CoCl3·5NH3 with AgNO3, how do I decide how many moles of AgCl form using Werner's primary and secondary valence?

I have been going through Werner's postulates for a few days and I can recite them, but I get stuck the moment a question asks me to apply them. The series I am looking at is the classic one in NCERT: CoCl3·6NH3, CoCl3·5NH3, CoCl3·4NH3, CoCl3·3NH3, with excess AgNO3 giving 3, 2, 1 and 0 moles of AgCl respectively.

What I understand: the metal has a primary valence, which is ionisable, and a secondary valence, which is not ionisable and is directed in space. What I do not understand is how to read the formula for CoCl3·5NH3 and place each chloride. NCERT writes it as $[Co(NH_3)_5Cl]Cl_2$ , so one chloride is inside the bracket and two are outside. Then the primary valence is 3, yet only two chlorides are ionisable? That feels contradictory.

Is it that a chloride inside the bracket can still count towards the primary valence while also acting as a ligand? And how is the secondary valence 6 in each of these four compounds? A clean way to count would help me because I keep second-guessing myself in tests.

7 answers & replies

[+]Ishita GoyalDropperSample· 21d ago

Yes, that's exactly the point. In $[Co(NH_3)_5Cl]Cl_2$ :

Primary valence is the oxidation state of cobalt, +3. It is satisfied by three Cl⁻ ions in total, but one of them is also sitting inside the coordination sphere. So that Cl⁻ is doing both jobs, satisfying a primary valence and a secondary valence.
Secondary valence is the coordination number, 6, satisfied by 5 NH3 + 1 Cl⁻.

Only the ions outside the bracket come off in solution and precipitate with Ag⁺. Two outside, so 2 moles AgCl. Count what is outside the square bracket and you have your answer.

[+]Rohan Sharma11thSample· 21d ago

i remember primary = 6 and secondary = 3 because the number of cl- is 3 so secondary is 3? 😅 is this right

[+]Nisha Verma12thSample· 21d ago

Rohan, other way round. Memory trick I use: Primary = Positive charge (oxidation state, ionisable, not directional). Secondary = Spatial (coordination number, non-ionisable, directed in space, gives geometry). So cobalt +3 means primary is 3 and octahedral six-coordination means secondary is 6.

[+][–]Rohan Sharma11thSample· 21d ago
oh i had them swapped lol. positive = primary, spatial = secondary. stealing this

[+]Farhan ShaikhDropperSample· 21d ago

The whole series in one table, learn this and you are safe:

Compound	Formula	Ions in solution	AgCl with excess AgNO3
CoCl3·6NH3	$[Co(NH_3)_6]Cl_3$	4	3 mol
CoCl3·5NH3	$[Co(NH_3)_5Cl]Cl_2$	3	2 mol
CoCl3·4NH3	$[Co(NH_3)_4Cl_2]Cl$	2	1 mol
CoCl3·3NH3	$[Co(NH_3)_3Cl_3]$	0	none

The last one is a neutral complex, no free chloride, so no precipitate and no conductivity. Werner measured molar conductivity to get the number of ions and matched it with the precipitation numbers.

[+][–]Zoya Khan12thSample· 21d agoAuthor
This table is the clearest thing I have seen on this. Thank you, I can finally see how the conductivity and the precipitate tell the same story.

[+]Harsh Gupta12thSample· 21d ago

Counting trap that got me in a test: coordination number is the number of donor atoms attached to the metal, not the number of ligands. $[Co(en)_3]^{3+}$ has only three ligands but CN is 6 since each en is bidentate. Same for $[Fe(C_2O_4)_3]^{3-}$ , CN = 6. If you count ligands you will be wrong and the examiner will smile.

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