Exercises — Werner's theory of coordination compounds

• Secondary valency (inside the bracket, the "hands"): the 6 $N{H}_3$ molecules. They never let go in water.
• Primary valency (outside the bracket, the "coins paid to friends"): the 3 $C{l}^{-}$.
• In water, only the outside ions swim free → 3 $C{l}^{-}$ ionize.
• So the compound breaks into $1+3=4$ ions: $[Co(N{H}_3{)}_6{]}^{3+}+3C{l}^{-}$.

• Primary valency $\to$ oxidation state of the metal (ionizable, non-directional charge balance).
• Secondary valency $\to$ coordination number (CN) (non-ionizable, directional / fixed in space). See Coordination number and geometry for how CN fixes the shape.

• Anything outside the bracket is a counter ion → the 4 $K^{+}$.
• Anything inside the bracket is a ligand → the 6 $C{N}^{-}$.
• So this salt gives $4K^{+}+[Fe(CN{)}_6{]}^{4-}=5$ ions in water.

Use $Q_{\text{complex}}=\text{OS}+\sum (\text{ligand charges})$, and read $Q_{\text{complex}}$ from the outside ions.

• (a) Outside = $2C{l}^{-}$, so complex ion is $+2$. Ligands: $5N{H}_3$ (each $0$) and $1C{l}^{-}$ ($-1$). $+2=\text{OS}+[5(0)+1(-1)]⟹\text{OS}=+2+1=+3.$
• (b) Outside = $2K^{+}$, so complex ion is $-2$. Ligands: $6C{l}^{-}$ ($-6$ total). $-2=\text{OS}+(-6)⟹\text{OS}=+4.$
• (c) No outside ions, so complex is neutral ($0$). Ligands: $4CO$ (each $0$). $0=\text{OS}+0⟹\text{OS}=0.$

• Inside the bracket: $4N{H}_3+2C{l}^{-}$ → the $2$ chlorides inside are ligands, locked by secondary valency.
• Outside: $1C{l}^{-}$ → this is the only free chloride.
• (a) Ions: $[Co(N{H}_3{)}_{4}C{l}_2{]}^{+}+C{l}^{-}=2$ ions.
• (b) Only the free $C{l}^{-}$ reacts with $A{g}^{+}$ → 1 mol AgCl per formula unit.

• Water ligands are neutral ($6\times 0=0$).
• Outside ions $=3C{l}^{-}$ → complex ion charge $=+3$. $+3=\text{OS}+0⟹\text{OS}=+3.✓$ Consistent: Cr is $+3$, CN $=6$, breaks into $4$ ions.

• Step 1 — how many Cl are free? $2$ mol AgCl means $2$ chlorides ionize → $2C{l}^{-}$ are outside.
• Step 2 — so how many Cl are ligands? Total Cl $=3$, free $=2$ → the remaining $1C{l}^{-}$ is inside (a ligand).
• Step 3 — check CN. Inside: $5N{H}_3+1C{l}^{-}=6$ donors → CN $=6$. ✓ matches Co(III).
• Step 4 — charge inside: $\text{OS}+\text{ligands}=+3+(-1)=+2$, so complex is $[Co(N{H}_3{)}_{5}Cl{]}^{2+}$.
• Formula: $[Co(N{H}_3{)}_{5}Cl]C{l}_2$, giving $1+2=3$ ions.

Count $1$ (the complex ion) $+$ (number of outside counter ions):

• $[Co(N{H}_3{)}_6]C{l}_3\to 1+3=4$ ions
• $[Co(N{H}_3{)}_{5}Cl]C{l}_2\to 1+2=3$ ions
• $[Co(N{H}_3{)}_{4}C{l}_2]Cl\to 1+1=2$ ions
• $[Co(N{H}_3{)}_{3}C{l}_3]\to 1+0=1$ particle (neutral, non-electrolyte) Ranking: $4>3>2>1$. More ions → higher molar conductivity, exactly Werner's measurement. See Conductivity and ionization of electrolytes.

• CN $=6$ but only $3N{H}_3$ are present → the metal needs 3 more ligands → all $3C{l}^{-}$ are pulled inside the bracket by secondary valency.
• Charge inside: $+3+3(-1)=0$ → the whole species is a neutral molecule, so there are no counter ions to ionize.
• No free $C{l}^{-}$ → $AgN{O}_3$ finds nothing to precipitate → $0$ mol AgCl.
• No ions → it does not conduct → molar conductivity $\approx 0$. This was Werner's "smoking gun".

