Chapter 5: Coordination Compounds

The number of moles of AgCl precipitated with excess $$\mathrm{AgNO_3}$$ equals the number of ionisable (free) chloride ions present outside the coordination sphere. Chloride ions bound to the metal inside the coordination sphere (acting as ligands) are not precipitated.

The secondary valence of the metal equals its coordination number — the total number of ligands directly bound to the metal inside the square brackets.

(i) $$\mathrm{PdCl_2 \cdot 4NH_3}$$: 2 mol AgCl $$\Rightarrow$$ 2 free $$\mathrm{Cl^-}$$ ions. So the compound is $$\mathrm{[Pd(NH_3)_4]Cl_2}$$. The four $$\mathrm{NH_3}$$ molecules lie inside the sphere $$\Rightarrow$$ secondary valence = 4.

(ii) $$\mathrm{NiCl_2 \cdot 6H_2O}$$: 2 mol AgCl $$\Rightarrow$$ 2 free $$\mathrm{Cl^-}$$ ions. So it is $$\mathrm{[Ni(H_2O)_6]Cl_2}$$. Six $$\mathrm{H_2O}$$ molecules lie inside the sphere $$\Rightarrow$$ secondary valence = 6.

(iii) $$\mathrm{PtCl_4 \cdot 2HCl}$$: 0 mol AgCl $$\Rightarrow$$ no free $$\mathrm{Cl^-}$$. All six chloride ions lie inside the sphere: $$\mathrm{H_2[PtCl_6]}$$ $$\Rightarrow$$ secondary valence = 6.

(iv) $$\mathrm{CoCl_3 \cdot 4NH_3}$$: 1 mol AgCl $$\Rightarrow$$ 1 free $$\mathrm{Cl^-}$$. So it is $$\mathrm{[Co(NH_3)_4Cl_2]Cl}$$, with 4 $$\mathrm{NH_3}$$ + 2 $$\mathrm{Cl^-}$$ inside the sphere $$\Rightarrow$$ secondary valence = 6.

(v) $$\mathrm{PtCl_2 \cdot 2NH_3}$$: 0 mol AgCl $$\Rightarrow$$ no free $$\mathrm{Cl^-}$$. All ligands lie inside: $$\mathrm{[Pt(NH_3)_2Cl_2]}$$, with 2 $$\mathrm{NH_3}$$ + 2 $$\mathrm{Cl^-}$$ $$\Rightarrow$$ secondary valence = 4.

In a tetrahedral complex the central metal lies at the centre and the four ligands occupy the four corners of a regular tetrahedron. Every corner of a tetrahedron is adjacent to every other corner, and all four positions are geometrically equivalent with respect to one another — there is no pair of positions that is uniquely 'opposite' (trans) and no pair that is uniquely 'adjacent' (cis).

Geometrical (cis–trans) isomerism requires that ligands can be placed in distinguishable relative positions. Since interchanging any two ligands of a tetrahedral complex produces a structure that can simply be rotated to coincide with the original, no distinct spatial arrangements arise.

Therefore tetrahedral complexes with two (or more) different types of unidentate ligands do not show geometrical isomerism.

The entity $$\mathrm{[Fe(NH_3)_2(CN)_4]^-}$$ is octahedral and is of the type $$\mathrm{[MA_4B_2]}$$, where $$\mathrm{A = CN^-}$$ and $$\mathrm{B = NH_3}$$. Such a complex shows two geometrical isomers, depending on the relative positions of the two $$\mathrm{NH_3}$$ ligands.

cis isomer: Draw Fe at the centre of an octahedron. Place the two $$\mathrm{NH_3}$$ ligands on two adjacent corners (at $$90^\circ$$ to each other) and the four $$\mathrm{CN^-}$$ ligands on the remaining four corners.

trans isomer: Place the two $$\mathrm{NH_3}$$ ligands on two opposite corners (at $$180^\circ$$, along one axis); the four $$\mathrm{CN^-}$$ ligands then occupy the four corners of the perpendicular square plane.

Each structure carries an overall charge of $$-1$$.

In $$\mathrm{[MnBr_4]^{2-}}$$ the bromide ligands carry a charge of $$4(-1) = -4$$ and the ion has charge $$-2$$, so manganese is in the $$+2$$ state. $$\mathrm{Mn^{2+}}$$ has the configuration $$3d^5$$.

The spin-only magnetic moment is related to the number of unpaired electrons $$n$$ by $$\mu = \sqrt{n(n+2)} \ \mathrm{BM}$$.

$$5.9 = \sqrt{n(n+2)} \Rightarrow n(n+2) = 34.8 \Rightarrow n = 5$$

So all five $$3d$$ electrons are unpaired. The coordination number is 4, so the geometry is either tetrahedral ($$sp^3$$) or square planar ($$dsp^2$$).

A square planar arrangement ($$dsp^2$$) would require one $$3d$$ orbital for hybridisation, which would force the $$d$$ electrons to pair up and reduce the number of unpaired electrons. Since all five electrons remain unpaired, no $$d$$ orbital is used in hybridisation; the hybridisation is $$sp^3$$.

Hence the complex ion is tetrahedral.

Step 1 : Write the formulae of the two salts clearly.
( i )  $$\mathrm{[Co(NH_3)_5Cl]SO_4}$$ ( ii )  $$\mathrm{[Co(NH_3)_5(SO_4)]Cl}$$

Step 2 : Show the species produced on dissolving each salt in water.

Step 3 : Devise simple qualitative tests for the free ions.

• Test for free sulphate  : add a few drops of $$\mathrm{BaCl_2}$$ solution. A dense white precipitate of $$\mathrm{BaSO_4}$$ confirms the presence of $$SO_4^{2-}$$ ions.
• Test for free chloride  : add a few drops of $$\mathrm{AgNO_3}$$ solution. A curdy white precipitate of $$\mathrm{AgCl}$$ confirms the presence of $$Cl^-$$ ions.

Step 4 : Carry out the tests on the two solutions.

Step 5 : Interpret the observations.
The two compounds contain the same atoms, but in aqueous solution they furnish different ions:

• Compound (i) gives $$SO_4^{2-}$$ ions but no $$Cl^-$$ ions.
• Compound (ii) gives $$Cl^-$$ ions but no $$SO_4^{2-}$$ ions.

Hence the sulphate ion acts as the counter-ion in (i) and as a coordinated ligand in (ii), while the chloride ion behaves oppositely. This exchange of a ligand with the counter-ion without any change in the chemical formula is the hallmark of ionisation isomerism.

Step 6 : State the conclusion.
Because the two complexes yield different ions in solution, the experimental evidence proves that $$\mathrm{[Co(NH_3)_5Cl]SO_4}$$ and $$\mathrm{[Co(NH_3)_5(SO_4)]Cl}$$ are ionisation isomers.

Step 1 : Oxidation state and ground-state electronic configuration of Ni

• Atomic number of Ni = 28 → ground state $$[\mathrm{Ar}]\,3d^{8}\,4s^{2}$$.
• In both complexes the metal is $$\mathrm{Ni^{2+}}$$ (overall charge $$-2$$ and each ligand is $$-1$$). Removing two 4s electrons gives

$$\mathrm{Ni^{2+}}:[\mathrm{Ar}]\,3d^{8}$$  (five 3d orbitals).

Step 2 : Filling of the five 3d orbitals in the free ion

Eight d-electrons distribute themselves in the five 3d orbitals according to Hund’s rule (three orbitals doubly occupied and two singly occupied):

Number of unpaired electrons in the free ion = 2.

Part A : $$\mathrm{[Ni(CN)_4]^{2-}}$$

1. Nature of ligand – $$\mathrm{CN^-}$$ is a strong-field ligand; its field exceeds the pairing energy, so the two unpaired d-electrons pair up.
2. Pairing inside the 3d set – the eight electrons now occupy only four of the five 3d orbitals:

All eight electrons are paired; the inner orbital $$d_{x^2-y^2}$$ is now vacant.

