In electrovalency, the oxidation number of the participating metal is always

Analysis of the Question

This question tests your understanding of electrovalency (also known as ionic bonding) and the behavior of metal atoms during chemical reactions. You need to identify what happens to the oxidation state of a metal when it forms an ionic bond.

Concept Explanation

Electrovalency occurs when a metal atom transfers one or more electrons to a non-metal atom.

• Metals have low ionization energies, meaning they easily lose their valence electrons to achieve a stable octet configuration (like a noble gas).
• When a neutral metal atom (which has an oxidation number of 0) loses electrons, it becomes a positively charged ion (a cation).
• Because the metal is losing negatively charged electrons, its oxidation number changes from 0 to a positive value (e.g., $Na \rightarrow Na^+ + e^-$).

In terms of oxidation numbers, moving from 0 to a positive value is mathematically an increase.

Step-by-Step Solution

1. Identify the starting state: A neutral metal atom in its elemental state has an oxidation number of 0.
2. Identify the process: In electrovalency, the metal loses electrons to form a cation.
3. Determine the change: When an atom loses electrons, it undergoes oxidation. Oxidation is defined as an increase in the oxidation number.
4. Evaluate the options: • Since the metal goes from 0 to a positive charge (e.g., $+1, +2, +3$), the oxidation number is increasing.

Conclusion

This aligns with the chemical principle that metals act as reducing agents by losing electrons, which results in an increase in their oxidation state.

Exam Tip: Remember the mnemonic OIL RIG: Oxidation Is Loss (of electrons/increase in oxidation number), Reduction Is Gain (of electrons/decrease in oxidation number). Since metals lose electrons in electrovalency, they are always oxidized, meaning their oxidation number increases.