Crystal Field Stabilization Energy (CFSE) — high-spin vs low-spin; spectrochemical series

3.4.9 · Chemistry › Coordination Chemistry

KYA ho raha hai (model)

Ek octahedral field mein (6 ligands $\pm x, \pm y, \pm z$ ke saath):

Set	Orbitals	Point karte hain...	Energy
$e_{g}$	$d_{z^{2}}, d_{x^{2} - y^{2}}$	ligands ki taraf	raised
$t_{2 g}$	$d_{x y}, d_{y z}, d_{x z}$	ligands ke beech	lowered

Unke beech ka energy gap $== Δ_{o} ==$ hai (octahedral splitting parameter, "10 Dq").

Orbital energies kaise derive karein (first principles)

Maano down-shift of each $t_{2 g}$ orbital $x$ ho (barycentre se neeche) aur up-shift of each $e_{g}$ orbital $y$ ho (barycentre se upar). Gap ki definition se:

$y - (-x) = \Delta_o \quad\Rightarrow\quad x + y = \Delta_o \tag{1}$

Barycentre conservation: total down = total up.

$3x = 2y \tag{2}$

Yeh step kyun? 3 orbitals har ek $x$ se girte hain, 2 orbitals har ek $y$ se uthte hain; gravity ka centre unchanged rehne ke liye dono balance karne chahiye.

(1) aur (2) solve karo: (2) se, $y = \frac{3}{2} x$ . (1) mein substitute karo:

$x + \frac{3}{2} x = Δ_{o} \Rightarrow \frac{5}{2} x = Δ_{o} \Rightarrow x = \frac{2}{5} Δ_{o}, y = \frac{3}{5} Δ_{o}$

High-spin vs Low-spin: competition

Yeh choice sirf $d^{4}$ – $d^{7}$ (octahedral) ke liye matter karti hai. $d^{1}$ – $d^{3}$ aur $d^{8}$ – $d^{10}$ ke liye filling forced hai — koi ambiguity nahi.

Figure — Crystal Field Stabilization Energy (CFSE) — high-spin vs low-spin; spectrochemical series

Spectrochemical Series ( $Δ_{o}$ kya set karta hai?)

$Δ_{o}$ higher oxidation state ke saath aur group mein neeche jaane par bhi badhta hai (3d < 4d < 5d), isliye 4d/5d complexes almost always low-spin hote hain.

Common mistake Common errors ko steel-man karna

(1) "Low-spin ka CFSE hamesha bada hota hai isliye woh hamesha zyada stable hai." Kyun sahi lagta hai: low-spin electrons ko deep $t_{2 g}$ mein pack karta hai, ek bada negative $Δ_{o}$ term deta hai. Fix: tumhe pairing-energy cost $+ m P$ subtract karni padegi. Net stability tab hi milti hai jab $Δ_{o} > P$ . $P$ bhool gaye toh tum har fluoride complex ke liye "low-spin predict" karoge — galat.

(2) Barycentre bhool ke $\pm \frac{1}{2} Δ_{o}$ use karna. Kyun sahi lagta hai: do sets hain, toh half-half symmetrically split ho jaata hai. Fix: 3 vs 2 orbitals hain, 1 vs 1 nahi. Balance force karta hai $- 0.4$ aur $+ 0.6$ , na ki $- 0.5/ + 0.5$ .

(3) Un electrons ke liye pairing energy count karna jo free ion mein already paired the. Fix: sirf woh extra pairs count karo jo ligand field force karta hai. $d^{6}$ HS ke liye, $m = 1$ hai (0 nahi), kyunki $t_{2 g}^{4}$ banana ek pair require karta hai hypothetical fully-spread arrangement ke zero pairs se pare — lekin consistently free-ion baseline se compare karo.

(4) Tetrahedral ke liye octahedral numbers use karna. Fix: tetrahedral splitting inverted hoti hai ( $e$ below $t_{2}$ ) aur $Δ_{t} \approx \frac{4}{9} Δ_{o}$ hota hai, isliye tetrahedral complexes essentially hamesha high-spin hote hain.

