Step 1: Identification of Coordination Entity Components

To determine the chemical formula of the coordination compound "Potassium tetrachloridopalladate(II)", we decompose the name into its constituent parts according to IUPAC nomenclature conventions:

• Counter-ion: "Potassium" indicates the presence of $K^+$ ions outside the coordination sphere.
• Central Metal Atom: "Palladate" with the suffix "-ate" indicates that the coordination entity is an anionic complex. The metal is Palladium ($Pd$).
• Ligands: "tetrachlorido" indicates four chloride ($Cl^-$) ligands.
• Oxidation State: "(II)" indicates the oxidation state of the central metal atom ($Pd^{2+}$).

Step 2: Determining the Charge of the Coordination Sphere

We calculate the net charge of the complex ion $[PdCl_4]^n$ using the oxidation state of the metal and the charge of the ligands [Per the Rule of Coordination Algebra]:

$Charge = (\text{Oxidation state of metal}) + (\text{Number of ligands} \times \text{Charge of each ligand})$

$Charge = (+2) + (4 \times -1)$

$Charge = +2 - 4 = -2$

Thus, the complex anion is $[PdCl_4]^{2-}$.

Step 3: Balancing the Formula via Electroneutrality

To form a neutral compound, the total positive charge from the potassium ions must balance the negative charge of the coordination entity [Per the Principle of Electroneutrality]:

Let $x$ be the number of $K^+$ ions:

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

$x = 2$

Therefore, two $K^+$ ions are required to balance one $[PdCl_4]^{2-}$ ion.

Step 4: Final Formula Construction

Combining the counter-ion and the coordination entity, we place the cation first followed by the complex ion within square brackets:

Formula: $K_2[PdCl_4]$

Visual Representation of the Coordination Geometry

Final Solution: K_2[PdCl_4]

Step 1: Identifying Electronic Configuration and Oxidation State

Both complexes feature a central iron atom in the $+2$ oxidation state. The ground state electronic configuration of neutral $\text{Fe}$ ($Z=26$) is $[Ar] 3d^6 4s^2$. Upon forming the $+2$ ion, the configuration becomes:

$\text{Fe}^{2+}: [Ar] 3d^6$

In both $[Fe(CN)_{6}]^{4-}$ and $[Fe(H_{2}O)_{6}]^{2+}$, the $3d$ subshell contains $6$ electrons. The difference in color arises from the nature of the ligands attached to the central metal ion.

Step 2: Analyzing Crystal Field Splitting Energy ($\Delta_o$)

According to the Crystal Field Theory (CFT), ligands split the degenerate $d$-orbitals of the metal ion into two sets: $t_{2g}$ (lower energy) and $e_g$ (higher energy). The magnitude of the crystal field splitting energy, denoted as $\Delta_o$, is determined by the spectrochemical series.

• Cyanide ($\text{CN}^-$): A strong-field ligand that causes large splitting.
• Water ($\text{H}_2\text{O}$): A weak-field ligand that causes small splitting.

[Per the Spectrochemical Series: $\text{H}_2\text{O} < \text{CN}^-$]

Step 3: Calculating $d-d$ Transition Energies

The color of a coordination compound arises from the absorption of specific wavelengths of visible light, which promote an electron from the lower $t_{2g}$ orbitals to the higher $e_g$ orbitals. The energy of the absorbed photon corresponds to $\Delta_o$.

The energy relationship is given by: $E = h\nu = \frac{hc}{\lambda}$

Since $\Delta_o(\text{CN}^-) > \Delta_o(\text{H}_2\text{O})$, the wavelength of light absorbed ($\lambda = \frac{hc}{\Delta_o}$) will be shorter for the cyanide complex compared to the aqua complex.

Step 4: Comparative Analysis Summary

The transition of electrons between $d$-orbitals (the $d-d$ transition) absorbs different portions of the white light spectrum due to the difference in $\Delta_o$. Consequently, the light transmitted or reflected—which determines the perceived color—is different for both compounds.

Final Solution: The different colors are due to the difference in the crystal field splitting energy ($\Delta_o$) caused by the nature of the ligands. $\text{CN}^-$ is a strong-field ligand causing large splitting (absorbing higher energy, shorter wavelength light), while $\text{H}_2\text{O}$ is a weak-field ligand causing smaller splitting (absorbing lower energy, longer wavelength light). This difference results in the absorption of light of different wavelengths, leading to different observed colors.

Step 1: Identification of Coordination Sphere Components

To determine the chemical formula, we dissect the IUPAC name "Potassium tri(oxalato)chromate(III)" into its constituent parts based on coordination nomenclature rules [Per IUPAC Recommendations for Inorganic Chemistry]:

• Central Metal Ion: "chromate" indicates that the complex ion is anionic. The metal is Chromium ($Cr$).
• Ligands: "tri(oxalato)" specifies three oxalate ions. Oxalate is an ambidentate chelating ligand denoted as $C_2O_4^{2-}$ (often abbreviated as $ox$).
• Counter Ion: "Potassium" ($K^+$) outside the coordination sphere balances the charge of the complex anion.
• Oxidation State: The Roman numeral (III) indicates the oxidation state of Chromium is $+3$.

