Problem 2

Heating a mixture of element A and fluorine (molar ratio 1 : 9, pressure ~1 MPa) to 900 °C gives three crystalline products B, C, D (all melting below 150 °C). The fluorine content of C is 36.7 % and of D is 46.5 % (by weight). Reaction of B with anhydrous

\ce{HOSO2F}

at -75 °C gives E + HF; X-ray diffraction maps of E show electron density maxima with values 52, 58, 104 and 350, roughly proportional to atomic numbers. 2.1 Assign these maxima to the elements in E (F, O, S, A) and determine the atomic number and identity of A. 2.2 Treating 450.0 mg of C with excess mercury liberates 53.25 cm

^3

of gas A at 101.0 kPa and 25 °C. Calculate the relative atomic mass of A and identify B, C, D and E. 2.3 Hydrolysis in water proceeds as follows: B hydrolyses to A,

\ce{O2}

and HF; C hydrolyses to A,

\ce{O2}

(in 4 : 3 molar ratio), aqueous

\ce{AO3}

and HF; D hydrolyses to aqueous

\ce{AO3}

and HF. Write the three balanced hydrolysis equations. 2.4 Quantitative hydrolysis of a mixture of B, C, D yields 60.2 cm

^3

of gas at 290 K and 100 kPa containing 40.0 %

\ce{O2}

by volume. The dissolved

\ce{AO3}

is titrated with 0.100 M

\ce{FeSO4}

, requiring 36.0 cm

^3

(

\ce{Fe^{2+}}

is oxidised to

\ce{Fe^{3+}}

and

\ce{AO3}

is reduced to A). Calculate the molar composition of the original mixture of B, C, D.

Step 3 of 5: Hydrolysis reactions

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\begin{aligned} \ce{XeF2 + H2O &-> Xe + 2 HF + 0.5 O2} \\ \ce{XeF4 + 2 H2O &-> \tfrac{2}{3} Xe + 4 HF + \tfrac{1}{3} XeO3 + 0.5 O2} \\ \ce{XeF6 + 3 H2O &-> XeO3 + 6 HF} \end{aligned}

Analysis

Xe(II) reduces water quantitatively to $\ce{O2}$ yielding $\ce{Xe^0}$ ; Xe(IV) disproportionates into $\ce{Xe^0}$ and $\ce{Xe^{VI}}$ (in a 2:1 ratio) while also evolving $\ce{O2}$ ( $n(\ce{Xe}):n(\ce{O2}) = 2/3 : 1/2 = 4:3$ ); Xe(VI) hydrolyses without redox directly into $\ce{XeO3}$ .