The Mistake Bank
Class 12 - Chapter 9: Coordination Compounds
Ligands approach, orbitals split. Make sure your logic doesn't split too. Master the inner complexities of transition metal complexes.
1. 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 out?
Student sees exactly 3 Chlorine atoms in the empirical formula.
Answer given: "3 moles of AgCl."
(They completely ignored the coordination sphere boundaries!)
Only the Primary Valency (Counter Ions) reacts!
Since there are only 4 $NH_3$ molecules available, two $Cl^-$ ions must step inside the coordination sphere to satisfy the secondary valency.
True structure: $[Co(NH_3)_4Cl_2]Cl$.
Only the single $Cl^-$ completely outside the square bracket is ionizable. Therefore, only 1 mole of $AgCl$ precipitates!
2. Nickel Complexes Hybridization
Valence Bond TheoryScenario: Determine the hybridization, shape, and magnetic nature of the Tetracyanonickelate(II) ion: $[Ni(CN)_4]^{2-}$.
Student counts 4 ligands and immediately assumes $sp^3$ hybridization, just like in $[NiCl_4]^{2-}$.
Answer given: "Tetrahedral, Paramagnetic."
Check the Ligand Field Strength!
The Cyanide ion ($CN^-$) is a Strong Field Ligand. It forces those two unpaired electrons to pair up against Hund's rule.
This frees up one inner 3d orbital. The hybridization uses this inner orbital: $d + s + p + p$.
Result: $dsp^2$ (Square Planar) and Diamagnetic (all paired).
3. IUPAC Naming of Anionic Complexes
NomenclatureScenario: Write the IUPAC name for the complex: $K_3[Fe(CN)_6]$.
Student names the metal normally as they would in simple salts.
Answer given: "Potassium hexacyanoiron(III)."
(Wrong Metal Name! You must check the charge of the coordination sphere.)
Anionic Complex Metals end in '-ate'!
In anionic complexes, the central metal's Latin name is used with the suffix -ate. Iron becomes Ferrate. (Copper becomes Cuprate, Silver becomes Argentate, etc.)
Correct Name: Potassium hexacyanoferrate(III).
4. Geometric Isomerism in Tetrahedral
IsomerismScenario: Can a tetrahedral complex of the type $MA_2B_2$ (e.g., $[Ni(CO)_2(PR_3)_2]$) exhibit Cis-Trans Geometrical Isomerism?
Student thinks: "It has two types of ligands, so I can draw them next to each other (cis) or opposite to each other (trans)."
Answer given: "Yes."
Tetrahedral positions are ALL Adjacent!
Because every position is equidistant and adjacent to every other position, there is no such thing as being "opposite" (trans) to another ligand.
Answer: No Geometrical Isomerism is possible in Tetrahedral complexes. (Only Square Planar $MA_2B_2$ shows GI).
5. Synergic Bonding & Bond Lengths
Metal CarbonylsScenario: Which complex possesses the longest Carbon-Oxygen ($C-O$) bond length? $[Mn(CO)_6]^+$ or $[Fe(CO)_4]^{2-}$?
Student confuses the Metal-Carbon bond with the Carbon-Oxygen bond. They assume that if the interaction is strong, everything shrinks.
Back Bonding WEAKENS the C-O bond!
$[Fe(CO)_4]^{2-}$ has a highly negative metal center ($Fe^{-2}$), meaning it aggressively back-donates electron density into CO. Adding electrons to an antibonding orbital lowers the C-O bond order, thereby increasing its bond length.
Answer: $[Fe(CO)_4]^{2-}$ has the longest C-O bond.
6. Crystal Field Splitting ($d^4$)
CFTScenario: Write the electronic configuration for a $d^4$ central metal ion in an Octahedral field where $\Delta_o > P$ (Splitting energy is greater than Pairing energy).
Student rigidly applies Hund's Rule: "All orbitals must be singly occupied before any pairing occurs."
They place 3 electrons in $t_{2g}$ and 1 electron in $e_g$.
Answer given: $t_{2g}^3 e_g^1$.
Energy decides! Splitting > Pairing.
It requires less energy for the 4th electron to pair up in the lower $t_{2g}$ level (paying the Pairing energy penalty) than to jump the massive gap to the $e_g$ level.
Correct Configuration: $t_{2g}^4 e_g^0$ (Low Spin complex).
7. Optical Isomerism in Square Planar
IsomerismScenario: Can a square planar complex of type $[M(AA)B_2]$ (containing a bidentate ligand) exhibit Optical Isomerism?
Student recalls that Octahedral complexes with bidentate ligands (like $[Co(en)_2Cl_2]^+$) show optical isomerism.
They assume the bidentate ligand makes the square planar complex asymmetric too.
Answer given: "Yes."
Square Planar complexes are NEVER Optically Active!
By definition, a Square Planar complex is flat. The molecular plane itself acts as a horizontal plane of symmetry, cutting through all atoms perfectly in half.
Answer: No, they are always achiral.
8. The Linkage Isomerism Trap
Structural IsomerismScenario: Identify the type of isomerism between $[Co(NH_3)_5(NO_2)]Cl_2$ and $[Co(NH_3)_5(ONO)]Cl_2$.
Student sees the ligand changing inside the bracket and confuses the terminology.
They often call it "Ionization Isomerism" or "Coordination Isomerism."
Look for Ambidentate Ligands!