Step 2: Determining the Charge of the Coordination Entity

We calculate the net charge ($x$) of the complex ion $[Cr(C_2O_4)_3]^x$ using the oxidation state of the central metal and the charge of the ligands:

$x = (\text{Oxidation state of } Cr) + 3 \times (\text{Charge of oxalate ion})$

$x = (+3) + 3 \times (-2)$ [Charge of oxalate is $-2$]

$x = 3 - 6 = -3$

Thus, the complex anion is $[Cr(C_2O_4)_3]^{3-}$.

Step 3: Balancing Charges and Constructing the Formula

To form a neutral compound, the total positive charge from the potassium ions must balance the $-3$ charge of the complex anion. Since each potassium ion carries a charge of $+1$:

$(+1) \times (\text{number of } K^+) + (-3) = 0$

Number of $K^+$ = $3$

Combining the counter ion and the complex entity, we obtain $K_3[Cr(C_2O_4)_3]$.

Visual Representation of the Complex Ion

Final Solution: $K_3[Cr(C_2O_4)_3]$

1. Theoretical Foundation: Werner’s Coordination Theory

Alfred Werner’s Coordination Theory (1893) provides the foundational framework for understanding the bonding in coordination compounds. It posits that metal atoms in coordination complexes exhibit two distinct types of valency:

• Primary Valency (Ionizable): Corresponds to the oxidation state of the metal. These satisfy the negative charge of the metal ion and are satisfied by negative ions.
• Secondary Valency (Non-ionizable): Corresponds to the coordination number (CN) of the metal. These are fixed for a metal ion and are satisfied by ligands (either neutral molecules or negative ions).

2. Postulates and Structural Implications

3. Visual Representation of Bonding

The following SVG illustrates the distinction between primary (dashed) and secondary (solid) valencies for a complex such as $[Co(NH_3)_6]Cl_3$.

4. Mathematical Verification of Valency

Consider the complex $[Co(NH_3)_6]Cl_3$. Per the rules of coordination chemistry [IUPAC Nomenclature and Oxidation State Rules]:

Let $x$ be the oxidation state of $Co$.

$x + 6(0) + 3(-1) = 0$ [Since $NH_3$ is neutral and $Cl$ has a charge of $-1$]

$x = +3$

• Primary Valency: $3$ (Satisfied by $3 Cl^-$ ions).
• Secondary Valency: $6$ (Determined by the number of $NH_3$ ligands coordinated to the $Co^{3+}$ center).

Final Solution: Werner’s postulates conclude that bonding in coordination compounds is governed by dual valency: primary valencies define the oxidation state and ionic character, while secondary valencies define the coordination number and stereochemical geometry of the complex.

Step 1: Analyzing the Coordination Sphere

The complex ion is $\left[Co(NH_{3})Cl(en)_{2}\right]^{2+}$.

• Central Metal: Cobalt ($Co^{3+}$), which is $d^6$ and typically octahedral in coordination number 6.
• Ligands:
  • One ammine ligand ($NH_3$), unidentate.
  • One chloride ligand ($Cl^-$), unidentate.
  • Two ethylenediamine ($en$) ligands, bidentate ($NH_2CH_2CH_2NH_2$).
• General Formula: $[M(AA)_2bc]$, where $AA$ is the bidentate ligand, and $b, c$ are unidentate ligands.

Step 2: Determining Geometrical Isomers

The complex exhibits two geometrical isomers based on the relative position of the two unidentate ligands ($NH_3$ and $Cl$) with respect to each other and the bidentate ligands.

Step 3: Evaluating Optical Activity

Optical activity requires the absence of a plane of symmetry (chirality). [Per the Criterion for Optical Activity].

• cis-Isomer: Does not possess a plane of symmetry. Therefore, the cis-isomer exists as a pair of enantiomers (non-superimposable mirror images).
• trans-Isomer: Possesses a plane of symmetry passing through the $Cl-Co-NH_3$ axis and the two $en$ rings. Therefore, it is achiral and optically inactive.

Step 4: Visualizing the Isomers

Step 5: Summary of Isomeric Forms

The total number of isomers identified is three:

1. $cis$-isomer (d-form): Optically active.
2. $cis$-isomer (l-form): Optically active (enantiomer of 1).
3. $trans$-isomer: Optically inactive.

Final Solution: The complex $[Co(NH_{3})Cl(en)_{2}]^{2+}$ exhibits two geometrical isomers (cis and trans), where the cis-isomer is further resolved into two optical enantiomers, resulting in a total of three distinct isomeric structures.

Theoretical Foundation: Crystal Field Theory (CFT)

In an isolated transition metal ion, all five $d$-orbitals ($d_{xy}$, $d_{yz}$, $d_{zx}$, $d_{x^2-y^2}$, and $d_{z^2}$) are degenerate, meaning they possess identical energy. When six ligands approach the metal ion in an octahedral arrangement along the Cartesian axes ($x, y, z$), the electrostatic field causes the degeneracy to split. Orbitals pointing directly at the ligands experience stronger electrostatic repulsion than those pointing between the axes [Per Crystal Field Theory].

