The coordination numbers of Co and Al in [Co(Cl)(en)2]Cl and K3[Al(C2O4)3] respectively are: (en = ethane-1,2-diamine)

🧠 CN = Donor-Atom Count, Period

Coordination number is never about how many ligands you can count — it's about how many donor atoms touch the metal. Each bidentate ligand contributes 2, each monodentate contributes 1. Multiply, then add.

🗺️ Two Spheres

$[\mathrm{Co(Cl)(en)_2}]\mathrm{Cl}$. Inside the bracket: 1 chlorido (1 donor) + 2 en (each $\times$ 2 donors). Total = $1 + 4 = 6$. The outer $\mathrm{Cl^-}$ is a counter-ion — ignore it.

$\mathrm{K_3[Al(C_2O_4)_3]}$. Inside the bracket: 3 oxalate (each bidentate, each $\times$ 2 donors). Total = $3 \times 2 = 6$.

So both metals have CN = 6 — option (4).

⚡ The "Donor Atom" Quick-Add

For each ligand, write a small superscript with its denticity, then sum:

• en$^{2}$, ox$^{2}$, EDTA$^{6}$, dien$^{3}$, $\mathrm{NH_3^{1}}$, $\mathrm{Cl^{-1}}$.

Quick-add the superscripts inside the bracket.

⚠️ The Outer Counter-Ion Trap

Students often add the lone outer $\mathrm{Cl^-}$ in $[\mathrm{Co(Cl)(en)_2}]\mathrm{Cl}$ and report CN = 7. The outer ion is outside the coordination sphere — it does not bind the metal. Always count what is inside the square brackets only.

$\boxed{\text{Answer: (4) 6 and 6}}$