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 2 of 3: Extract

k_S

and

k_Y

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k_{obs} = k_S + k_Y[\text{py}]:\quad k_Y = \dfrac{7.20 - 3.45}{0.122 - 0.061} \times 10^{-4} = 5.8 \times 10^{-3}\ \text{dm}^3\,\text{mol}^{-1}\,\text{s}^{-1};\quad k_S \approx 0\ \text{s}^{-1}

Analysis

A plot of $k_{obs}$ vs $[\text{py}]$ is linear through the origin: the slope gives $k_Y = 5.8 \times 10^{-3}$ dm $^3$ mol $^{-1}$ s $^{-1}$ and the intercept $k_S \approx 0$ s $^{-1}$ , meaning the solvent-assisted path is negligible; at $[\text{py}] = 0.10$ M essentially all product forms by direct substitution.

Common pitfall. Keep the units distinct:

k_S

in s

^{-1}

(first order) and

k_Y

in dm

^3

mol

^{-1}

^{-1}

(second order).