Step 1: Classification of Energy Levels

• $e_g$ orbitals: The $d_{x^2-y^2}$ and $d_{z^2}$ orbitals point directly toward the ligands (along the axes). These experience higher repulsion and are destabilized to a higher energy level.
• $t_{2g}$ orbitals: The $d_{xy}$, $d_{yz}$, and $d_{zx}$ orbitals point between the axes. These experience lower repulsion and are stabilized to a lower energy level.

Step 2: Mathematical Representation of Energy Split

The total energy separation between the two sets is defined as the crystal field splitting energy, denoted as $\Delta_o$.

• The $e_g$ set is elevated by $+0.6\Delta_o$ above the barycenter.
• The $t_{2g}$ set is lowered by $-0.4\Delta_o$ below the barycenter.

Step 3: Visual Representation of Orbital Splitting

Step 4: Summary Table of Orbital Energy

Final Solution: In an octahedral field, the five degenerate d-orbitals split into two sets: a triply degenerate lower-energy $t_{2g}$ set and a doubly degenerate higher-energy $e_g$ set, separated by an energy magnitude of $\Delta_o$.

Step 1: Analyzing the Coordination Entity

The complex ion given is $[PtCl_{2}(en)_{2}]^{2+}$. We first identify the coordination geometry and ligand arrangement:

• Central Metal: Platinum ($Pt^{4+}$).
• Coordination Number: $6$ (Octahedral geometry).
• Ligands: Two chloride ions ($Cl^-$) and two ethylenediamine ($en$) molecules. Since $en$ is a bidentate ligand, it occupies two coordination sites each ($2 \times 2 = 4$).
• Geometric Isomers: The complex exhibits cis-trans isomerism. Only the cis-isomer is optically active, as the trans-isomer possesses a plane of symmetry [Per the principle of molecular chirality in coordination chemistry].

Step 2: Defining the Optical Isomers

The cis-isomer of $[PtCl_{2}(en)_{2}]^{2+}$ lacks a plane of symmetry and a center of inversion, existing as a pair of non-superimposable mirror images (enantiomers), labeled as $\Delta$ (delta) and $\Lambda$ (lambda) isomers.

Step 3: Visualization of Geometric Structures

Step 4: Theoretical Justification

The chirality arises due to the helical arrangement of the bidentate ethylenediamine ligands around the $Pt$ center. By applying the Cahn-Ingold-Prelog (CIP) priority rules for coordination compounds:

• In the $\Delta$ isomer, the skew lines formed by the $en$ ligands describe a right-handed helix.
• In the $\Lambda$ isomer, the skew lines formed by the $en$ ligands describe a left-handed helix [By the IUPAC nomenclature rules for chiral coordination compounds].

The trans-isomer is excluded from optical isomerism because it contains a plane of symmetry passing through the $Cl-Pt-Cl$ axis, making it achiral.

Final Solution: The optical isomers of $[PtCl_{2}(en)_{2}]^{2+}$ are the pair of enantiomers derived from the cis-geometric configuration, represented as the $\Delta$ and $\Lambda$ forms which are non-superimposable mirror images of each other.

Step 1: Nomenclature and Oxidation State Analysis

The complex is $K[Cr(H_2O)_2(C_2O_4)_2] \cdot 3H_2O$.

• IUPAC Name: Potassium diaquadioxalatochromate(III) trihydrate.
• Oxidation State Calculation: Let the oxidation state of $Cr$ be $x$.
  $[+1] + x + 2(0) + 2(-2) = 0$ [Since $H_2O$ is neutral and oxalate $C_2O_4^{2-}$ has a charge of $-2$]
  $1 + x - 4 = 0 \implies x = +3$.

Step 2: Coordination Number and Electronic Configuration

• Coordination Number (CN): The ligands are two $H_2O$ (monodentate) and two $C_2O_4^{2-}$ (bidentate).
  $CN = (2 \times 1) + (2 \times 2) = 2 + 4 = 6$. [Per the coordination theory, the number of donor atoms directly bonded to the central metal atom].
• Electronic Configuration of $Cr^{3+}$:
  Ground state $Cr$ ($Z=24$): $[Ar] 3d^5 4s^1$
  $Cr^{3+}$ state: $[Ar] 3d^3 4s^0$

Step 3: Stereochemistry and Geometric Representation

Given $CN = 6$, the geometry is octahedral. Since it contains two bidentate ligands, it exhibits geometrical isomerism (cis/trans) and optical isomerism.

Step 4: Magnetic Moment Calculation

The configuration of $Cr^{3+}$ is $t_{2g}^3 e_g^0$. There are 3 unpaired electrons ($n=3$).

Using the spin-only magnetic moment formula $\mu = \sqrt{n(n+2)} \mu_B$:
$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \mu_B$. [Per Hund's Rule and Pauli Exclusion Principle].

Final Solution: The complex is Potassium diaquadioxalatochromate(III) trihydrate. It possesses an octahedral geometry with a $d^3$ configuration, a coordination number of 6, and a magnetic moment of 3.87 BM.

