electrons without any pairs in the 3d orbitals get paired with each other.

Nickel (Z = 28) ‘s external electronic configuration present in the ground state can be defined as 3d⁸4s². Meanwhile, Nickel in this available structure lies in a + 2 oxidation state. Nickel reaches this desired state by opting to lose two of its 4s-electrons. This ends up being Ni²⁺ ions with external electronic combinations of 3d⁸ orbitals around them. As we know, the CP ion is a weak link of ligand, it doesn’t possess the desired ability to create any such electron pairing module.

5.6

Ans – According to the subject matter, the chemical is a tetrahedral combination since it consists of four ligands. Because Cl- is an intense field ligand, it won’t induce the 3d electrons to couple.
Consequently, sp³ hybridization will occur. Because of the unpaired nature of 2 electrons, [NiCl₄]^2- is a paramagnetic compound. Ni is located in the zero oxidation condition in [Ni(CO)₄], meaning that its chemical structure is 3d⁸ 4s². It is provided the one that follows:

CO, however, is a potent field ligand. This results in the coupling of unpaired 3d electrons. Additionally, it causes the 4s to migrate towards the 3d orbital, which causes sp³ to hybridize. Since there aren’t any unpaired electrons in existence, [Ni(CO)₄] is diamagnetic in this scenario.

5.7

Ans – Fe is the main metal ion in the connections that are provided, and in both situations, its oxidation state is +3. It is going to be in a d⁵ arrangement as a result.

CN couples with unpaired electrons because it is an extremely powerful field ligand. As a result, the d-orbital contains just a single unattached electron.

Therefore, the force of magnetic pull may be computed as follows:

H₂O fails to couple with unpaired electrons because the substance is a fragile field ligand. The d-orbital has 5 sets of electrons which are unpaired.

Thus, the force of magnetic attraction may be computed as follows:

As a result, the combined state of [ Fe(CN)₆ ]³⁺ is slightly paramagnetic, whereas [ Fe(H₂O)₆ ]³⁺ is substantially paramagnetic. 

5.8

Ans – Co is present in the +3 state of oxidation in [Co(NH₃)₆]³⁺, which results in an arrangement of Co to be d⁶.

NH₃ pairs electrons that are unpaired because it is an exceptionally potent field ligand. The result of this process will be d²sp³ hybridization. The following is seen in the sections that follow.

It represents an interior orbital group as a result. The oxidation state of Ni occurs in the +2 oxidation state in [Ni(NH₃)₆]²⁺, which results in an arrangement of Ni as that of d⁸.

NH₃ pairs electrons that are unpaired because it is a strong field ligand. Only one d-orbital remains, therefore it is unable to construct the d²sp³. It consequently acts as an insignificant field ligand. Consequently, sp³d² hybridization will occur. The reason for this is displayed in the following order:

Hence, this type of structure is defined as an external orbital arrangement.

5.9

Ans – The compound in question has emerged as [Pt(CN)₄]^2-. Platinum is the main metal ion associated with this combination because it is in the oxidation state +2. It creates a flat square architecture. This indicates that dsp² hybridization is going to happen. Thus, 5d⁸ is the electrical composition Pt (+2).

Every single one of the electrons that are unpaired in the molecule will couple as a result of the powerful magnetic field ligand CN^–. As a result, there won’t be any electrons and no pairings.

5.10

Ans – The arrangement of electronics of Mn in [Mn(H₂O)₆]²⁺ is d⁵, which indicates that it stands associated with the +2 oxidation state. It has an octahedral crystal pattern. An inadequate field ligand comprises water. Consequently, the arrangement of electrons in [Mn(H₂O)₆]²⁺ is expected to be t²g³ eg².
The electronic structure of Mn in [Mn(CN)₆]^2- is d⁵, which implies that it belongs located in the +2 oxidation state. It has an octahedral crystal configuration. A potent field ligand contains cyanide. As a result, the electron configuration in [Mn(CN)₆]^2- tends to be considered as t²g⁵ eg⁰.

