Which of the following complex to posesses d2sp3 hybridustion? (a) [Ni(NH3)6]2+ (b) (CoF6}3- (c)[Co(NH3)6]3+ (d) (FeF6]3-

Key Theory

1. Oxidation State and d-Electron Count
   • First find the oxidation state of the metal to know its electronic configuration.
2. Ligand Strength (Spectro-chemical Series)
   • Strong-field ligands (CN$^-$, CO, NH$_3$, en) usually cause pairing of d-electrons, favouring low-spin inner-orbital (d$^2$sp$^3$) octahedral complexes.
   • Weak-field ligands (F$^-$, Cl$^-$, H$_2$O) do NOT cause pairing → high-spin outer-orbital (sp$^3$d$^2$).
3. Valence-Bond Hybridisation
   • Six coordinate bonds can be made either by d$^2$sp$^3$ (inner) or sp$^3$d$^2$ (outer) hybrid orbitals.

Step-by-Step Reasoning for Each Complex

Hence, the only d$^2$sp$^3$ complex in the list is [Co(NH$_3$)$_6$]$^{3+}$.

Imagine Lego Blocks!

Suppose every metal ion is like a Lego base and the ligands (NH$_3$, F$^-$ etc.) are Lego sticks that want to plug in. The metal needs special holes (orbitals) to let six sticks fit neatly in an octahedral pattern (like putting six pencils around a ball).

Two ways to arrange the holes:

1. Inner-layer holes: Use two of the already-existing d-holes plus one s-hole and three p-holes. This mix is called d$^2$sp$^3$.
2. Outer-layer holes: Keep those inner d-holes busy and instead pull in two bigger, outer d-holes (4d, 5d …). That mix is sp$^3$d$^2$.

Whether the metal chooses the inner or outer holes depends on how bossy the ligand is:

• Strong / medium ligands (like NH$_3$, CN$^-$) shout, "Buddy, pair up your electrons and clear two d-holes for me!" → inner (d$^2$sp$^3$)
• Weak ligands (like F$^-$, Cl$^-$) are polite and never force pairing → outer (sp$^3$d$^2$)

Looking at the options, only [Co(NH$_3$)$_6$]$^{3+}$ has a medium-strong ligand (NH$_3$) and therefore pushes cobalt to use the inner d-orbitals: d$^2$sp$^3$.

Step-by-Step Solution

1. Identify oxidation states

   • Ni in [Ni(NH$_3$)$_6$]$^{2+}$: $x + 6(0) = +2 \Rightarrow x = +2$ → Ni$^{2+}$ ($d^8$).
   • Co in [CoF$_6$]$^{3-}$: $x + 6(-1) = -3 \Rightarrow x = +3$ → Co$^{3+}$ ($d^6$).
   • Co in [Co(NH$_3$)$_6$]$^{3+}$: $x + 6(0) = +3 \Rightarrow x = +3$ → Co$^{3+}$ ($d^6$).
   • Fe in [FeF$_6$]$^{3-}$: $x + 6(-1) = -3 \Rightarrow x = +3$ → Fe$^{3+}$ ($d^5$).

2. Check ligand strength using the spectro-chemical series:
   $\text{CN}^- > \text{NH}_3 > \text{H}_2O > \text{F}^- > \text{Cl}^-$

3. Apply VBT rules

   Ni$^{2+}$, $d^8$, ligand NH$_3$ (medium): Pairing energy is high for Ni$^{2+}$; electrons stay unpaired in 3d, so 4s, 4p, 4d are used → $sp^3d^2$

   Co$^{3+}$, $d^6$, ligand F$^-$ (weak): No pairing, so 4s, 4p, 4d → $sp^3d^2$

   Co$^{3+}$, $d^6$, ligand NH$_3$ (medium-strong): Electrons pair up in 3d, freeing two 3d orbitals, combine with 4s and three 4p → $d^2sp^3$

   Fe$^{3+}$, $d^5$, ligand F$^-$ (weak): No pairing, outer orbitals → $sp^3d^2$

4. Conclude

   $\boxed{\text{Only }[\text{Co(NH}_3)_6]^{3+}\text{ is }d^2sp^3}$