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How do the electron arrangements in Bohr-Rutherford diagrams for helium, neon, and argon explain the
chemical properties of
these elements?



Answer :

In Bohr-Rutherford diagrams, the electron arrangements for helium, neon, and argon explain their chemical properties based on their electron configuration in the outer energy level.

1. Helium (He) has two electrons in its outermost energy level. This makes it stable because it has a full outer shell (like the noble gases), which makes helium unreactive or inert.

2. Neon (Ne) has eight electrons in its outermost energy level, also known as a full valence shell. This full shell gives neon stability, making it unreactive like helium.

3. Argon (Ar) has eight electrons in its outermost energy level as well, resulting in a full valence shell. This full shell provides argon with stability, making it unreactive similar to helium and neon.

The stability of these elements due to their full valence shells explains why helium, neon, and argon are classified as noble gases and have very low reactivity. The electron arrangements in their Bohr-Rutherford diagrams show that these elements have achieved a stable configuration, making them unlikely to form chemical bonds with other elements easily.

Therefore, the electron arrangements in Bohr-Rutherford diagrams for helium, neon, and argon help explain their chemical properties by demonstrating their stable configuration with full valence shells, leading to low reactivity and the characteristics of noble gases.

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