Q.Give valence bond description for the bonding in the complex [VCl4]- . Draw box diagrams for free metal ion. Which hybrid orbitals are used by the metal ? State the number of unpaired electrons.

[Co(NH3)6]3+ has Co3+ (d6) surrounded by the strong-field ligand NH3; the d-electrons pair up (low spin), freeing two inner 3d orbitals for d2sp3 hybridisation — this makes it an inner orbital, low-spin, diamagnetic complex.

Co3+ has configuration [Ar]3d6. NH3 is a strong-field ligand (high in the spectrochemical series), so it causes the d-electrons to pair up rather than spread out (low-spin arrangement):

t2g6 eg0 — all six electrons paired in three lower orbitals, leaving two of the five 3d orbitals completely empty.

These two empty 3d orbitals, combined with the 4s and three 4p orbitals, give d2sp3 hybridisation, which uses INNER (n−1) d orbitals — hence 'inner orbital complex' (also called low-spin or spin-paired complex). Since all electrons are paired, the complex is diamagnetic, not paramagnetic (ruling out option d), and it is not an outer-orbital (sp3d2, using 4d) complex (ruling out option b), nor high-spin (ruling out option c).

dsp2 hybridisation gives a square planar geometry; it occurs when a strong-field ligand forces pairing of the d8 metal ion's electrons, as happens with CN- and Ni2+.

In $[\text{Ni(CN)}_4]^{2-}$, nickel is present as Ni2+, a d8 ion ([Ar] 3d8). CN- is a strong field ligand and causes the unpaired 3d electrons to pair up, freeing one 3d orbital. This vacant 3d orbital, together with one 4s and two 4p orbitals, undergoes dsp2 hybridisation, giving a square planar, diamagnetic complex.

By contrast, $[\text{Ni(CO)}_4]$ has Ni(0), d10, and is sp3 hybridised (tetrahedral); $[\text{MnCl}_4]^{2-}$ and most Cu2+ complexes here do not match this classic dsp2 example.

$Fe^{2+}$ ($3d^6$) with the weak-field ligand $H_2O$ forms a high-spin octahedral complex using outer $4s,4p,4d$ orbitals — i.e. $sp^3d^2$ hybridisation.

Iron has atomic number 26: $Fe = [Ar]3d^64s^2$. On forming $Fe^{2+}$, the two $4s$ electrons are lost first:

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

$H_2O$ is a weak-field ligand (low in the spectrochemical series), so it is not strong enough to pair up the $3d^6$ electrons into the lower three orbitals. Instead, the six 3d electrons remain spread across all five 3d orbitals in a high-spin arrangement ($t_{2g}^4 e_g^2$ in crystal-field language), leaving the $3d$ subshell too occupied/spread to participate directly in hybrid bonding orbitals. The metal ion must instead use its empty outer $4s$, three $4p$, and two $4d$ orbitals to accept the six ligand lone pairs, giving:

$sp^3d^2 \text{ hybridisation (using outer } 4d \text{ orbitals)}$

This is called an outer-orbital (ionic) octahedral complex, and being high-spin with 4 unpaired electrons, $[Fe(H_2O)_6]^{2+}$ is strongly paramagnetic.

(Contrast this with a strong-field ligand like $CN^-$, as in $[Fe(CN)_6]^{4-}$: there, all six $3d$ electrons pair up in the lower $t_{2g}$ set, freeing the two remaining $3d$ orbitals for inner-orbital $d^2sp^3$ hybridisation — a low-spin complex, diamagnetic since every electron is paired.)

F⁻ is a weak-field ligand, so it does not pair up Co(III)'s d-electrons, giving an outer-orbital octahedral complex with sp³d² hybridisation.

In $[CoF_6]^{3-}$, cobalt is in the +3 oxidation state, with configuration $Co^{3+}: [Ar]3d^6$.

$F^-$ is a weak-field ligand (low in the spectrochemical series), so it is unable to force pairing of the $3d^6$ electrons into the lower $t_{2g}$ orbitals. The complex therefore remains high-spin, and the outer $4d$ (in this case, the $ns,np,nd$ set is $4s, 4p, 4d$) orbitals are used for bonding rather than the inner $3d$ orbitals — this is an outer orbital (sp³d²) octahedral complex.

Hybridisation: $4s + 4p^3 + 4d^2 = sp^3d^2$, giving 6 equivalent hybrid orbitals directed octahedrally, each accepting a lone pair from a fluoride ligand.

Square planar geometry arises from dsp$^2$ hybridisation of the central metal atom's orbitals.

In a square planar complex (e.g. [Ni(CN)$_4$]$^{2-}$, [PtCl$_4$]$^{2-}$), the metal uses one $d$ orbital ($d_{x^2-y^2}$), one $s$ orbital, and two $p$ orbitals ($p_x$, $p_y$) to form four equivalent hybrid orbitals arranged at 90° to each other in a plane — this is dsp$^2$ hybridisation. It typically occurs with strong-field ligands causing electron pairing, freeing a $d$ orbital for hybridisation (common for $d^8$ metal ions like Ni$^{2+}$, Pd$^{2+}$, Pt$^{2+}$).

(For comparison: tetrahedral complexes are sp$^3$, and octahedral complexes are sp$^3$d$^2$ or d$^2$sp$^3$.)

Using the spin-only formula $\mu=\sqrt{n(n+2)}$ BM, $\mu=1.73$ BM requires exactly $n=1$ unpaired electron; only $[Cu(NH_3)_4]^{2+}$ (Cu$^{2+}$, $d^9$, always 1 unpaired electron) matches.

