Intuition The one-line ideaA metal ion has empty orbitals . Ligands carry lone pairs . The metal offers a set of hybridized empty orbitals , the ligands donate their lone pairs into them (coordinate bonds), and the geometry is just the shape of the hybrid set . Whether the metal uses its inner d d d ( n − 1 ) d (n-1)d ( n − 1 ) d outer d d d n d nd n d magnetism and the name inner vs outer orbital complex .
Definition Valence Bond Theory (VBT) of complexesThe metal–ligand bond is a coordinate (dative) covalent bond formed when a filled ligand orbital (lone pair) overlaps with a vacant hybrid orbital of the central metal ion. The number of empty hybrid orbitals = coordination number ; their geometry = the shape of the complex.
The key questions VBT answers:
Geometry → from hybridization type.Magnetic moment → from number of unpaired electrons.Inner vs outer → which d d d
C.N.
Hybridization
Geometry
Which d d d
2
s p sp s p linear
none
4
s p 3 sp^3 s p 3 tetrahedral
none → outer
4
d s p 2 dsp^2 d s p 2 square planar
( n − 1 ) d (n-1)d ( n − 1 ) d inner
6
s p 3 d 2 sp^3d^2 s p 3 d 2 octahedral
n d nd n d outer
6
d 2 s p 3 d^2sp^3 d 2 s p 3 octahedral
( n − 1 ) d (n-1)d ( n − 1 ) d inner
Intuition Inner vs outer in plain words
Inner orbital complex = the metal sacrifices (pairs up) its electrons to free up the inner ( n − 1 ) d (n-1)d ( n − 1 ) d d 2 s p 3 d^2sp^3 d 2 s p 3 Outer orbital complex = the metal keeps electrons unpaired and instead uses outer n d nd n d s p 3 d 2 sp^3d^2 s p 3 d 2
A magnetic moment arises from unpaired electron spins . Each electron has spin quantum number s = 1 2 s=\tfrac12 s = 2 1 n n n S = n ⋅ 1 2 = n 2 S = n\cdot\tfrac12 = \tfrac{n}{2} S = n ⋅ 2 1 = 2 n
The magnitude of the spin angular momentum (in QM) is S ( S + 1 ) ℏ \sqrt{S(S+1)}\,\hbar S ( S + 1 ) ℏ S = n 2 S=\tfrac{n}{2} S = 2 n
S ( S + 1 ) = n 2 ( n 2 + 1 ) = n 2 ⋅ n + 2 2 = n ( n + 2 ) 4 S(S+1) = \frac{n}{2}\left(\frac{n}{2}+1\right) = \frac{n}{2}\cdot\frac{n+2}{2} = \frac{n(n+2)}{4} S ( S + 1 ) = 2 n ( 2 n + 1 ) = 2 n ⋅ 2 n + 2 = 4 n ( n + 2 )
So S ( S + 1 ) = 1 2 n ( n + 2 ) \sqrt{S(S+1)} = \tfrac12\sqrt{n(n+2)} S ( S + 1 ) = 2 1 n ( n + 2 ) μ \mu μ g ≈ 2 g\approx2 g ≈ 2 1 2 \tfrac12 2 1
n n n 0
1
2
3
4
5
μ \mu μ 0
1.73
2.83
3.87
4.90
5.92
Intuition WHY we can run thisbackwards
Experiments measure μ \mu μ μ \mu μ n n n n n n deduce inner vs outer geometry. VBT is predictive both ways.
[ Fe ( CN ) 6 ] 4 − [\text{Fe}(\text{CN})_6]^{4-} [ Fe ( CN ) 6 ] 4 − Step — oxidation state: CN − \text{CN}^- CN − − 1 -1 − 1 − 4 -4 − 4 + 2 +2 + 2 Why? charge balance: x + 6 ( − 1 ) = − 4 x + 6(-1) = -4 x + 6 ( − 1 ) = − 4 Step — config: Fe 2 + = [ Ar ] 3 d 6 \text{Fe}^{2+}=[\text{Ar}]3d^6 Fe 2 + = [ Ar ] 3 d 6 Why? Fe is [ Ar ] 3 d 6 4 s 2 [\text{Ar}]3d^64s^2 [ Ar ] 3 d 6 4 s 2 4 s 4s 4 s Step — ligand: CN − \text{CN}^- CN − strong field → forces pairing. Why? it's high in the spectrochemical series.
