Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic con

Exercises · 5.24

Q. Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex:

(i)

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

(ii)

[Co(NH_3)_5Cl]Cl_2

(iii)

[CrCl_3(py)_3]

(iv)

Cs[FeCl_4]

(v)

K_4[Mn(CN)_6]

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Concept understanding — Crystal Field Splitting

Crystal Field Splitting: From Intuition to Precision

Imagine you are a negatively charged electron sitting on a metal ion. All around you, the space is perfectly spherical — every direction feels the same. Your energy depends only on how far you are from the nucleus, not on which way you face.

Now imagine that six negative ions (or the negative ends of polar molecules) march in from the $x$ , $y$ , and $z$ axes and stop close to you. Suddenly, the space around you is no longer uniform. If you try to move straight toward one of these approaching ions, you feel a strong repulsion — that path costs extra energy. If you move between the axes (say, along a diagonal), you feel less repulsion because you are farther from the incoming charges.

This is the core intuition: when ligands approach a metal ion, they break the spherical symmetry of the space around the metal. Different directions in space are no longer equivalent. Electrons in orbitals that point directly at the ligands get pushed up in energy; electrons in orbitals that point between the ligands stay lower.

The Precise Statement

d

For the most common geometry — octahedral — here is what happens:

Six ligands sit at the corners of an octahedron, along the $+x$ , $-x$ , $+y$ , $-y$ , $+z$ , $-z$ axes.
The $d_{x^2-y^2}$ and $d_{z^2}$ orbitals point their lobes directly along these axes. These are the $e_g$ set. They feel maximum repulsion → higher energy.
The $d_{xy}$ , $d_{xz}$ , and $d_{yz}$ orbitals point their lobes between the axes (into the octahedral faces). These are the $t_{2g}$ set. They feel less repulsion → lower energy.

The energy gap between these two sets is denoted by $\Delta_o$ (or $10\,Dq$ ). The $t_{2g}$ set drops by $0.4\Delta_o$ and the $e_g$ set rises by $0.6\Delta_o$ , keeping the average energy unchanged (the "barycentre" rule).

Note

The labels $e_g$ and $t_{2g}$ come from group theory — they describe how the orbitals transform under the symmetry operations of an octahedron. You do not need to memorise the derivation, but the notation is standard in every exam.

Why This Matters

Crystal field splitting explains three things you will see repeatedly:

Colour — electrons can jump from $t_{2g}$ to $e_g$ by absorbing visible light. The gap $\Delta_o$ determines the colour you see.
Magnetism — if $\Delta_o$ is large, electrons pair up in the lower $t_{2g}$ set (low spin). If $\Delta_o$ is small, electrons spread out (high spin). This changes the number of unpaired electrons. …

Why this formula?

Crystal Field Splitting: Why the Energy Splitting Occurs

Crystal Field Theory (CFT) explains how the d-orbitals of a transition metal ion split in energy when placed in an electrostatic field created by surrounding ligands (anions or polar molecules). The key result is that five degenerate d-orbitals split into two or more sets with different energies. Let's understand why this happens.

1. The Starting Point: Degenerate d-Orbitals

In a free transition metal ion (no ligands), all five d-orbitals have the same energy (degenerate). Their shapes are:

$d_{xy}$ , $d_{xz}$ , $d_{yz}$ — lobes lie between the x, y, z axes (called $t_{2g}$ set in octahedral symmetry)
$d_{x^2-y^2}$ , $d_{z^2}$ — lobes point directly along the x, y, z axes (called $e_g$ set)

Key idea: The spatial orientation of each orbital determines how it interacts with approaching ligands.

2. The Octahedral Case: Why $e_g$ Orbitals Are Higher in Energy

Imagine six ligands approaching along the +x, –x, +y, –y, +z, –z axes (octahedral geometry).

What happens to $d_{x^2-y^2}$ and $d_{z^2}$ ?

Their lobes point directly at the ligands.
The negatively charged ligands repel the electron density in these orbitals.
This repulsion raises the energy of these orbitals — they become less stable (higher energy).

What happens to $d_{xy}$ , $d_{xz}$ , $d_{yz}$ ?

Their lobes point between the axes (e.g., $d_{xy}$ lobes lie in the xy-plane but at 45° to x and y).
They avoid the ligands — less repulsion.
Their energy is lower than the $e_g$ set.

