3.4.8
Ek metal ion ke paanch d point negative charges maan ke chalte hain), kuch d
Ek aisa model jo metal–ligand bond ko purely electrostatic (ionic) interaction maanta hai. Ligands ko point negative charges (ya point dipoles) ki tarah model kiya jaata hai. Yeh theory explain karti hai ki jab ligands approach karte hain toh metal ke d splitting kyun hoti hai.
Humein iska zaroorat KYU hai? Valence Bond Theory colour , magnetism trends, ya kyun kuch complexes high-spin aur kuch low-spin hain — yeh explain nahi kar sakti thi. CFT ek hi number se teeno predict karta hai: splitting energy Δ
Paanch d
Orbitals
Lobes kahan point karte hain...
d x 2 − y 2 , d z 2 axes ke saath (x , y , z
d x y , d y z , d x z axes ke beech mein
Intuition Yeh dono kyun alag split hote hainEk octahedron mein, 6 ligands ± x , ± y , ± z . Toh:
d x 2 − y 2 , d z 2 strong repulsion feel hoti hai → energy badhti hai . Is set ko == e g == d x y , d y z , d x z ghatti hai . Is set ko == t 2 g ==
Dono ke beech ka gap Δ oc t 10 D q
Barycentre rule derive karna (first principles se).
Jab ligands approach karte hain, paanch orbitals ki average (weighted) energy constant rehni chahiye — energy conserve hoti hai; hum sirf redistribute karte hain. Yahi barycentre / centre-of-gravity rule hai.
Maano t 2 g x e g y y − ( − x ) = Δ oc t
3 ( − x ) + 2 ( + y ) = 0 and x + y = Δ oc t
Kyun? t 2 g 3 orbitals hain, e g 2 ; barycentre se total displacement ka sum zero hona chahiye.
Solve karo: pehle equation se, 3 x = 2 y x = Δ − y 3 ( Δ − y ) = 2 y ⇒ 3Δ = 5 y ⇒ y = 5 3 Δ , x = 5 2 Δ
Intuition Order KYUN FLIP hota haiEk tetrahedron mein 4 ligands hote hain, aur khaas baat yeh hai ki unme se koi bhi axis par nahi baithe hote — woh axes ke beech mein baithe hote hain (socho ek cube ke alternate corners). Ab:
t 2 d x y , d y z , d x z zyada karib point karte hain → raised .e d x 2 − y 2 , d z 2 zyada door point karte hain → lowered .
Labels aur order octahedral ke mukable inverted hain, aur koi g
Δ t e t
Sirf 4 ligands hain (6 ke badle) → kam total repulsion.
Koi bhi orbital seedha ligand ki taraf point nahi karta → effect aur bhi weak hai.
Ek geometric/electrostatic calculation se yeh famous factor milta hai:
Kyunki Δ t e t almost hamesha pairing energy P . Toh tetrahedral complexes essentially hamesha high-spin hote hain — electrons pair karne ke liye kabhi itna energy gap nahi hota.
Agar Δ oc t < P weak field → pehle sab orbitals mein single fill karo → high-spin (max unpaired).
Agar Δ oc t > P strong field → pehle t 2 g low-spin (min unpaired).
Yeh sirf d 4 d 7 d 1 , d 2 , d 3 , d 8 , d 9 , d 10
d 6 2 + d 6 3 d 6 4 s 2 2 + 4 s 2 3 d 6
[ FeF 6 ] 4 − − weak field ligand hai → Δ oc t < P t 2 g 4 e g 2 4 unpaired → paramagnetic.
Yeh step kyun? Weak field matlab electrons pair hone se pehle spread out ho jaate hain.[ Fe(CN) 6 ] 4 − − strong field ligand hai → Δ oc t > P t 2 g 6 e g 0 0 unpaired → diamagnetic.
