Correct Option: (a)

Explanation: According to **Werner's Theory**, only the ions present outside the square brackets (the **coordination sphere**) are ionisable in aqueous solution and can react with reagents like silver nitrate to form precipitates. Since 1 mole of the complex yields exactly 2 moles of solid $\text{AgCl}$, there must be **2 free chloride ($\text{Cl}^{-}$) ions** outside the coordination sphere:

Ambidentate Ligand: A monodentate ligand which contains **two different donor atoms** and has the ability to coordinate to the central metal atom/ion through either of the two atoms is called an ambidentate ligand.

Example: The nitrite ion ($\text{NO}_2^{-}$):

• It can coordinate to the metal via the **nitrogen** atom: $\text{M} \leftarrow \text{NO}_2$ (nitro-N complex)
• It can coordinate via the **oxygen** atom: $\text{M} \leftarrow \text{ONO}$ (nitro-O complex)

Another classic example is the thiocyanate ion ($\text{SCN}^{-}$ which coordinates via S, and $\text{NCS}^{-}$ which coordinates via N).

Step 1: Calculate the Coordination Number (C.N.).

Coordination number is the total number of coordinate sigma bonds formed between ligands and the central metal ion.

• Ethylenediamine ($\text{en}$) is a **bidentate ligand** (each $\text{en}$ molecule donates 2 electron pairs). So, $2 \text{ bidentate ligands} \times 2 = 4$ bonds.
• Chloride ($\text{Cl}^{-}$) is a **monodentate ligand** (each $\text{Cl}$ donates 1 pair). So, $2 \text{ monodentate ligands} \times 1 = 2$ bonds.
• $\text{Total C.N.} = 4 + 2 = 6$ (Octahedral structure).

Step 2: Determine the Oxidation State of Platinum ($x$).

Let the oxidation state of $\text{Pt}$ be $x$. Ethylenediamine ($\text{en}$) is a neutral ligand (charge = 0), and chloride has a charge of $-1$:

Answer: The coordination number of platinum is $6$ and its oxidation state is $+4$ (written as $\text{Pt(IV)}$).

Chelation: When a di- or polydentate ligand uses two or more donor atoms simultaneously to bind to a single central metal atom/ion, it forms a cyclic ring-like structure. This process is called chelation, and the ligand is known as a chelating ligand.

Chelate Effect: The thermodynamic stability of a chelated complex is significantly greater than that of an analogous complex containing only monodentate ligands. This stabilization is called the chelate effect.

Comparison & Stability:

• $[\text{Co}(\text{en})_3]^{3+}$ is **much more stable** than $[\text{Co}(\text{NH}_3)_6]^{3+}$.
• Reason: While ammonia is a monodentate ligand, ethylenediamine ($\text{en}$) is bidentate. Binding of three $\text{en}$ ligands forms three stable five-membered rings with Cobalt. From an entropy perspective, the displacement of coordinated monodentate water molecules by bidentate chelates results in an overall increase in the number of free particles in solution ($\Delta S \gt 0$), rendering the chelated complex highly stable.

(i) $[\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2$:

• Cation is inside the coordination sphere. We list the ligands in alphabetical order: five ammine molecules ("pentaammine") and one chloride ("chlorido").
• Let $x$ be the oxidation state of Cobalt: $x + 5(0) + 1(-1) + 2(-1) = 0 \implies x = +3$.
• Since the complex ion is cationic, the metal name remains **cobalt**.
• IUPAC Name: Pentaamminechloridocobalt(III) chloride

(ii) $\text{K}_3[\text{Fe}(\text{CN})_6]$:

• The cation is potassium (written first). The anion is the complex sphere containing six cyano groups ("hexacyanido").
• Let $x$ be the oxidation state of Iron: $3(+1) + x + 6(-1) = 0 \implies x = +3$.
• Since the complex sphere is anionic, the metal name suffix changes to **-ate** (ferrate).
• IUPAC Name: Potassium hexacyanidoferrate(III)

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

• The central metal is Chromium ($\text{Cr}$).
• "tris(ethane-1,2-diamine)" implies three bidentate ethylenediamine ($\text{en}$) ligands: $(\text{en})_3$.
• The oxidation state of chromium is $+3$. Since $\text{en}$ is neutral, the coordination sphere charge is $+3$: $[\text{Cr}(\text{en})_3]^{3+}$.
• To balance the $+3$ sphere, we require 3 chloride counter-ions ($\text{Cl}^{-}$).
• Formula: $[\text{Cr}(\text{en})_3]\text{Cl}_3$

(ii) Ammineaquadichloridoplatinum(II):

