Class 12 Chemistry Chapter 5 Coordination Compounds

Key Concept: Weak vs Strong Ligands

d) It does not distinguish between weak and strong ligands.
[Solution Description] Valence Bond Theory cannot distinguish between weak and strong ligands, as it does not account for differences in ligand field strength that influence electronic configurations.

Key Concept: Ligand Type

b) Bidentate
[Solution Description] Ethylenediamine (NH$_2$CH$_2$CH$_2$NH$_2$) has two donor nitrogen atoms that can coordinate with a metal center, making it a bidentate ligand. A bidentate ligand can form two bonds with the central atom/ion.

Key Concept: Structural Differences

c) $[Co(NH_3)_5(H_2O)](NO_3)_2\cdot H_2O$
[Solution Description] A solvate isomer differs in the arrangement of solvent molecules, either directly bonded to the metal center or as free molecules. In this case, we have $[Co(NH_3)_5(NO_3)](NO_3)$, where the nitrate ion is outside the coordination sphere. For its solvate isomer, one possible structure could be $[Co(NH_3)_5(H_2O)](NO_3)_2 \cdot H_2O$, where water replaces the nitrate ion inside the coordination sphere.

Key Concept: Optical Activity

b) Cis-isomer
[Solution Description] Only the cis-isomer of the complex [PtCl$_2$(en)$_2$]$^{2+}$ shows optical activity as it can exist in two non-superimposable mirror image forms.

Key Concept: Geometrical Isomerism

c) $[\text{Pt(NH}_3)_2\text{Cl}_2]$

[Solution Description] The complex $[\text{Pt(NH}_3)_2\text{Cl}_2]$ can exhibit geometrical isomerism because it has a square planar structure, allowing for cis and trans configurations. For example, if both $\text{NH}_3$ groups are adjacent, it is a cis isomer; if they are opposite, it’s a trans isomer.

Key Concept: Nomenclature Rules

b) Tetraamminedichloridopalladium(II)

[Solution Description] According to nomenclature rules, the name begins with the ligands followed by the metal. The ligands are named alphabetically. Here, we have ammine (NH$_3$) and chloride (Cl). Therefore, the name will start as ‘Tetraammine’ because there are four ammines, followed by ‘dichlorido’ for the two chlorides. Finally, add the metal palladium and its oxidation state which can be +2 here. So, the name becomes Tetraamminedichloridopalladium(II).

Key Concept: Geometry Prediction

b) Tetrahedral
[Solution Description] To determine the geometry, we need to consider the coordination number which is 4 for [Zn(NH$_3$)$_4$]$^{2+}$. This coordination number typically leads to a tetrahedral or square planar geometry. However, zinc(II) complexes with ammine ligands usually adopt a tetrahedral geometry, so the geometry is tetrahedral.

Key Concept: Ambidentate Ligands

b) SCN$^-$
[Solution Description]
Ambidentate ligands can attach to the central metal atom/ion through two different atoms but not simultaneously. NO$_2^-$ is an example of an ambidentate ligand, as it can bind through either nitrogen or oxygen.

Key Concept: Valence Identification

b) Primary valence = 3; Secondary valence = 6
[Solution Description] The primary valence corresponds to the charge on the complex ion, which is satisfied by negative ions like SO$_4^{2-}$. In this case, the cobalt ion has a +3 oxidation state, so the primary valence is 3. Secondary valences refer to the coordination number, which is determined by the ligands attached to the metal through coordinate bonds. Here Co is coordinated with 5 NH$_3$ molecules and 1 Cl$^-$ ion, providing a total of 6 coordination sites.

Key Concept: Structure Prediction

b) Tetrahedral with $sp^3$ hybridization
[Solution Description] According to VBT, $[NiCl_4]^{2-}$ is formed by hybridization involving $sp^3$, resulting in a tetrahedral geometry. Nickel in this compound has an electron configuration of $3d^8 4s^0$. Chloride ions are weak field ligands and do not cause pairing of electrons in the d-orbitals. Thus, the hybridization involves four orbitals: one s and three p, forming $sp^3$ hybrid orbitals with a tetrahedral arrangement.