3. Hybridisation – the vacant inner $$3d_{x^2-y^2}$$, the $$4s$$ and two $$4p$$ orbitals hybridise to give four $$dsp^{2}$$ hybrid orbitals lying in one plane (square-planar geometry).
4. Bond formation – each $$\mathrm{CN^-}$$ donates a lone pair to one $$dsp^{2}$$ orbital, forming four σ-bonds. All electrons are paired → the ion is diamagnetic.

Result: $$\mathrm{[Ni(CN)_4]^{2-}}$$ is an inner-orbital ($$dsp^{2}$$), square-planar, diamagnetic complex.

Part B : $$\mathrm{[NiCl_4]^{2-}}$$

1. Nature of ligand – $$\mathrm{Cl^-}$$ is a weak-field ligand; its field is smaller than the pairing energy, so the two unpaired d-electrons do not pair.
2. 3d configuration remains as in the free ion – still two unpaired electrons.
3. Hybridisation – no inner 3d orbital is vacant; therefore one $$4s$$ and three $$4p$$ orbitals hybridise to give four $$sp^{3}$$ hybrid orbitals pointing to the corners of a tetrahedron.
4. Bond formation – each $$\mathrm{Cl^-}$$ donates a lone pair to one $$sp^{3}$$ hybrid orbital. The two unpaired 3d electrons remain unpaired, so the complex is paramagnetic ($$\mu = \sqrt{2(2+2)} \approx 2.83\,\mathrm{BM}$$).

Result: $$\mathrm{[NiCl_4]^{2-}}$$ is an outer-orbital ($$sp^{3}$$), tetrahedral, paramagnetic complex with 2 unpaired electrons.

Conclusion

• Strong-field $$\mathrm{CN^-}$$ causes pairing, leaving an inner 3d orbital empty; the ion is square-planar ($$dsp^{2}$$) and diamagnetic.
• Weak-field $$\mathrm{Cl^-}$$ cannot cause pairing; the ion uses outer-orbital $$sp^{3}$$ hybridisation, remains tetrahedral and paramagnetic with two unpaired electrons.

Step 1 : Electronic configuration of free Ni atom

Atomic number 28 → $$\mathrm{Ni}:[\mathrm{Ar}]\,3d^{8}\,4s^{2}$$.

Step 2 : Oxidation state and d-electron count in each complex

• For $$\mathrm{[NiCl_4]^{2-}}$$: $$x + 4(-1) = -2 \Rightarrow x = +2$$. Hence Ni is $$\mathrm{Ni^{2+}}$$ with configuration $$[\mathrm{Ar}]\,3d^{8}\,4s^{0}$$.
• For $$\mathrm{[Ni(CO)_4]}$$: CO is neutral, so the oxidation state of Ni is 0. The metal retains its ground-state configuration $$[\mathrm{Ar}]\,3d^{8}\,4s^{2}$$.

Step 3 : Bonding in $$\mathrm{[NiCl_4]^{2-}}$$ (VBT)

• $$\mathrm{Cl^-}$$ is a weak-field ligand and does not force pairing of the 3d electrons of $$\mathrm{Ni^{2+}}$$.
• Distribution of the eight 3d electrons (Hund’s rule):

• No inner d-orbital is vacant, so Ni(II) uses the empty $$4s$$ and three $$4p$$ orbitals to make four $$sp^{3}$$ hybrid orbitals pointing towards the corners of a tetrahedron.
• Each $$\mathrm{Cl^-}$$ donates its lone pair to one $$sp^{3}$$ hybrid orbital. The two unpaired 3d electrons remain unpaired, so the complex is paramagnetic ($$\mu = \sqrt{2(2+2)} \approx 2.83\,\mathrm{BM}$$).

Step 4 : Bonding in $$\mathrm{[Ni(CO)_4]}$$ (VBT)

• $$\mathrm{CO}$$ is a very strong-field $$\pi$$-acceptor ligand. Its field is large enough to push the two 4s electrons into the 3d set and pair every electron there:
    $$\mathrm{Ni^{0}}\;(3d^{8}\,4s^{2}) \;\longrightarrow\; 3d^{10}\,4s^{0}$$.
• All ten d-electrons are now paired. The empty $$4s$$ and three $$4p$$ orbitals hybridise to give four $$sp^{3}$$ hybrid orbitals directed tetrahedrally.
• Each CO donates a lone pair into one $$sp^{3}$$ hybrid orbital. With no unpaired electron left, the complex is diamagnetic.

Step 5 : Conclusion

Both complexes are tetrahedral ($$sp^{3}$$) but their magnetic behaviour differs because of the ligand-field strength. The weak-field $$\mathrm{Cl^-}$$ leaves the two unpaired 3d electrons of $$\mathrm{Ni^{2+}}$$ intact → paramagnetic; the very strong-field CO pairs every electron in $$\mathrm{Ni^{0}}$$ to give $$3d^{10}$$ → diamagnetic.

Step 1 : Oxidation state and d‑electron count of the metal
For both complexes the metal is iron in the +3 oxidation state.

• Atomic number of Fe = 26
  Electronic configuration = $$[\mathrm{Ar}]\,3d^{6}\,4s^{2}$$
• For $$\mathrm{Fe^{3+}}$$ three electrons are removed (first the two 4s and one 3d):
  $$\mathrm{Fe^{3+}}:[Ar]\,3d^{5}$$  ⇒  $$d^{5}$$ system.

Step 2 : Octahedral crystal-field splitting
Both species are octahedral (six ligands around the metal) so the five d orbitals split into

• Lower-energy set: $$t_{2g}(d_{xy},d_{yz},d_{xz})$$
• Higher-energy set: $$e_{g}(d_{x^{2}-y^{2}},d_{z^{2}})$$

Step 3 : Relative magnitudes of crystal-field splitting (Δ_o) and pairing energy (P)

Step 4 : Electron arrangements

• High-spin $$d^{5}$$ (weak field, $$[\mathrm{Fe(H_{2}O)_{6}]}^{3+}$$)
  Because $$\Delta_{o} < P$$, electrons occupy all five d-orbitals singly before any pairing occurs:
  $$t_{2g}^{3}\,e_{g}^{2}$$ ⇒ 5 unpaired electrons.
• Low-spin $$d^{5}$$ (strong field, $$[\mathrm{Fe(CN)_{6}]}^{3-}$$)
  Here $$\Delta_{o} > P$$, so electrons first pair in the lower $$t_{2g}$$ set:
  $$t_{2g}^{5}\,e_{g}^{0}$$ ⇒ 1 unpaired electron.

Step 5 : Magnetic moments
The spin-only magnetic moment is given by $$\mu = \sqrt{n(n+2)}\,\text{B.M.}$$ where $$n$$ is the number of unpaired electrons.

• For $$[\mathrm{Fe(H_{2}O)_{6}]}^{3+}$$: $$n = 5 \Rightarrow \mu = \sqrt{5(5+2)} \approx 5.92\,\text{B.M.}$$ (strongly paramagnetic).
• For $$[\mathrm{Fe(CN)_{6}]}^{3-}$$: $$n = 1 \Rightarrow \mu = \sqrt{1(1+2)} \approx 1.73\,\text{B.M.}$$ (weakly paramagnetic).

Step 6 : Conclusion
Because $$\mathrm{H_{2}O}$$ is a weak-field ligand it produces a high-spin $$d^{5}$$ configuration with five unpaired electrons, leading to strong paramagnetism. In contrast, the strong-field ligand $$\mathrm{CN^{-}}$$ forces electron pairing to give a low-spin $$d^{5}$$ configuration with only one unpaired electron, so $$[\mathrm{Fe(CN)_{6}]}^{3-}$$ is only weakly paramagnetic.

Valence-bond approach for an octahedral complex
For six ligands the central metal ion can use either

• the inner set of orbitals: $$3d,\,3d,\,4s,\,4p_x,\,4p_y,\,4p_z \;\Rightarrow\; d^{2}sp^{3}$$ (an inner-orbital complex), or
• the outer set of orbitals: $$4s,\,4p_x,\,4p_y,\,4p_z,\,4d,\,4d \;\Rightarrow\; sp^{3}d^{2}$$ (an outer-orbital complex).

Which choice is made depends on (a) the d-electron count of the metal ion and (b) how strongly the ligand field can pair those d electrons.