• The central metal is Platinum ($\text{Pt}$).
• The ligands are one ammine ($\text{NH}_3$), one aqua ($\text{H}_2\text{O}$), and two chlorido ($\text{Cl}_2$).
• Let's calculate the overall sphere charge with Pt(II): $+2 (\text{Pt}) + 0 (\text{NH}_3) + 0 (\text{H}_2\text{O}) + 2(-1) (\text{Cl}) = 0$. The complex is neutral.
• Formula: $[\text{Pt}(\text{NH}_3)(\text{H}_2\text{O})\text{Cl}_2]$

Ionization Isomerism: This isomerism arises when a coordination compound yields different ions in solution, despite having the exact same chemical composition. This occurs because an ionisable counter-ion acts as a ligand in one isomer, displacing a coordinated ligand into the outer sphere.

Isomeric Formulas:

• Isomer A: $[\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Br}$ (red-violet)
• Isomer B: $[\text{Co}(\text{NH}_3)_5\text{Br}]\text{SO}_4$ (red)

Chemical Distinction:

1. Treating **Isomer A** with an aqueous barium chloride ($\text{BaCl}_2$) solution yields **no precipitate**, but treating it with silver nitrate ($\text{AgNO}_3$) yields a **pale yellow precipitate** of silver bromide ($\text{AgBr}$), confirming free $\text{Br}^{-}$ ions:
   $$ [\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Br}(aq) + \text{AgNO}_3(aq) \rightarrow \text{AgBr}(s)\downarrow + [\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{NO}_3(aq) $$
2. Treating **Isomer B** with aqueous silver nitrate ($\text{AgNO}_3$) yields **no precipitate** (since $\text{Br}$ is bound), but treating it with barium chloride ($\text{BaCl}_2$) yields a **dense white precipitate** of barium sulfate ($\text{BaSO}_4$), confirming free $\text{SO}_4^{2-}$ ions:
   $$ [\text{Co}(\text{NH}_3)_5\text{Br}]\text{SO}_4(aq) + \text{BaCl}_2(aq) \rightarrow \text{BaSO}_4(s)\downarrow + [\text{Co}(\text{NH}_3)_5\text{Br}]\text{Cl}(aq) $$

Geometrical Structures:

• $cis$-isomer: Identical ligands are located adjacent (at $90^\circ$ angles) to each other:
  $$ \text{Cl} - \text{Pt} - \text{Cl} \quad \text{and} \quad \text{NH}_3 - \text{Pt} - \text{NH}_3 \quad \text{arranged adjacently} $$
• $trans$-isomer: Identical ligands are located opposite (at $180^\circ$ angles) to each other:
  $$ \text{Cl} - \text{Pt} - \text{NH}_3 \quad \text{arranged linearly} $$

Why Tetrahedral Complexes Cannot Show Geometrical Isomerism:

In a tetrahedral geometry (bond angles of $109.5^\circ$), **all four coordination positions are completely equivalent** and equidistant relative to one another in three-dimensional space. Since any arrangement of ligands A and B leads to identical spatial separation, no distinct adjacent ($cis$) or opposite ($trans$) isomer configurations can exist.

Optical Isomerism: Optical isomers (known as enantiomers) are coordination compounds that are **non-superimposable mirror images** of one another. They differ only in their ability to rotate plane-polarized light—the dextrorotatory ($d$) isomer rotates light to the right, and the levorotatory ($l$) isomer rotates it to the left.

Structures of $[\text{Co}(\text{en})_3]^{3+}$ Enantiomers:

These octahedral structures possess no plane of symmetry because of the three bidentate ethylenediamine ($\text{en}$) rings:

• The **$d$-isomer** (dextro) has a right-handed helical spiral of chelate loops.
• The **$l$-isomer** (levo) is the exact mirror image, with a left-handed helical spiral. When placed on top of each other, they cannot be aligned perfectly (non-superimposable).

(i) Linkage Isomerism:

• Type: Linkage Isomerism.
• Difference: This occurs because of the presence of the ambidentate nitrite ligand. In $[\text{Co}(\text{NH}_3)_5(\text{NO}_2)]\text{Cl}_2$, the ligand coordinates through the **nitrogen atom** ($\text{-NO}_2$), making the complex yellow-brown. In $[\text{Co}(\text{NH}_3)_5(\text{ONO})]\text{Cl}_2$, it coordinates through the **oxygen atom** ($\text{-ONO}$), making the complex red and highly sensitive to acids.

(ii) Coordination Isomerism:

• Type: Coordination Isomerism.
• Difference: This arises in salts containing both complex cationic and complex anionic spheres. The isomers differ by the **interchange of ligands** between the two metal centers. In the first isomer, ammonia ligands coordinate with cobalt and cyano ligands coordinate with chromium. In the second isomer, the ligands are swapped completely.