Step 1: Identification of Coordination Entity and Counter Ions

The chemical formula provided is $[Ni(NH_{3})_{6}]Cl_{2}$. In this structure, the coordination sphere is represented by $[Ni(NH_{3})_{6}]^{2+}$, and the chloride ions ($Cl^{-}$) serve as the counter ions [Per the definition of Coordination Compounds, where the species inside the square brackets constitutes the primary coordination entity].

Step 2: Calculation of the Oxidation State of the Central Metal Ion

To determine the systematic name, we must first establish the oxidation state of the central nickel ($Ni$) atom. Let the oxidation state of $Ni$ be $x$.

• The ligand $NH_{3}$ (ammine) is a neutral molecule, meaning its charge is $0$ [Per IUPAC nomenclature rules for neutral ligands].
• The counter ion is $Cl^{-}$, which has a charge of $-1$. Since there are two chloride ions, the total charge from counter ions is $2 \times (-1) = -2$.
• To maintain electrical neutrality of the total compound, the charge on the coordination entity $[Ni(NH_{3})_{6}]$ must be $+2$.

Equation: $x + 6(0) = +2$
Result: $x = +2$. Thus, the oxidation state of Nickel is represented as $Ni(II)$.

Step 3: IUPAC Nomenclature Rules Application

We apply the following IUPAC naming protocols:

1. Ligands: The ligand $NH_{3}$ is named "ammine". Since there are six of them, the prefix "hexa" is used [Per IUPAC rule: prefixes like di-, tri-, tetra-, penta-, hexa- are used for stoichiometric counts].
2. Metal Center: Because the coordination entity is a cation (the complex part is positive), the name of the metal remains "nickel" without any suffix modification [Per IUPAC rule: if the complex ion is cationic, the metal name is unchanged].
3. Oxidation State: The oxidation number must be included in parentheses as a Roman numeral immediately following the metal name: $(II)$.
4. Counter Ion: The chloride ions are named simply as "chloride". The stoichiometric prefix is not required for the counter ion [Per IUPAC rule: the naming of the counter ion is independent of its quantity].

Visual Representation of the Coordination Geometry

Synthesis of the Systematic Name

Combining the components: Hexaammine + nickel(II) + chloride.

Final Solution: Hexaamminenickel(II) chloride

Step 1: Electronic Configuration and Coordination Environment

The central metal ion in the complex $[Ti(H_{2}O)_{6}]^{3+}$ is Titanium in the $+3$ oxidation state. The ground state configuration of a neutral Titanium atom ($Z=22$) is $[Ar] 3d^2 4s^2$. Upon losing three electrons to form $Ti^{3+}$, the configuration becomes:

$Ti^{3+} = [Ar] 3d^1$

[Per the Aufbau Principle and Hund's Rule, the remaining electron resides in one of the five degenerate $d$-orbitals.]

Step 2: Crystal Field Splitting in Octahedral Geometry

As six $H_2O$ ligands approach the $Ti^{3+}$ ion in an octahedral arrangement, the degeneracy of the $d$-orbitals is lifted due to the electrostatic repulsion between the ligand lone pairs and the metal $d$-electrons. The $d$-orbitals split into two distinct energy levels:

• $t_{2g}$ set: Lower energy ($d_{xy}, d_{yz}, d_{zx}$), containing 3 degenerate orbitals.
• $e_g$ set: Higher energy ($d_{x^2-y^2}, d_{z^2}$), containing 2 degenerate orbitals.

Step 3: d-d Transition Mechanism

The single electron occupies the lower energy $t_{2g}$ orbital. When the complex absorbs light from the white light spectrum, the energy of the photon corresponds to the energy gap ($\Delta_o$) between the $t_{2g}$ and $e_g$ levels [Per the Bohr Frequency Condition: $\Delta E = h\nu = \frac{hc}{\lambda}$].

The electron is promoted from the $t_{2g}$ level to the $e_g$ level. This process is known as a $d-d$ transition. For $[Ti(H_{2}O)_{6}]^{3+}$, the energy absorbed corresponds to the yellow-green portion of the visible spectrum.

Step 4: Complementary Color Observation

The color observed by the human eye is the complementary color of the light absorbed. According to the color wheel of visible light:

Since the complex absorbs the yellow-green light to facilitate the excitation of the $d$-electron, the transmitted or reflected light is perceived as violet.

Final Solution: The violet color of $[Ti(H_{2}O)_{6}]^{3+}$ arises from the excitation of the single $3d$ electron from the $t_{2g}$ level to the $e_g$ level ($d-d$ transition) upon the absorption of yellow-green light, resulting in the observation of the complementary violet color.

1. Classification of Isomerism in Coordination Compounds

Coordination isomers are compounds with identical molecular formulas but different arrangements of atoms or ligands. They are broadly categorized into two major divisions: Structural Isomerism and Stereoisomerism.