5.11 (Currently removed from NCERT textbook)

Ans – As we all know, the total stability constant that was provided is depicted as (β₄) = 2.1 × 10¹³.
The detachment of the whole compound in the equilibrium constant is the inverse of the total stability constant. This information is given in the ones that follow:

Exercise Questions with Solutions of Class 12 Chemistry Chapter 5 – Coordination Compounds

5.1

Ans – A Swiss-based scientist, Alfred Werner (1866-1919) was the very first to develop the framework of coordination compounds. He synthesized and characterized an enormous amount of coordination particles, and then examined their physical & chemical behaviour via standard methods of investigation. Werner postulated the existence of a primary & secondary polarity in metal ions.

Several ligands—ions or molecules containing a few pairs of electrons that are potentially exchanged with the metal—are joined to a central metal atom or ion to form a metal group.

In 1823, he proposed the above hypothesis to explain the isomeric characteristics of ammonia & cobalt chloride as well as the structure & creation of complex or coordination substances. Arguably the most significant characteristic of metallic substances is their ability to act like Lewis acids & build clusters with a range of bases that act as Lewis acids.

Being known as the founder of coordination chemistry, Werner was the very first inorganic chemist to win the Nobel Prize for Chemistry in 1913.

The 1st hypothesis to describe the characteristics of linking in coordination compounds was Werner’s hypothesis. The primary tenets of Werner’s concept are as follows:
(i) Coordinated compound elements have 2 different kinds of valencies such as primary & secondary valencies.
(ii) Ionizable main valencies are ions that remain negative. It shows the path that is dotted.
(iii) The metallic ion oxidation frequency is equal to the primary charge. Both neutral ions & non-ionizing secondary valencies are satisfied. It’s represented with a solid path.
(vi) The number for the coordination of metal ions is referred to as the secondary polarity.
(v) In the area allotted to a certain topology of the coordination complex, those valencies extend in a designated vector.

5.2

Ans – A double salt also referred to as Mohr’s salt is created when mixtures of FeSO₄ & (NH₄)₂SO₄ are combined in a 1:1 molar proportion. The chemical equation for it is FeSO₄.(NH₄)₂SO₄.6H₂O. The sodium chloride breaks down in aqueous solution in the following manner:

Every particle of the ions and other substances comprising Fe²⁺ ions are tested in the resulting solution. Conversely, a complicated substance [Cu(NH₃)₄]SO₄ is created once CuSO₄ and NH₃ are combined in solution with a molecular concentration of 1: 4. Cu2+ ions won’t provide their distinctive assessments as Fe²⁺ ions do given that they are component of the complex structure (contained in double brackets).

5.3

Ans – The following explains each of them. For example,

(i) Coordinating entity
A coordination entity is made up of a core metal atom or particle coupled to a certain amount of ligands, which are ions or molecules.
For instance, check the below equations:
[Ni(NH₃)₆]²⁺, [Fe(CN)₆]⁴⁺ can be denoted as cationic complexes.
[PtCl₄]^2-, [Ag(CN)₂]^– are known as anionic complexes.
[Ni(CO)₄], [Co(NH₃)₄Cl₂ ] are called neutral complexes.

(ii) Ligand
The uncharged molecules or cancelled ions that envelop the metal atom in a coordinated substance are known as ligands. These ligands typically have two electrons in the valence state, no less than one of which is unused, and are polarized. H₂O, CN^–, NH₃, CO, and so on are examples of ligands.

(iii) Coordination number
The overall number of ligands (impartial or unfavourable ions) attached to the central-metal atoms in the coordinated ring is known as the quantity of coordination that surrounds the central-metal molecule. It is renowned for its ligancy as well.
For instance, the core metal ion in K₂[PtCl₆] is platinum, whereas the ligands are chloride particles. The coordination coefficient is six as it comprises six ligands.

(iv) Coordination polyhedron
The dimensional structure of the ligands that are effectively connected to the predominant metal ion in the supporting field can potentially be characterized as coordinated polyhedrons around the fundamental atom.
A square planar, for instance, is a coordinating polyhedron.

(v) Homoleptic compound
These constitute complexes in which just one kind of donor function is connected to the metal ion. For instance, a homoleptic compound is [PtCl₆]².

(vi) Heteroleptic group
These arrangements are those in which the metal ion is connected to many donor groups. A particular instance of a heteroleptic group is [Co(NH₃)₄Cl₂]⁺.