Key Concept: Electroplating

c) $[Ag(CN)_2]^–$
[Solution Description] Electroplating involves depositing a layer of metal onto a surface using an electric current. Coordination compounds like $[Ag(CN)_2]^–$ are used because they provide a source of silver ions in solution, essential for uniform and controlled deposition during electroplating.

Key Concept: Tetrahedral Complexes

a) Due to identical spatial arrangement
[Solution Description] Tetrahedral complexes do not exhibit geometrical isomerism because all four positions are equivalent due to their identical spatial arrangement, thus making any potential isomer indistinguishable.

Key Concept: Complex Nomenclature, Isomerism in Coordination Compounds

c) Diammine-dibromo-dichloroplatinate(IV)
[Solution Description]
The given complex formula corresponds to an anionic complex ion. In naming, we begin with the ligands listed in alphabetical order followed by the metal. The prefixes denote the number of each type of ligand. Here:

– Ammonia (ammine) is neutral: $2 \rightarrow diammine$.
– Chloride is negative: $2 \rightarrow dichloro$
– Bromide is negative: $2 \rightarrow dibromo$.

Platinum, being the central metal in its anionic form, is named as ‘platinate’. Thus, the full name is derived as follows:

“Diammine-dibromo-dichloroplatinate(IV)” because Pt is in the +4 oxidation state (each halogen contributes -1).

Due to the possibility of arranging these ligands ($NH_3$, $Cl^-$, $Br^-$) around Pt in different spatial arrangements, it can exhibit geometric isomerism (cis/trans). Optical isomerism arises when cis-form creates chiral centers.

Hence, the accurate IUPAC name is Diammine-dibromo-dichloroplatinate(IV).

Key Concept: Magnetic Property

a) Paramagnetic
[Solution Description] The complex [CoF$_6$]$^{3-}$ has cobalt in a +3 oxidation state, resulting in a 3$d^6$ electron configuration. Fluoride is a weak field ligand, leading to no pairing of electrons within the $t_{2g}$ set. Therefore, unpaired electrons remain, making [CoF$_6$]$^{3-}$ paramagnetic.

Key Concept: Coordination Number and Shape

b) A coordination number of 4 can result in both tetrahedral or square planar shapes.
[Solution Description] Generally, a coordination number of 4 can lead to either a tetrahedral or square planar shape. A coordination number of 6 usually leads to an octahedral shape.

Key Concept: Application

a) $[Co(NH_3)_6][Cr(CN)_6]$
[Solution Description]
Coordination isomerism requires two different metal ions with interchangeable ligands. The given complex must have this property. Among the options, $[Co(NH_3)_6][Cr(CN)_6]$ has cobalt and chromium as different metal centers and suitable ligand sets that can be interchanged.

Key Concept: Ligand Naming

c) Carbonyl
[Solution Description] The ligand CO is named as “carbonyl” in coordination chemistry due to its neutral nature and specific historical naming convention.

Key Concept: Basic Concepts

b) It forms complexes with metal ions
[Solution Description] EDTA acts as a chelating agent, forming stable complexes with calcium and magnesium ions present in hard water, allowing for their quantitative analysis through titration.

Key Concept: Colour Explanation

d) $[Co(NH3)6]3+$
[Solution Description] A complex appears yellow if it absorbs light in the violet region of the spectrum. For example, $[Co(NH3)6]3+$ can absorb violet light, resulting in a yellow appearance due to the d-d transition that corresponds to the absorbed wavelength.

Key Concept: Isomer Comparison

c) Color
[Solution Description] Linkage isomers have the same chemical formula and number of ligands but may vary in physical properties such as color due to different atom binding sites which affect the electronic structure.

Key Concept: Magnetic Properties

b) $[NiCl_4]^{2-}$
[Solution Description]
To determine if a complex is paramagnetic, we must look at the electron configuration and see if there are any unpaired electrons. For tetrahedral complexes:
– $[Ni(CO)_4]$: Ni in this complex is in zero oxidation state with the electron configuration $[Ar]3d^{10}$, which means all electrons are paired (diamagnetic).
– $[NiCl_4]^{2-}$: Ni is in +2 oxidation state, with an electron configuration of $[Ar]3d^{8}$, having two unpaired electrons (paramagnetic).
– $[CuCl_4]^{2-}$: Cu is in +2 oxidation state, with an electron configuration of $[Ar]3d^{9}$, having one unpaired electron (paramagnetic).
– $[PtCl_4]^{2-}$: Pt is typically square planar and diamagnetic.

