# Click here to see the question of chemistry. A light metal M,which is good conductor of electricity,forms a layer when it is exposed to air.The formulae of the layer is M203.It is amphoteric...

Click here to see the question of chemistry.

A light metal M,which is good conductor of electricity,forms a layer when it is exposed to air.The formulae of the layer is M203.It is amphoteric oxide.

Identify M and show its amphoteric nature by writing chemical equations.

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the oxide is Al203 and the answer can be easily fou nd by the way given by 'edobro'.

the amphoiteric nature if aluminium oxide can be shown by the following equations

rxn with acid:

Al2O3 + 6HCl -----> 2AlCl3 + 3H20

rxn with base:

Al2O3 + 6NaOH -----> 2Na3AlO3 + 3H2O

** M is ALUMINIUM.**

**its chemical equation is **

**aluminium oxide is Al2O3**

There's a smiple way to solve this problem.First of all check out the various Elements that are good conductors eleminate all the non-conductive elemtes like gases and half-metals( A **half-metal** is any substance that acts as a conductor to electrons of one spin orientation, but as an insulator to those of the opposite orientation.search in wikipedia for more informations on half-metals).

2)Second Step is to take a more logical wiev on your given formula it is M203, So the first thing to take in concider is to Use the valence rule, first Oxygen on your formula has a valence number of -2 in some cases -1 but for now it has a valence number of -2, since we have three atoms => 3*(-2)= -6 now we have to find the other part of the formula M2, we can use X instead of M ==> X*2-6=0 sine the atom should be eletroneutral the answer is x=3 So the element which we are searching for has a valence number of +3 (this element also might have diferent valence numbers but in this formula it has a valence of +3) now we are left to find the final answer.Good Condutors That have a valence number of +3 are elements lieke : Aluminum,Titanium,Vandanium etc, wich also have a valence number of +3,the key is we are searching for a good conducter these elements are silver, copper, gold, and aluminum.

Since we are searching for an amphoteric oxide aluminum would come to our first solution, because Al2O3 is amphoteric.

**so the answer M=Aluminum (Al).**