2. Detailed Examples and Structural Analysis

A. Structural Isomerism

• Linkage Isomerism: Occurs when an ambidentate ligand binds via different donor atoms.
  Example: $[Co(NH_3)_5(NO_2)]Cl_2$ (Nitro isomer) and $[Co(NH_3)_5(ONO)]Cl_2$ (Nitrito isomer).
• Coordination Isomerism: Involves redistribution of ligands between two metal centers.
  Example: $[Co(NH_3)_6][Cr(CN)_6]$ and $[Cr(NH_3)_6][Co(CN)_6]$.
• Ionization Isomerism: Occurs when the counter ion acts as a potential ligand.
  Example: $[Co(NH_3)_5SO_4]Br$ and $[Co(NH_3)_5Br]SO_4$.
• Solvate Isomerism: Specifically relates to the coordination of $H_2O$.
  Example: $[Cr(H_2O)_6]Cl_3$ (violet) vs $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$ (blue-green).

B. Stereoisomerism

Stereoisomers have the same connectivity but differ in spatial orientation [Per the VSEPR Theory].

• Geometrical Isomerism: Square planar complexes like $[Pt(NH_3)_2Cl_2]$ exist as *cis* (same groups on adjacent sides) and *trans* (same groups on opposite sides).
• Optical Isomerism: Occurs in chiral complexes, such as $[Co(en)_3]^{3+}$. These exist as non-superimposable mirror images known as dextro ($d$) and laevo ($l$) forms.

Final Solution: Coordination compounds exhibit structural isomerism (Linkage, Coordination, Ionization, Solvate) and stereoisomerism (Geometrical and Optical). Each type arises from differences in connectivity or spatial orientation of ligands around the central metal ion as dictated by the coordination geometry.

Step 1: IUPAC Nomenclature

To name the coordination compound $K_{4}[Mn(CN)_{6}]$, we follow the IUPAC guidelines for coordination entities:

• Cation: Potassium ($K^+$).
• Ligand: Hexa-cyano (six $CN^-$ ligands).
• Central Metal: Manganese, acting as an anion within the complex, hence named "Manganate".
• Oxidation State Calculation: Let $x$ be the oxidation state of $Mn$. $4(+1) + x + 6(-1) = 0 \Rightarrow x - 2 = 0 \Rightarrow x = +2$. [Per the principle of electro-neutrality]
• Final Name: Potassium hexacyanomanganate(II).

Step 2: Electronic Configuration & Coordination Number

• Central Metal Ion: $Mn^{2+}$.
• Atomic Number ($Z$): $25$.
• Ground State Configuration ($Mn$): $[Ar] 3d^5 4s^2$.
• Configuration of $Mn^{2+}$: $[Ar] 3d^5 4s^0$.
• Coordination Number (C.N.): The complex contains 6 monodentate cyanide ligands bonded to the central atom. Therefore, C.N. = 6.

Step 3: Stereochemistry & Magnetic Properties

The complex adopts an octahedral geometry due to the coordination number of 6. Cyanide ($CN^-$) is a strong field ligand, which induces spin-pairing in the $d$-orbitals. [Per Crystal Field Theory]

Step 4: Calculation of Magnetic Moment

The magnetic moment ($\mu$) is calculated using the spin-only formula: $\mu = \sqrt{n(n+2)} \text{ B.M.}$

Since the number of unpaired electrons ($n$) = 1:

$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732 \text{ B.M.}$ [Per the Bohr-van Leeuwen theorem]

Final Solution: The IUPAC name is Potassium hexacyanomanganate(II). The oxidation state is +2, coordination number is 6, electronic configuration is $[Ar] 3d^5$, stereochemistry is octahedral, and the magnetic moment is 1.732 B.M.

Step 1: Analyzing the IUPAC Name Components

To determine the chemical formula of the coordination compound Hexaammineplatinum(IV), we decompose the nomenclature into its constituent parts:

• Ligand: "Hexaammine" indicates six ($6$) molecules of the neutral ligand ammonia ($\text{NH}_3$).
• Central Metal Ion: "Platinum" is the metal core, with the symbol $\text{Pt}$.
• Oxidation State: "(IV)" denotes that the platinum ion has an oxidation state of $+4$.

Step 2: Determining the Charge of the Coordination Entity

The total charge of a coordination entity is calculated by the sum of the oxidation state of the central metal atom and the charges of the coordinating ligands [Per Coordination Chemistry Rules].

Let $x$ be the total charge of the coordination entity:

$x = (\text{Oxidation state of Pt}) + (\text{Number of ligands} \times \text{Charge of } \text{NH}_3)$

Given that $\text{NH}_3$ is a neutral ligand with a charge of $0$:

$x = (+4) + (6 \times 0)$

$x = +4$

Step 3: Constructing the Chemical Formula

By convention, the central metal atom is written first, followed by the ligands in alphabetical order (only $\text{NH}_3$ is present here), enclosed within square brackets. The net charge is indicated as a superscript outside the closing bracket [According to IUPAC Recommendations for Coordination Nomenclature].

Formula = $[\text{Pt}(\text{NH}_3)_6]^{4+}$

Visual Representation of the Coordination Geometry

Final Solution: $[\text{Pt}(\text{NH}_3)_6]^{4+}$

Step 1: Identification of Coordination Sphere Components

To determine the oxidation state of the central metal ion (Cobalt, $\text{Co}$), we must identify the individual charges of all ligands coordinated to the metal within the coordination sphere $[ \text{Co}(\text{H}_2\text{O})(\text{CN})(\text{en})_2 ]^{2+}$.

