Werner's theory of coordination compounds: Visual walkthrough

See also: the parent topic, Conductivity and ionization of electrolytes, Coordination number and geometry, Ligands — classification (mono/poly-dentate, chelate).

Step 1 — Meet the raw ingredients (no theory yet)

WHAT. We are handed a bottle. On the label: it was made by mixing cobalt trichloride ( $C o C l_{3}$ ) with five molecules of ammonia ( $N H_{3}$ ). Nothing about brackets, charges, or shape — just the recipe $C o C l_{3} \cdot 5 N H_{3}$ .

WHY start here. Werner's whole method is forensic: we only get to use facts we can measure. So far the only fact is "how many of each atom". The dot $\cdot$ in $C o C l_{3} \cdot 5 N H_{3}$ is an honest confession — it means "these stuck together and we don't yet know how." Our job is to replace that dot with structure.

PICTURE. Look at the figure: a central cobalt ball ( $C o$ ), three chlorine balls ( $C l$ ), and five ammonia molecules ( $N H_{3}$ ) — all just floating, unattached. The red $C o$ is the character whose secrets we will uncover.

Figure — Werner's theory of coordination compounds

Step 2 — Two kinds of "sticking" (Werner's one big idea)

WHAT. Werner claims the metal binds other atoms in two completely different ways. We name them and give each a picture.

WHY two kinds. A single kind of bonding cannot explain the facts we will meet in Step 4 (some chlorines react with silver, some don't). When one rule fails to sort your data into groups, you need a second rule. Werner's second rule is: there is an inner grip and an outer grip.

PICTURE. The figure shows two concentric zones around $C o$ :

an inner ring of fixed hooks — the secondary valency — that hold ligands tightly and in fixed directions;
an outer cloud of charge — the primary valency — balanced by ions that can drift away.

Step 3 — Pin down the two fixed numbers

WHAT. We look up (or recall) the two constants for cobalt(III): $oxidation state + 3 coordination number, CN 6$

Reading the equation term-by-term. The $+ 3$ is the primary valency — the size of the outer charge, so it will demand $3$ units of $- 1$ charge outside to stay neutral. The $6$ is the secondary valency — the exact number of inner hooks, no more, no fewer. Cobalt(III) always shows CN $= 6$ ; that rigidity is what makes the deduction possible.

WHY these matter now. Everything downstream is bookkeeping against these two numbers. Six hooks must be filled. A $+ 3$ core must be neutralised.

PICTURE. Six empty hooks drawn around $C o$ (all currently vacant, in red), plus a " $+ 3$ " badge on the metal for the outer charge.

Recall Where do these numbers come from?

Question: Is CN $= 6$ a guess? ::: No — Werner fixed it by counting isomers and precipitates across the whole $C o C l_{3} \cdot n N H_{3}$ series; $6$ is the value that makes every row consistent.

Step 4 — The measurement that cracks the case

WHAT. We drop silver nitrate ( $A g N O_{3}$ ) into the dissolved sample. Silver grabs only free chloride: $A g^{+} + C l^{-} \to A g C l ↓$ (a white solid falls out). We measure how much falls. Result: only $2$ of the $3$ chlorines precipitate.

WHY this test. $A g N O_{3}$ is a detector for free ions. A chloride locked on an inner hook is not free, so silver cannot reach it. So the count of precipitate = the count of outside (primary) chlorides. This single number splits the three $C l$ into two groups.

PICTURE. Two chlorides shown drifting free (they meet $A g^{+}$ and drop as red $A g C l$ ); one chloride stays clamped on an inner hook, untouched.

Step 5 — Fill the six hooks (the deduction)

WHAT. We now assign every atom to a zone.

$5 \times N H_{3}$ are neutral molecules with lone pairs → they land on $5$ hooks.
Hooks required $= 6$ , filled so far $= 5$ → one hook is still empty.
We have exactly one chlorine unaccounted for (the one silver couldn't catch). It must sit on the last hook.

WHY the leftover $C l$ goes inside. Step 4 proved one chlorine is not free. "Not free" means "clamped by the inner grip" means "on a hook" means it is a ligand. Geometry agrees: we needed one more hook filled, and here is one more binder. Two facts, one conclusion.

PICTURE. All six hooks now occupied: five by $N H_{3}$ , one by the red inner $C l$ . The other two $C l$ float outside as free ions.

Step 6 — Balance the charge (find the bracket's charge)

WHAT. Add up the charges inside the bracket: $Q_{complex} = C o^{3 +} (+ 3) + 5 N H_{3} 5 \times (0) + inner C l^{-} 1 \times (- 1) = + 2$

Reading it term-by-term.

$+ 3$ — the cobalt's oxidation state (Step 3), a positive core.
$5 \times 0$ — each ammonia is neutral, so five of them add nothing to the charge.
$1 \times (- 1)$ — the inner chloride carries $- 1$ ; it partly cancels the metal's $+ 3$ .
Total $+ 3 - 1 = + 2$ : the whole bracket is a cation (net positive) of charge $+ 2$ .

WHY. The atoms inside are physically bonded but still carry their charges; the bracket's net charge is just their sum. This number decides how many outside ions we need.

PICTURE. A balance scale: $+ 3$ on one pan, $- 1$ (inner $C l$ ) partly lifting it, leaving a red " $+ 2$ " reading on the whole bracket.

