Question

Lowest oxidation number of an atom in a compound \( A_2B \) is -2. The number of electrons in its valence shell is:

Answer 1

Correct Answer: 6

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.