Recall Feynman: 12-saal ke bache ko explain karo

Imagine karo paanch bachche identical kursiyon par baithe hain ( $d$ -orbitals). Ab cheh grumpy neighbours (ligands) aate hain aur dhakka dete hain. Do bachche jinki kursiyan neighbours ke saamne hain, upar oonchi stools par push ho jaate hain (uncomfortable, $e_{g}$ ); teen bachche jo neighbours ke beech mein baithe hain woh aaram wale low cushions mein dhasak jaate hain ( $t_{2 g}$ ). Bachche aaram wale cushions prefer karte hain, isliye group pehle se zyada khush hai (zyada stable) — yeh "saved happiness" hi CFSE hai. Ab: ek chhaatha bachcha aata hai. Kya woh kisi ke saath aaram wali cushion par squeeze kare (annoying = pairing energy $P$ ) ya ek akeli oonchi stool par chadhe (height cost = $Δ_{o}$ )? Agar stool bahut oonchi hai toh woh squeeze karta hai (low-spin); agar stool neechi hai toh chadh jaata hai (high-spin). CO jaise bossy neighbours bahut oonchi stools banate hain → sab squeeze karte hain.

Flashcards

CFSE ka full form kya hai aur yeh kya measure karta hai?

Crystal Field Stabilization Energy; woh net energy jisse ek complex stabilize hota hai kyunki

d

-electrons preferentially lower-energy split orbitals occupy karte hain (unsplit barycentre ke relative).

Octahedral field mein kaun se orbitals

t_{2 g}

banate hain aur kaun se

e_{g}

t_{2 g} = d_{x y}, d_{y z}, d_{x z}

(ligands ke beech point karte hain, lower);

e_{g} = d_{z^{2}}, d_{x^{2} - y^{2}}

(ligands ki taraf point karte hain, higher).

t_{2 g}

aur

e_{g}

ki energies barycentre ke relative derive karo.

Maano down-shift

x

, up-shift

y

. Gap:

x + y = Δ_{o}

; barycentre:

3 x = 2 y

. Solve karo →

x = 0.4 Δ_{o}

y = 0.6 Δ_{o}

. Toh

t_{2 g} = - 0.4 Δ_{o}

e_{g} = + 0.6 Δ_{o}

High-spin vs low-spin kya determine karta hai?

Δ_{o}

ko pairing energy

P

se compare karo.

Δ_{o} < P

→ high-spin (weak field);

Δ_{o} > P

→ low-spin (strong field).

Kis

d^{n}

count ke liye HS/LS distinction exist karti hai (octahedral)?

d^{4}

d^{7}

tak. (

d^{1}

–

d^{3}

d^{8}

–

d^{10}

forced hain.)

High-spin

d^{6}

ka CFSE (

P

ignore karke)?

t_{2 g}^{4} e_{g}^{2}

(- 0.4 \times 4 + 0.6 \times 2) Δ_{o} = - 0.4 Δ_{o}

Low-spin

d^{6}

ka CFSE (

P

ignore karke)?

t_{2 g}^{6}

- 0.4 \times 6 Δ_{o} = - 2.4 Δ_{o}

[Co(NH_{3})_{6}]^{3 +}

ka diamagnetic hona low-spin kyun prove karta hai?

Diamagnetic = koi unpaired electrons nahi =

t_{2 g}^{6} e_{g}^{0}

, jo require karta hai

Δ_{o} > P

(low-spin).

Inhe field strength se order karo: CN⁻, H₂O, I⁻.

I⁻ < H₂O < CN⁻ (weak → strong).

CO, neutral hone ke bawajood F⁻ se stronger-field ligand kyun hai?

CO ke paas empty π* orbitals hain jo metal

t_{2 g}

density accept karte hain (π-backbonding),

t_{2 g}

neeche aata hai aur

Δ_{o}

badhta hai; F⁻ ek π-donor hai jo ulta karta hai.

Δ_{t}

Δ_{o}

se compare karo, aur tetrahedral complexes high-spin kyun hote hain?

Δ_{t} \approx \frac{4}{9} Δ_{o}

; yeh itna chhota hai ki

Δ_{t} < P

almost hamesha hota hai, isliye electrons spread out ho jaate hain (high-spin).

Oxidation state aur metal row

Δ_{o}

ko kaise affect karte hain?

Δ_{o}

higher oxidation state aur group mein neeche jaane par badhta hai (3d<4d<5d), isliye 4d/5d complexes usually low-spin hote hain.

Connections

Crystal Field Theory — Octahedral vs Tetrahedral Splitting
Magnetic Properties of Coordination Compounds (spin-only formula)
Colour of Transition-Metal Complexes (d-d transitions, $\Delta_o$ and λ)
Jahn–Teller Distortion
Pairing Energy and Hund's Rule
Ligand Field Theory & π-backbonding
Stability & Kinetic Inertness (Cr³⁺, Co³⁺)