Step — pair up: d 6 d^6 d 6 ↑ ↓ ↑ ↓ ↑ ↓ _ _ \uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \_\ \_ ↑↓ ↑↓ ↑↓ _ _ two empty 3 d 3d 3 d orbitals freed.
Step — hybridize: two 3 d 3d 3 d 4 s 4s 4 s 4 p 4p 4 p d 2 s p 3 d^2sp^3 d 2 s p 3 octahedral, inner orbital .
Step — magnetism: n = 0 ⇒ μ = 0 n=0 \Rightarrow \mu=0 n = 0 ⇒ μ = 0 diamagnetic .
[ FeF 6 ] 3 − [\text{FeF}_6]^{3-} [ FeF 6 ] 3 − Oxidation: x + 6 ( − 1 ) = − 3 ⇒ x+6(-1)=-3 \Rightarrow x + 6 ( − 1 ) = − 3 ⇒ 3 + ^{3+} 3 + Config: Fe 3 + = 3 d 5 \text{Fe}^{3+}=3d^5 Fe 3 + = 3 d 5 Ligand: F − \text{F}^- F − weak field → no pairing. Why? low in series.
Distribution: d 5 d^5 d 5 ↑ ↑ ↑ ↑ ↑ \uparrow\ \uparrow\ \uparrow\ \uparrow\ \uparrow ↑ ↑ ↑ ↑ ↑ no empty inner 3 d 3d 3 d .
Hybridize: must use outer orbitals: 4 s , 4 p x , 4 p y , 4 p z , 4 d x 2 − y 2 , 4 d z 2 4s,4p_x,4p_y,4p_z,4d_{x^2-y^2},4d_{z^2} 4 s , 4 p x , 4 p y , 4 p z , 4 d x 2 − y 2 , 4 d z 2 s p 3 d 2 sp^3d^2 s p 3 d 2 outer .
Magnetism: n = 5 ⇒ μ = 5 ⋅ 7 = 35 = 5.92 n=5 \Rightarrow \mu=\sqrt{5\cdot7}=\sqrt{35}=5.92 n = 5 ⇒ μ = 5 ⋅ 7 = 35 = 5.92
[ Ni ( CN ) 4 ] 2 − [\text{Ni}(\text{CN})_4]^{2-} [ Ni ( CN ) 4 ] 2 − [ NiCl 4 ] 2 − [\text{NiCl}_4]^{2-} [ NiCl 4 ] 2 − Ni 2 + = 3 d 8 \text{Ni}^{2+}=3d^8 Ni 2 + = 3 d 8
With CN − \text{CN}^- CN − d 8 d^8 d 8 ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ _ \uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \_ ↑↓ ↑↓ ↑↓ ↑↓ _ one 3 d 3d 3 d → d s p 2 dsp^2 d s p 2 square planar , n = 0 n=0 n = 0 diamagnetic .
With Cl − \text{Cl}^- Cl − n = 2 n=2 n = 2 s p 3 sp^3 s p 3 tetrahedral, paramagnetic (μ = 2.83 \mu=2.83 μ = 2.83 Why this matters: same metal, same C.N. (4), but ligand strength changes shape and magnetism .
