I am reading this wikipedia article that I don't understand. What I don't understand is:

suppose we have two elements $X$ and $Y$ having oxidation numbers $x$ and $y$ respectively. Can we prove that the compound formed will be $X_yY_x$.

I tried to understand this method as: I think this method comes from the fact of charge neutrality. Suppose the compound to be formed has formula $X_{y'}Y_{x'}$. The charge on $y'$ number of $X$ elements is $xy'$. The charge on $x'$ number of $Y$ elements is $x'y$. Now by charge neutrality we should have : $$|x|y'=x'|y|$$
this implies $\dfrac{|x|}{|y|}=\dfrac{x'}{y'}$.

But I could not further understand why the we divide $x$ and $y$ with their greatest common factor to find the required solution.

I have also found some contradiction examples to this Crisscross rule. Example is : Solid-state-semiconductor $GaAs$, in this Gallium has valancy $3$ and $As$ has valancy $5$.
The main points on which I want answers are:

  • What is the logic behind this method, how to prove this method?
  • Who, how and when developed this method?
  • Is this method only applicable to ionic compounds?
  • Why compounds like $GaAs$ do not follow this rule, how charge neutrality is preserved in these compounds. I think these don't follow this rule because these are not ionic compounds, if yes then how the chemical formula of compounds like solid state $GaAs$ is known(or decided) and how it is justified, how one atom of $Ga$ balances one atom of $As$ ,their valencies are different, how they are bonded(covalently I guess) together?
  • $\begingroup$ I think you mean the oxidation number for N, As, and P are +5, not -5... en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements I've never heard of this method, but it's just simple math. You know that the net charge of a neutral ionic molecule is 0: For example, say you have Al(+3) and O (-2). You end up with 2 Al and 3 O because +3*2 = -2*3 = 6; 6 is the least common multiple of 2 and 3. You can do this with any binary ionic molecule. $\endgroup$
    – rch
    Mar 16, 2014 at 7:34
  • $\begingroup$ @rch We could also have $Al_4O_6$ why always take the least common multiple? Also in case of $GaAs$ shouldn't we have $Ga_5As_3$. $\endgroup$
    – user31782
    Mar 16, 2014 at 7:41
  • $\begingroup$ Oh I see what you mean. Yeah, this sure does sound like something a high school teacher thought of to make this easier to teach. $\endgroup$
    – rch
    Mar 16, 2014 at 10:21

3 Answers 3


What is the logic behind this method, how to prove this method?

The logic of the method is in fact trivial. Suppose you have to write a compound knowing his oxidation numbers : $$\ce{xA^{a} + yB^{b} -> A_{x}B_{y}}$$

Where $a$ and $b$ are oxidation numbers that have opposite sings, $A$, $B$ chemical species and $x$, $y$ stoichiometric coefficient $ \in \mathbb{N}$. We can start from this assumption: if a compound is neutral the algebraic sum of the oxidation numbers of his atoms must be zero ( point .4 of current IUPAC definition of the oxidation state) so we have to find when:

$$ ax+by=0 \space [1] $$

But we assumed that $a$ and $b$ have opposite sings so we can write:

$$ ax - by=0$$

This is a single equation with two variables so you can't solve it easily because it can have several solutions. However when $\textbf{x=b}$ and $\textbf{y=a}$ the equation is always solved, because:

$$ ab - ab=0$$ And so you can always write a compound formula that satisfy equation 1 using this formula: $$\ce{xA^{a} + yB^{b} -> A_{b}B_{a}}$$

This works for $\ce{Al^{+3} , O^{-2}}$: $$\ce{xAl^{+3} + yO^{-2} -> Al_{2}O_{3}}$$ but what's about $\ce{Ti^{+4} , O^{-2}}$?

This turns out $\ce{Ti2O4}$ and not $\ce{TiO2}$ we need another step, we need to divide the subscripts by their greatest common factor gcf(a,b) because we have found another solution of the equation 1 with higher coefficient. In this case the gcf(2,4) is 2 and dividing the stoichiometric coefficient by two we get $\ce{TiO2}$.

Who, how and when developed this method?

I've found it the first time in Chemistry and Human Affairs by William Evans Price, George Howard Bruce, World Book Company, 1949. I don't know who is the inventor. I think you can easily deduce the crisscross method simply looking at many compounds and their oxidation number! Is not so hard to imagine!

Is this method only applicable to ionic compounds?

No, it works for every compounds that can be describe with oxidation numbers. Oxidation numbers are the real charge in ionic compounds and a formal (or imaginary) charge in compounds with covalent bonds, look at $TiO_2$ example.

Why compounds like $GaAs$ do not follow this rule?