5.4

Ans – A coordination mixture is created when a molecule or ion known as a ligand donates two electrons to a dominant metal atom or ion in order to create a coordinated covalent connection. In order to determine the framework, reaction time, and characteristics of coordination compounds, ligands are essential. Here is an explanation of different types of ligands:

(i) Unidentate ligand
The kind of molecule that possesses the capacity to create a coordinate relationship with a single donor atom is known as unknown. Unidentate ligands are molecules that have a single donor region. Cl^– and NH₃, for instance, are not the same ligand.

(ii) Bidentate ligand
A bidentate ligand consists of 2 donor atoms, which indicates that two atoms with the capacity equipped to donate their single pair of electrons tend to be available. Bidentate ligands are compounds that offer two pairs of electrons to a metal atom. For instance, the oxalate ion & ethane-1,2-diamine. All of these are listed down below:

(iii) Ambidentate ligand
Ambidentate ligands are binding molecules that have the ability to bind to the primary metal atom via 2 distinct atoms. For instance, the SCN & NO₂ groups. The nitrogen & oxygen atoms are capable of attacking the NO₂ association. The nitrogen and sulphur atoms can assault the SCN molecule.

5.5

Ans – (i) Assume that Co has a coordination number of x.
As a result, we can write:
x + 0 + (-1) + 2 (0) = +2
⇒ x – 1 = +2
⇒ x = +3
Thus, cobalt has a coordination number of +3.

(ii) Assume that Pt has a coordination number of x.
As a result, we can write:
x + 4 (-1) = -2
⇒ x = +2
Thus, platinum has a coordination number of +2.

(iii) Assume that Cr has a coordination number of x.
As a result, we can write:
x + 0 + 3 (-1) = 0
⇒ x – 3 = 0
⇒ x = +3
Thus, chromium has a coordination number of +3.

(iv) Assume that Co has a coordination number of x.
As a result, we can write:
x + 2 (-1) + 0 = +1
⇒ x – 2 = +1
⇒ x = +3
Thus, cobalt has a coordination number of +3.

(v) Assume that Fe has a coordination number of x.
As a result, we can write:
x + 6 (-1) = -3
⇒ x – 6 = -3
⇒ x = +3
Thus, iron has a coordination number of +3.

5.6

Ans – (a) [Zn(OH)₄]^2-
(b) [Pt(NH₃)₆]⁴⁺
(c) K₂[PdCl₄]
(d) [Cu(Br)₄]^2-
(e) [CO(NH₃)₆]₂ (SO₄)₃
(f) K₂[Ni(CN)₄]
(g) K₃ [Cr(OX)₃]
(h) [CO(NH₃)₅ONO]²⁺
(i) [Pt(NH₃)₂Cl₂]
(j) [CO(NH₃)₅NO₂]²⁺

5.7

Ans – (i) Hexaammine cobalt (III) chloride.
(ii) Diammine chlorido (methylamine) platinum (II) chloride.
(iii) Hexaaquatitanium (III) ion.
(iv) Tetraammine chlorido nitrito-N-cobalt (IV) chloride.
(v) Hexaaquamanganese (II) ion.
(vi) Tetrachloridonickelate (II) ion.
(vii) Hexaammine nickel (II) chloride.
(viii) Tris (ethane -1,2-diamine) cobalt (III) ion.
(ix) Tetra carbonyl nickel (0).

5.8

Ans – When two or more substances exhibit an identical chemical composition but distinct fundamental formulae and characteristics, this is known as isomerism. The primary cause of such instances is due to disparate spatial or structural configurations. Now, let us examine all the different forms of isomerism. Presented below is a discussion of them with illustrations.