The Splitting Pattern

\Delta_{\text{oct}} = E(e_g) - E(t_{2g})

Where:

$E(e_g)$ = energy of $d_{x^2-y^2}$ and $d_{z^2}$ (higher)
$E(t_{2g})$ = energy of $d_{xy}$ , $d_{xz}$ , $d_{yz}$ (lower)
$\Delta_{\text{oct}}$ is called the crystal field splitting energy (CFSE)

Why the name? The $e_g$ orbitals are "doubly degenerate" (2 orbitals), $t_{2g}$ are "triply degenerate" (3 orbitals). The letters come from group theory symmetry labels.

3. The Energy Conservation Rule

The total energy of all five d-orbitals must remain constant (no energy is created or destroyed). So:

The center of gravity (average energy) of the split set equals the original degenerate energy.
For octahedral splitting:
- 2 $e_g$ orbitals go up by $+0.6\Delta_{\text{oct}}$ each
- 3 $t_{2g}$ orbitals go down by $-0.4\Delta_{\text{oct}}$ each

Check:

2 \times (+0.6\Delta) + 3 \times (-0.4\Delta) = 1.2\Delta - 1.2\Delta = 0

This conservation of energy is a fundamental constraint — the splitting is not arbitrary.

4. The Tetrahedral Case: Why It's Opposite and Smaller

In a tetrahedral complex, four ligands approach from alternate corners of a cube. The axes are different:

The $d_{xy}$ , $d_{xz}$ , $d_{yz}$ orbitals now point closer to the ligands (more repulsion).
The $d_{x^2-y^2}$ and $d_{z^2}$ orbitals point away from ligands (less repulsion).

Result:

$e$ set ( $d_{x^2-y^2}$ , $d_{z^2}$ ) — lower energy
$t_2$ set ( $d_{xy}$ , $d_{xz}$ , $d_{yz}$ ) — higher energy

The splitting is inverted compared to octahedral.

Magnitude:

\Delta_{\text{tet}} \approx \frac{4}{9} \Delta_{\text{oct}}

Why smaller?

Only 4 ligands (vs. 6) → less total repulsion.
Ligands are not directly along axes → weaker interaction. …

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Concept: Crystal Field Splitting – the arrangement of ligands around a central metal ion splits the d-orbital energies, determining electronic configuration, magnetic moment, and stereochemistry.

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

IUPAC name: Potassium diaquadioxalatochromate(III) trihydrate
Oxidation state: Cr = +3 (K⁺, 2×H₂O neutral, 2×C₂O₄²⁻ → +1 + x + 0 + 2(−2) = 0 ⇒ x = +3)
Electronic configuration: Cr³⁺ = $[Ar]3d^3$
Coordination number: 6 (2 H₂O + 2 bidentate oxalate = 6 donor atoms)
Stereochemistry: Octahedral; oxalate is bidentate, so cis/trans isomers possible — here, the complex is likely cis due to chelate ring stability.
Magnetic moment: 3 unpaired electrons → $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ BM

(ii) $[Co(NH_3)_5Cl]Cl_2$

IUPAC name: Pentaamminechloridocobalt(III) chloride
Oxidation state: Co = +3 (5 NH₃ neutral, Cl⁻ ligand, 2 Cl⁻ counterions → x + 0 − 1 − 2 = 0 ⇒ x = +3)
Electronic configuration: Co³⁺ = $[Ar]3d^6$
Coordination number: 6
Stereochemistry: Octahedral; no geometrical isomerism (only one Cl ligand).
Magnetic moment: Strong-field NH₃ causes low-spin $t_{2g}^6$ → 0 unpaired electrons → $\mu = 0$ BM

(iii) $[CrCl_3(py)_3]$

IUPAC name: Trichloridotris(pyridine)chromium(III)
Oxidation state: Cr = +3 (3 Cl⁻, 3 pyridine neutral → x − 3 = 0 ⇒ x = +3)
Electronic configuration: Cr³⁺ = $[Ar]3d^3$
Coordination number: 6
Stereochemistry: Octahedral; facial (fac) and meridional (mer) isomers possible — both exist.
Magnetic moment: 3 unpaired electrons → $\mu \approx 3.87$ BM

(iv) $Cs[FeCl_4]$

IUPAC name: Caesium tetrachloridoferrate(III)
Oxidation state: Fe = +3 (Cs⁺, 4 Cl⁻ → +1 + x − 4 = 0 ⇒ x = +3) …

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This problem asks for the systematic IUPAC name, oxidation state, electronic configuration, coordination number, stereochemistry, and magnetic moment for five coordination complexes. The key is to determine the metal’s oxidation state from the overall charge and ligand charges, then use crystal field theory to predict the d-electron count, geometry, and magnetic behavior. For each complex, the final answer is given in a compact table below.