Step 1: Determine oxidation state and electronic configuration of $\text{Fe}$.

Let oxidation state of iron be $x$: $x + 6(-1) = -4 \implies x = +2$.

Step 2: Account for the effect of the ligand.

Cyanide ($\text{CN}^{-}$) is a **strong field ligand**. Under its influence, the 6 electrons in the $3d$ subshell are forced to pair up against Hund's rule, leaving two $3d$ orbitals empty:

Step 3: Hybridisation and orbital overlap.

To accommodate 6 coordinate bonds from the ligands, the central iron ion hybridises its two empty $3d$ orbitals, one $4s$ orbital, and three $4p$ orbitals:

Step 4: Conclusions.

• Geometry: Octahedral
• Magnetic Behavior: Diamagnetic (since all electrons are fully paired, $n = 0$)
• Spin Type: Low-spin complex (or inner orbital octahedral complex)

Step 1: Define central ion parameters.

In both complexes, cobalt is in the $+3$ oxidation state ($\text{Co}^{3+}$):

Step 2: Analyze $[\text{Co}(\text{NH}_3)_6]^{3+}$.

• $\text{NH}_3$ behaves as a **strong field ligand** in this complex. It forces the six $3d$ electrons of $\text{Co}^{3+}$ to pair up: $(\uparrow\downarrow) (\uparrow\downarrow) (\uparrow\downarrow)$, freeing up two $3d$ orbitals.
• These two $3d$, one $4s$, and three $4p$ orbitals undergo **$d^2sp^3$ hybridisation**.
• Since there are no unpaired electrons, the complex is **diamagnetic**. It is an **inner orbital complex**.

Step 3: Analyze $[\text{CoF}_6]^{3-}$.

• Fluoride ($\text{F}^{-}$) is a **weak field ligand** and cannot force electrons to pair. The six $3d$ electrons remain in their high-spin distribution: $(\uparrow\downarrow) (\uparrow) (\uparrow) (\uparrow) (\uparrow)$, leaving no vacant $3d$ orbitals.
• The ion must utilize outer shell orbitals ($4s, 4p, 4d$) to accommodate ligand electrons, undergoing **$sp^3d^2$ hybridisation**.
• With 4 unpaired electrons, the complex is highly **paramagnetic** and is classified as an **outer orbital complex**.

Summary Comparison:

Step 1: Identify nickel's oxidation state in both complexes.

In both complexes, nickel is in the $+2$ oxidation state ($\text{Ni}^{2+}$):

Step 2: Analyze $[\text{Ni}(\text{CN})_4]^{2-}$.

• Cyanide ($\text{CN}^{-}$) is a **strong field ligand**. It forces the eight $3d$ electrons to pair up: $(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)$, leaving one $3d$ orbital completely empty.
• This empty $3d$, one $4s$, and two $4p$ orbitals undergo **$dsp^2$ hybridisation**.
• Geometry: Square Planar
• Magnetic behavior: Diamagnetic (no unpaired electrons)

Step 3: Analyze $[\text{NiCl}_4]^{2-}$.

• Chloride ($\text{Cl}^{-}$) is a **weak field ligand**. It cannot force electron pairing, so the electrons remain distributed as: $(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow)(\uparrow)$ with 2 unpaired electrons.
• One $4s$ and three $4p$ orbitals hybridise to form **$sp^3$ hybridisation**.
• Geometry: Tetrahedral
• Magnetic behavior: Paramagnetic (with 2 unpaired electrons)

Correct Option: (a)

Explanation: In $[\text{Ni(CO)}_4]$, nickel is in the $0$ oxidation state with the configuration $[\text{Ar}] 3d^8 4s^2$. Since carbon monoxide ($\text{CO}$) is an exceptionally strong field ligand, it forces the two electrons in the $4s$ orbital to pair with the eight electrons in the $3d$ orbitals. This shifts nickel's configuration to $3d^{10}$, making it highly stable and diamagnetic. The empty $4s$ and three $4p$ orbitals undergo $sp^3$ hybridisation, yielding a **tetrahedral** geometry.

