Question

The value of $n$ in the molecular formula $B{e_n}A{l_2}S{i_6}{O_{18}}$ is:

Answer 1

Hint: The number of electrons present in the valence shell of the atoms that are free to lose in a gaseous or liquid state are known as valence electrons. Whenever an atom loses/gains electrons, it gets oxidized/ reduced accordingly and attains nearest noble gas configuration. The overall oxidation number is balanced such that the total charge on the compound remains neutral.

Complete step by step answer:
In the given compound, we can see that the compound is overall neutral. This means that the oxidation states of the individual atoms in the compound balance each other. The sum of the oxidation states of the individual atoms will be zero.
The oxidation states of the atoms in the given compound are:
$Be = + 2$
$Al = + 3$
$Si = + 4$
$O = - 2$
Now, we need to determine the oxidation number of the atoms.
Oxidation number = Number of atoms x Oxidation state of each atom
In the given compound, the oxidation number can be written mathematically as:
$2 \times n + 2 \times ( + 3) + 6 \times ( + 4) + 18 \times ( - 2) = 0$
Here, number of boron atoms = 2
Number of aluminum atoms = 2
Number of silicon atoms = 6
Number of oxygen atoms = 18
Now solving the above equation, we get:
$ \Rightarrow 2n + 6 + 24 - 36 = 0$
$ \Rightarrow n = 3$
Hence, the total number of boron atoms present in the compound is 3.

Note:
The increase in the oxidation state of an atom, through a chemical reaction, is known as an oxidation. The decrease in the oxidation state is known as reduction. Reactions like these involve the formal transfer of electrons. This means that there is a net gain in electrons during reduction and a net loss of electrons during oxidation. For pure elements, the oxidation state is zero or the ionic charge on the element will remain zero.