[ Co ( NH 3 ) 6 ] 3 + [\text{Co}(\text{NH}_3)_6]^{3+} [ Co ( NH 3 ) 6 ] 3 + 3 + = 3 d 6 ^{3+}=3d^6 3 + = 3 d 6 NH 3 \text{NH}_3 NH 3 t t t ↑ ↓ ↑ ↓ ↑ ↓ \uparrow\downarrow\,\uparrow\downarrow\,\uparrow\downarrow ↑↓ ↑↓ ↑↓ 3 d 3d 3 d d 2 s p 3 d^2sp^3 d 2 s p 3 n = 0 n=0 n = 0
Recall Forecast:[ CoF 6 ] 3 − [\text{CoF}_6]^{3-} [ CoF 6 ] 3 − 3 + = 3 d 6 ^{3+}=3d^6 3 + = 3 d 6 F − \text{F}^- F − d 6 = ↑ ↓ ↑ ↑ ↑ ↑ d^6 = \uparrow\downarrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow d 6 =↑↓ ↑ ↑ ↑ ↑ n = 4 n=4 n = 4 s p 3 d 2 sp^3d^2 s p 3 d 2 paramagnetic μ = 4 ⋅ 6 = 4.90 \mu=\sqrt{4\cdot6}=4.90 μ = 4 ⋅ 6 = 4.90 n = 0 n=0 n = 0 same ion, ligand decides everything.
d d d 4 s 4s 4 s Why it feels right: 4 s 4s 4 s 3 d 3d 3 d Fix: for cations, 4 s 4s 4 s Fe 2 + \text{Fe}^{2+} Fe 2 + 3 d 6 3d^6 3 d 6 not 3 d 4 4 s 2 3d^44s^2 3 d 4 4 s 2 3 d 3d 3 d 4 s 4s 4 s
Common mistake "Strong ligand always means low spin / inner orbital."Why it feels right: strong = pairing = inner, usually true.
Fix: pairing only frees inner d d d are enough electrons to pair into a smaller set. For d 1 d^1 d 1 d 3 d^3 d 3 d 4 d^4 d 4 d 7 d^7 d 7
Common mistake "VBT explains colour of complexes."Why it feels right: VBT explains geometry and magnetism so well you expect colour too.
Fix: VBT is silent on colour and on why some ligands are strong/weak (no orbital-splitting energy). That needs CFT/Crystal Field Theory .
s p 3 d 2 sp^3d^2 s p 3 d 2 d 2 s p 3 d^2sp^3 d 2 s p 3 Fix: Same shape (octahedral) but different d d d : d 2 s p 3 d^2sp^3 d 2 s p 3 ( n − 1 ) d (n-1)d ( n − 1 ) d s p 3 d 2 sp^3d^2 s p 3 d 2 n d nd n d
Mnemonic Remember the magnetism formula & inner/outer"n n n n n n μ \mu μ → n ( n + 2 ) \sqrt{n(n+2)} n ( n + 2 ) Inner = INNer = d d d (d 2 s p 3 d^2sp^3 d 2 s p 3 Outer = s p 3 d 2 sp^3d^2 s p 3 d 2 d d d
"STrong ligand → STacks (pairs) electrons → STays low spin."
Recall Feynman: explain to a 12-year-oldImagine the metal is a hotel with empty rooms (orbitals). Ligands are guests carrying gifts (lone pairs) who only move into empty rooms. The hotel must arrange its rooms into a neat shape (square, triangle-pyramid, octahedron) before guests arrive — that arranging is "hybridization." Some pushy guests (strong ligands like CN⁻) make the hotel's own residents share rooms (pair up) so more fancy inner rooms open — that's an inner complex , and because everyone's paired, the hotel isn't "magnetic." Polite guests (weak ligands like F⁻) don't force sharing, so the hotel opens extra outer rooms instead, leaves residents un-paired, and the building acts magnetic. Count the lonely (unpaired) residents, plug into n ( n + 2 ) \sqrt{n(n+2)} n ( n + 2 )
What does VBT say a metal–ligand bond is? A coordinate (dative) covalent bond: ligand lone pair donated into a vacant hybrid orbital of the metal.
Spin-only magnetic moment formula? μ = n ( n + 2 ) \mu=\sqrt{n(n+2)} μ = n ( n + 2 ) BM,
n n n = number of unpaired electrons.
Hybridization & geometry for inner-orbital octahedral complex? d 2 s p 3 d^2sp^3 d 2 s p 3 , octahedral, uses inner
( n − 1 ) d (n-1)d ( n − 1 ) d orbitals, low spin.
Hybridization for outer-orbital octahedral complex? s p 3 d 2 sp^3d^2 s p 3 d 2 , octahedral, uses outer
n d nd n d orbitals, high spin.