Let’s work through each complex one by one. The approach is always the same: find the oxidation state of the metal, then its d-electron configuration, then the coordination number (which tells you geometry), and finally use the ligand field strength to decide whether the complex is high-spin or low-spin — that determines the magnetic moment.

1. Complex (i): $K[Cr(H_2O)_2(C_2O_4)_2] \cdot 3H_2O$

Step 1: Oxidation state of Cr.

The complex ion is $[Cr(H_2O)_2(C_2O_4)_2]^-$ (since K⁺ is +1, the anion must be –1). Water is neutral, oxalate ( $C_2O_4^{2-}$ ) is –2 each. Let Cr oxidation state be $x$ :

$x + 2(0) + 2(-2) = -1 \implies x - 4 = -1 \implies x = +3$ .

So Cr is in +3 state.

Step 2: Electronic configuration.

Cr (atomic number 24): $[Ar] 3d^5 4s^1$ . For Cr³⁺, remove three electrons (the 4s electron and two 3d electrons): $[Ar] 3d^3$ . So $d^3$ configuration.

Step 3: Coordination number.

Water is monodentate (one donor oxygen atom) and oxalate ( $C_2O_4^{2-}$ ) is bidentate (two donor oxygen atoms). So total donor atoms: 2 from the two water molecules + 2 × 2 from the two oxalate ligands = 6. Coordination number = 6. Geometry: octahedral.

Step 4: Stereochemistry.

For an octahedral complex with two monodentate and two bidentate ligands, the possible isomers are cis and trans. Here, the two water molecules can be adjacent (cis) or opposite (trans). The given formula doesn’t specify which, but typically such complexes are cis because oxalate is a chelating ligand that prefers to occupy adjacent sites. We’ll note both possibilities, but the common form is cis.

Step 5: Magnetic moment.

For $d^3$ in an octahedral field, the crystal field splitting puts three electrons in the $t_{2g}$ orbitals (all unpaired). No pairing possible because $t_{2g}$ holds three electrons singly. So number of unpaired electrons $n = 3$ . Magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{3 \times 5} = \sqrt{15} \approx 3.87$ BM.

Watch out

A common mistake is to think Cr³⁺ is $d^4$ or to forget that oxalate is bidentate — that would mess up the coordination number. Always count donor atoms, not just ligand molecules.

IUPAC name: Potassium diaquadioxalatochromate(III) trihydrate.

(Note: “diaqua” for two water, “dioxalato” for two oxalate ligands — following the same simple-prefix convention NCERT itself uses for oxalato, e.g. “trioxalatochromate(III)” for [Cr(C2O4)3]3– — “chromate(III)” for the metal in –ate form, and “trihydrate” for the three water molecules outside the coordination sphere.)

2. Complex (ii): $[Co(NH_3)_5Cl]Cl_2$

Step 1: Oxidation state of Co.

The complex ion is $[Co(NH_3)_5Cl]^{2+}$ (because two Cl⁻ outside). NH₃ is neutral, Cl⁻ is –1. Let Co be $x$ :

$x + 5(0) + (-1) = +2 \implies x - 1 = +2 \implies x = +3$ .

So Co is +3.

Step 2: Electronic configuration.

Co (atomic number 27): $[Ar] 3d^7 4s^2$ . For Co³⁺, remove three electrons: $[Ar] 3d^6$ . So $d^6$ configuration.

Step 3: Coordination number.

Five NH₃ (monodentate) + one Cl⁻ (monodentate) = 6 donor atoms. Coordination number = 6, octahedral.

Step 4: Stereochemistry.

With five identical ligands and one different, the only isomerism possible is the position of Cl — but since all NH₃ are equivalent, there’s no cis/trans isomerism here. The complex is simply octahedral with one Cl ligand.