Geometrical isomerism:

Stereoisomers with a distinct configuration of bands or atoms surrounding pairs of bonds are known as geometrical isomers. Cis-trans isomerism is another name for geometrical isomerism. Geometrical isomerism is common in heteroleptic compounds. The style is caused by the different geometric configurations of the ligands. For instance:

Optical isomerism:

When two compounds have exactly identical structural & molecular formulas but are unable to be combined with one another, this is known as optical isomerism. Optical isomers are merely mimic representations of one another. This type of isomerism is seen in asymmetric substances. Isomers cannot be superimposable. Rather, they are mirrored versions of one another. For instance:

Linkage isomerism:

In chemistry, linkage isomerism is a kind of isomerism that happens whenever substances that coordinate have an identical component but change in the connection between a ligand & a metal atom. Such isomerism happens whenever the coordinated ring contains an ambidentate ligand. For instance, [Co(NH₃)₅(NO₂)] (NO₃)₂ & [Co(NH₃)₅(ONO)] (NO₃)₂

Coordination isomerism:

Whenever the ligands connected to a metal ion in a substance of coordination molecules alter, coordinated isomerism—a kind of structural isomerism—occurs. These ligands of different metal ions in a compound are switched between cationic and anionic organizations, Coordination isomerism takes place. For instance, [Co(NH₃)₆] [Cr(CN)₆] & [Cr(NH₃)₆] [Co(CN)₆].

Ionization isomerism:

Whenever a substance has a comparable structure but generates distinct ions in a liquid state, this is known as ionization isomerism. It takes place when ions move back and forth within a coordination field. This type of isomerism occurs when the counter-ion takes over the role of the ligand within the coordination sphere.

Ionization isomers are therefore referred to as combinations since they share an identical composition nevertheless, whenever suspended in water, they produce different ions. For instance, [Co(NH₃)₅(NO₂)] (NO₃)₂ & [Co(NH₃)₅(NO₃)] (NO₃)(NO₂).

Solvate isomerism:

Compounds with a comparable substance but different solvent molecular attachments to metal ions are known as solvent isomerism. This phenomenon is known as hydrate isomerism whenever the solvent being used is water solvent.

Solvent isomers vary in that the chemical solvent molecular structure is either in the crystalline lattice as an independent solvent agent or is directly attached to the metal ion. For example, [Cr(H₂O)₆] Cl₃ & [Cr(H₂O)₅Cl] (H₂O)Cl₂.

5.9

Ans – (i) For this complex, no geometrical isomers can occur, as there is only a single type of ligand present, which is a bidentate ligand.

(ii) For this compound, two geometrical isomers can be formed: the Facial form and the Meridional form. These are shown below:

5.10

Ans – (i)

(ii)

(iii) The below structure is of the cis-form

The below structure is of the trans-form

5.11

Ans – (i)

(ii)

(iii)

5.12

Ans – Three isomers that are possible will be:

No one will exhibit optical isomers from the provided isomers. It is rare for tetrahedral compounds to undergo optical isomerization. This only occurs in the presence of asymmetric chelating agents.

5.13

Ans – Because of the [Cu(H₂O)₄]²⁺ ions, the water-soluble CuSO₄ solution is [Cu(H₂O)₄]SO₄, a substance with a colour that is blue.

(i) The addition of KF causes the less powerful H₂O ligands to be substituted with F^– ligands, resulting in the green deposit [CuF₄]^2- ions.

(ii) The weakened H₂O ligands are replaced by Cl^– ligands upon the addition of KCl, resulting in the formation of the vivid green [CuCl₄]^2- ion.

5.14

Ans – Cuprous cyanide and cyanogen air are produced when the first cupric cyanide breaks down. The mixture is created when too much potassium cyanide combines with cuprous cyanide K₃[Cu(CN)₄],

Therefore, [Cu(CN)₄]^3- is the coordinated molecule that was created in the resultant process mentioned as stated earlier. The structure of the complex ion is extremely persistent and cannot break apart or ionize to produce Cu²⁺ ions because CN^– is a powerful ligand. As a result, none of the precipitates containing H₂S forms.

CuSO₄(aq) + 4KCN(aq) → K₂[Cu(CN)₄](aq) + K₂SO₄(aq) 
i.e., [Cu(H₂O)₄]²⁺ + 4CN^– → [Cu(CN)₄]^2- + 4H₂O 

5.15

Ans – (i) Fe is within the +2 oxidation state in [Fe(CN)₆]⁴⁻, which ends up in the arrangement of Fe to be d⁶.

Because it acts as an extremely powerful magnetic ligand, CN^– pairs electrons that are unpaired. The result of this will be d²sp³ hybridization. Everything is displayed in the one that follows:

Consequently, the resulting structure is diamagnetic and has a structure that is octahedral.