(i) $$\mathrm{[Co(NH_3)_6]^{3+}}$$

1. Cobalt atomic number 27. $$\mathrm{Co^{3+}}:[\mathrm{Ar}]\,3d^{6}$$.
2. Distribution of the six 3d electrons in the free ion (Hund’s rule):

There are four unpaired electrons in the free ion.

3. Although $$\mathrm{NH_3}$$ is only a moderate-field ligand, the high +3 charge on $$\mathrm{Co^{3+}}$$ makes the effective ligand field strong enough to force pairing of the 3d electrons.
4. After pairing the six electrons occupy only three of the five 3d orbitals:

Two inner 3d orbitals are now vacant.

5. These two vacant 3d orbitals together with $$4s$$ and three $$4p$$ orbitals form six $$d^{2}sp^{3}$$ hybrid orbitals → inner-orbital octahedral complex.
6. All electrons are paired → complex is diamagnetic ($$\mu \approx 0\,\mathrm{BM}$$).

(ii) $$\mathrm{[Ni(NH_3)_6]^{2+}}$$

1. Nickel atomic number 28. $$\mathrm{Ni^{2+}}:[\mathrm{Ar}]\,3d^{8}$$.
2. Distribution of the eight 3d electrons (Hund’s rule):

Two unpaired electrons are present.

3. $$\mathrm{NH_3}$$ alone is not a strong enough ligand to force further pairing on the +2 metal ion, and any pairing in $$d^{8}$$ would still leave two electrons in the higher set, so no inner 3d orbital can be made vacant.
4. To accommodate six lone pairs from the ligands, $$\mathrm{Ni^{2+}}$$ must use the next available set $$4s,\,4p_x,\,4p_y,\,4p_z,\,4d,\,4d$$ → $$sp^{3}d^{2}$$ hybridisation.
5. Because the d-orbitals used come from the outer (fourth) shell, $$\mathrm{[Ni(NH_3)_6]^{2+}}$$ is an outer-orbital octahedral complex.
6. Two unpaired electrons remain → complex is paramagnetic ($$\mu \approx 2.83\,\mathrm{BM}$$).

Conclusion
The high +3 charge on Co(III) lets $$\mathrm{NH_3}$$ pair the six 3d electrons, freeing two inner 3d orbitals for $$d^{2}sp^{3}$$ hybridisation. In the +2 Ni complex no inner 3d orbital can be made empty, so $$\mathrm{Ni^{2+}}$$ has to use outer $$4d$$ orbitals to form $$sp^{3}d^{2}$$ hybrids. Hence $$\mathrm{[Co(NH_3)_6]^{3+}}$$ is an inner-orbital complex while $$\mathrm{[Ni(NH_3)_6]^{2+}}$$ is an outer-orbital complex.

Step 1 — Oxidation state of platinum

Let the oxidation state of Pt be $$x$$:

$$x + 4(-1) = -2 \;\Rightarrow\; x = +2$$

So the metal is $$\mathrm{Pt(II)}$$.

Step 2 — d-electron count of $$\mathrm{Pt(II)}$$

Pt (Z = 78): $$[\mathrm{Xe}]\,4f^{14}\,5d^9\,6s^1$$. Removing two electrons (the 6s electron and one 5d) gives $$[\mathrm{Xe}]\,4f^{14}\,5d^8$$, i.e. a $$d^8$$ system.

Step 3 — Geometry and ligand field

• $$\mathrm{CN^-}$$ is a strong-field ligand.
• For a $$d^8$$ ion in a strong field the most stable arrangement is square planar ($$dsp^2$$ hybridisation).

Step 4 — Square-planar crystal-field splitting

When the four $$\mathrm{CN^-}$$ ligands approach Pt in a square-planar geometry the five d orbitals split into four levels. The accepted energy order, from lowest to highest, is:

$$d_{xz}\,=\,d_{yz}\;\lt\;d_{z^2}\;\lt\;d_{xy}\;\lt\;d_{x^2-y^2}$$

Equivalently, using HTML-safe symbols: $$d_{xz} = d_{yz}$$ < $$d_{z^2}$$ < $$d_{xy}$$ < $$d_{x^2-y^2}$$.

The degenerate $$d_{xz}/d_{yz}$$ pair lies at the bottom; $$d_{z^2}$$ comes next (its torus lies in the plane of the ligands but the bulk of its electron density is along the perpendicular $$z$$ axis, so it is only mildly destabilised); $$d_{xy}$$ sits well above because its lobes point between the in-plane ligands; and the strongly destabilised $$d_{x^2-y^2}$$, whose lobes point straight at the four ligands, lies highest of all. The gap between $$d_{xy}$$ and $$d_{x^2-y^2}$$ is large.

Step 5 — Filling the eight d electrons

Step 6 — Unpaired electrons

All occupied orbitals are fully paired, so $$\mathrm{[Pt(CN)_4]^{2-}}$$ has 0 unpaired electrons and is diamagnetic.

Step 1 : Write the electronic configuration of the metal ion

Manganese in the +2 oxidation state has configuration

$$\mathrm{[Ar] 3d^5 4s^0} \;\;(d^5).$$

Hence both complexes contain a central metal ion with a d⁵ set of electrons.

Step 2 : Recall octahedral crystal-field splitting

In an octahedral field the five degenerate d orbitals split into

• three lower-energy $$t_{2g}(d_{xy},d_{yz},d_{xz})$$ orbitals, and
• two higher-energy $$e_g(d_{z^2},d_{x^2-y^2})$$ orbitals.

The energy gap is the octahedral crystal-field splitting parameter $$\Delta_o$$.

Step 3 : Compare the ligand field strengths

• $$\mathrm{H_2O}$$ is a weak-field (high-spin) ligand. For Mn(II) $$\Delta_o$$ is smaller than the electron-pairing energy $$P$$ (energy needed to force two electrons into the same orbital).
• $$\mathrm{CN^-}$$ is a strong-field (low-spin) ligand. Here $$\Delta_o$$ is larger than $$P$$.

Step 4 : Populate the d orbitals for both situations

(a) Hexaaquomanganese(II), $$[\mathrm{Mn(H_2O)_6}]^{2+}$$

Because $$\Delta_o < P$$, electrons avoid pairing and occupy the higher $$e_g$$ level before pairing in $$t_{2g}$$. The filling order is

$$t_{2g}^3e_g^2$$

Diagram to draw: three $$t_{2g}$$ boxes each with a single ↑; two $$e_g$$ boxes each with a single ↑.

Number of unpaired electrons = 5.

(b) Hexacyanomanganate(II), $$[\mathrm{Mn(CN)_6}]^{4-}$$

Because $$\Delta_o > P$$, electrons pair up in the lower $$t_{2g}$$ set before any enter $$e_g$$. The filling order is

$$t_{2g}^5e_g^0$$

Diagram to draw: first $$t_{2g}$$ box has ↑↓, second $$t_{2g}$$ box ↑↓, third $$t_{2g}$$ box ↑.

Number of unpaired electrons = 1.

Step 5 : Explain the observed magnetic behaviour

Magnetic moment $$\mu$$ depends on the number (n) of unpaired electrons: $$\mu=\sqrt{n(n+2)}\,\mathrm{BM}$$.

• Hexaaquo ion: $$n=5 \Rightarrow \mu\approx5.92\,\mathrm{BM}$$ (strongly paramagnetic).
• Hexacyano ion: $$n=1 \Rightarrow \mu\approx1.73\,\mathrm{BM}$$ (weakly paramagnetic).

Thus Crystal Field Theory accounts for the presence of five unpaired electrons in $$[\mathrm{Mn(H_2O)_6}]^{2+}$$ and only one unpaired electron in $$[\mathrm{Mn(CN)_6}]^{4-}$$.

Werner’s explanation of bonding in coordination compounds

Alfred Werner (1893) interpreted the bonding pattern of transition-metal complexes through a set of postulates that account for all observed properties (molar conductivity, precipitation behaviour, isomerism, etc.). The postulates and the way they explain bonding are set out below.