Step 5: Magnetic moment.

NH₃ is a strong field ligand (high in spectrochemical series), so for $d^6$ in an octahedral field, strong field leads to low-spin configuration: all six electrons pair up in $t_{2g}$ ( $t_{2g}^6$ ), leaving $e_g$ empty. Number of unpaired electrons $n = 0$ . Magnetic moment $\mu = 0$ BM (diamagnetic).

Tip

For $d^6$ octahedral complexes, the magnetic moment is a quick giveaway: if the complex is low-spin (strong field), $\mu = 0$ ; if high-spin (weak field), $\mu = \sqrt{4 \times 6} \approx 4.90$ BM. Here, NH₃ is strong enough to cause pairing.

IUPAC name: Pentaamminechloridocobalt(III) chloride.

(Note: “pentaammine” for five NH₃, “chlorido” for Cl ligand, “cobalt(III)” for the metal, and “chloride” for the counterions.)

3. Complex (iii): $[CrCl_3(py)_3]$

Step 1: Oxidation state of Cr.

The complex is neutral. py (pyridine) is neutral, Cl⁻ is –1 each. Let Cr be $x$ :

$x + 3(-1) + 3(0) = 0 \implies x - 3 = 0 \implies x = +3$ .

So Cr is +3 again.

Step 2: Electronic configuration.

Cr³⁺: $[Ar] 3d^3$ , same as in (i).

Step 3: Coordination number.

Three Cl⁻ (monodentate) + three py (monodentate) = 6 donor atoms. Coordination number = 6, octahedral.

Step 4: Stereochemistry.

Here we have three identical Cl and three identical py ligands. This gives two possible geometric isomers: facial (fac) where three Cl occupy one face of the octahedron, and meridional (mer) where three Cl lie in a plane. Both are possible, but the complex is often isolated as the fac isomer because it is more stable. We’ll note both.

Step 5: Magnetic moment.

Again $d^3$ , so $n = 3$ , $\mu = \sqrt{15} \approx 3.87$ BM. (pyridine is a moderate field ligand, but for $d^3$ there’s no high-spin/low-spin choice — always three unpaired electrons.)

IUPAC name: Trichloridotris(pyridine)chromium(III).

(Or: fac-trichloridotris(pyridine)chromium(III) if specifying stereochemistry.)

4. Complex (iv): $Cs[FeCl_4]$

Step 1: Oxidation state of Fe.

The complex ion is $[FeCl_4]^-$ (Cs⁺ is +1, so anion is –1). Cl⁻ is –1 each. Let Fe be $x$ :

$x + 4(-1) = -1 \implies x - 4 = -1 \implies x = +3$ .

So Fe is +3.

Step 2: Electronic configuration.

Fe (atomic number 26): $[Ar] 3d^6 4s^2$ . For Fe³⁺, remove three electrons: $[Ar] 3d^5$ . So $d^5$ configuration.

Step 3: Coordination number.

Four Cl⁻ ligands, each monodentate. Coordination number = 4. Geometry: tetrahedral (since Cl⁻ is a weak field ligand and Fe³⁺ is $d^5$ , tetrahedral is favored over square planar).

Step 4: Stereochemistry.

Tetrahedral complexes with four identical ligands have no geometric isomerism. So no stereoisomers.

Step 5: Magnetic moment.

For tetrahedral $d^5$ , the crystal field splitting is smaller than octahedral, and the configuration is always high-spin (because pairing energy is larger than splitting). The five electrons occupy all five d-orbitals singly: $e^2 t_2^3$ . Number of unpaired electrons $n = 5$ . Magnetic moment $\mu = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92$ BM. …

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Excellent. Here is the Crystal Field Splitting method applied to each complex. The method is Crystal Field Theory (CFT) Analysis, which involves determining the geometry, oxidation state, d-electron count, and then the splitting pattern (high-spin vs low-spin) to find the magnetic moment.

Method: Crystal Field Theory (CFT) Analysis

Steps:

Determine the Oxidation State of the central metal ion.
Determine the Coordination Number (CN) and Geometry (e.g., octahedral, tetrahedral).
Determine the d-electron configuration ( $d^n$ ) of the metal ion.
Identify the ligand field strength (strong field = low spin, weak field = high spin for $d^4$ to $d^7$ ).
Draw the Crystal Field Splitting diagram (e.g., $t_{2g}$ and $e_g$ for octahedral; $e$ and $t_2$ for tetrahedral).
Calculate the Magnetic Moment using the spin-only formula:

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

where $n$ = number of unpaired electrons.

