Problem 2

Square-planar Pt(II) complexes substitute with retention of stereochemistry, and the rate of substitution of X by Y depends on the ligand trans to X (trans effect):

\ce{CN-} > \ce{H-} > \ce{NO2-}, \ce{I-} > \ce{Br-}, \ce{Cl-} > \text{pyridine}, \ce{NH3}, \ce{OH-}, \ce{H2O}

. (a) Draw all stereoisomers of square-planar

\ce{Pt(py)(NH3)BrCl}

(py = pyridine). (b) Write reaction schemes, via intermediates, to prepare each stereoisomer of

\ce{[Pt(NH3)(NO2)Cl2]-}

in water starting from

\ce{PtCl4^{2-}}

\ce{NH3}

\ce{NO2-}

, exploiting the trans effect. (c) Substitution follows

\mathrm{Rate} = k_S[\ce{ML3X}] + k_Y[\ce{Y}][\ce{ML3X}]

; for

[\ce{py}] \gg [\ce{Pt}]

\mathrm{Rate} = k_{obs}[\ce{Pt}]

. For the displacement of

\ce{Cl-}

by pyridine in methanol at 25 °C:

[\text{py}]

= 0.122, 0.061, 0.030 mol dm

^{-3}

give

k_{obs}

7.20 \times 10^{-4}

3.45 \times 10^{-4}

1.75 \times 10^{-4}

^{-1}

. Find

k_S

and

k_Y

and decide which pathway dominates at

[\text{py}] = 0.10

mol dm

^{-3}

. (d) A gold nanoparticle of diameter 13 nm carries 90 oligonucleotide groups, 98 % bound to a Pt(IV) complex; a 1.0 cm

^3

vessel contains

1.0 \times 10^{-6}

mol dm

^{-3}

in Pt (density of Au = 19.3 g cm

^{-3}

). Calculate the masses of Pt and Au used.

Step 3 of 3: Nanoparticle masses

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m(\mathrm{Pt}) = 10^{-6} \times 10^{-3} \times 195.1 = 2.0 \times 10^{-7}\ \text{g};\quad N_{np} = \dfrac{6.0 \times 10^{14}}{90 \times 0.98} = 6.8 \times 10^{12};\quad m(\mathrm{Au}) = N_{np} \times \tfrac{4}{3}\pi (6.5 \times 10^{-7})^3 \times 19.3 = 1.5 \times 10^{-4}\ \text{g}

Analysis

$n(\mathrm{Pt}) = 10^{-9}$ mol $= 6.0 \times 10^{14}$ atoms; with $88$ Pt atoms per nanoparticle there are $6.8 \times 10^{12}$ particles. Each particle of radius $6.5 \times 10^{-7}$ cm contains $1.2 \times 10^{-18}$ cm $^3$ of gold ( $2.3 \times 10^{-17}$ g), so total gold mass is $1.5 \times 10^{-4}$ g.