The problem asks for the number of valence electrons of an atom B, given that its lowest possible oxidation number in a compound A₂B is -2.
Concept Used:
The oxidation number (or oxidation state) of an atom in a compound represents its degree of oxidation. For a non-metal element, the lowest (most negative) possible oxidation number is related to the number of electrons it needs to gain to achieve a stable electron configuration, typically an octet (8 electrons in its valence shell).
The relationship between the minimum oxidation number and the number of valence electrons (\(V_e\)) for a main group element is given by the formula:
\[
\text{Lowest Oxidation Number} = V_e - 8
\]
This formula arises because the lowest oxidation state is achieved when the atom gains enough electrons to complete its octet.
Step-by-Step Solution:
Step 1: Identify the given information.
We are given that the lowest oxidation number of atom B is -2.
\[
\text{Lowest Oxidation Number of B} = -2
\]
Step 2: Apply the formula relating the lowest oxidation number to the number of valence electrons.
Let \(V_e\) be the number of electrons in the valence shell of atom B. Using the formula:
\[
\text{Lowest Oxidation Number} = V_e - 8
\]
Step 3: Substitute the given value and solve for \(V_e\).
\[
-2 = V_e - 8
\]
Final Computation & Result:
To find the number of valence electrons, we rearrange the equation:
\[
V_e = 8 - 2
\]
\[
V_e = 6
\]
This means that atom B has 6 electrons in its valence shell. Such elements belong to Group 16 of the periodic table (e.g., Oxygen, Sulfur), and their most common negative oxidation state is indeed -2.
The number of electrons in its valence shell is 6.
