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I had a question in my exam:

Which of the following can be the $n$-factor of $\ce{P4O6}?$

A. 8
B. 9
C. 10
D. 12

I went for options A and B as in cold water $1$ mole of $\ce{P4O6}$ on hydrolysis give $4$ moles of $\ce{H3PO3}$, whereas in hot water it gives $3$ moles of $\ce{H3PO4}$ and $1$ mole of $\ce{PH3}$, thus getting $n$-factor $8$ and $9,$ respectively.

However, the answer given is A, B and C. How can the $n$-factor be $10?$

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  • $\begingroup$ Assuming the n-factor is a measure of dissociation, what happens when you dissolve (for instance) H3PO4 in water? $\endgroup$ – sjb-2812 Nov 18 at 14:36
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    $\begingroup$ Whoever designed this question doesn't understand the equivalent weight concept. Equivalent weight is dependent on a reaction. So the key question is n-factor for which reaction? $\endgroup$ – M. Farooq Nov 18 at 14:41
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    $\begingroup$ @sjb-2812, n factor is very old concept of equivalent weights which is still taught in India. It has nothing to do with dissociation. $\endgroup$ – M. Farooq Nov 18 at 14:42
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    $\begingroup$ Anyway it is necessary that Zenix tells us what is the n-factor, for him. $\endgroup$ – Maurice Nov 18 at 16:54
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As M. Farooq pointed in the comments, the concept of $n$-factor is applied to the reactions a given compound participates in, namely of the following types:

  1. acid-base, neutralization (in a context of Arrhenius approach, i.e. the amount of transported hydronium ions);
  2. redox (total change in oxidation state per mole of the substance);
  3. precipitation and double decomposition reactions.

This concept is so rarely used these days that I actually had to look this up to refresh my memory.

Apparently, $\ce{P4O6}$ doesn't have any protons to transfer, so the first category is irrelevant. I also don't think that the question counts basicity of the acids that can be formed by hydrolysis or decomposition of $\ce{P4O6}$ in water, so I doubt your reasoning is relevant to this question.

Besides, in hot water there is also elementary phosphorous among the products, and the stoichiometry is going to be quite different and won't justify $n$-factor $9.$ However, since the decomposition of $\ce{P4O6}$ in hot water is in fact a redox reaction:

$$\ce{6 \overset{+3}{P}_4O6(s) + 24 H2O(l) ->[Δ] 8 \overset{0}{P}(s, red) + 15 H3\overset{+5}{P}O4(aq) + \overset{-3}{P}H3(g)}\tag{R1}$$

$$n\text{-factor}(\ce{P4O6}) = \frac{8 × |3- 0| + 15 × |3 - 5| + 1 × |3 - (-3)|}{6} = 10\tag{1}$$

which explains answer C.

Oxidation with oxygen, another redox reaction, explains answer A:

$$\ce{\overset{+3}{P}_4O6(s) + 2 O2(g) ->[\pu{50 - 120 °C}] \overset{+5}{P}_4O10(s)}\tag{R2}$$

$$n\text{-factor}(\ce{P4O6}) = 4 × |3 - 5| = 8\tag{2}$$

Another redox reaction, for example, thermal decomposition in vacuum, could also demonstrate another $n$-factor:

$$\ce{4 \overset{+3}{P}_4O6(s) + 2 O2(g) ->[\pu{210 - 250 °C}] 3 \overset{+3}{P}_2\overset{+5}{P}_2O8 + 4 \overset{0}{P}(s,red)}\tag{R3}$$

$$n\text{-factor}(\ce{P4O6}) = \frac{3 × 2 × |3 - 5| + 4 × |3 - 0|}{4} = 6\tag{3}$$

And so on, so it looks like the question requires you to come up with the examples of the reactions that would be impossible for the listed $n$-factors. I think this is an unjust choice for an exam as oxy-compounds of phosphorous have rich chemistry and I still cannot come up with the suitable process for the $n$-factor $9$.

Notes

The reactions are adapted from [1, pp. 167]; they can also be found in Russian and German Wikipedia articles on phosphorous(III) oxide.

References

  1. R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)
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