Hybridization for square planar vs tetrahedral 4-coordinate? Square planar =
d s p 2 dsp^2 d s p 2 (inner
d d d ); tetrahedral =
s p 3 sp^3 s p 3 .
Which electrons are removed first when forming a transition-metal cation? The
n s ns n s electrons, before
( n − 1 ) d (n-1)d ( n − 1 ) d .
d-config of Fe 2 + \text{Fe}^{2+} Fe 2 + Fe 3 + \text{Fe}^{3+} Fe 3 + 3 d 6 3d^6 3 d 6 and
3 d 5 3d^5 3 d 5 .
Magnetism of [ Fe ( CN ) 6 ] 4 − [\text{Fe}(\text{CN})_6]^{4-} [ Fe ( CN ) 6 ] 4 − n = 0 n=0 n = 0 , diamagnetic (
d 2 s p 3 d^2sp^3 d 2 s p 3 , inner).
Magnetism of [ FeF 6 ] 3 − [\text{FeF}_6]^{3-} [ FeF 6 ] 3 − n = 5 n=5 n = 5 ,
μ = 5.92 \mu=5.92 μ = 5.92 BM, paramagnetic (
s p 3 d 2 sp^3d^2 s p 3 d 2 , outer).
Why is [ Ni ( CN ) 4 ] 2 − [\text{Ni}(\text{CN})_4]^{2-} [ Ni ( CN ) 4 ] 2 − [ NiCl 4 ] 2 − [\text{NiCl}_4]^{2-} [ NiCl 4 ] 2 − CN⁻ is strong → pairs
d 8 d^8 d 8 → frees a
3 d 3d 3 d →
d s p 2 dsp^2 d s p 2 square planar; Cl⁻ weak → no pairing →
s p 3 sp^3 s p 3 tetrahedral.
Two major failures of VBT? Cannot explain colour (spectra) and cannot predict why ligands are strong/weak; ignores orbital splitting energies.
μ \mu μ n = 3 n=3 n = 3 3 ⋅ 5 = 15 = 3.87 \sqrt{3\cdot5}=\sqrt{15}=3.87 3 ⋅ 5 = 15 = 3.87 BM.
Strong-field vs weak-field effect on spin? Strong field → pairing → low spin/inner; weak field → no pairing → high spin/outer.
Number of hybrid orbitals
mu = sqrt of n times n plus 2
Intuition Hinglish mein samjhoDekho, VBT ka funda simple hai: metal ion ke paas kuch khaali orbitals hote hain, aur ligands apne lone pair leke aate hain. Ligand apna lone pair metal ke khaali hybrid orbital mein daal deta hai — yahi coordinate bond hai. Pehle metal ka oxidation state nikaalo, fir uska d d d d n d^n d n 4 s 4s 4 s pehle nikalte hain, isliye Fe 2 + \text{Fe}^{2+} Fe 2 + 3 d 6 3d^6 3 d 6 3 d 4 4 s 2 3d^44s^2 3 d 4 4 s 2
Ab inner vs outer ka khel ligand strength pe depend karta hai. Strong ligand (jaise CN⁻, NH₃) metal ke electrons ko pair kara deta hai, jisse andar wale ( n − 1 ) d (n-1)d ( n − 1 ) d d 2 s p 3 d^2sp^3 d 2 s p 3 inner orbital, low spin complex hai. Weak ligand (jaise F⁻, Cl⁻) pairing force nahi karta, to metal ko bahar wale n d nd n d orbital use karne padte hain → s p 3 d 2 sp^3d^2 s p 3 d 2 outer orbital, high spin complex.
Magnetism samajhna easy hai: jitne unpaired electrons (n n n μ = n ( n + 2 ) \mu=\sqrt{n(n+2)} μ = n ( n + 2 ) n = 0 n=0 n = 0 μ \mu μ n n n
Ek important baat — VBT geometry aur magnetism to bataa deta hai, lekin colour kyun hota hai ya ligand strong/weak kyun hai ye explain nahi kar paata. Uske liye Crystal Field Theory padhni padti hai. Isliye VBT ko ek solid pehla tool samjho, complete picture ke liye CFT zaruri hai.
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