(ii) Fe lies within the +3 oxidation condition in [Fe(F)₆]^3-, which results in an arrangement of Fe to be d⁵.

Electron coupling is not induced by F^– because the latter is a weak magnetic ligand. Consequently, sp³d² hybridization is going to happen. The reason for this is displayed below:

This structure is paramagnetic & has a geometry called an octahedral.

(iii) Co exists in the +3 state of oxidization in [Co(C₂O₄)₃]^3-, which results in the physical form of Co as that of d⁶.

Oxalate pairs electrons that are unpaired because it is an extremely powerful potential ligand. The result of this process will constitute d²sp³ hybridization. The reason behind this is displayed below:

Consequently, the resulting structure is diamagnetic and has an octahedral geometry.

(iv) The structure and arrangement of Co is going to be d⁶ in [CoF₆]^3-since the oxidation process of Co corresponds to the oxidation state of +3.

Electron collaboration is not induced by F^– since it is a moderate space ligand. Consequently, sp³d² hybridization will occur. The underlying structure is paramagnetic and has a structure that is octahedral.

5.16

Ans –

5.17

Ans – An ordered collection of metallic ions & ligands arranged according to their respective oxidation state and endurance is called a spectrochemical series. The design is employed to determine the minimal or high-spin nature of a coordination compound.
The accusation brought on by the ion of metal & the magnetic field generated by the ligand determines the magnetic field of the crystal breaking down, or ∆₀. Certain ligands can generate powerful fields, which at first will end up in a significant separation.
Meanwhile, other sorts of ligands may generate moderate fields, which will lead to a minor separation of d-orbitals. As seen further down, ligands may generally be stacked in a sequence that improves electromagnetic field strength in the following sequence:
I^– < Br^– < SCN^– < Cl^– < S^2- < F^– < OH^– < C₂O₄^2- < H₂O < NCS^– < edta^4- < NH₃ < en < CN^– < CO

5.18

Ans – Being formulated in 1929, crystal field theory views the relationship between an ion of metal & ligand as solely electrostatic, with the ligands being seen as points of charge close to the core atom’s orbital coordinates. Crystal field theory governs how these ligands communicate with the core metal atom or ion. Subsequently, it’s been referred to as ligand field theory.
These d-orbitals separate into 2 distinct sets, one with a greater potential and the other set having a smaller power whenever the ligands get closer to an intermediate metal ion. Crystal field splitting energy (Δ₀ to supply octahedral force) is the amount of power that differentiates the two distinct pairs of orbitals. Strong spin compounds are formed when the 4th electron occupies one of the two e°g orbitals, resulting in the arrangement t³_2ge¹_g when Δ₀ < P (pairing potential).
These ligands are known to be fragile field ligands because Δ0 < P. Weak spin complexes are formed when the 4^th electron couples together with a single of the two t_2g orbital structures, resulting in an arrangement known as t⁴_2ge¹_g if Δ₀ > P. Robust field ligands are compounds with a coefficient Δ₀ > P.

5.19

Ans – The arrangement of Cr is going to be d³ in [Cr(NH₃)₆]³⁺ since the oxidized form of Cr is in the oxidation state of +3.

NH₃ is not involved in electron coupling because the substance is a poor field ligand. The result of this process happens to be d²sp³ hybridization.

The structure is paramagnetic & has a structure that is octahedral. The chemical structure of Ni is expected to be d8 in [Ni(CN)₄]^2- since the oxidation state of Ni corresponds to the +2 oxidation state.

Because it functions as an effective magnetic ligand, CN^– pairs electrons that are unpaired. The following will result in the integration of dsp². The reason for that is displayed below:

As a result, the framework’s topology is diamagnetic in addition to the squared plane.

5.20

Ans – [Ni(H₂O)₆]²⁺ has 2 electrons with no pairing that fail to couple when the weakly charged H₂O ligand is present. Ni exists within the + 2 oxidation state and has a 3d⁸ electronic arrangement. The subject is coloured as a result. The d-d evolution absorbs the red spectrum, while green light is released as a supplementary illumination.
Additionally, Ni has a 3d⁸ electronic structure and exists in the +2 oxidation state in [Ni(CN)₄]^2-. However, the 2 electrons that are not paired in the third orbitals couple up when an extremely powerful ligand, CN^–, is present. Therefore, there isn’t a single electron that is unpaired. It is transparent and as a result, it remains colourless.