Thus, both $[NiCl_4]^{2-}$ and $[CuCl_4]^{2-}$ are paramagnetic, but since this asks for a single answer among given options, based on typical examples, $[NiCl_4]^{2-}$ is considered here.

Key Concept: Complex Analysis, Advanced Reasoning

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that $[Fe(CN)_5(NO)]^{2-}$ is a heteroleptic complex. In this complex, iron (Fe) is coordinated to five cyanide (CN) ions and one nitrosyl (NO) ligand. Since there are two different types of donor groups (CN and NO), it qualifies as a heteroleptic complex, making the assertion true.

The reason mentions that heteroleptic complexes have multiple types of ligands, which can influence the electronic properties of the central metal atom or ion. This statement is also correct. Different ligands can affect factors such as the energy levels, stability, and reactivity of the complex by altering the electron density around the metal center.

Although both the assertion and reason are true, the reason directly explains why having multiple types of ligands makes the complex heteroleptic because they provide varied coordination environments affecting the metal’s properties. Thus, the reason correctly explains the assertion.

Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Basic Concept

a) Sigma ($\sigma$)
[Solution Description] In metal carbonyls, the ligand donates electron density to the metal using a sigma ($\sigma$) bond.

Key Concept: Coordination Number and Hybridization, Coordination Number and Electronic Configuration

c) $sp^3d^2$; 40
[Solution Description] The complex [Zn(en)$_3$]$^{2+}$ indicates Zn is bonded to three ethylenediamine molecules, each contributing two donor atoms, giving a coordination number of 6. This implies an octahedral geometry requiring $sp^3d^2$ hybridization. To calculate EAN:
– Atomic number of Zn = 30
– Electrons lost (Zn$^{2+}$) = 2
– Ligand electrons provided = 6 $\times$ 2 = 12
So, EAN = 30 – 2 + 12 = 40.

Key Concept: Oxidation Numbers

b) +3

[Solution Description] In the complex K$_3$[Cr(CN)$_6$], potassium (K) has an oxidation state of +1. There are 3 K atoms, contributing a total charge of +3. Cyanide (CN$^-$) as a ligand carries a -1 charge each, with 6 such CN$^-$ ions bringing a total charge of -6. Let x be the oxidation number of Cr. The sum of oxidation states in the entire neutral complex is zero:

$3(+1) + x + 6(-1) = 0$

Simplifying,

$3 + x – 6 = 0$

$x = 6 – 3$

$x = +3$

Therefore, the oxidation number of Cr is +3.

Key Concept: Simple Nomenclature

b) Dichlorodiammineplatinum(II)

[Solution Description] The IUPAC name for $[Pt(NH_3)_2Cl_2]$ follows this sequence: naming ligands first (ammine), then the central metal with its oxidation state in Roman numerals (platinum(II)), followed by the type of ligands (dichloride).

Key Concept: Conceptual Understanding

b) Due to ambidentate ligands
[Solution Description] Linkage isomerism in coordination compounds occurs due to the presence of ambidentate ligands that can bind through more than one atom, such as thiocyanate (NCS$^-$) which can bind through nitrogen or sulfur.

Key Concept: Electron Pairing Energy, Transition Metal Complexes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To analyze the problem, consider a $d^6$ octahedral complex. The presence of strong field ligands indicates that $\Delta_o > P$, where $\Delta_o$ is the crystal field splitting energy and $P$ is the electron pairing energy. This implies that electrons will pair up in the lower $$t_{2g}$$ orbitals rather than occupy higher $e_g$ orbitals. As a result, all six electrons in the $d$-orbitals are paired in the $t_{2g}$ set ($t_{2g}^6 e_g^0$), resulting in no unpaired electrons, which makes the complex diamagnetic. Therefore, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Basic Definition

d) $[Cu(NH_3)_4]$

[Solution Description] The coordination sphere consists of the central metal atom/ion and the ligands attached to it, all enclosed within square brackets. In the given complex $[Cu(NH_3)_4]SO_4$, the part inside the square brackets is Cu along with four $NH_3$ ligands. Thus, the coordination sphere is $[Cu(NH_3)_4]$.