Yeh step kyun? Bada gap pair karna e g
d 5 t 2 g 3 e g 2 CFSE = ( 3 ) ( − 0.4Δ ) + ( 2 ) ( + 0.6Δ ) = − 1.2Δ + 1.2Δ = 0 Yeh step kyun? Har orbital ki energy ko uske electron count se multiply karo aur sum karo. High-spin d 5 zero hota hai — yahi key reason hai ki aise ions (jaise Mn2 + 3 +
d 6 t 2 g 6 e g 0 P CFSE = 6 ( − 0.4Δ ) + 0 = − 2.4 Δ oc t ( + pairing terms ) Yeh step kyun? Badi − 2.4Δ d 6 [ Co(NH 3 ) 6 ] 3 + [ Fe(CN) 6 ] 4 −
Intuition Complexes coloured KYUN hote hainEk photon jiska energy Δ oc t t 2 g e g d–d transition hai). Woh wavelength absorb hoti hai; hum complementary colour dekhte hain. Bada Δ λ Δ oc t = h ν = λ h c
Ligands us Δ
I − < B r − < S 2 − < C l − < F − < O H − < H 2 O < N H 3 < e n < N O 2 − < C N − < C O
Mnemonic Spectrochemical series"I Bet Silly Clowns From Old Hospitals Need Energetic Nice Cool Cars"
I⁻, Br⁻, S²⁻, Cl⁻, F⁻, OH⁻, H₂O, NH₃, en, NO₂⁻, CN⁻, CO
Common mistake "Tetrahedrale t 2 g Kyun sahi lagta hai: Octahedral mein e g t 2 g g Fix: Subscript g gerade ) ka matlab hai structure mein ek centre of inversion hai. Tetrahedron mein koi centre of inversion nahi hota, toh hum sirf e t 2
Common mistake "Tetrahedral complexes low-spin ho sakte hain agar ligand strong ho."Kyun sahi lagta hai: Octahedral low-spin strong ligands ke saath hota hai, toh tetrahedral mein bhi ho sakta hai.
Fix: Δ t e t = 9 4 Δ oc t < P hamesha high-spin treat kiya jaata hai.
e g Kyun sahi lagta hai: "Beech mein = ligands se bachte hain = stable = low."
Fix: Octahedral mein, e g d x 2 − y 2 , d z 2 saath point karte hain = ligands ki taraf = raised. Tum ise tetrahedral case se confuse kar rahe ho jahan geometry flip hoti hai.
Common mistake "Hum jo colour dekhte hain wahi absorb hota hai."Kyun sahi lagta hai: Naive "object IS its colour."
Fix: Hum woh colour dekhte hain jo absorb hue colour ka complementary hota hai. Red absorb karo → green dikhega.
Recall Feynman: ek 12-saal ke bachche ko samjhaoSocho metal ion ke d Δ e g 4/9
Kaun sa model M–L bonds ko point-charge electrostatic interactions maanta hai? Crystal Field Theory (CFT)
Octahedral fields mein kaun sa d-orbital set raised hota hai aur kyun? e g d x 2 − y 2 , d z 2
Barycentre ke relative t 2 g e g t 2 g = − 0.4 Δ oc t e g = + 0.6 Δ oc t
Δ t e t Δ oc t Δ t e t = 9 4 Δ oc t ≈ 0.44 Δ oc t
Tetrahedral complexes almost hamesha high-spin KYUN hote hain? Δ t e t Δ oc t Δ t e t < P
Tetrahedral labels mein 'g' subscript KYUN nahi hota? Tetrahedron mein centre of inversion nahi hota (centrosymmetric nahi hai).
Low-spin (octahedral) ke liye condition kya hai? Strong field, Δ oc t > P
Low-spin d 6 t 2 g 6 e g 0
High-spin d 5 0
Complexes mein colour KISSE aata hai? d–d electronic transition (t 2 g → e g λ E = h c / λ = Δ
Sabse strong aur sabse weak common field ligand kaun sa hai? Strongest: CO (aur CN⁻); weakest: I⁻.
Electrostatic ionic interaction
Colour, magnetism, spin state