5.21

Ans – Fe has a d⁶ geometry and is at the +2 oxidation state within the two compounds. The design possesses 4 unpaired electrons, which are classified according to this subject. Distinct comparative places in the spectrochemical sequence are occupied by ligands such as H₂O & CN^– ion.
The crystal field splitting energy (∆₀) varies between them. It goes without saying that they collect energy absorbed by the naked light spectrum that corresponds to various wavelengths or vibrations. The emitted colours and (VIBGYOR) are likewise not the same. This indicates the reality that compounds’ colours in substances might vary.

5.22

Ans – The metallic carbon covalent (M – C) in metallic carbonyl has two distinct σ & π -bond characteristics. The atom-based orbital shapes of the metallic element and the carbon monoxide C-atom intersect in the manner that follows to create a connection:

(a) Whenever an atom of any unoccupied d-orbital crosses across with an orbital that contains one individual electron pair on the carbon monoxide C-atom (: C = O:), the initial σ-bond involving the metal and carbon is created.
(b) The electrons that emerge from the occupied d-orbitals of a transitioning metallic atom or ion are returned to the anti-bonding π-orbitals of carbon monoxide, in conjunction with the σ-bond in metallic carbonyl. The metal-based ligand link is stabilized as a result. The remainder of this chart illustrates the two distinct ideas mentioned in the past.

5.23

Ans – (i) K₃[Co(C₂O₄)₃] => [CO(C₂0₄)₃]^3-. x + 3 (-2) = -3.
Degree of oxidation, x = +3 Since C₂O₄^2- is bidentate, the coordinating factor also happens to be 6.
In the circumstance of Co+3, every single electron is coupled.

(ii) cis – [Cr(en)2Cl₂] + Cl^–
x + 0 – 2 = +1
The degree of oxidation tends to be x = +3
Given that “en” is bidentate, the assigned degree is 6. Cr³⁺ is a paramagnetic cfi instance.

(iii) (NH₄)₂[COF₄] = (NH₄)₂²⁺[COF₄]^2-
x – 4 = -2.
Four will be the coordination element.
Oxidative state remains x = + 2
Co²⁺ is a paramagnetic d⁵ instance.

(iv) [Mn(H₂0)₆]²⁺S0₄^2-
x+0f+2
Therefore, the degree of oxidation, x- + 2
Six is the coordination ratio.
Mn⁺² is a paramagnetic d⁵ instance.

5.24

Ans –

5.25

Ans – The level of significance of collaboration among the two different species that determine the state of equilibrium is known as the level of stability of a coordination complex (aq). The connective stability is expressed quantitatively by the equilibrium constant’s strength (the durability or creation). It is a measurement of how strongly the metallic ion & the substance’s ligands are connected.
A normal equilibrium event occurs when an intricate system forms in solution using this combination. This can be shown as that of M + 4L ⇌ ML₄. The complex’s durability is gauged by the reaction’s optimum frequency.
The stabilization constant could also be described as the equilibrium constant associated with a reversal response. This event is another name for the equilibrium constant. It is likewise possible to write the aforesaid complex’s growth through the following procedures:

The formula for the stability parameter becomes β₄ = K₁K₂K₃K₄.
The more powerful the metal-ligand connection, the higher the degree of stability constant. The following factors will determine the level of stability of the compound since it consists of:

1. Metallic character.
2. Metallic oxidation state.
3. Ligand nature, such as chelating ligands forming robust compounds.
4. The stability will increase with the ligand’s fundamental potential.

5.26

Ans – The enhanced equilibrium of a metallic complex that results from a polydentate ligand binding to a metal ion is known as the chelate phenomenon. In contrast to unidentate ligands, the combination is more durable because the ligand creates a circular configuration surrounding the metal ion. The aforementioned term additionally stands for coordination complex stabilization factor.
The chelate phenomenon happens whenever the donation atoms in a bidentate or polydentate ligand are arranged so that, upon coordinating through the primary metal ion, a 5 or 6-membered loop is created. For instance,

5.27

Ans – (i) When it comes to biological structures, coordinating molecules are crucial. Chlorophyll, the chemical compound that is in charge of the photosynthesis process, is a magnesium-based coordinated molecule. Iron coordinates to form haemoglobin, the red hue 1 of blood that carries oxygen in the air.
Cobalt is a coordinating component of B12 vitamin, and it contains cyanocobalamine, the anti-pernicious anemia agent. Proteins such as carbonic anhydrase and carboxypeptidase A are examples of additional biologically significant molecules that include coordinated metallic ions (stimulants that facilitate biological processes).

