cfse in d7 tetrahedral

Key Ideas Needed

1. Crystal-field splitting in tetrahedral complexes
   The five d-orbitals split into:

   • Lower energy set: $e$ (2 orbitals) at –0.6 $\Delta_t$ each.
   • Higher energy set: $t_2$ (3 orbitals) at +0.4 $\Delta_t$ each. The values are chosen so that the overall centre of gravity remains the same:
     $2(-0.6 \Delta_t) + 3(+0.4 \Delta_t) = 0$

2. High-spin vs low-spin
   In tetrahedral geometry $\Delta_t$ is small ( $\Delta_t \approx \frac{4}{9}\,\Delta_o$ ), always smaller than typical pairing energy $P$. Therefore tetrahedral complexes are almost always high-spin; we fill all orbitals singly before pairing.

3. Electron distribution for d(^{7})

   • First 5 electrons: one in each of the five d-orbitals (2 in $e$, 3 in $t_2$).
   • 6th & 7th electrons: must pair; they choose the lower set ($e$) because that costs $P$ but saves 0.6 $\Delta_t$ each. Resulting occupancies: $e^4\;t_2^3$

4. Crystal-field stabilisation energy (CFSE)
   $\text{CFSE} = (n_e)(-0.6\,\Delta_t) + (n_{t_2})(+0.4\,\Delta_t)$ Here $n_e = 4$ and $n_{t_2}=3$.

   $\text{CFSE} = 4(-0.6\,\Delta_t) + 3(+0.4\,\Delta_t) = -2.4\,\Delta_t + 1.2\,\Delta_t = -1.2\,\Delta_t$

5. Interpretation
   Negative value means the complex is stabilised by 1.2 units of the tetrahedral splitting energy relative to the hypothetical unsplit state.

Imagine 5 "rooms" and 7 kids

• A metal ion has 5 d-rooms for its electrons (kids).
• In a tetrahedral house these rooms split into two floors:
  • Ground floor (e set) – just 2 rooms, a little lower.
  • First floor (t8 set) – 3 rooms, a little higher.
• The lift between the floors needs energy called $\Delta_t$.
• Electrons are like playful kids – they first take every empty room single-single before sharing (Hund’s rule) because sharing costs the pairing energy (P).

For 7 kids (d8):

1. First put one kid in each of the 2 ground-floor rooms (\rightarrow) 2 kids.
2. Next use the 3 first-floor rooms one each (\rightarrow) total 5 kids.
3. Now all 5 rooms have one kid each; the 6th and 7th kids must share. They prefer the cheaper ground floor, so those two pair up there.

Finally you have 4 kids down (e) and 3 kids up (t8).

Every kid down saves 0.6 units, every kid up costs 0.4 units. Do the maths and you get a net saving (CFSE) of –1.2 (\Delta_t).

Step-by-step Calculation of CFSE for d$^{7}$ Tetrahedral Complex

1. Write the splitting pattern

   $\text{Lower } e: -0.6\,\Delta_t\quad (2\text{ orbitals})$ $\text{Upper } t_2: +0.4\,\Delta_t\quad (3\text{ orbitals})$

2. Place 7 d–electrons (high spin)

   • Electrons 1–2 $\rightarrow$ fill each $e$ orbital singly.
   • Electrons 3–5 $\rightarrow$ fill each $t_2$ orbital singly.
   • Electron 6 $\rightarrow$ pair in one $e$ orbital (lower energy).
   • Electron 7 $\rightarrow$ pair in the other $e$ orbital.

   Occupancy obtained: $e^4\;t_2^3$

3. Count electrons in each set
   $n_e = 4,\;\; n_{t_2}=3$

4. Apply the CFSE formula

   \text{CFSE} &= n_e(-0.6\,\Delta_t) + n_{t_2}(+0.4\,\Delta_t) \\ &= 4(-0.6\,\Delta_t) + 3(+0.4\,\Delta_t) \\ &= -2.4\,\Delta_t + 1.2\,\Delta_t \\ &= \boxed{-1.2\,\Delta_t} \end{aligned}$$

5. Express relative to octahedral splitting (optional)
   Since $\Delta_t = \frac{4}{9}\,\Delta_o$:
   $\text{CFSE} = -1.2\left(\frac{4}{9}\,\Delta_o\right) = -\frac{16}{15}\,\Delta_o \approx -1.07\,\Delta_o$

Hence the crystal field stabilisation energy for a high-spin d$^{7}$ tetrahedral complex is –1.2 (\Delta_t).