Is not right if we assume that we are dealing with $Ga(+3)$ and $As(-3)$ the equation is this: $$\ce{xGa^{+3} + yAs^{-3} -> Ga_{3}As_{3}}$$ if we divide the subscripts by the gcf(3,3) that is 3, we get $GaAs$. So this method will be always valid until you can describe the compound effectively with oxidation numbers, unfortunately not all the oxidation numbers are so easy to determine.

In fact I don't find the criss cross method useful I think is better to calculate the least common multiple of $a$ and $b$ and then divide it by $a$ for $x$ and $b$ for $y$. I think this photo, from J. Chem. Educ., 2012, 89 (11), pp 1436–1438, summarize the best way to do it: enter image description here


The general solution of |a|x=|b|y s.t. x,y∈ℕ is x=n|a| and y=n|b| where n∈ℕ, that is we can use x=2|a|,3|a|...n|a| and similarly y=2|b|,3|b|...n|b| then why use x=|a| and y=|b| only?

(rearranged after the advices in the comment of mannaia) In fact you can write $Na_{2}Cl_{2}$ without breaking any charge neutrality or valence rule. NaCl is a crystal with a lattice and you can't consider $NaCl$ like a stand alone unit, what matter is the charge balance, the proportion between positive and negative ions. Maybe if you are dealing with a crystal made of 23456 atoms of $Na^{+}$ and 23456 atoms of $Cl^{-}$ you could write $Na_{23456}Cl_{23456}$ this is in fact a solution of the equation and is not theoretically wrong however if you are talking about a generic crystal of NaCl you should write the ratio between $Na^{+}$ and $Cl^{-}$ ions it has no sense to choose randomly one of the solutions because has stated mannaia: "there's no need to choose a larger unit, whereas with a smaller one you can fully explain how the crystal is formed". Mathematically speaking we write the ratio to Lowest Terms.

Why always take the "gcf"?

In fact we don't take the "gcf" we divide for the "gcf" because this is the best way to get the ratio in the lowest terms.

Furthermore as explained in the $TiO_{2}$ example. $$ A=\{(b,a),(2b,2a),(3b,3a),\ldots\}. $$ Is a set of solutions of the equation but not all the solutions. If you have a=6 b=-4 you have the solution (2,3) too, here is the plot of the solutions in $\mathbb{N}^{2}$. enter image description here

For compounds with covalent bond is different, because you are dealing with different stand alone units called molecule. For example if you have $C^{-4}$ and $H^{+1}$ the solutions not divide by gcf of the equation have no sense. For example $C_{2}H_{8}$ doesn't exist due the valence of the atoms involved.

  • $\begingroup$ 1. "However when x=b and y=x the equation is always solved, because:": Did you mean y=b? 2. The correct equation is infact $|a|x=|b|y$. The general solution of $|a|x=|b|y\ \ s.t.\ \ x,y \in \mathbb{N}$ is $x=n|a|$ and $y=|n|b|$ where $n \in \mathbb{N}$, that is we can use $x=2|a|,3|a|...n|a|$ and similarily $y=2|b|,3|b|...n|b|$ then why use $x=|a|$ and $y=|b|$ only? Why always take the "gcf"? 3. Note: In solid state $GaAs$, $As$ does not have -3 oxidation state. So the last part of your answer is technically incorrect. $\endgroup$
    – user31782
    Mar 20, 2014 at 14:06
  • 2
    $\begingroup$ @Anupam 1. yes i've correct the typo. 2. Yes! I've simplify it to reduce the length of my answer but if you want I will add it when I've more time to improve it 3. "I've said if we assume" in fact I don't like very much talk of oxidation number or charge for this compound however normally is arsenide has charge -3 see here and here. $\endgroup$
    – G M
    Mar 20, 2014 at 15:20
  • 1
    $\begingroup$ @GM I wouldn't say that it's just a mathematical convention...where is the chemistry behind ? From the chemistry point of view, $\ce{Na2Cl2}$ is definitely not correct: because there's no need to choose a larger unit, whereas with a smaller one you can fully explain how the crystal is formed (3D repetition of the least unit, according to rigorous symmetry rules). $\endgroup$
    – mannaia
    Mar 24, 2014 at 14:39
  • $\begingroup$ @mannaia I agree with you what I mean is that NaCl is not it self a stand alone unit. And Na2Cl2 do not brake any stoichiometric or valence rule is only not the right way to describe it. However I will edit the answer to make it more clear thank for the comment. $\endgroup$
    – G M
    Mar 24, 2014 at 14:57

Does this also work with compounds other than only ionic compounds? Yes - take a look at this : in water hydrogen has a ($\ce{+I}$) partially positive oxidation state and oxygen has a ($\ce{-II}$) partially negative oxidation state. So if you work out the rule then you get : $\ce{H2O1}$ (and you can leave the 1 away so you get $\ce{H2O}$).

