What are the decay products of HTO?

Question

Tritium, symbol $\ce{T}$, is hydrogen with a mass of $3\ \mathrm{amu}$. It is radioactive and undergoes $\beta$ decay. Which of the following could be the only products after a quantity of $\ce{HTO}$ undergoes decay?

1. $\ce{HeOH}$
2. $\ce{H2O}$, $\ce{O2}$ and $\ce{He}$
3. $\ce{H2O}$, $\ce{H2}$ and $\ce{He}$

The answer is 2 only, however I am finding it difficult to understand how I could have worked this out. Tritium undergoes beta decay, so: $$\ce{^3_1H+ -> ^3_2He^{2+} + e- + \bar {\nu}_e}$$

I had assumed it wouldn't be possible for the electron emitted in beta decay to become an orbital electron. I have seen the following question: What does HTO decay into?, but it doesn't answer my specific query. I could state that the helium ions produced gain 2 electrons from the oxygen ions to become helium, with these oxygen atoms then reacting to form $\ce{O2}$ and some of the hydrogen combining with oxygen to form water, leading to answer 2, but this feels somewhat arbitrary. What approach could be taken to solve this problem?

Answer 1

I don’t consider your working to be arbitrary at all. Let’s first consider the β decay including the associated $\ce{OH}$ part of $\ce{HTO}$:

$$\ce{_1^3THO -> _2^3He^2+ + OH- + e-^ + \bar{\nu_e}}\tag{1}$$

As you suggested, the electron created in equation (1) won’t stick around, it is emitted into the environment, leaving us with a very reactive $\ce{He^2+}$ ion. Obviously, that is a strong oxidant as it wants to get reduced itself. The closest electron sources are $\ce{OH-}$ and $\ce{HTO}$, so we can consider a number of other reactions. However, it makes sense to organise it in a simple redox reaction. The reduced component is obviously going to be $\ce{He^2+}$ which only leaves $\ce{HTO}$ being the oxidised component — and since both hydrogen and tritium are in the $\mathrm{+I}$ oxidation state in $\ce{HTO}$, it must be oxygen which is oxidised to either peroxide or elemental oxygen. Indeed, peroxides are not stable and typically disproportionate into oxygen and water anyway, so we might as well just consider the formation of oxygen. Here goes:

$$\textbf{Reduction:}\\ \begin{gather}\ce{He^2+ + 2 e- -> He^{\pm 0}}\tag{2}\end{gather}$$ $$\textbf{Oxidation:}\\ \begin{align}\ce{2HTO^{-II} &-> O2^{\pm0} + 4 e-}\tag{3.1}\\ \ce{2 HTO & -> O2 + 4 e- + 2 H+ + 2 T+}\tag{3.2}\end{align}$$ $$\textbf{Redox:}\\ \begin{gather}\ce{2 He^2+ + 2 HTO -> 2 He + O2 + 2 H+ + 2 T+}\tag{4}\end{gather}$$

Equation (3) and (4) would be similar if I had chosen to oxidise only to $\ce{H2O2}$ and then would have been followed by an equation (5) to show the disproportionation of hydrogen peroxide:

$$\ce{2 H2O2 -> O2 + 2 H2O}\tag{5}$$

In the end, you will arrive more or less at the same substances.

Note that to end up with hydrogen gas, you would have needed a reductive environment capable of reducing $\ce{H^{+I}}$ to $\ce{H2^{\pm0}}$ — not the case if you have a severly electron-desiring ion like $\ce{He^2+}$ around.

Answer 2

The answer really can't be deduced by an understanding of the various chemical reactions, but I think we can all agree that answer (1) is obviously wrong. So the answer must be either (2) or (3).

Since the decay of tritium is consuming a species of hydrogen the solution will be hydrogen deficient, so (3) is out.

Thus (2) is the answer and it is more from logical deduction rather than an understanding of all of the chemical equations.