Step 2: Mathematical Formulation of the Oxidation State

Let the oxidation number of Cobalt be denoted as $x$. According to the algebraic sum rule for coordination complexes, the sum of the oxidation numbers of all atoms in the complex ion must equal the net charge of the coordination entity [Per IUPAC Nomenclature Rules for Coordination Compounds].

The equation is structured as follows:

$x + [1 \times (\text{charge of } \text{H}_2\text{O})] + [1 \times (\text{charge of } \text{CN}^-)] + [2 \times (\text{charge of } \text{en})] = \text{Net Charge}$

Step 3: Algebraic Calculation

Substituting the values derived in Step 1 into the equation:

$x + [1 \times (0)] + [1 \times (-1)] + [2 \times (0)] = +2$

Solving for $x$:

$x + 0 - 1 + 0 = +2$

$x - 1 = +2$

$x = +2 + 1$

$x = +3$

Step 4: Geometric Contextualization

The complex is an octahedral species where Cobalt acts as the central metal atom. The oxidation state $+3$ indicates that Cobalt has lost three electrons to form the complex, typically resulting in a $d^6$ configuration for the $\text{Co}^{3+}$ ion.

Final Solution: The oxidation number of the metal (Cobalt) is $+3$.

1. Role in Biological Systems

Coordination compounds are essential for the survival of organisms as they serve as prosthetic groups in various enzymes and proteins [Per Biochemistry principles].

• Chlorophyll: A coordination complex of Magnesium ($Mg^{2+}$) with a porphyrin ring. It is crucial for photosynthesis, capturing light energy [Referenced via Photosynthetic pigment theory].
• Hemoglobin: A coordination complex of Iron ($Fe^{2+}$) with protoporphyrin. It functions as an oxygen carrier in the bloodstream [By the principle of reversible ligand binding].
• Vitamin $B_{12}$ (Cyanocobalamin): A complex where Cobalt ($Co^{3+}$) is coordinated with a corrin ring system, essential for red blood cell formation.

2. Role in Medicinal Chemistry

Coordination compounds are employed as therapeutic agents, particularly in oncology and heavy metal toxicity treatment [Per Pharmacology protocols].

• Cisplatin: The complex cis-$[Pt(NH_3)_2Cl_2]$ is a potent anti-cancer drug. It binds to DNA, inhibiting replication in rapidly dividing cancer cells [By the mechanism of DNA cross-linking].
• EDTA Therapy: Disodium salt of EDTA (Ethylenediaminetetraacetic acid) is used to treat lead or mercury poisoning by forming stable, water-soluble chelate complexes that are excreted in urine [Per Chelation Therapy].

3. Role in Analytical Chemistry

Coordination chemistry facilitates the qualitative and quantitative determination of metal ions through colorimetric changes or precipitation [Per Analytical Chemistry axioms].

• EDTA Titrations: EDTA is used as a titrant to determine the hardness of water by forming complexes with $Ca^{2+}$ and $Mg^{2+}$ ions [By Complexometric titration principle].
• Dimethylglyoxime (DMG): Forms a distinct red precipitate with Nickel ($Ni^{2+}$) ions in the presence of ammonium hydroxide, acting as a specific spot test for qualitative identification.

4. Role in Extraction/Metallurgy of Metals

Ligands are used to selectively dissolve metals from ores, a process known as hydrometallurgical extraction [Per Extractive Metallurgy standards].

• Gold and Silver Extraction: Gold is extracted by treating the native ore with dilute $NaCN$ solution in the presence of air, forming a soluble cyano-complex:
  $4Au(s) + 8CN^-(aq) + 2H_2O(aq) + O_2(g) \rightarrow 4[Au(CN)_2]^-(aq) + 4OH^-(aq)$
  The gold is subsequently recovered by reduction with Zinc [By the MacArthur-Forrest process].
• Mond Process (Nickel): Nickel is purified by reaction with $CO$ to form a volatile complex, $[Ni(CO)_4]$, which is then thermally decomposed to obtain pure metal.

Final Solution: Coordination compounds are indispensable to biological systems (e.g., Hemoglobin), medicinal applications (e.g., Cisplatin for cancer), analytical detection (e.g., EDTA for water analysis), and industrial metallurgy (e.g., Cyanide process for gold extraction).

Step 1: Nomenclature and Oxidation State Determination

The complex is formulated as $[Co(NH_{3})_{5}Cl]Cl_{2}$.

• IUPAC Name: Pentaamminechloridocobalt(III) chloride [Per IUPAC nomenclature rules for coordination compounds, ligands are listed alphabetically: ammine before chlorido].
• Oxidation State ($x$):
  Let the oxidation state of $Co$ be $x$.
  $x + 5(0) + 1(-1) + 2(-1) = 0$
  $x - 1 - 2 = 0 \Rightarrow x = +3$
  The Cobalt ion is in the $+3$ oxidation state ($Co^{3+}$).