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

IUPAC Name: Potassium diaquadioxalatochromate(III) trihydrate
Oxidation State: $+3$ (K is +1, oxalate is -2 each, water is neutral; $1 + x + 0 + 2(-2) = 0 \Rightarrow x = +3$ )
Electronic Configuration: $Cr^{3+}$ is $[Ar]3d^3$
Coordination Number: 6 (2 water + 2 bidentate oxalates = 6 donor atoms)
Stereochemistry: Octahedral
Ligand Field: $H_2O$ and $C_2O_4^{2-}$ are weak field ligands.
CFT Diagram (Octahedral, $d^3$ ): All three electrons go into the $t_{2g}$ $t_{2 g}$ orbitals (Hund's rule).
- Unpaired electrons: 3
Magnetic Moment:

\mu = \sqrt{3(3+2)} = \sqrt{15} \approx \mathbf{3.87 \text{ BM}}

(ii) $[Co(NH_3)_5Cl]Cl_2$

IUPAC Name: Pentaamminechloridocobalt(III) chloride
Oxidation State: $+3$ (5 NH $_3$ are neutral, Cl inside coordination sphere is -1, 2 Cl outside are -2; $x + 0 - 1 - 2 = 0 \Rightarrow x = +3$ )
Electronic Configuration: $Co^{3+}$ is $[Ar]3d^6$
Coordination Number: 6
Stereochemistry: Octahedral
Ligand Field: NH $_3$ is a strong field ligand (causes pairing).
CFT Diagram (Octahedral, $d^6$ , Low Spin): All 6 electrons pair up in the $t_{2g}$ $t_{2 g}$ orbitals.
- Unpaired electrons: 0
Magnetic Moment:

\mu = \sqrt{0(0+2)} = \mathbf{0 \text{ BM}} \text{ (Diamagnetic)}

(iii) $[CrCl_3(py)_3]$

IUPAC Name: Trichloridotris(pyridine)chromium(III)
Oxidation State: $+3$ (3 Cl are -3, pyridine is neutral; $x - 3 = 0 \Rightarrow x = +3$ )
Electronic Configuration: $Cr^{3+}$ is $[Ar]3d^3$
Coordination Number: 6
Stereochemistry: Octahedral (can exist as facial and meridional isomers, but the question asks for stereochemistry in general).
Ligand Field: Cl $^-$ and pyridine are weak to intermediate field ligands. For $d^3$ , there is no high-spin/low-spin choice.
CFT Diagram (Octahedral, $d^3$ ): Three unpaired electrons in $t_{2g}$ $t_{2 g}$ .
- Unpaired electrons: 3
Magnetic Moment: …

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Here is a breakdown of the common mistakes students make with Crystal Field Splitting and coordination compound problems, specifically for the question you provided.

1. Mistake: Forgetting the "Outside" Water in Hydrated Salts

The Error: In complex (i) $K[Cr(H_2O)_2(C_2O_4)_2] \cdot 3H_2O$ , students often count the 3 water molecules as ligands inside the coordination sphere. This leads to a wrong coordination number (e.g., 7 instead of 6) and a wrong IUPAC name.

Why it happens: The dot ( $\cdot$ ) indicates water of crystallization (hydration), not coordination. These water molecules are outside the square brackets.

How to Avoid:

Rule: Anything inside the square brackets [ ] is the coordination sphere. Anything outside is a counter-ion or water of crystallization.
Action: For naming, write the complex part first, then add the hydrated water as a separate word.
Correct Approach: The complex is [Cr(H2O)2(C2O4)2]^-. The coordination number is 6 (2 water + 2 oxalate, each oxalate is bidentate).

2. Mistake: Incorrect Oxidation State Calculation (Especially with Oxalate)

The Error: In complex (i), students forget that oxalate ( $C_2O_4^{2-}$ ) is a -2 ligand. They might treat it as neutral or -1, leading to a wrong oxidation state for Cr.