Check each complex: (a) $[CoCl_6]^{4-}$ — overall charge $-4$ with 6 $Cl^-$ ($-6$) gives Co oxidation state $+2$, so $Co^{2+}$ is $3d^7$; $Cl^-$ is a weak-field ligand so this is high-spin, giving 3 unpaired electrons ($\mu=\sqrt{3\times5}=3.87$ BM).

(b) $TiCl_4$ — Ti is $+4$, so $Ti^{4+}$ is $3d^0$, no unpaired electrons ($\mu=0$, diamagnetic).

(c) $[Cu(NH_3)_4]^{2+}$ — $Cu^{2+}$ is $3d^9$; irrespective of ligand field, a $d^9$ configuration always has exactly one unpaired electron, giving $\mu=\sqrt{1\times3}=\sqrt3=1.73$ BM.

(d) $[Ni(CN)_4]^{2-}$ — $Ni^{2+}$ is $3d^8$; $CN^-$ is a strong-field ligand, forcing a square-planar, low-spin, diamagnetic arrangement ($\mu=0$). So only $[Cu(NH_3)_4]^{2+}$ gives the 1.73 BM moment.

Cl⁻ is a weak field ligand, so Ni²⁺'s d-electrons stay unpaired and the complex is sp³ hybridised (tetrahedral).

In [NiCl₄]²⁻, nickel is in the +2 oxidation state: Ni²⁺ = [Ar] 3d⁸ (8 electrons in five 3d orbitals: ↑↓ ↑↓ ↑ ↑ ↑ across the orbitals, i.e. 2 unpaired electrons in the free ion).

Cl⁻ is a weak field ligand (low in the spectrochemical series), so it cannot force pairing of the 3d electrons into fewer orbitals. Since the 3d orbitals remain unavailable for bonding (all are singly/doubly occupied and none are left empty for hybridisation), nickel uses one 4s and three 4p orbitals to bond with the four Cl⁻ ligands.

This gives sp³ hybridisation, a tetrahedral geometry, and the complex is paramagnetic (2 unpaired electrons remain in the 3d orbitals).

Four equivalent hybrid orbitals directed tetrahedrally come from mixing one s and three p orbitals — sp$^3$ hybridisation — option (c).

In valence bond theory (VBT), the geometry of a coordination complex is linked to the type of hybrid orbitals the central metal atom/ion uses to bond with its ligands:

• sp$^3$ hybridisation (1 s + 3 p orbitals) → 4 equivalent orbitals directed towards the corners of a regular tetrahedron → tetrahedral complex, e.g. [NiCl$_4$]$^{2-}$, [Ni(CO)$_4$].
• dsp$^2$ hybridisation → square planar geometry, e.g. [Ni(CN)$_4$]$^{2-}$.
• d$^2$sp$^3$/sp$^3$d$^2$ hybridisation → octahedral geometry. So a tetrahedral complex ion corresponds to sp$^3$ hybridisation.

F- is a weak-field ligand, so [FeF6]3- is an outer-orbital (sp3d2, octahedral) high-spin complex.

$sp^3d^2$ hybridisation corresponds to an outer-orbital, high-spin octahedral complex, which occurs with weak-field ligands that cannot force electron pairing. $F^-$ is a weak-field ligand, so $[FeF_6]^{3-}$ uses the outer 4d orbitals along with 4s and 4p, giving $sp^3d^2$ hybridisation. $[Fe(CN)_6]^{4-}$ and $[Co(NH_3)_6]^{3+}$ use strong/moderately strong field ligands that cause pairing, giving inner-orbital $d^2sp^3$ complexes, while $[Ni(CN)_4]^{2-}$ is square planar ($dsp^2$).

$[FeF_6]^{4-}$ contains high-spin $Fe^{2+}$ ($d^6$) with 4 unpaired electrons, giving a spin-only magnetic moment of $\sqrt{24}\approx4.90$ BM.

Since the six $F^-$ ligands each carry charge $-1$ and the overall complex ion charge is $-4$, the iron must be $Fe^{2+}$: $x + 6(-1) = -4 \Rightarrow x=+2$. $Fe^{2+}$ has the configuration $[Ar]3d^6$. $F^-$ is a weak-field ligand (low in the spectrochemical series), so it does not force electron pairing, and the complex adopts the high-spin octahedral configuration $t_{2g}^4 e_g^2$, which has 4 unpaired electrons. Using the spin-only formula $\mu=\sqrt{n(n+2)}$ BM with $n=4$: $\mu=\sqrt{4\times6}=\sqrt{24}=4.899\approx4.90$ BM.

As a weak-field ligand, F⁻ leaves Fe's d electrons unpaired, forcing the complex to use outer d orbitals — sp³d² hybridisation (an outer-orbital complex).

In $[FeF_6]^{3-}$, iron is in the +3 oxidation state (since each F⁻ contributes −1 and the overall complex ion charge is −3): Fe³⁺ has the configuration $3d^5$.

F⁻ is a weak-field ligand in the spectrochemical series, meaning it cannot supply enough crystal-field splitting energy to force pairing of the 5 d electrons into fewer orbitals. So all 5 d electrons remain unpaired, occupying the inner 3d orbitals fully as singly-filled — this leaves no empty inner (3d) orbitals available for bonding.

Since the inner d orbitals are unavailable, the complex must use the outer d orbitals (4d) along with 4s and 4p to accommodate the six ligand pairs, giving $sp^3d^2$ hybridisation. This is called an outer-orbital (or high-spin) octahedral complex, and it is paramagnetic (5 unpaired electrons).

(By contrast, a strong-field ligand like CN⁻ would pair up the d electrons and allow use of the inner 3d orbitals — giving $d^2sp^3$, an inner-orbital/low-spin complex.)