1. Existence of two kinds of valency
     • Primary valency (PV) corresponds to the usual oxidation state of the metal, is ionisable and non-directional.
     • Secondary valency (SV) is equal to the coordination number of the metal, is non-ionisable and directional.
2. Satisfaction of the valencies
     • PV is satisfied exclusively by negative ions.
     • SV is satisfied by neutral molecules or by anions (called ligands). These donor groups remain directly bound to the metal, forming what Werner wrote inside square brackets.
3. Fixed number and definite spatial arrangement of SV
   For a given metal in a given oxidation state the number of SV remains constant; furthermore, because SV are directional they are arranged in characteristic geometries (octahedral when SV = 6, tetrahedral or square-planar when SV = 4, etc.).
4. Dissociation behaviour in solution
   When the compound dissolves, only the ions associated with PV are set free; groups bound through SV stay attached to the metal and therefore do not react with typical ionic reagents (e.g. $$\mathrm{AgNO_3}$$).

Example 1 : $$\mathrm{[Co(NH_3)_6]Cl_3}$$

• Cobalt(III) possesses three PV, hence three chloride ions lie outside the coordination sphere:

$$\mathrm{[Co(NH_3)_6]^{3+} + 3\,Cl^{-}}$$

• It has six SV, so six $$\mathrm{NH_3}$$ molecules are directly bonded to the metal, giving an octahedral structure.
• In aqueous solution the salt gives four ions and shows a conductivity corresponding to a 3:1 electrolyte, exactly as postulated.

Example 2 : $$\mathrm{[PtCl_2(NH_3)_2]}$$

• Pt(II): two PV satisfied by two chloride ions inside the coordination sphere.
• Four SV are fulfilled by two $$\mathrm{Cl^{-}}$$ and two $$\mathrm{NH_3}$$, arranged at the corners of a square plane (directionality of SV).
• No chloride is available for precipitation with $$\mathrm{AgNO_3}$$, confirming that the chlorides are held via SV and are therefore non-ionisable.

Support from isomerism

The compound $$\mathrm{[CoCl_2(NH_3)_4]Cl}$$ exists in two isomeric forms, cis and trans. Such isomerism can arise only if the six SV adopt an octahedral arrangement around Co(III), once again illustrating the directional nature of SV.

Summary of the bonding picture

According to Werner, a coordination compound consists of a coordination sphere—the central metal ion plus the ligands bound through its fixed number of secondary valencies—surrounded by counter-ions that satisfy the metal’s primary valency. Only the latter dissociate in solution. The secondary valencies are directional, imparting definite geometries to the complexes.

Connection with modern view  Primary valency is now called the metal’s formal charge, while secondary valencies correspond to metal–ligand coordinate (dative) bonds. Werner’s empirical postulates therefore anticipated the electronic model developed later.

When equimolar aqueous solutions of $$\mathrm{FeSO_4}$$ and $$\mathrm{(NH_4)_2SO_4}$$ are mixed they crystallise as Mohr’s salt, $$\mathrm{FeSO_4\cdot(NH_4)_2SO_4\cdot6H_2O}$$.

Mohr’s salt is a double salt. In water it dissociates completely into its simple constituent ions:

$$\mathrm{FeSO_4\cdot(NH_4)_2SO_4\cdot6H_2O \;\xrightarrow{H_2O}\; Fe^{2+} + 2\,SO_4^{2-} + 2\,NH_4^{+} + 6\,H_2O}$$

Because free $$\mathrm{Fe^{2+}}$$ ions are produced, the solution gives all the usual qualitative tests for ferrous ions (e.g. a dark blue colour with $$\mathrm{K_3[Fe(CN)_6]}$$).

Conversely, treating $$\mathrm{CuSO_4}$$ with four molar equivalents of aqueous ammonia gives the deep-blue complex salt:

$$\mathrm{CuSO_4 + 4NH_3 \rightarrow [Cu(NH_3)_4]SO_4}$$

This is a coordination compound. In water it ionises only as

$$\mathrm{[Cu(NH_3)_4]SO_4 \xrightarrow{H_2O} [Cu(NH_3)_4]^{2+} + SO_4^{2-}}$$

The copper remains locked inside the coordination sphere; no free $$\mathrm{Cu^{2+}}$$ ions are present. Hence reagents that detect $$\mathrm{Cu^{2+}}$$ (such as $$\mathrm{H_2S}$$ or $$\mathrm{K_4[Fe(CN)_6]}$$) do not react and the usual test is negative.

Therefore

• Mixture I (Mohr’s salt) behaves as a double salt → releases free $$\mathrm{Fe^{2+}}$$ → test is obtained.
• Mixture II forms the complex ion $$\mathrm{[Cu(NH_3)_4]^{2+}}$$ → no free $$\mathrm{Cu^{2+}}$$ → test is not obtained.

Definitions and explanations

1. Coordination entity
   A coordination entity (or complex) is the species formed when a central metal atom/ion is bonded to a fixed number of ions or molecules (ligands) through coordinate (dative) bonds. The entire unit is written inside square brackets with its overall charge outside.
   Examples
   • $$\mathrm{[Fe(CN)_6]^{3-}}$$ – central ion $$\mathrm{Fe^{3+}}$$ with six cyanide ligands.
   • $$\mathrm{[CoCl_3(NH_3)_3]}$$ – central ion $$\mathrm{Co^{3+}}$$ with three chloride and three ammine ligands.
2. Ligand
   A ligand is an ion or molecule that possesses at least one lone pair of electrons which it can donate to the central metal ion to form a coordinate bond. Ligands may be neutral or charged and are classified as monodentate, bidentate, polydentate, etc., depending on the number of donor atoms.
   Examples
   • $$\mathrm{NH_3}$$ (ammine) – a neutral monodentate ligand donating the lone pair on nitrogen.
   • $$\mathrm{Cl^-}$$ (chlorido) – a monodentate anionic ligand donating a lone pair on chlorine.
3. Coordination number (CN)
   The coordination number is the total number of ligand donor atoms directly bonded to the central metal ion (it counts donor atoms, not ligands — one bidentate ligand contributes 2).
   Examples
   • $$\mathrm{[PtCl_6]^{2-}}$$ – six chloride donor atoms; CN = 6.
   • $$\mathrm{[Ni(CO)_4]}$$ – four carbonyl donor atoms; CN = 4.
4. Coordination polyhedron
   The coordination polyhedron is the spatial arrangement (geometry) of the donor atoms around the central metal ion. Typical geometries are octahedral, tetrahedral and square-planar.
   Examples
   • $$\mathrm{[Co(NH_3)_6]^{3+}}$$ – six donor atoms at the corners of a regular octahedron (octahedral polyhedron).
   • $$\mathrm{[PtCl_4]^{2-}}$$ – four donor atoms at the corners of a square around Pt (square-planar polyhedron).
5. Homoleptic complexes
   Homoleptic complexes contain only one kind of ligand around the central metal ion.
   Examples
   • $$\mathrm{[Ni(CO)_4]}$$ – only carbonyl ligands.
   • $$\mathrm{[Cu(NH_3)_4]^{2+}}$$ – only ammine ligands.
6. Heteroleptic complexes
   Heteroleptic complexes contain two or more different kinds of ligands around the central metal ion.
   Examples
   • $$\mathrm{[Co(NH_3)_4Cl_2]^+}$$ – ammine and chlorido ligands.
   • $$\mathrm{[Pt(NH_3)_2Cl_2]}$$ (cis- or trans-platin) – ammine and chlorido ligands.

Step 1  Recall the meaning of denticity
In any coordination compound a ligand is an ion or molecule that donates one or more lone‐pair electrons to the central metal ion, forming a coordinate (dative) bond. The number of donor atoms of a single ligand that actually attach to the metal at the same time is called its denticity (from the Latin dens, tooth). According to this number ligands are classified as follows.

Step 2  Unidentate (monodentate) ligands
A ligand that possesses only one donor atom and therefore can form only one coordinate bond with the metal centre at a given time is called unidentate (or monodentate).
Examples, each furnishing a single lone pair:

• Halide ion $$\mathrm{Cl^-}$$  (donor atom Cl)
• Ammonia $$\mathrm{NH_3}$$  (donor atom N)

Step 3  Didentate (bidentate) ligands
A ligand that contains two separate donor atoms and can attach to the same metal ion through both of them simultaneously is termed didentate (bidentate). When both donor atoms bind, a five- or six-membered ring called a chelate ring is produced.