• Ligand budget: CN $=6$. Available anions: $4C{l}^{-}$ (from $PtC{l}_4$) $+2C{l}^{-}$ (from $2KCl$) $=6$. All six fit inside → $[PtC{l}_6]$.
• Charge inside: $+4+6(-1)=-2$ → complex ion is $[PtC{l}_6{]}^{2-}$.
• Counter ions: the $2K^{+}$ must sit outside to balance $-2$.
• Formula: $K_2[PtC{l}_6]$ → ions $=1+2=3$ ($2K^{+}+[PtC{l}_6{]}^{2-}$).
• AgCl: every $C{l}^{-}$ is a ligand → 0 mol precipitate. ✓

• Octahedron: the two $Cl$ ligands are either adjacent (cis, $90^{\circ }$ apart) or opposite (trans, $180^{\circ }$ apart) → exactly 2 isomers. See figure.
• Planar hexagon: the two $Cl$ could be at positions $1,2$ / $1,3$ / $1,4$ → 3 isomers.
• Werner isolated 2 bottles → nature agrees with the octahedron. This is how counting isomers deduced the 3D shape without ever seeing an atom. More on this in Isomerism in coordination compounds (cis-trans, optical).

• CN counts donor atoms, not molecules. Each en gives $2$ donors → $3\times 2=6$ → CN $=6$.
• Charge: en is neutral, outside $=3C{l}^{-}$ → complex is $+3$. $+3=\text{OS}+0⟹\text{OS}=+3.$
• Ions: $[Co(en{)}_3{]}^{3+}+3C{l}^{-}=4$ ions. See Ligands — classification (mono/poly-dentate, chelate).

• 3 mol AgCl → $3C{l}^{-}$ are outside (ionizable) → all three chlorides are counter ions.
• 4 ions confirms: $1$ complex ion $+3C{l}^{-}$. ✓
• So no $C{l}^{-}$ inside → the $6$ CN slots are filled by the 6 $H_{2}O$ ligands.
• Charge inside: $+3+6(0)=+3$ → $[Cr(H_{2}O{)}_6{]}^{3+}$.
• Formula: $[Cr(H_{2}O{)}_6]C{l}_3$.
• Valencies: secondary valency (CN $=6$) = the $6$ water ligands; primary valency ($=+3$, oxidation state) = balanced by the $3$ outside $C{l}^{-}$.

The ion that is free (gives the test) sits outside; the ion locked inside is a ligand.

• X: $S{O}_4^{2-}$ is free (outside), $B{r}^{-}$ is the ligand (inside). Inside: $5N{H}_3+1B{r}^{-}$ → charge $+3+(-1)=+2$ → $[Co(N{H}_3{)}_{5}Br{]}^{2+}$. Counter ion: $S{O}_4^{2-}$ ($-2$) balances $+2$. → $[\mathbf{Co}(\mathbf{N}{\mathbf{H}}_3{)}_5\mathbf{Br}]\mathbf{S}{\mathbf{O}}_4$.
• Y: $B{r}^{-}$ is free (outside), $S{O}_4^{2-}$ is the ligand (inside). Inside: $5N{H}_3+1S{O}_4^{2-}$ → charge $+3+(-2)=+1$ → $[Co(N{H}_3{)}_{5}S{O}_4{]}^{+}$. Counter ion: $B{r}^{-}$ ($-1$) balances $+1$. → $[\mathbf{Co}(\mathbf{N}{\mathbf{H}}_3{)}_5\mathbf{S}{\mathbf{O}}_4]\mathbf{Br}$. These are ionization isomers — same atoms, different partition inside/outside the bracket.

• Shape / CN / formula / conductivity / isomers → Werner (the skeleton). ✓
• WHY a particular CN, bonding, hybridisation, magnetism → Valence Bond Theory (VBT) of complexes.
• Colour and the energy of d-orbital splitting → Crystal Field Theory and colour of complexes. Werner gave the frame; the electronic theories add the electrons.

Sort every atom into inside the bracket (ligand, secondary valency, never ionizes) or outside (counter ion, primary valency, ionizes); then charge and particle counts fall out of $Q_{\text{complex}}=\text{OS}+\sum (\text{ligand charges})$.