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Topic 10: Ligand Exchange and Stability in d-Block Complexes

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Total cards: 19•Content version: v4•Public deck•Updated: 8/1/2026

Card preview (19)

#1
Front (Question)

Most d-block metals react with acids to … The metal is … which is …

Back (Answer)

liberate H2 and form a metal salt oxidised favourable

#2
Front (Question)

delta G (J) =

Back (Answer)

-nFE n = moles of electrons transferred per mole of reaction F = 96485 C mol^-1 (Faradays Constant) E = Energy (V) -(RT)lnK

#3
Front (Question)

Requirements for a thermodynamically favourable reaction

Back (Answer)

For the overall reaction E > 0 G < 0

#4
Front (Question)

Strongly polarising, metal-aqua cations readily … to form … As the metals charge increases, … so … may be lost to produce a …

Back (Answer)

lose a proton an acidic solution and a Metal-Hydroxide species It becomes even more polarising 2 protons may be lost Oxide ligand

#5
Front (Question)

hydroxide and oxo ligands can lead to …which have higher …

Back (Answer)

hydroxy and oxo bridged species nuclearity

#6
Front (Question)

Solvated metal cations will react with … to replace the … but sometimes …

Back (Answer)

other donor ligands co-ordinated water ligands not all water ligands are displaced

#7
Front (Question)

… charged ligands are required around a central metal ion than neutral ligands.

Back (Answer)

fewer

#8
Front (Question)

1st row transition metals coordinated to chloride or iodide ions tend to be … due to …

Back (Answer)

tetrahedral packing (space around metal)

#9
Front (Question)

K1 (equilibrium constant) = K2 = K3 =

Back (Answer)

[M(L)(H2O)5]^n+/[M(H2O)6]^n+[L] [M(L)2(H2O)4]^n+/[M(L)(H2O)5]^n+[L] etc

#10
Front (Question)

beta 6 = beta 2 =

Back (Answer)

K1 x K2 x K3 x K4 x K5 x K6 K1 x K2

#11
Front (Question)

log beta 6 =

Back (Answer)

logK1 + logK2 + logK3 + logK4 + logK5 + logK6

#12
Front (Question)

Trend in Km is … but can deviate due to …

Back (Answer)

K1 > K2 … >K6 changes in geometry or Jahn-Teller effects

#13
Front (Question)

Why does the value of K1 to K6 steadily decrease for ammonia complexes

Back (Answer)

The ammonia ligand can replace another ammonia ligand instead of a. H2O (fewer water molecules to replace)

#14
Front (Question)

To exchange a ligand, an octahedral complex must go via a … so the …

Back (Answer)

5 or 7 coordinate intermediate CFSE is very small

#15
Front (Question)

Why are d^6 low complexes kinetically inert to ligand exchange?

Back (Answer)

It has a very large CFSE in the octahedral so there is a large CFAE energy barrier to ligand exchange

#16
Front (Question)

Why can d^10 rapidly exchange ligands

Back (Answer)

There’s a very small CFAE barrier to exchange ligands so it can rapidly exchange ligands

#17
Front (Question)

Do monodentate or bidentate ligands bind more strongly? Why?

Back (Answer)

bidentate If the bond attaching the monodentate ligand is broken, it will likely be swept away If one bond attaching the bidentate ligand is broken, the other bond holds it near the metal so will likely re coordinate so both bonds must be broken at the same time Entropy - Monodentate: number of species remains same so little effect Bidenate: increases number of species so favours chelate effect Enthalpy: Monodenate: lone pairs repel bidenate: this repulsion already built in

#18
Front (Question)

Perfect chelate ring The larger the ring, …

Back (Answer)

5-membered ring the less stable as held not as close to metal ion

#19
Front (Question)

What do E values tell you about the likelihood of oxidation or reduction?

Back (Answer)

The more positive, the more likely reduction is (oxidation less likely) The more negative, the less likely reduction is (oxidation more likely) Low positive = both fairly stable