Step 2: Electronic Configuration

• Atomic number of $Co (Z=27)$: $[Ar] 3d^{7} 4s^{2}$
• Electronic configuration of $Co^{3+}$: Remove 3 electrons (two from $4s$, one from $3d$): $[Ar] 3d^{6}$

Step 3: Coordination Number and Geometry

• Coordination Number: The central metal atom is bonded to 5 $NH_{3}$ molecules (monodentate) and 1 $Cl^{-}$ ion (monodentate). Total = $5 + 1 = 6$.
• Geometry: A coordination number of 6 corresponds to an octahedral geometry.

Step 4: Stereochemistry and Magnetic Moment

• Stereochemistry: The complex $[Co(NH_{3})_{5}Cl]^{2+}$ is of the type $[MA_{5}B]$. This type of complex does not exhibit geometrical isomerism (cis-trans) nor optical isomerism, as there is no arrangement that creates non-superimposable mirror images or distinct spatial orientations for the single $Cl$ ligand.
• Magnetic Moment ($μ$):
  $NH_{3}$ is a strong field ligand, but for $Co^{3+}$ ($d^{6}$), even with weak ligands, it typically forms low-spin complexes. However, in this specific octahedral field, all electrons are paired in $t_{2g}$ orbitals.
  Configuration: $t_{2g}^{6}, e_{g}^{0}$.
  Number of unpaired electrons ($n$) = 0.
  Magnetic Moment formula: $μ = \sqrt{n(n+2)} \text{ B.M.}$
  $μ = \sqrt{0(0+2)} = 0 \text{ B.M.}$
  The complex is diamagnetic.

Final Solution: The complex is Pentaamminechloridocobalt(III) chloride. The oxidation state is +3, electronic configuration is $3d^{6}$, coordination number is 6, it has no stereoisomers, and the magnetic moment is 0 B.M. (diamagnetic).

Step 1: Analyzing the Formation of Mohr's Salt ($FeSO_{4} \cdot (NH_{4})_{2}SO_{4} \cdot 6H_{2}O$)

When an aqueous solution of ferrous sulfate ($FeSO_{4}$) is mixed with ammonium sulfate ($(NH_{4})_{2}SO_{4}$) in a 1:1 molar ratio, the resulting compound is a double salt known as Mohr's salt. The chemical reaction is defined as:

$FeSO_{4}(aq) + (NH_{4})_{2}SO_{4}(aq) + 6H_{2}O(l) \rightarrow FeSO_{4} \cdot (NH_{4})_{2}SO_{4} \cdot 6H_{2}O(s)$

[Per the definition of double salts, these compounds exist only in the solid state and dissociate completely into their constituent simple ions when dissolved in water.]

In aqueous solution, the dissociation follows:

$[FeSO_{4} \cdot (NH_{4})_{2}SO_{4} \cdot 6H_{2}O](aq) \rightarrow Fe^{2+}(aq) + 2NH_{4}^{+}(aq) + 2SO_{4}^{2-}(aq) + 6H_{2}O(l)$

Because the $Fe^{2+}$ ions are released into the solution as free, hydrated ions, they remain available to react with chemical reagents (e.g., potassium ferricyanide), thereby giving a positive qualitative test for iron(II).

Step 2: Analyzing the Formation of Tetraamminecopper(II) Sulfate

When an aqueous solution of copper(II) sulfate ($CuSO_{4}$) is mixed with aqueous ammonia ($NH_{3}$) in a 1:4 molar ratio, a coordination complex is formed:

$CuSO_{4}(aq) + 4NH_{3}(aq) \rightarrow [Cu(NH_{3})_{4}]SO_{4}(aq)$

[Per Werner’s Coordination Theory, ligands coordinate to the central metal ion via dative covalent bonds.]

In this coordination entity, the $Cu^{2+}$ ion is surrounded by four $NH_{3}$ ligands in a square planar geometry. The coordination entity $[Cu(NH_{3})_{4}]^{2+}$ is highly stable and does not dissociate into free $Cu^{2+}$ ions in an aqueous medium. The dissociation equilibrium is defined as:

$[Cu(NH_{3})_{4}]SO_{4}(aq) \rightarrow [Cu(NH_{3})_{4}]^{2+}(aq) + SO_{4}^{2-}(aq)$

Since the concentration of free $Cu^{2+}(aq)$ ions in the solution is negligible (governed by the instability constant $K_{inst}$ of the complex), the solution fails to yield the characteristic qualitative tests associated with free copper(II) ions.

Step 3: Comparative Analysis

Final Solution: Mohr's salt is a double salt that dissociates completely in water, releasing free $Fe^{2+}$ ions which react to give their characteristic test. Conversely, $[Cu(NH_{3})_{4}]SO_{4}$ is a coordination complex that retains its identity in solution, sequestering the $Cu^{2+}$ ion within the stable coordination entity, thereby preventing it from reacting during standard qualitative testing.

Step 1: Identifying the Coordination Entity

The given coordination compound is $[Pt(NH_{3})_{2}Cl(NH_{2}CH_{3})]Cl$. This is a coordination sphere (cationic) followed by a counter-ion (anion).