(ii) The application of chelate treatment in therapeutic Chemistry is gaining attention. The management of issues brought on by hazardous levels of metallic substances in plant and animal ecosystems serves as a prime instance.
Therefore, through the creation of coordination molecules, the chelating binding substances D-penicillamine and desferrioxime B eliminate excessive iron and copper from the body. Lead toxicity can be treated using EDTA. Certain platinum coordination chemicals efficiently stop tumor development. Several instances involve ds-platin & its derivatives.

(iii) Numerous qualitative as well as quantitative chemical investigations make inquiries regarding coordination molecules. The identification and estimate of metallic ions using both conventional and instrumental approaches of examination are based on the well-known colour responses that they produce with various ligands, particularly chelating ligands, due to the production of coordination units. EDTA, DMG (dimethylglyoxime), a-nitroso-β-naphthol, cupron, and others are illustrations of these types of compounds.

(iv) Complex synthesis is used in several significant metal mining methods, such as the ones for the recovery of gold and silver. For instance, in an aqueous state, gold and cyanide interact to create the coordination phenomenon [Au(CN)₂]^– when oxygen and water are present. Zinc is capable of being added to this solvent to break down the gold that remains in metal condition.

5.28

Ans – The coordination number of cobalt is 6
⇒ the complex will be [Co(NH₃)₆]Cl₂
When ionization occurs, it produces three different ions. The following reaction illustrates this process:

5.29

Ans – Number of unpaired electrons in [Cr(H₂O)₆]³⁺ = 3. Hence, magnetic moment can be calculated as,

Number of unpaired electrons [Fe(H₂0)₆]²⁺ = 4. Hence, magnetic moment can be calculated as,

Number of unpaired electrons [Zn(H₂0)₆]²⁺ = 0. Hence the magnetic moment will be 0(since n = 0).
Thus, (ii) [Fe(H₂0)₆]²⁺ has highest value of magnetic moment.

5.30

Ans – Fe appears within the +3 state of oxidation throughout all the complexes that are provided. Since C₂0₄^2- is a bidentate binding ligand, it creates chelate loops; as a consequence, (iii) is perhaps the most persistent compound of the ones listed in the question.

5.31

Ans – Boosting field abilities or the ligands’ CFSE levels derived from the spectro-chemical series, correspond to the following sequence as the metallic ion is secured: H₂O < NH₃ < NO₂^–. Consequently, the following is the sequence of the energy collected for stimulation:
[Ni(H₂O₆)]²⁺ < [Ni(NH₃)₆]²⁺ < [Ni(NO₂)₆]^4-
Since E = hc/λ, the wavelengths that are absorbed will be in reverse order.

Related Study Resources of Chapter 5 – Coordination Compounds

Students can use the links below to get extra study materials for Class 12 Chemistry Chapter 5: Coordination Compounds

Sl No.	Related Links
1	Class 12 Chemistry Chapter 5 Coordination Compounds – Important Questions

NCERT Solutions for Class 12 Coordination Compounds

Coordination Compounds is a crucial chapter in Class 12 Chemistry, covering topics like Werner’s theory, coordination number, ligand types, and crystal field theory. Our NCERT Solutions for Class 12 Chemistry Chapter 5 provide step-by-step explanations, helping students understand complex topics with ease. These solutions are designed as per the latest NCERT syllabus, making them ideal for board exam preparation and competitive exams like JEE and NEET.

With our Class 12 Chemistry Chapter 5 NCERT Solutions, students can master topics such as isomerism in coordination compounds, valence bond theory, and the stability of complexes. The detailed answers help in grasping reaction mechanisms and electronic configurations effectively. Practicing these solutions ensures better conceptual clarity and enhances problem-solving skills, making it easier to tackle exam questions with confidence.