Answer :
The correct classification of the three elements is as follows:
- Fe (Iron) is a metal due to its excellent conductivity, luster, malleability, and ductility.
- Te (Tellurium) is a metalloid because it exhibits properties intermediate between metals and nonmetals, with its semiconductor behavior.
- S (Sulfur) is a nonmetal because it lacks the characteristic properties of metals, being a poor conductor and brittle solid at room temperature.
1. Fe (Iron): Metal
Iron (Fe) is classified as a metal. Metals are typically good conductors of electricity and heat, have a lustrous appearance, and are malleable and ductile. Iron is a good conductor of electricity and heat, and it exhibits metallic luster. It is also malleable, which means it can be easily hammered into thin sheets, and ductile, meaning it can be drawn into wires. These properties are characteristic of metals, making iron fall into this category.
2. Te (Tellurium): Metalloid
Tellurium (Te) is classified as a metalloid. Metalloids have properties that are intermediate between metals and nonmetals. They can exhibit characteristics of both groups.
Tellurium is a semiconductor, which means its electrical conductivity is between that of metals and insulators (nonmetals). While it can conduct electricity under certain conditions, it is not as efficient as typical metals. Additionally, tellurium shares some chemical properties with nonmetals. Hence, it is considered a metalloid.
3. S (Sulfur): Nonmetal
Sulfur (S) is classified as a nonmetal. Nonmetals are generally poor conductors of electricity and heat, lack metallic luster, and are not malleable or ductile.
Sulfur is a nonmetal because it does not conduct electricity well, and it does not possess the typical properties of metals. It is a yellow, brittle solid at room temperature, and its properties are more similar to those of other nonmetals.
In summary, Fe (Iron) is a metal , Te (Tellurium) is a metalloid and S (Sulfur) is a nonmetal.
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