Problem 1

Nitrogen trifluoride is a surprisingly stable compound that was first prepared by the melt electrolysis of a mixture of ammonium fluoride and hydrogen fluoride. (a) At which electrode does

\ce{NF3}

form? Write the balanced half-reaction for its formation. (b) The related compounds

\ce{NH2F}

and

\ce{NHF2}

are very unstable and form as side products. Which of

\ce{NF3}

\ce{NHF2}

\ce{NH2F}

is expected to condense at the lowest temperature? (c) The N–F bond lengths in these molecules are 136, 140 and 142 pm; assign them using a simple electrostatic (partial-charge) model. (d) When

\ce{NHF2}

is bubbled through a solution of KF in HF, a binary nitrogen–fluorine compound is obtained as a mixture of two geometric isomers; write a balanced equation for its formation. (e)

\ce{NF4+}

salts form from

\ce{NF3}

and

\ce{F2}

in the presence of a suitable reagent — propose one and write the equation. (f)

\ce{NF4+}

hydrolyzes to

\ce{NF3}

and

\ce{O2}

, but less

\ce{O2}

than expected is often obtained; write the hydrolysis equation and a possible side reaction lowering the

\ce{O2}

\ce{NF3}

ratio. (g) A tetrafluoroammonium salt (a candidate solid rocket fuel) contains 65.6% fluorine by mass, all released as

\ce{NF3}

and

\ce{F2}

on heating, with

n(\ce{F2}) = 2.5\,n(\ce{NF3})

. Determine the formula of the salt.

Step 6 of 6: Deduce the salt formula

← Previous Next →

x{:}y = 1{:}4 \text{ in } (\ce{NF4})_x\ce{AF}_y;\quad 65.6\%\,\text{F} \Rightarrow A = \ce{Xe} \Rightarrow (\ce{NF4})_2\ce{XeF8}

Analysis

Each $\ce{NF4+}$ releases one $\ce{NF3}$ ; the observed $n(\ce{F2}):n(\ce{NF3}) = 2.5:1$ means the anion supplies 1.5 $\ce{F2}$ per $\ce{NF4+}$ ... Counting released F atoms: per $\ce{NF4+}$ , $\ce{NF3}$ carries 3 F and 2.5 $\ce{F2}$ carries 5 F, total 8, so the anion $\ce{AF}_y^{x-}$ must satisfy the charge and F balance, giving $x:y = 1:4$ , i.e. $\ce{(NF4)AF4}$ or $\ce{(NF4)2AF8}$ . The F mass fraction 65.6% fixes $M_A \approx 131$ — xenon — so the salt is $\ce{(NF4)2XeF8}$ (65.9% F).

Common pitfall. Do not forget that the anion itself contains releasable fluorine — the 2.5:1 ratio is what reveals it; skipping it leads to trying to fit a fluoride-free anion.