Why it happens: Oxalate is a common bidentate ligand, but its charge is often overlooked.

How to Avoid:

Memorize Ligand Charges: Oxalate = -2, Water = 0, Ammonia = 0, Chloride = -1, Cyanide = -1, Pyridine (py) = 0.
Step-by-step calculation:
- Complex (i): $K^+$ is +1. The complex ion is $[Cr(H_2O)_2(C_2O_4)_2]^-$ .
- Let Cr oxidation state = $x$ .
- $x + 2(0) + 2(-2) = -1 \implies x - 4 = -1 \implies x = +3$ .
- Result: Cr is in +3 oxidation state.

3. Mistake: Confusing Coordination Number with Number of Ligands

The Error: In complex (i), students say coordination number = 4 (because there are 4 ligand molecules: 2 water + 2 oxalate). They forget that oxalate is bidentate (donates 2 electron pairs).

Why it happens: They count "things" instead of "donor atoms."

How to Avoid:

Definition: Coordination number = number of donor atoms directly bonded to the metal.
Action: For each ligand, ask: "How many atoms does it use to bond?"
- Water ( $H_2O$ ): Monodentate (1 donor atom). 2 waters = 2 donor atoms.
- Oxalate ( $C_2O_4^{2-}$ ): Bidentate (2 donor atoms). 2 oxalates = 4 donor atoms.
- Total: 2 + 4 = 6. Coordination number = 6.

4. Mistake: Wrong Electronic Configuration (Ignoring Crystal Field)

The Error: For complex (v) $K_4[Mn(CN)_6]$ , students write the configuration for a free $Mn^{2+}$ ion ( $d^5$ ) as $t_{2g}^3 e_g^2$ (high spin). They forget that CN⁻ is a strong field ligand.

Why it happens: They apply the "half-filled stability" rule blindly without considering the ligand's strength.

How to Avoid:

Step 1: Find oxidation state. $K_4$ = +4. CN⁻ = -1 each (total -6). So Mn must be +2 ( $d^5$ ).
Step 2: Identify ligand. CN⁻ is a strong field ligand (causes large splitting).
Step 3: Apply pairing. Strong field → electrons pair up in $t_{2g}$ before going to $e_g$ .
Correct Configuration: $t_{2g}^5 e_g^0$ (low spin).
Magnetic Moment: $\mu = \sqrt{n(n+2)}$ where $n$ = number of unpaired electrons. Here, $n=1$ . $\mu = \sqrt{1(3)} = \sqrt{3} \approx 1.73$ BM.

5. Mistake: Ignoring Stereochemistry (Geometrical Isomerism)

The Error: For complex (iii) $[CrCl_3(py)_3]$ , students just write "octahedral" and stop. They miss the fac-mer isomerism.

Why it happens: They think all $[MA_3B_3]$ complexes are the same.

How to Avoid:

Recognize the formula: $[M(A)_3(B)_3]$ is a classic case of geometrical isomerism in octahedral complexes.
Two possibilities:
- Facial (fac): Three identical ligands occupy the corners of one face of the octahedron.
- Meridional (mer): Three identical ligands lie in a plane that goes through the metal (a meridian).
Action: For $[CrCl_3(py)_3]$ , both fac and mer isomers are possible. You must state this.

6. Mistake: Forgetting to Specify the Geometry for Coordination Number 4 …

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Q.Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex:

1. Complex (i): K[Cr(H2O)2(C2O4)2]⋅3H2OK[Cr(H_2O)_2(C_2O_4)_2] \cdot 3H_2OK[Cr(H2​O)2​(C2​O4​)2​]⋅3H2​O

2. Complex (ii): [Co(NH3)5Cl]Cl2[Co(NH_3)_5Cl]Cl_2[Co(NH3​)5​Cl]Cl2​

3. Complex (iii): [CrCl3(py)3][CrCl_3(py)_3][CrCl3​(py)3​]

4. Complex (iv): Cs[FeCl4]Cs[FeCl_4]Cs[FeCl4​]

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1. Complex (i): $K[Cr(H_2O)_2(C_2O_4)_2] \cdot 3H_2O$

2. Complex (ii): $[Co(NH_3)_5Cl]Cl_2$

3. Complex (iii): $[CrCl_3(py)_3]$

4. Complex (iv): $Cs[FeCl_4]$

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