Sodium fluoride (NaF) contains an ionic bond due to the transfer of electrons between the sodium (Na) atom and the fluoride (F) atom, leading to the formation of charged ions. Here's a detailed explanation of the process:
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Electron Configuration:
- Sodium has an atomic number of 11, and its electron configuration is \(1s^2 2s^2 2p^6 3s^1\). Sodium has one electron in its outermost shell (the 3s shell), which it can easily lose to achieve a stable electronic configuration similar to that of neon (a noble gas).
- Fluorine has an atomic number of 9, with the electron configuration \(1s^2 2s^2 2p^5\). Fluorine has seven electrons in its outer shell and needs one more electron to complete its octet and achieve a stable configuration like neon.
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Electron Transfer:
- When sodium and fluorine come into contact, sodium tends to lose its one valence electron, becoming a positively charged ion (Na⁺).
- This loss results in a full outer shell for sodium, which is energetically favorable.
- Meanwhile, the fluoride ion (F⁻) is formed when fluorine gains that lost electron, achieving a full outer shell with eight electrons and becoming negatively charged.
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Formation of Ions:
- The result of this electron transfer is the creation of Na⁺ and F⁻ ions.
- Sodium ion (Na⁺) has a positive charge of +1, and the fluoride ion (F⁻) has a negative charge of -1.
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Ionic Bonding:
- The oppositely charged ions (Na⁺ and F⁻) are attracted to each other due to electrostatic forces. This attraction forms an ionic bond, resulting in the compound sodium fluoride.
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Crystal Lattice Structure:
- In solid form, sodium fluoride arranges itself in a crystal lattice structure where each sodium ion is surrounded by fluoride ions and vice versa, maximizing the attraction between the opposite charges and stabilizing the compound.
In summary, sodium fluoride exhibits an ionic bond due to the electrostatic attraction between the positively charged sodium ions and negatively charged fluoride ions formed through electron transfer.