Comparison:

• $\text{Sc}^{3+}$ Ion: The electronic configuration of Scandium is $[\text{Ar}] 3d^1 4s^2$, so the trivalent ion is $\text{Sc}^{3+} = [\text{Ar}] 3d^0$. Because it contains **zero $d$-electrons** ($d^0$), there are no unpaired electrons (making it **diamagnetic**) and no $d-d$ transitions can take place (making it **colorless**).
• $\text{Ti}^{3+}$ Ion: The electronic configuration of Titanium is $[\text{Ar}] 3d^2 4s^2$, so the trivalent ion is $\text{Ti}^{3+} = [\text{Ar}] 3d^1$. It contains **one unpaired electron** ($n = 1$). This unpaired electron makes the complex **paramagnetic**, and its transition from lower energy $d$-orbitals to higher ones ($d-d$ transition) absorbs yellow-green light, making the complex appear deeply **violet**.

Primary Assumptions of CFT:

1. The bonding between the central metal ion and ligands is assumed to be **purely electrostatic** (ionic) in nature.
2. Ligands are treated as **point charges** (if anionic) or as **point dipoles** (if neutral).
3. The five $d$-orbitals in a free isolated gaseous metal ion are degenerate (possess identical energy).

Octahedral Splitting Mechanism:

In an octahedral complex, the six ligands approach the central metal ion along the three Cartesian coordinate axes ($x, y,$ and $z$).

• The lobes of the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals (called the **$e_g$ set**) point directly along these axes, causing them to experience significant electrostatic repulsion and rise in energy.
• The lobes of the $d_{xy}, d_{yz},$ and $d_{zx}$ orbitals (called the **$t_{2g}$ set**) lie in between the axes. They experience less repulsion, and their energy drops relative to the average energy level (the barycentre).
• This splitting of the degenerate orbitals into $t_{2g}$ (lower energy) and $e_g$ (higher energy) sets is called **crystal field splitting ($\Delta_o$)**.

(A) When $\Delta_o \gt P$ (Strong Field / Low Spin):

• The energy required to split the orbitals ($\Delta_o$) is greater than the pairing energy ($P$) required to force electrons into the same orbital.
• The fourth electron preferentially pairs up in the lower-energy $t_{2g}$ level rather than jumping to the higher $e_g$ level.
• Configuration: $t_{2g}^4 e_g^0$ (unpaired electrons $n = 2$).

(B) When $\Delta_o \lt P$ (Weak Field / High Spin):

• The splitting energy ($\Delta_o$) is less than the pairing energy ($P$).
• The fourth electron will jump to the higher-energy $e_g$ sub-level according to Hund's rule, preventing pairing in the lower level.
• Configuration: $t_{2g}^3 e_g^1$ (unpaired electrons $n = 4$).

Step 1: Determine the configuration of the manganese ion ($\text{Mn}^{2+}$).

Step 2: Distribute electrons in an octahedral field (high-spin).

Since water ($\text{H}_2\text{O}$) is a **weak field ligand**, the splitting energy is less than the pairing energy ($\Delta_o \lt P$). The five electrons are distributed singly across both levels:

This yields $n = 5$ unpaired electrons.

Step 3: Calculate the spin-only magnetic moment ($\mu_s$).

Answer: The crystal field configuration is **$t_{2g}^3 e_g^2$** and the spin-only magnetic moment is **$5.92\text{ B.M.}$**

Concept of Synergic Bonding:

In metal carbonyl complexes (such as $[\text{Ni(CO)}_4]$), bonding between the carbon monoxide ($\text{CO}$) ligand and the metal involves two simultaneous steps:

1. Sigma ($\sigma$) Bond Formation: The carbon atom of the carbonyl ligand donates its lone pair of electrons into a vacant $d$-orbital of the metal ion: $\text{M} \leftarrow \text{C}\equiv\text{O}$.
2. Pi ($\pi$) Backbonding: Simultaneously, a filled non-bonding $d$-orbital of the metal overlaps with the empty **anti-bonding $\pi^{*}$ molecular orbital** of the carbon monoxide ligand, transferring electron density back from the metal to the ligand: $\text{M} \rightarrow \text{C}\equiv\text{O}$.

Synergic Effect: This bidirectional electron transfer is self-reinforcing. The $\sigma$-bond donation increases electron density on the metal, which in turn facilitates stronger $\pi$-backdonation. Conversely, the $\pi$-backdonation decreases electron density on the metal, facilitating stronger $\sigma$-bonding. This highly stable interaction is called **synergic bonding**, and it significantly strengthens the overall metal-carbon bond.

Spectrochemical Series: It is an experimentally determined series in which common ligands are arranged in the increasing order of their **crystal field splitting energy ($\Delta_o$)** power.

Ligands that produce small crystal field splitting are classified as weak field ligands, whereas those that produce large splitting are classified as strong field ligands.

Increasing Order of Splitting Power:

Where chloride ($\text{Cl}^{-}$) is the weakest ligand, and cyanide ($\text{CN}^{-}$) is the strongest among the group.