• Oxalate ion $$\mathrm{C_2O_4^{2-}}$$: the two O atoms bearing negative charge each donate a lone pair.
• Ethane-1,2-diamine (en) $$\mathrm{NH_2CH_2CH_2NH_2}$$: the two N atoms each supply a lone pair.

Step 4  Ambidentate ligands
Certain monodentate ligands possess two different potential donor atoms, but only one of them can bind to the metal ion at a time. Such ligands are called ambidentate. Depending on experimental conditions they may coordinate through either donor atom, giving linkage isomers.

• Nitrite ion $$\mathrm{NO_2^-}$$ — can bind through N (-NO₂) or through an O atom (-ONO).
• Thiocyanate ion $$\mathrm{SCN^-}$$ — can bind through S (-SCN) or through N (-NCS).

Thus, unidentate ligands donate one electron pair from one donor atom, didentate ligands donate two electron pairs simultaneously from two donor atoms, and ambidentate ligands have two possible donor atoms but utilise only one at a time.

Background : In coordination chemistry the central metal atom/ion and its surrounding ligands can be put together in more than one way, or can occupy different spatial positions; the resulting compounds are called isomers. All the observed isomerisms are gathered under two broad heads – structural and stereoisomerism.

1. Structural isomerism (formula of the complex actually changes)

• (a) Ionisation isomerism – interchange between a ligand in the coordination sphere and a counter-ion produces different ions in solution.
  Example : $$[\mathrm{Co(NH_3)_5SO_4}]\,\mathrm{Br}$$ (gives Br^– in solution) and $$[\mathrm{Co(NH_3)_5Br}]\,\mathrm{SO_4}$$ (gives SO₄^2–).
• (b) Hydrate (solvate) isomerism – water (or another solvent) may act either as a coordinated ligand or as a mere crystal-lattice molecule.
  Example : $$[\mathrm{Cr(H_2O)_6}]\mathrm{Cl_3}$$ (all six H₂O inside the sphere), $$[\mathrm{Cr(H_2O)_5Cl}]\mathrm{Cl_2}\!\cdot\!\mathrm{H_2O}$$ and $$[\mathrm{Cr(H_2O)_4Cl_2}]\mathrm{Cl}\!\cdot\!2\mathrm{H_2O}$$.
• (c) Linkage isomerism – an ambidentate ligand bonds through different donor atoms.
  Example : $$[\mathrm{Co(NH_3)_5(NO_2)}]^{2+}$$ (nitro, M–N) and $$[\mathrm{Co(NH_3)_5(ONO)}]^{2+}$$ (nitrito, M–O).
• (d) Coordination isomerism – possible when both cation and anion are coordination entities; ligands are exchanged between the two metal centres.
  Example : $$[\mathrm{Co(NH_3)_6}][\mathrm{Cr(CN)_6}]$$ and $$[\mathrm{Cr(NH_3)_6}][\mathrm{Co(CN)_6}]$$.

2. Stereoisomerism (formula is the same; only spatial arrangement differs)

• (a) Geometrical isomerism – ligands occupy different relative positions like cis/trans, fac/mer etc.
  Example : square-planar $$[\mathrm{Pt(NH_3)_2Cl_2}]$$ exists as cis-platin (anticancer drug) and trans-platin.
• (b) Optical isomerism – nonsuperimposable mirror images (enantiomers) rotate plane-polarised light in opposite directions.
  Example : octahedral $$[\mathrm{Cr(en)_3}]^{3+}$$ (where en = ethylenediamine) has d- and l- forms.

Thus, coordination compounds can show four kinds of structural and two kinds of stereoisomerism, each illustrated above with a representative pair (or set) of isomers.

Step 1 • Identify the stereochemistry of the complex
The central metal ion is $$\mathrm{Pt^{II}}$$ (a d⁸ ion). Complexes of the type $$\mathrm{Pt^{II}L_4}$$ are almost invariably square-planar. Hence $$\mathrm{[Pt(NH_3)(Br)(Cl)(py)]}$$ must be treated as a square-planar system having four different monodentate ligands.

Step 2 • Strategy for enumerating geometrical isomers
For a square-planar complex with four different ligands (denoted $$\mathrm{A},\,\mathrm{B},\,\mathrm{C},\,\mathrm{D}$$) the only symmetry operation that leaves the molecule unchanged is a rotation of 90° or 180° about the perpendicular axis through the metal. Because of this, two arrangements are considered identical if one can be brought to the other by such a rotation in the plane. The trans pair (the two ligands lying opposite each other) therefore uniquely identifies a geometrical isomer.

Step 3 • List the possible trans pairs
Choose one ligand as a reference, say $$\mathrm{NH_3}$$. The ligand placed opposite to it (trans to it) can be:

• $$\mathrm{Br^-}$$
• $$\mathrm{Cl^-}$$
• $$\mathrm{py}$$ (pyridine)

Thus there are three distinct trans combinations and hence three geometrical isomers.

Step 4 • Draw / describe every isomer

1. Isomer I: $$\mathrm{NH_3}$$ trans $$\mathrm{Br^-}$$, with $$\mathrm{Cl^-}$$ cis to $$\mathrm{NH_3}$$ and $$\mathrm{py}$$ cis to $$\mathrm{Br^-}$$.
   Describe a diagram: draw a square with Pt at the centre; place $$\mathrm{NH_3}$$ at the top, $$\mathrm{Br^-}$$ at the bottom, $$\mathrm{Cl^-}$$ at the left, $$\mathrm{py}$$ at the right.
2. Isomer II: $$\mathrm{NH_3}$$ trans $$\mathrm{Cl^-}$$.
   Describe a diagram: top $$\mathrm{NH_3}$$, bottom $$\mathrm{Cl^-}$$, left $$\mathrm{Br^-}$$, right $$\mathrm{py}$$.
3. Isomer III: $$\mathrm{NH_3}$$ trans $$\mathrm{py}$$.
   Describe a diagram: top $$\mathrm{NH_3}$$, bottom $$\mathrm{py}$$, left $$\mathrm{Br^-}$$, right $$\mathrm{Cl^-}$$.

Any other arrangement is merely a 90° rotation of one of these and is therefore not new.

Step 5 • Check for optical activity
All three isomers are square-planar; the molecule itself is the plane of symmetry. Consequently each structure possesses at least one symmetry element (a mirror plane in the molecular plane) and therefore cannot be chiral. Hence none of the geometrical isomers can show optical isomerism.

Conclusion
There are three geometrical (trans) isomers of $$\mathrm{[Pt(NH_3)(Br)(Cl)(py)]}$$ and zero of them exhibit optical isomerism.

The ionic species present in an aqueous solution of copper(II) sulphate is

$$\mathrm{Cu^{2+} + SO_4^{2-}}$$

When an excess of aqueous $$\mathrm{KCN}$$ is added, the following sequence of steps takes place.

1. Initial interaction (precipitation)
   

   $$\mathrm{Cu^{2+} + 2\,CN^- \;\longrightarrow\; Cu(CN)_2 \;(white\, ppt)}$$

   The white precipitate $$\mathrm{Cu(CN)_2}$$ is not stable; it immediately undergoes internal redox change:

   $$\mathrm{2\,Cu(CN)_2 \;\longrightarrow\; 2\,CuCN\,(white\, ppt) + (CN)_2\,(g)}$$

   Thus, insoluble $$\mathrm{CuCN}$$ containing copper in the +1 oxidation state is obtained.

