Mannaia, chemistry.stackexchange.com/questions/9317/….

Note that the text also writes:"...one of the four covalent bond is formed by the $As$ atom contributing two electron( coordinate-bond), while the $Ga$ atom contribute none...". This implies that $As$ on an average does not have $-3$ oxidation state.

If GaAs crystal contains two different type of bonds then will the electrons corresponding to the two different types of bonds have different band gaps. I am familiar with the 1.46eV band gap energy of $GaAs$. What is the band gap energy of the electrons in the coordinate bond?

@Anupam the fourth bond does not differ from the others. They are equivalent: that's why a diamond-like structure characterizes such a crystal.

If all the bonds are equivalent the coordinate bond should be identical with the usual covalent bond. Shouldn't all the equivalent bonds cause same oxidation states? How can two identical types of bonds be different in some way. I have another doubt(Excuse me for this , I am not in touch with chemistry from some years) Would the hybradysation occurre? If yes then all the bonds will be exactly equivalent, that is the electron pair in the coordinate bond will be exactly at same distance from $As$ as the usual covalent bond(in which both As and Ga share 1-1 electron).

@Anupam I think the questions you rise, though very interesting, would hardly receive an exhaustive answer here. Hybridization can be used to explain why the $4$ bonds are equivalent. Further, we can say that electrons will be closer to $\ce{As}$ being the latter more electronegative and this will explain $-3$ oxidation state. Of course, this is a very simplified description of a complex crystal and was originally proposed to show that the crisscross method can work as a shortcut to obtain the chemical formula of gallium arsenide.

It appears your answer is right. $As$ should have $-3$ oxidation state. I am quite confused with the the book that I have already mentioned in which $As$ makes 5 bonds. Let's come to my present question. Does this method works for purely covalent compounds? If the diff. b/w electronnegativities of the constituent atoms is very less then would it be approriate to talk about oxidation states? Would we use valancy in that case?

@Anupam That's a good point, worth opening a new question. I like the definition given to the oxidation number (or state) by en.wikipedia.org/wiki/Oxidation_state: "hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic". That means that you can use oxidation numbers also for slightly polarized covalent bonds. However, oxidation numbers should never be assimilated to formal charges, especially when the difference in electronegativity is so small. You could do that for $\ce{NaCl}$, certainly not for $\ce{GaAs}$.

This is the perfect answer to the contradiction. I think the OP is just a bit confused (or may be haste like me) in the difference between Oxidation State and Valency. Oxidation state is dependent on the valency to a limited amount, and more dependent on the electronic (orbital) configuration of elements, especially in p and d block elements.

Could you explain why we divide the oxidation numbers with their gcf.

@Anupam The so-called unit cell for such a crystal contains only two atoms (one $\ce{Ga}$ and one $\ce{As}$). Applying the symmetry and periodicity rules (specific to $\ce{GaAs}$) to such a unit, you will manage to obtain the crystal in the three dimensions. In other words, the unit cell (obtained by dividing the oxydation numbers by their gcf) contains enough information to explain the organization of the atoms within our crystal. We don't need more ... writing $\ce{Ga3As3}$ would simply be redundant.

I am asking in general for all compounds like molecules. While applying criss-cross rule for any compound we divide the oxidation numbers with their gcf, why?

@Anupam because, you want the least unit that is sufficient to explain the macroscopic properties of a given compound. You may call it molecule, unit cell, atom: all these are the invisible "bricks" that will help to build the macroscopic "wall"...

@Anupam The least unit for $\ce{H2O2}$ is $\ce{H2O2}$, not $\ce{OH}$. There is a covalent bond between the oxygen atoms, that keeps together the two $\ce{OH}$ pseudo-units. If you consider $\ce{OH}$ as the truly least unit for $\ce{H2O2}$, you will miss that covalent bond that is crucial to explain the macroscopic behavior of such a compound (e.g. its strong oxidizing power).

How criss-cross rule applies to $H_2O_2$?

Quoting book, Chemistry the Easy Way: "As a general rule, reduce the subscript numbers in the final formula to their lowest term. There are, however, certain exceptions, such $\ce{H2O2}$. For these exceptions, you must know more specific information about the compound" ... (covalent bond between oxygen atoms).

How about $C_2H_2$ any many others like this. We cannot apply criss-cross rule in genral. Suppose we know we have some compound made up of $C$ and $H$ then we will apply criss cross rule to get $CH_4$! which is can be wrong if the compound $C_2H_2$. So this method should be true only for ionic compounds not colvalent compounds.

@Anupam It seems to me that for water, it works. Anyway, at some point we should forget crisscross and focus on chemistry.

What I want to say is that I think criss-cross method is only applicable to ionic compounds only.

@Anupam water is what ? ionic ?

I meant criss-cross method is true for all ionic compounds but not for all covalent compounds( For some covalent compounds this method works).