Key Concept: Bond Formation

b) Carbon
[Solution Description] The M-C sigma bond in metal carbonyls is formed by the donation of a lone pair of electrons from the carbonyl carbon into a vacant orbital of the metal.

Key Concept: Ligand Types

a) $Cl^{-}$

[Solution Description] A monodentate ligand has one donor atom that coordinates to the central metal atom/ion. $Cl^{-}$ is an example of a monodentate ligand because it can only donate through its single available site.

Key Concept: Unpaired Electrons

a) 0
[Solution Description]
To solve this, we need to determine the electronic configuration of $\mathrm{Fe^{2+}}$. Iron ($ \mathrm{Fe}$) is in group 8 of the periodic table with an atomic number of 26. Thus, its electronic configuration is $[Ar]\,3d^6\,4s^2$. In $\mathrm{Fe^{2+}}$, it loses two electrons, so the configuration becomes $3d^6$. The CN$^{-}$ ligand is a strong field ligand causing pairing of the electrons. Therefore, all six 3d electrons will pair up, resulting in zero unpaired electrons.

Key Concept: Basic Definition

c) +3
[Solution Description]
The complex $[Co(NH_3)_6]^{3+}$ consists of cobalt (Co) as the central metal ion and ammonia ($NH_3$) as neutral ligands. Since $NH_3$ does not contribute to the charge, the entire charge on the complex $[Co(NH_3)_6]^{3+}$ must come from the oxidation state of Co. Therefore, the oxidation number of Co is +3.

Key Concept:  Medicinal Chemistry

c) Desferrioxamine B

[Solution Description] Chelate therapy for removing excess iron often utilizes the compound desferrioxamine B. This ligand forms strong complexes with iron, effectively reducing its levels in the bloodstream through excretion.

Key Concept: Coordination Sphere Identification

c) [AlF$_6$]
[Solution Description] The coordination sphere is identified as the part enclosed within square brackets. For Na$_3$[AlF$_6$], the coordination sphere is [AlF$_6$].

Key Concept: Crystal Field Theory Basics

d) d-d transition
[Solution Description] Crystal field theory states that the colour in coordination compounds is caused by d-d transitions. Electrons are excited from lower energy d orbitals ($t_{2g}$) to higher energy d orbitals ($e_g$).

Key Concept: Basic Naming

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that the given complex is correctly named according to IUPAC nomenclature rules. Here, the ammonia ligands are prefixed with “tetra-” indicating four NH$_3$ groups, followed by the name of the metal ion copper with its oxidation state in Roman numerals as (II), and then the sulfate as the counter-ion. According to IUPAC rules, when naming ionic coordination compounds, the complete name of the cation comes before the anion. Thus, both the Assertion and Reason are true, and the Reason correctly explains the assertion.

Key Concept: Charge Calculation

b) -3

[Solution Description]

In this complex, chromium is the central metal atom. The oxalate ion ($C_2O_4^{2-}$) has a charge of -2 and acts as a bidentate ligand, meaning each oxalate ion coordinates twice to the metal center. With three oxalate ligands, the total negative charge contributed by the ligands is $3 \times -2 = -6$. To determine the charge on the overall complex ion, we need to consider the oxidation state of chromium, which typically is +3 in such complexes to balance out the ligand charges. Therefore, the charge on the entire complex ion is calculated as follows:

$n = (+3) + 3(-2)$

Simplifying gives:

$n = 3 – 6 = -3$

Thus, the charge on the complex ion is -3.

Key Concept: Conceptual Understanding

a) The counter ion is a potential ligand
[Solution Description] Ionisation isomerism arises when the counter ion in a complex acts as a potential ligand that can replace another ligand within the coordination sphere, leading to different ions in solution.

Key Concept: Oxidation Number

c) +3

[Solution Description] For $[Co(NH_3)_5Cl]^{2+}$, we calculate the oxidation state of Co by assigning known charges: $NH_3$ (neutral), Cl$^-$ (-1). The overall charge is +2:

Let x be the oxidation number of Co:

$x + 5(0) – 1 = +2$

Solving for x gives:

$x = +2 + 1 = +3$

Hence, the oxidation number of Co is +3.

Key Concept: Simple Identification

b) 2
[Solution Description] Didentate ligands are defined as ligands that have two donor atoms capable of binding to a central metal ion simultaneously.