2. Dissolution of the precipitate in excess cyanide
   

   The precipitated $$\mathrm{CuCN}$$ now reacts with the excess $$\mathrm{CN^-}$$ ions present in the solution to give a soluble complex anion:

   $$\mathrm{CuCN + 3\,CN^- \;\longrightarrow\; [Cu(CN)_4]^{3-}}$$

   In ionic form (taking the potassium counter-ions into account):

   $$\mathrm{CuCN + 3\,KCN \;\longrightarrow\; K_3[Cu(CN)_4]}$$

Hence, the coordination entity formed is

$$\boxed{\mathrm{[Cu(CN)_4]^{3-}}}$$

Absence of a copper(II) sulphide precipitate on passing $$\mathrm{H_2S}$$

Passing $$\mathrm{H_2S(g)}$$ through the above solution supplies sulphide ions:

$$\mathrm{H_2S \rightleftharpoons 2\,H^+ + S^{2-}}$$

For precipitation to occur, the ionic product $$[\mathrm{Cu^{2+}}][\mathrm{S^{2-}}]$$ (or for $$\mathrm{Cu^{+}}$$, $$[\mathrm{Cu^+}]^2[\mathrm{S^{2-}}]$$) must exceed the solubility-product $$K_{sp}$$ of copper sulphide. In our solution almost all the copper is locked up inside the very stable complex ion $$\mathrm{[Cu(CN)_4]^{3-}}$$, whose formation constant $$\beta_f$$ is extremely high:

$$\mathrm{Cu^+ + 4\,CN^- \rightleftharpoons [Cu(CN)_4]^{3-}}\qquad \beta_f \approx 10^{25}$$

Consequently, the equilibrium concentration of free $$\mathrm{Cu^{2+}}$$ (or $$\mathrm{Cu^+}$$) in the solution becomes vanishingly small. The ionic product with $$\mathrm{S^{2-}}$$ never reaches $$K_{sp}(\mathrm{CuS})$$, so

no precipitate of copper sulphide is obtained.

In short, the strong complex-forming ability of cyanide keeps copper in solution as $$\mathrm{[Cu(CN)_4]^{3-}}$$, preventing any reaction with sulphide ions.

In an octahedral environment six ligands approach the central metal ion along the x, y and z axes (the positive and negative directions of each axis). Because the lobes of two of the five d-orbitals point exactly along these axes, while the lobes of the other three lie in between the axes, the originally degenerate set of five d-orbitals splits into two energy levels.

Which orbitals go where?

• The $$d_{x^2 - y^2}$$ and $$d_{z^2}$$ orbitals have their electron density directly on the Cartesian axes. They experience stronger repulsion from the six ligand fields and therefore rise in energy. This doubly-degenerate pair is called the $$e_g$$ set.
• The $$d_{xy}$$, $$d_{yz}$$ and $$d_{xz}$$ orbitals lie between the axes. They encounter less repulsion and become the lower-energy, triply-degenerate $$t_{2g}$$ set.

Energy separation

The difference in energy between the two sets is denoted $$\Delta_\mathrm{o}$$ (the octahedral crystal-field splitting energy). The dotted horizontal line drawn midway between the two sets is the barycentre, the original mean energy of the unsplit d-orbitals.

How to draw the diagram

1. Start with a single horizontal line and write “five degenerate d-orbitals”.
2. Draw a vertical arrow upward labelled “apply octahedral crystal field”.
3. Above the arrow, draw two horizontal lines side-by-side (same height) to represent the $$e_g$$ set. Label each small box $$d_{x^2 - y^2}$$ and $$d_{z^2}$$. Write $$e_g$$ slightly to the right of the pair.
4. Below these, draw three horizontal lines (same height, slightly lower) to represent the $$t_{2g}$$ set. Label the boxes $$d_{xy}$$, $$d_{yz}$$ and $$d_{xz}$$. Write $$t_{2g}$$ to the right.
5. Draw a dotted horizontal line exactly midway between the $$e_g$$ and $$t_{2g}$$ levels; label it “barycentre”.
6. Indicate the energy gap between the two sets by a double-headed vertical arrow; write $$\Delta_\mathrm{o}$$ beside it.

That completed sketch is the required figure for the splitting of $$d$$-orbitals in an octahedral crystal field.

Step 1 : What is the spectrochemical series?

When a transition-metal ion is placed in a ligand field its five degenerate d orbitals split into sub-sets. For an octahedral field the splitting is into the lower $$t_{2g}$$ set and the upper $$e_g$$ set; the energy gap is the crystal-field splitting energy $$\Delta_o$$. Experimental measurement of $$\Delta_o$$ from the electronic absorption spectra of complexes containing the same metal ion but different ligands shows that different ligands produce very different splittings. Arranging the common ligands in the order of increasing field strength (i.e. increasing $$\Delta_o$$) gives the spectrochemical series:

$$\mathrm{I^-}\;\lt\;\mathrm{Br^-}\;\lt\;\mathrm{SCN^-}\;\lt\;\mathrm{Cl^-}\;\lt\;\mathrm{S^{2-}}\;\lt\;\mathrm{F^-}\;\lt\;\mathrm{OH^-}\;\lt\;\mathrm{C_2O_4^{2-}}\;\lt\;\mathrm{H_2O}\;\lt\;\mathrm{NCS^-}\;\lt\;\mathrm{edta^{4-}}\;\lt\;\mathrm{NH_3}\;\lt\;\mathrm{en}\;\lt\;\mathrm{CN^-}\;\lt\;\mathrm{CO}$$

(Note: $$\mathrm{SCN^-}$$ and $$\mathrm{NCS^-}$$ are the same thiocyanate ion acting as linkage isomers; donation through the softer S atom gives a weaker field, whereas donation through the harder N atom gives a stronger field, so the two appear at different positions in the series.)

The series is termed spectrochemical because it is constructed from spectral (absorption) data. It allows the field strength of any ligand to be ranked qualitatively against others.

Step 2 : Weak-field vs. strong-field ligands

The decisive criterion that separates the two classes is the magnitude of $$\Delta_o$$ compared with the electron-pairing energy $$P$$:

• If $$\Delta_o\;\lt\;P$$, the ligand is a weak-field ligand and the complex is high-spin.
• If $$\Delta_o\;\gt\;P$$, the ligand is a strong-field ligand and the complex is low-spin.

Thus the spectrochemical series places every common ligand on a single field-strength scale, and the position of a ligand on this scale (relative to the pairing energy $$P$$) decides whether it behaves as a weak-field or a strong-field ligand and hence whether the resulting complex is high-spin or low-spin.

1.  Crystal-field splitting energy (CFSE)

In an isolated gaseous transition-metal ion the five d orbitals are degenerate (same energy). When the ion is placed in a symmetrical field of surrounding ligands, the electrostatic repulsion between ligand lone pairs and metal d electrons removes this degeneracy.

• For an octahedral arrangement, the three orbitals that lie between the Cartesian axes — $$\{d_{xy},\,d_{yz},\,d_{zx}\}$$, the $$t_{2g}$$ set — experience less repulsion and are lowered in energy.
• The two orbitals that point along the axes — $$\{d_{x^{2}-y^{2}},\,d_{z^{2}}\}$$, the $$e_g$$ set — experience stronger repulsion and are raised in energy.

The energy gap between the centroids of these two groups is the crystal-field splitting energy, denoted $$\Delta_o$$ for an octahedral complex.

2.  How the magnitude of $$\Delta_o$$ decides the d-electron arrangement

While filling electrons into the split orbitals, the ion has to balance two competing energies:

1. Pairing energy $$P$$: the extra energy needed to place a second electron in an already-occupied (lower-energy) orbital.
2. Promotion energy $$\Delta_o$$: the energy needed to promote an electron from the lower $$t_{2g}$$ set to the upper $$e_g$$ set.

Therefore the actual ground-state configuration of the metal ion is governed by the competition between $$\Delta_o$$ and $$P$$. Ligands near the strong-field end of the spectrochemical series (CN⁻, CO, NH₃) produce a large $$\Delta_o$$ and low-spin complexes, whereas weak-field ligands (I⁻, Br⁻, F⁻, H₂O) give a small $$\Delta_o$$ and high-spin configurations.