• Central Metal Atom/Ion: $Pt$ (Platinum)
• Ligands in the coordination sphere:
  • $NH_{3}$: Ammine (Neutral, Charge = 0)
  • $Cl$: Chloro (Anionic, Charge = $-1$)
  • $NH_{2}CH_{3}$: Methanamine (Neutral, Charge = 0)
• Counter-ion: $Cl^{-}$ (Chloride)

Step 2: Calculation of Oxidation State of Platinum ($Pt$)

Let the oxidation state of $Pt$ be $x$. The sum of the charges of all ligands and the central metal must equal the total charge on the coordination sphere. Since the counter-ion is $Cl^{-}$, the coordination sphere $[Pt(NH_{3})_{2}Cl(NH_{2}CH_{3})]$ must have a net charge of $+1$.

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

$x - 1 = +1$

$x = +2$

[Per the rules of oxidation state determination in coordination compounds]

Step 3: Determining the Alphabetical Order of Ligands

According to IUPAC norms, ligands are named in alphabetical order regardless of the number of ligands:

• Ammine ($NH_{3}$): Begins with 'a'
• Chloro ($Cl^{-}$): Begins with 'c'
• Methanamine ($NH_{2}CH_{3}$): Begins with 'm'

The sequence is: Ammine, Chloro, Methanamine. Since there are two $NH_{3}$ groups, we use the prefix 'di-'.

Step 4: Constructing the IUPAC Name

Following IUPAC nomenclature for coordination compounds (prefix for ligands + name of central metal + oxidation state in Roman numerals + name of counter-ion):

• Prefixes: Diammine
• Ligand order: Diamminechloromethanamineplatinum(II)
• Counter-ion: chloride

Step 5: Final Verification

Applying the standard rules for cationic complexes:

1. Name the ligands alphabetically: Diammine, Chloro, Methanamine.
2. Identify metal: Platinum.
3. Identify oxidation state: (II).
4. Identify counter-ion: Chloride.

Resulting name: Diamminechloromethanamineplatinum(II) chloride

Final Solution: Diamminechloromethanamineplatinum(II) chloride

Step 1: Identification of Coordination Entity Components

The complex ion is given as $[NiCl_{4}]^{2-}$. The components are defined as follows:

• Central Metal Atom/Ion: Nickel ($Ni$).
• Ligands: Four chloride ions ($Cl^{-}$). The prefix for four is "tetra-". Since the ligand is anionic, the suffix "-ide" is changed to "-o", resulting in "chlorido".
• Coordination Entity Charge: The complex is an anion (indicated by the overall charge of $-2$).

Step 2: Calculation of Oxidation State of Nickel

Let $x$ be the oxidation state of Nickel ($Ni$).

The oxidation state of a chloride ion ($Cl^{-}$) is $-1$. [Per IUPAC Nomenclature rules for anionic ligands].

The sum of oxidation states in a complex ion equals the total charge of the complex:

$x + 4(-1) = -2$ [Per the Principle of Algebraic Sum of Oxidation States in a Coordination Complex]

$x - 4 = -2$

$x = +2$

The oxidation state of Nickel is $II$.

Step 3: Application of IUPAC Naming Conventions

According to IUPAC recommendations for coordination compounds:

1. Ligand Naming: List ligands alphabetically. We have four chlorido ligands: "tetrachlorido".
2. Metal Naming: Because the complex ion is anionic ($[NiCl_{4}]^{2-}$), the central metal must end in the suffix "-ate". Thus, "Nickel" becomes "Nickelate".
3. Oxidation State: The oxidation state is indicated by Roman numerals in parentheses immediately following the name of the metal.
4. Overall Structure: Name of ligands + Name of central metal (with suffix) + (Oxidation state) + "ion".

Step 4: Systematic Construction of the Name

Combining the identified parts:

• Ligand portion: tetrachlorido
• Metal portion: nickelate(II)
• Suffix for ion: ion

Final Solution: tetrachloridonickelate(II) ion

1. Theoretical Foundation and Coordination Geometry

The complex $[Co(NH_{3})_{2}Cl_{2}(en)]^{+}$ is an octahedral complex of the type $[M(AA)(a)_{2}(b)_{2}]$, where $M = Co^{3+}$, $(AA) = ethylenediamine$ (a bidentate ligand), $a = NH_{3}$, and $b = Cl^{-}$. The coordination number is 6.

2. Geometrical Isomerism Analysis

Geometrical isomers arise from the relative orientation of the monodentate ligands ($Cl$ and $NH_{3}$) in the octahedral field. We classify these based on the position of the chloride ligands:

• Cis-isomer: The two $Cl$ ligands are adjacent to each other ($90^\circ$ angle).
• Trans-isomer: The two $Cl$ ligands are opposite to each other ($180^\circ$ angle).

3. Optical Isomerism Analysis

Optical activity requires the absence of a plane of symmetry (chirality).

• The trans-isomer possesses a plane of symmetry passing through the $Cl-Co-Cl$ axis and the $en$ ligand plane, rendering it optically inactive.
• The cis-isomer exists in two enantiomeric forms (non-superimposable mirror images) due to the lack of a center or plane of symmetry, rendering it optically active.

4. Geometric Representations (SVG)

5. Summary of Isomeric Forms

Final Solution: The complex exhibits three isomers: one optically inactive trans-isomer and a pair of optically active (d and l) cis-enantiomers.