Other examples include :

$\ce{HCl}$ : ox.state (H) = $\ce{+I}$ and ox.state (Cl) = $\ce{-I}$ ... so... : H 1 Cl 1 ->remove ('1')-> HCl

This rule even works on certain, more complicated molecules :

$\ce{H2SO4}$ : ox.state(H) = $\ce{+I}$ and ox.state(SO4) = $\ce{-II}$ ..... so... : H 2 SO4 1 ->remove('1')-> H2SO4

$\ce{KMnO4}$ : ox.state(K) = $\ce{+I}$ and ox.state(MnO4) = $\ce{-I}$ .....so.... : K 1 MnO4 1 -> remove ('1') -> KMnO4

(of course all the charges mentioned above are partial(+/-) charges, because they are covalently bonded molecules)

And why the greatest common factor is taken... :

Take a look at calcium oxide. Here you have one calcium ($\ce{+II}$) ion and one oxygen ($\ce{-II}$) ion. When the rule is applied you get $\ce{Ca2O2}$. Now, you can divide both indexes by 2 to get $\ce{CaO}$. But why? Look at $\ce{CaO}$ : Here the one calcium atom has lost 2 electrons to achieve a stable noble gas configuration. And oxygen has gained 2 electrons to achieve a stable noble gas configuration. Both atoms are 'happy' and there is no need for another calcium atom or oxygen atom. So $\ce{CaO}$ is enough.

This is the bonding in $\ce{GaAs}$, I found this online: (source)

So the crisscross method wouldn't work with semiconducting materials. However it does (as mentioned above) work for certain normal molecules such as H2O, HCl,... these semiconducting materials just are exceptions to the rule...

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – jonsca
    May 18, 2018 at 21:48

I think the contradiction you mention in your question is only apparent and is due to a misunderstanding around the word valency, which is by itself a rather ambiguous term (valency vs oxidation state).

According to Wikipedia, valency yields the number of bonds an element tends to have. For instance, oxygen tends to form $2$ chemical bonds: its valency is then $2$.

Valency is different from valence electrons: oxygen has $6$ valence electrons: a valence electron is an electron that can participate in the formation of a chemical bond.

If we consider arsenic, its valency is $3$ and not $5$, as you write in your question. $5$ is the number of valence electrons. Of course, this value of $3$ is based on the octet rule, that's just a rule (that often works) not a law (that should always work).

Considering $\ce{GaAs}$, $4$ bonds are actually formed between $\ce{As}$ and $\ce{Ga}$, as discussed in the text given by LievenB in his answer. One should more precisely say that each $\ce{As}$ atom is bonded to $4$ $\ce{Ga}$ atoms and each $\ce{Ga}$ atom is bonded to $4$ $\ce{As}$ atoms, in a diamond-like blende structure. This is discussed in the book Fundamentals of Solid State Engineering of M. Razieghi, here below:

enter image description here

The text also says that $\ce{As}$ attracts more strongly electrons. It adds also that for the fourth bond, $\ce{As}$ contributes two electrons, while $\ce{Ga}$ contributes none.

Now, knowing that each $\ce{As}$ is bonded to $4$ $\ce{Ga}$ atoms. Let's consider a given $\ce{As}$ atom.

  • That $\ce{As}$ atom and one $\ce{Ga}$ will share $2$ electrons to form a covalent bond: these $2$ electrons are closer to $\ce{As}$: oxidation number will become $-1$.

  • The same $\ce{As}$ atom and a second $\ce{Ga}$ will share $2$ electrons to form a second covalent bond: these $2$ electrons are again closer to $\ce{As}$: oxidation number will become $-2$.

  • Again the same $\ce{As}$ atom and a third $\ce{Ga}$ will share $2$ electrons to form a third covalent bond: these $2$ electrons are again closer to $\ce{As}$: oxidation number will become $-3$.

  • The last bond, the fourth, will built up with the $2$ remaining valence electrons of our $\ce{As}$. Like the others, those $2$ electrons will stay closer to the $\ce{As}$: as a consequence, no change in the oxidation will further occur, since those $2$ electrons are both coming from $\ce{As}$ and not from $\ce{Ga}$.

Such a reasoning would suggest that $\ce{As}$ has an oxidation number of $-3$ and $\ce{Ga}$ of $+3$ (again Wikipedia).

Now, coming back to the so-called CrissCross method: if you apply the definition you quote in your question:

suppose we have two elements $X$ and $Y$ having oxidation numbers x and y respectively. Can we prove that the compound formed will be $X_{y}Y_{x}$.

and you use oxydation number $+3$ for $\ce{Ga}$ and $-3$ for $\ce{As}$, the logic formula $\ce{GaAs}$ is finally obtained.

For other arsenides, this method will not be easy to apply, since the oxidation number may not be clear for the elements involved, especially if one has no idea about the structure by which atoms are organized.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – jonsca
    May 18, 2018 at 21:47

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