Step 1  Oxidation state and d-electron count

• For $$\mathrm{[Cr(NH_3)_6]^{3+}}$$: the ligands $$\mathrm{NH_3}$$ are neutral.
    ⇒ Charge on metal $$= +3$$, i.e. the ion is $$\mathrm{Cr^{3+}}$$.
• Atomic number of Cr = 24 ⇒ ground-state configuration of the atom = $$[\mathrm{Ar}]\,3d^{5}4s^{1}$$.
    Remove three electrons for $$\mathrm{Cr^{3+}}$$ ⇒ $$[\mathrm{Ar}]\,3d^{3}$$ (so it is a d³ ion).
• For $$\mathrm{[Ni(CN)_4]^{2-}}$$: each $$\mathrm{CN^-}$$ carries –1 charge, overall charge –2.
    ⇒ Charge on metal $$= +2$$, i.e. the ion is $$\mathrm{Ni^{2+}}$$.
• Atomic number of Ni = 28 ⇒ atom = $$[\mathrm{Ar}]\,3d^{8}4s^{2}$$.
    Remove two electrons for $$\mathrm{Ni^{2+}}$$ ⇒ $$[\mathrm{Ar}]\,3d^{8}$$ (a d⁸ ion).

Step 2  Nature of the ligand and geometry adopted

• $$\mathrm{NH_3}$$ is a moderate-field ligand. In an octahedral site it generally does not cause pairing of electrons.
• $$\mathrm{CN^-}$$ is a strong-field ligand and favours maximum pairing. For a d⁸ ion it usually forces a square-planar arrangement (hybridisation $$dsp^2$$) rather than tetrahedral.

Step 3  Crystal-field splitting and electronic configuration

1. $$\mathrm{[Cr(NH_3)_6]^{3+}}$$ (octahedral)
     Splitting pattern: $$t_{2g}$$ (lower) and $$e_g$$ (higher).
     d³ distribution (no pairing forced): $$t_{2g}^{3}e_g^{0}$$.
     Number of unpaired electrons $$n = 3$$.
     Therefore the complex is paramagnetic.
2. $$\mathrm{[Ni(CN)_4]^{2-}}$$ (square planar)
     For a square-planar field the d-levels descend in the order
     $$d_{x^2-y^2}$$ > $$d_{xy}$$ > $$d_{z^2}$$ > $$d_{xz}=d_{yz}$$.
     With a strong ligand, the eight electrons occupy the four lowest orbitals pairwise:
     $$d_{xz}^{2}d_{yz}^{2}d_{z^2}^{2}d_{xy}^{2}d_{x^2-y^2}^{0}$$.
     Number of unpaired electrons $$n = 0$$.
     Hence the complex is diamagnetic.

Step 4  Conclusion

Because the octahedral d³ ion in $$\mathrm{[Cr(NH_3)_6]^{3+}}$$ retains three unpaired electrons, it shows paramagnetism, whereas the square-planar d⁸ ion in $$\mathrm{[Ni(CN)_4]^{2-}}$$ has all electrons paired, making the complex diamagnetic.

Step 1 : Electronic configuration of the central ion

Atomic number of nickel = 28; $$\mathrm{Ni}:[\mathrm{Ar}]\,3d^{8}\,4s^{2}$$. For the +2 oxidation state the two 4s electrons are removed, leaving $$\mathrm{Ni^{2+}}:[\mathrm{Ar}]\,3d^{8}$$ in both complexes.

Step 2 : $$\mathrm{[Ni(H_2O)_6]^{2+}}$$ — octahedral, weak field

• Six water ligands give an octahedral geometry.
• For a $$d^{8}$$ ion in an octahedral field there is only one possible electron arrangement: $$t_{2g}^{6}\,e_g^{2}$$ — two unpaired electrons regardless of field strength.
• $$\mathrm{H_2O}$$ is a weak-field ligand, so the crystal-field splitting $$\Delta_o$$ is small. The d–d transition $$t_{2g}\,\rightarrow\,e_g$$ requires absorption of a relatively low-energy (long-wavelength) photon, which falls in the red/orange region of visible light. The complementary colour transmitted by the solution is green, accounting for the observed green colour.

Step 3 : $$\mathrm{[Ni(CN)_4]^{2-}}$$ — square-planar, strong field

• With four strong-field $$\mathrm{CN^-}$$ ligands a $$d^{8}$$ Ni(II) ion adopts a square-planar geometry ($$dsp^{2}$$ hybridisation) instead of tetrahedral, because square-planar coordination is more stable for $$d^{8}$$ with strong-field ligands.
• The square-planar splitting is much larger than the octahedral $$\Delta_o$$, especially the gap between the $$d_{xy}$$ and the highest $$d_{x^{2}-y^{2}}$$ orbital (the LUMO).
• The eight d-electrons fill the four lower-energy orbitals as $$d_{xz}^{2}\,d_{yz}^{2}\,d_{z^{2}}^{2}\,d_{xy}^{2}\,d_{x^{2}-y^{2}}^{0}$$ — all paired (diamagnetic).
• The lowest d–d transition ($$d_{xy}\,\rightarrow\,d_{x^{2}-y^{2}}$$) requires so much energy that the absorption falls in the ultraviolet region. No visible light is absorbed, so the solution looks colourless.

Step 4 : Conclusion

Although both complexes contain Ni(II), they differ in geometry and in the field strength of their ligands. The small octahedral splitting produced by weak-field $$\mathrm{H_2O}$$ leads to absorption of red light and a green colour; the very large square-planar splitting produced by strong-field $$\mathrm{CN^-}$$ shifts the lowest d–d transition into the UV, so $$\mathrm{[Ni(CN)_4]^{2-}}$$ appears colourless.

Step 1 – Oxidation state of iron in each complex

• For $$\mathrm{[Fe(CN)_6]^{4-}}$$: each $$\mathrm{CN^-}$$ is $$-1$$, so $$x + 6(-1) = -4 \Rightarrow x = +2$$.
• For $$\mathrm{[Fe(H_2O)_6]^{2+}}$$: water is neutral, so $$x = +2$$.

Iron is $$\mathrm{Fe^{2+}}$$ ($$d^{6}$$) in both species.

Step 2 – Octahedral crystal-field splitting for $$d^{6}$$

In an octahedral field the five 3d orbitals split into the lower-energy $$t_{2g}$$ set and the higher-energy $$e_g$$ set, separated by the crystal-field splitting energy $$\Delta_o$$.

Step 3 – Ligand-field strengths

• $$\mathrm{CN^-}$$ is a strong-field ligand (right end of the spectrochemical series).
• $$\mathrm{H_2O}$$ is a weak-field ligand.

Hence $$\Delta_o(\mathrm{[Fe(CN)_6]^{4-}}) \gg \Delta_o(\mathrm{[Fe(H_2O)_6]^{2+}})$$.

Step 4 – Electronic configurations

Step 5 – Consequence for colour

• The visible colour of a coordination complex arises from d–d transitions of energy $$\Delta_o$$.
• A large $$\Delta_o$$ (strong field) means higher-energy (shorter-wavelength) light is absorbed; the colour observed is the complement of that shorter-wavelength colour.
• A small $$\Delta_o$$ (weak field) absorbs lower-energy (longer-wavelength) light, so the complementary colour observed is different.

Because $$\Delta_o$$ — and hence the wavelength absorbed — differs sharply between $$\mathrm{CN^-}$$ and $$\mathrm{H_2O}$$, the two complexes display different colours even though iron is in the same oxidation state. This is a direct consequence of their different positions in the spectrochemical series.

Objective: To explain how a metal atom or ion is bound to the carbonyl ligand $$\mathrm{CO}$$ in metal-carbonyl complexes such as $$\mathrm{[Ni(CO)_4]}$$, $$\mathrm{[Fe(CO)_5]}$$ and $$\mathrm{[Cr(CO)_6]}$$.

1. Relevant features of the CO molecule and the metal

• The carbon monoxide molecule has a lone pair of electrons on the carbon atom (its highest-energy filled $$\sigma$$ orbital) and two degenerate empty $$\pi^{*}$$ antibonding orbitals.
• The transition metal possesses partially filled $$(n-1)d$$ orbitals and empty $$ns$$ and $$np$$ orbitals that can both donate and accept electron density.

2. Two complementary orbital interactions

1. $$\sigma$$-donation (forward donation): the filled lone-pair orbital of carbon donates electron density into an empty metal orbital of appropriate symmetry (an $$s$$, $$p_z$$ or $$d_{z^{2}}$$ orbital), forming a metal–carbon $$\sigma$$ bond.
2. $$\pi$$-back donation (back bonding): a filled metal $$d_\pi$$ orbital ($$d_{xz}$$ or $$d_{yz}$$) overlaps sideways with an empty $$\pi^{*}$$ orbital on the coordinated CO. Electron density flows from the metal back to the ligand, forming a metal–carbon $$\pi$$ bond and simultaneously weakening the internal C–O bond.

