Which of the following is the correct Lewis structure of sodium fluoride (NaF)?
Na-F:
Na-F:
Na+, [:Ë:]-
The Correct Answer and Explanation is :
The correct Lewis structure for sodium fluoride (NaF) is:
Na⁺ [:F:]-
Here’s an explanation:
1. Sodium (Na) and Fluorine (F) atomic structure:
- Sodium (Na) is a metal with an atomic number of 11, meaning it has 11 electrons. In its neutral state, it has an electron configuration of 1s² 2s² 2p⁶ 3s¹. Sodium tends to lose one electron to achieve a stable noble gas configuration (1s² 2s² 2p⁶), thus forming a Na⁺ ion with a charge of +1.
- Fluorine (F) is a nonmetal with an atomic number of 9, so it has 9 electrons. Fluorine needs one more electron to fill its valence shell (making it stable with 8 electrons in its outermost shell). By gaining an electron, it becomes F⁻, a negatively charged ion.
2. Formation of NaF:
- Sodium (Na) donates its single valence electron to fluorine (F), forming Na⁺ (a cation) and F⁻ (an anion). This ionic bond forms due to the electrostatic attraction between the oppositely charged ions.
3. Lewis Structure:
- In the Lewis structure of NaF, sodium is represented by Na⁺, with no valence electrons shown because it has lost its lone electron.
- Fluorine is represented by [:F:]⁻, where the colon symbols (:) represent the lone pairs of electrons in fluorine’s valence shell. Fluorine has a total of 8 electrons in its valence shell, consisting of the 7 electrons it had originally plus the 1 electron it gained from sodium, thus making it F⁻.
4. Ionic Bonding:
- The ionic bond is the result of the electrostatic attraction between Na⁺ and F⁻, with sodium losing an electron and fluorine gaining one. This transfer of electrons creates oppositely charged ions that are held together by this strong attraction.
Therefore, the correct Lewis structure for sodium fluoride (NaF) is Na⁺ [:F:]-, which reflects the ionic nature of the compound.