The Mistake Bank
Class 12 - Chapter 9: Coordination Compounds
Ligands approach, orbitals split. Make sure your logic doesn't split too.
Werner's Theory (Precipitation)
StoichiometryScenario: 1 mole of $CoCl_3 \cdot 4NH_3$ is treated with excess $AgNO_3$. How many moles of $AgCl$ precipitate?
Student sees 3 Cl atoms.
Answer: "3 moles of AgCl."
(Ignores the coordination sphere!)
Find Secondary Valency!
Cobalt usually has Coordination Number = 6.
Structure: $[Co(NH_3)_4Cl_2]Cl$
Only the Cl outside the square bracket is ionizable.
Answer: 1 mole of AgCl.
Nickel Complexes Hybridization
Valence Bond TheoryScenario: Determine hybridization and shape of $[Ni(CN)_4]^{2-}$.
Student counts 4 ligands and assumes Tetrahedral ($sp^3$) like $[NiCl_4]^{2-}$.
Check Ligand Strength!
$CN^-$ is a Strong Field Ligand.
It forces pairing of electrons in the 3d orbital.
This leaves one 'd' orbital empty for hybridization.
Result: $dsp^2$ (Square Planar) and Diamagnetic.
IUPAC Naming of Anions
NomenclatureScenario: Name the complex: $K_3[Fe(CN)_6]$.
Student names metal as usual.
Answer: "Potassium hexacyanoiron(III)."
(Wrong Metal Name!)
Anionic Complex = 'ate'!
Since the complex ion is Anionic (negatively charged part), the metal name must end in -ate.
Iron becomes Ferrate.
Answer: Potassium hexacyanoferrate(III).
Geometric Isomerism (Tetrahedral)
IsomerismScenario: Does a tetrahedral complex $MA_2B_2$ show Geometrical Isomerism?
Student tries to draw Cis and Trans structures.
Answer: "Yes."
All positions are Adjacent!
In a perfect tetrahedron ($sp^3$), all bond angles are $109.5^\circ$. Every position is adjacent to every other position.
There is no "opposite" side.
Answer: No GI in Tetrahedral complexes. (Square planar $MA_2B_2$ shows GI).
Synergic Bonding Strength
Metal CarbonylsScenario: Which has the longest C-O bond length? $[Mn(CO)_6]^+$ or $[Fe(CO)_4]^{2-}$?
Student confuses M-C bond with C-O bond.
Thinks strong attraction means shorter bonds everywhere.
Back Bonding weakens C-O!
More negative charge on Metal $\rightarrow$ Stronger back donation into CO's antibonding orbital.
This increases M-C bond order but decreases C-O bond order.
Lower Bond Order = Longer Bond Length.
Answer: $[Fe(CO)_4]^{2-}$ (Most negative metal).
Crystal Field Splitting ($d^4$)
CFTScenario: Write config for $d^4$ ion if $\Delta_o > P$ (Octahedral field).
Student fills singly first (Hund's Rule): $t_{2g}^3 e_g^1$.
(Ignores the splitting energy!)
Splitting > Pairing Energy
If $\Delta_o > P$ (Strong Field), it is energetically favorable to pair up in the lower level than to jump to the higher level.
The 4th electron pairs up in $t_{2g}$.
Answer: $t_{2g}^4 e_g^0$.
Confess Your Sins!
"Isomers and Hybridization... did you get tangled in the coordination sphere?"
Did one of these catch you? Or do you have a different horror story from your last exam?
Scroll down to the comments section below and tell us:
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