- Via Nitrogen: $-NO_2$ (Nitro form)
- Via Oxygen: $-ONO$ (Nitrito form)
Because the point of attachment (linkage) changes, this is specifically called Linkage Isomerism.
9. Metal Carbonyl Oxidation State
CalculationsScenario: Calculate the oxidation state of Nickel in Tetracarbonylnickel: $[Ni(CO)_4]$.
Student confuses the Carbonyl ligand ($CO$) with the Oxide ligand ($O^{2-}$), or thinks Carbon donates a charge.
They calculate: $x + 4(-1) = 0 \implies \mathbf{+4}$.
Carbon Monoxide is a NEUTRAL Ligand!
$$ x + 4(0) = 0 $$ $$ \mathbf{x = 0} $$
Metals in homoleptic carbonyl complexes almost always exist in a remarkably stable Zero (0) Oxidation State!
10. The Brown Ring Complex Trap
Exceptional StatesScenario: Determine the oxidation state of Iron in the Brown Ring test complex: $[Fe(H_2O)_5(NO)]SO_4$.
Student knows sulfate is $-2$, making the complex sphere $+2$.
They assume $NO$ (Nitric oxide) is a neutral ligand like water.
Calculation: $x + 5(0) + 1(0) = +2 \implies \mathbf{Fe \text{ is } +2}$.
NO acts as Nitrosonium ($NO^+$)!
Calculation: $x + 5(0) + 1(+1) = +2$
$$ x + 1 = 2 \implies \mathbf{x = +1} $$
Iron is in a highly unusual +1 Oxidation State in the Brown Ring Complex!
11. The Chelate Effect Driving Force
ThermodynamicsScenario: Why is $[Cu(en)_2]^{2+}$ immensely more stable than $[Cu(NH_3)_4]^{2+}$ in aqueous solution?
Student focuses entirely on bond strength: "Ethylenediamine forms stronger coordinate covalent bonds than Ammonia."
(Bond enthalpy $\Delta H$ is actually quite similar for both M-N bonds!)
It is Entropy ($\Delta S$) Driven!
Notice the particles: 3 molecules on the left produce 5 molecules on the right.
Because multidentate (chelating) ligands release multiple water molecules for every one ligand attached, the randomness (Entropy, $\Delta S$) of the system increases massively. This makes $\Delta G$ highly negative, driving incredible stability.
12. The Illusion of Color
Physical PropertiesScenario: Predict the color of the aqueous $[Ti(H_2O)_6]^{4+}$ ion.
Student identifies Titanium as a transition metal.
They remember: "Transition metal complexes exhibit beautiful colors due to d-d transitions."
Answer given: "Colored (e.g., purple/blue)."
Check for d-electrons first!
Titanium's ground state is $[Ar] 3d^2 4s^2$. When it becomes $Ti^{4+}$, it loses all four valence electrons, resulting in a $3d^0$ configuration.
Because there are zero electrons available to jump, no light is absorbed. The complex is strictly Colorless!
13. Facial and Meridional (Fac-Mer)
IsomerismScenario: Which type of geometrical isomerism is exhibited by the Octahedral complex $[Co(NH_3)_3(NO_2)_3]$?
Student tries to apply Cis/Trans labeling, assuming "all next to each other" is Cis and "opposite" is Trans.
Answer given: "Cis and Trans isomers."
$MA_3B_3$ exclusively forms Fac/Mer Isomers!
- Facial (fac): Three identical ligands occupy one triangular face of the octahedron (all are at $90^\circ$ to each other).
- Meridional (mer): Three identical ligands form an arc around the meridian of the octahedron (one pair is at $180^\circ$).
14. Magnetic Moment Calculation
FormulasScenario: Calculate the spin-only magnetic moment ($\mu$) for the complex $K_4[Fe(CN)_6]$.
Student calculates the oxidation state of Fe as +2.
They use the formula $\mu = \sqrt{n(n+2)}$ but plug in the oxidation state ($n=2$) or the total d-electrons ($d^6 \implies n=6$).
Answer given: $\sqrt{2(4)} = 2.82 \text{ BM}$ or $\sqrt{6(8)} = 6.9 \text{ BM}$.
'n' is the number of UNPAIRED electrons!
2. $CN^-$ is a Strong Field Ligand. It completely forces all 6 electrons to pair up in the lower $t_{2g}$ level.
3. Number of Unpaired Electrons ($n$) = 0.
$$ \mu = \sqrt{0(0+2)} = \mathbf{0 \text{ BM}} $$ The complex is completely diamagnetic!
15. Inner vs Outer Orbital Complex
HybridizationScenario: Determine the hybridization of $[CoF_6]^{3-}$ vs $[Co(NH_3)_6]^{3+}$.
Student assumes that since both are octahedral Co(III) complexes, they must have the exact same $d^2sp^3$ hybridization.
Weak Field = Outer ($4d$), Strong Field = Inner ($3d$)!
- With $NH_3$ (Strong Field): Electrons pair up. Two inner $3d$ orbitals become empty. Hybridization is $\mathbf{d^2sp^3}$ (Inner Orbital Complex, Low Spin).
- With $F^-$ (Weak Field): Electrons DO NOT pair up. Inner 3d is full. The metal must borrow empty orbitals from the next shell up ($4d$). Hybridization is $\mathbf{sp^3d^2}$ (Outer Orbital Complex, High Spin).
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:
Live classes starting on E Acad Sutra
ReplyDelete