3. Synergic character

• The two interactions reinforce each other: stronger ligand-to-metal $$\sigma$$ donation increases the electron density on the metal, which enhances $$\pi$$ back donation; stronger back donation removes the surplus electron density built up on the metal, allowing still more $$\sigma$$ donation.
• This co-operative mechanism is called synergic bonding.

4. Consequences and experimental evidence

5. Pictorial representation

Draw the metal centre on the left with its filled $$d_{xz}$$/$$d_{yz}$$ orbitals and CO on the right with its filled $$\sigma$$ lone pair on C and empty $$\pi^{*}$$ lobes. Use arrows to indicate (a) $$\sigma$$ donation from ligand to metal and (b) $$\pi$$ back donation from metal to ligand.

6. Net picture

• The M–C bond has a strong $$\sigma$$ component reinforced by a $$\pi$$ component; it has partial multiple-bond character.
• The internal C–O bond is correspondingly weakened relative to free CO.
• The overall interaction is termed synergic $$\sigma$$–$$\pi$$ bonding.

One-line conclusion: bonding in metal carbonyls is a synergic combination of ligand-to-metal $$\sigma$$ donation and metal-to-ligand $$\pi$$ back donation, which together produce a strong (partial-multiple-bond) M–C linkage, a weakened C–O bond and a metal centre stabilised in a low oxidation state with an 18-electron count.

Step 1 – Oxidation state and d-electron configuration
Atomic number of Ti = 22. Atom: $$[\mathrm{Ar}]\,3d^{2}\,4s^{2}$$. In $$\mathrm{[Ti(H_2O)_6]^{3+}}$$ titanium is in the +3 oxidation state, so three electrons are removed (first the two 4s and then one 3d), leaving

$$\mathrm{Ti^{3+}}:\;[\mathrm{Ar}]\,3d^{1}$$   ($$d^{1}$$ system).

Step 2 – Nature of the ligand and geometry
$$\mathrm{H_2O}$$ is a weak-field ligand. With six water molecules the complex is octahedral.

Step 3 – Crystal-field splitting
In an octahedral field the five d orbitals split into a lower triply degenerate set $$t_{2g}(d_{xy},\,d_{yz},\,d_{zx})$$ and an upper doubly degenerate set $$e_g(d_{x^{2}-y^{2}},\,d_{z^{2}})$$ separated by the crystal-field splitting energy $$\Delta_o$$.

The single d-electron in $$\mathrm{Ti^{3+}}$$ occupies the lowest orbital:

Step 4 – d–d transition and absorption of light
On exposure to white light the lone d-electron can be promoted from $$t_{2g}$$ to $$e_g$$:

$$\Delta_o = h\nu = \dfrac{hc}{\lambda}$$

For $$\mathrm{[Ti(H_2O)_6]^{3+}}$$ the absorption maximum lies near $$\lambda \approx 500\,\mathrm{nm}$$ (about 20,300 cm$${}^{-1}$$) — in the green/yellow region of the visible spectrum.

Step 5 – Relation between absorbed and observed colour
The colour perceived is the complement of the colour absorbed. The complement of green-yellow is violet, so the aqueous solution looks violet.

Conclusion
The single d-electron in the octahedral $$d^{1}$$ ion $$\mathrm{[Ti(H_2O)_6]^{3+}}$$ absorbs green-yellow light (~500 nm) for the $$t_{2g}\,\rightarrow\,e_g$$ transition; the transmitted/reflected light is therefore rich in the complementary violet hue, giving the complex its characteristic violet colour.

The term chelation comes from the Greek for a crab's claw. A chelate is a coordination compound in which the metal ion is bound to a single ligand through two or more donor atoms, producing one or more five- or six-membered rings.

Consider a metal ion $$\mathrm{M^{n+}}$$ that can be surrounded either by six monodentate ammine ligands or by three bidentate ethane-1,2-diamine (en) ligands. Starting from the aquated ion, the two substitution equilibria are:

(a) with monodentate ammonia

$$\mathrm{[M(H_2O)_6]^{n+}\;+\;6\,NH_3 \;\rightleftharpoons\;[M(NH_3)_6]^{n+}\;+\;6\,H_2O}$$

Particle count: $$7 \to 7$$   ($$\Delta n = 0$$)

(b) with bidentate ethane-1,2-diamine (chelation)

$$\mathrm{[M(H_2O)_6]^{n+}\;+\;3\,en \;\rightleftharpoons\;[M(en)_3]^{n+}\;+\;6\,H_2O}$$

Particle count: $$4 \to 7$$   ($$\Delta n = +3$$)

• The chelation reaction produces three more independent particles than the reactants supply; the ammonia reaction produces no net change in the number of particles.
• The larger positive entropy change in the chelation reaction ($$\Delta S > 0$$) makes its Gibbs free-energy change $$\Delta G = \Delta H - T\Delta S$$ more negative, so its equilibrium constant is much larger than the corresponding monodentate reaction.

This pronounced increase in stability of a complex when a polydentate ligand replaces an equivalent number of similar monodentate ligands is called the chelate effect.

Quantitative illustration (nickel(II))

• Overall formation constant for the non-chelated complex: $$\beta_6\,\mathrm{[Ni(NH_3)_6]^{2+}} \approx 10^{8}$$.
• Overall formation constant for the chelated complex: $$\beta_3\,\mathrm{[Ni(en)_3]^{2+}} \approx 10^{18}$$.

Because $$\beta_3 \gg \beta_6$$, $$\mathrm{[Ni(en)_3]^{2+}}$$ is far more stable than $$\mathrm{[Ni(NH_3)_6]^{2+}}$$ — a clear manifestation of the chelate effect. The same phenomenon is even more pronounced with polydentate ligands such as EDTA⁴⁻.

The three complexes

$$\mathrm{[Ni(H_2O)_6]^{2+}},\;\mathrm{[Ni(NH_3)_6]^{2+}},\;\mathrm{[Ni(NO_2)_6]^{4-}}$$

all contain nickel in the +2 oxidation state:

• For $$\mathrm{[Ni(NO_2)_6]^{4-}}$$ each $$\mathrm{NO_2^-}$$ ligand carries a $$-1$$ charge, so $$x + 6(-1) = -4 \Rightarrow x = +2$$.
• For the two cationic complexes the ligands are neutral, so $$x = +2$$.

Each species is therefore an octahedral $$d^8$$ Ni(II) complex. The wavelength of the visible absorption depends only on the crystal-field splitting energy $$\Delta_o$$ produced by the ligand:

$$E = \Delta_o = \dfrac{hc}{\lambda}$$

so a larger $$\Delta_o$$ corresponds to a higher-energy d–d transition and therefore a shorter wavelength of absorption.

The relevant section of the spectrochemical series (weak field → strong field) is

$$\mathrm{H_2O}\;\lt\;\mathrm{NH_3}\;\lt\;\mathrm{NO_2^-}$$

Hence:

• $$\mathrm{H_2O}$$ produces the smallest $$\Delta_o$$ (longest wavelength of absorption).
• $$\mathrm{NH_3}$$ produces an intermediate $$\Delta_o$$.
• $$\mathrm{NO_2^-}$$ produces the largest $$\Delta_o$$ (shortest wavelength of absorption).

The energies of the d–d transitions therefore satisfy

$$E(\mathrm{[Ni(NO_2)_6]^{4-}})\;\gt\;E(\mathrm{[Ni(NH_3)_6]^{2+}})\;\gt\;E(\mathrm{[Ni(H_2O)_6]^{2+}})$$

and the corresponding wavelengths of absorption follow the reverse order:

$$\boxed{\lambda(\mathrm{[Ni(H_2O)_6]^{2+}})\;\gt\;\lambda(\mathrm{[Ni(NH_3)_6]^{2+}})\;\gt\;\lambda(\mathrm{[Ni(NO_2)_6]^{4-}})}$$