Question

The electronic configuration of a metal M is \[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{1}}\]. The formula of its oxides will be
A. MO
B.\[{{M}_{2}}O\]
C.\[{{M}_{2}}{{O}_{3}}\]
D. \[M{{O}_{2}}\]

Answer 1

Hint: Sodium is a chemical element having the symbol Na.
Its atomic number is 11.
It is a smooth, silvery-white very reactive metal. It is an alkali metal and lies at the top of group 1 of the periodic table.

Complete step by step solution:Here in this question, we are given a metal M which has 11 electrons. We have to find out the formula of its oxide.
To find out the formula of its oxide we have to find out the valency of the metal.
This is an ionic compound.
Oxide ion is the anion here and has a -2 charge.
The given metal will lose electrons to form a cation. Here the given metal has one electron in the outermost shell which it will lose to form an M+ cation.
We know that the compound must be electrically neutral and we have to decide how many of each ion are required for the positive and negative charges to nullify each other.
The charge on Na metal is +1.
The charge on O is -2.
So, we need two\[N{{a}^{+}}\] ions for one\[{{O}^{2-}}\] ion as the compound is neutral in nature.
So, the formula becomes \[N{{a}_{2}}O\].
So, the formula is \[{{M}_{2}}O\].

So, option B is correct.

Additional Information: Sodium oxide is an ionic compound having the formula \[N{{a}_{2}}O\]. It is utilized in ceramics and glasses. It is a white solid.
Sodium oxide is created by the reaction of sodium with sodium hydroxide, sodium peroxide, or sodium nitrite.
\[2NaOH+2Na\to 2N{{a}_{2}}O+{{H}_{2}}\]
NaOH is contaminated with water so higher amounts of sodium are used, the excess of which is distilled from the crude product.

Note: Another way to write the right formula for an ionic compound is to utilize the crisscross method.
In this method, the numerical value of each of the ion charges crosses over which becomes the subscript of the other ion.
Signs of the charges are not included in this process.