Recall Why do the ammonias contribute zero?

Question: If $N H_{3}$ bonds to the metal, why does it add $0$ to the charge? ::: It is a neutral molecule; it donates a lone pair to form the bond but carries no net ionic charge, so $5 \times 0 = 0$ .

Step 7 — Add counter-ions, write the full formula

WHAT. The whole compound must be neutral overall. The bracket is $+ 2$ , so we need $- 2$ of counter-ions outside. Two free chlorides ( $2 \times C l^{-}$ ) give exactly $- 2$ . And indeed Step 4 measured $2$ free chlorides. Everything closes.

$[C o (N H_{3})_{5} C l]^{+ 2} + 2 C l^{-} ⟶ [C o (N H_{3})_{5} C l] C l_{2}$

Reading it term-by-term. The bracket $[]^{+ 2}$ is the tight inner package (Step 6). The $2 C l^{-}$ are the free swimmers (Step 4). Written together with no dot, they form the true formula $[C o (N H_{3})_{5} C l] C l_{2}$ — the mystery dot of Step 1 is gone.

PICTURE. The finished ion: the bracketed $+ 2$ complex at the centre, two red free $C l^{-}$ counter-ions outside. In water it splits into exactly 3 particles.

Step 8 — Edge & degenerate cases (never leave a scenario unshown)

WHAT. Slide the number of ammonias and re-run the same machine. Two extremes test the logic:

Recipe	Free $C l$ (silver test)	Inner $C l$	Bracket charge	Formula	Ions
$C o C l_{3} \cdot 6 N H_{3}$	$3$	$0$	$+ 3$	$[C o (N H_{3})_{6}] C l_{3}$	$4$
$C o C l_{3} \cdot 5 N H_{3}$	$2$	$1$	$+ 2$	$[C o (N H_{3})_{5} C l] C l_{2}$	$3$
$C o C l_{3} \cdot 4 N H_{3}$	$1$	$2$	$+ 1$	$[C o (N H_{3})_{4} C l_{2}] C l$	$2$
$C o C l_{3} \cdot 3 N H_{3}$	$0$	$3$	$0$	$[C o (N H_{3})_{3} C l_{3}]$	$1$ (neutral)

WHY these two extremes matter.

Top row (6 $N H_{3}$ ): all six hooks taken by ammonia → zero inner chloride → all three $C l$ outside → silver precipitates all $3$ . Charge $+ 3 + 0 = + 3$ , needs $3 C l^{-}$ outside.
Bottom row (3 $N H_{3}$ ): three ammonias fill three hooks; the remaining three hooks must take chloride → all three $C l$ inside. Charge $+ 3 + 3 (- 1) = 0$ → the bracket is neutral, a non-electrolyte: no free ions, no precipitate, conductivity $\approx 0$ . This "silent" complex is the degenerate case — the same equation with the counter-ion count hitting zero.

PICTURE. Two mini-diagrams side by side: left = six ammonia hooks, three free red $C l$ ; right = three inner red $C l$ filling hooks, nothing outside, " $0$ " precipitate.

The one-picture summary

Everything at once: the metal's two fixed numbers ( $+ 3$ and CN $6$ ) feed a sorting machine; the silver test splits chlorides into inner (locked, ligands) and outer (free, counter-ions); charge balance gives the bracket charge; neutrality gives the counter-ion count; and the final bracketed formula plus its ion count drops out — the parent table's every row is just this one loop run again.

Recall Feynman retelling (explain the whole walkthrough to a friend)

Imagine cobalt is a person with six hands (that never let go) and a wallet with 3 coins ( $+ 3$ ). We toss in 5 ammonia toys and 3 chloride toys. The five ammonias each take a hand. That leaves one free hand — one chloride has to grab it, so it's held tight (a ligand). The other two chlorides can't find a hand, so they wander off as free swimmers. Now count the wallet: the metal owes $+ 3$ , but the one chloride it's holding chips in $- 1$ , so the tight package is only $+ 2$ . To be neutral overall, exactly two free chloride swimmers hang around outside → $[C o (N H_{3})_{5} C l] C l_{2}$ . When we drop in the silver detective, it can only arrest the two swimmers — the held chloride is safe in a hand. Two arrests = two precipitates. And since the package plus two swimmers makes three loose particles in water, the conductivity agrees too. Same story, different toy counts, gives every other row.

Flashcards

In the silver-nitrate test on a cobalt-ammine, what exactly does the precipitate count equal?

The number of free/outside-bracket chlorides (primary valency), not the total chlorines.

For

C o C l_{3} \cdot 5 N H_{3}

, why does one chloride sit inside the bracket?

Co(III) needs CN

= 6

; five

N H_{3}

fill five hooks, so the sixth hook takes a chloride ligand (also confirmed: only 2 of 3 Cl are free).

How is the bracket (complex-ion) charge computed?

Oxidation state of metal

+

sum of ligand charges; e.g.

+ 3 + 5 (0) + 1 (- 1) = + 2

Why is

[C o (N H_{3})_{3} C l_{3}]

a non-electrolyte giving no precipitate?

All three Cl are inner ligands; bracket charge

= + 3 + 3 (- 1) = 0

; zero free ions, so zero conductivity and zero AgCl.