Question

A metal M forms an oxide having the formula $ {M_2}{O_3} $ . It belongs to the third period. Write the atomic number and valency of the metal.

Answer 1

Hint: The number of electrons released or gained by an element while forming a chemical bond is termed as its valency. The compound given above is a metal oxide and we have to find its atomic number, valency and to find these we will also evaluate its electronic configuration. The complete answer of this question is given in the below section in brief.

Complete answer:
As it is already mentioned in the question that it belongs to the third period, therefore, the period number=number of shells (n).
Therefore, n $ = 3 $
Now we will find its valency. The formula of the compound is given in the question. Let M be x.
The valency of oxygen is $ - 2 $ .
Now, $ 2x - 6 = 0 $
So $ x = 3 $
Therefore, its valency is $ 3 $ . Now we know that it is a metal, therefore, its valence shell contains $ 3 $ electrons. So its electronic configuration is $ 2,8,3 $ .
Therefore, its atomic number $ = 13 $
Hence the metal is Aluminum $ \left( {Al} \right) $ with valency equal to $ 3 $ .

Note:
 As we look into the trends of the modern periodic table we see that the density of aluminum is lower than those of other common metals. When it is exposed to air, it forms a protective oxide layer with oxygen due to its greater affinity towards oxygen. It is silver colored metal and is a good reflector of light. We also use aluminum foil papers in our home to store food.