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Why Cu , Ag , Au form most of the compound in 2 , 1 & 3 oxidation state respectively ?
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Why Cu , Ag , Au form most of the compound in 2 , 1 & 3 oxidation s...
Cu (copper), Ag (silver), and Au (gold) are all transition metals. Transition metals have unique properties that make them prone to forming compounds.

One reason why Cu, Ag, and Au form compounds easily is their electron configuration. They have partially filled d-orbitals, which allows them to readily gain or lose electrons to achieve a stable electron configuration. This makes them good candidates for forming compounds with other elements.

Additionally, Cu, Ag, and Au have relatively low electronegativity values. Electronegativity is a measure of an atom's ability to attract electrons in a chemical bond. Transition metals tend to have lower electronegativity values compared to non-metals, which means they are more likely to lose electrons and form positive ions in compounds.

Another factor is the size of the atoms. Cu, Ag, and Au have relatively large atomic radii compared to other transition metals. This larger size allows for greater flexibility in bonding with other elements, increasing the likelihood of compound formation.

Overall, the combination of their electron configuration, low electronegativity, and larger atomic radii make Cu, Ag, and Au more prone to forming compounds with other elements.
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Why Cu , Ag , Au form most of the compound in 2 , 1 & 3 oxidation state respectively ?
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Why Cu , Ag , Au form most of the compound in 2 , 1 & 3 oxidation state respectively ? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Why Cu , Ag , Au form most of the compound in 2 , 1 & 3 oxidation state respectively ? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Why Cu , Ag , Au form most of the compound in 2 , 1 & 3 oxidation state respectively ?.
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