Why is F- bigger than Na+?
The fluoride ion (F-) is larger than the sodium ion (Na+) primarily due to the difference in their nuclear charge and its effect on electron attraction, despite both ions having the same number of electrons.Understanding Ionic Size
The size of an ion, known as its ionic radius, is determined by the balance between the attractive force of the protons in the nucleus and the repulsive forces among the electrons. When comparing ions that have the same number of electrons (isoelectronic species), the key factor becomes the number of protons.The Role of Effective Nuclear Charge (Zeff)
Effective nuclear charge (Zeff) is the net positive charge experienced by an electron in a polyelectronic atom. It is influenced by the actual nuclear charge (number of protons) and the shielding effect of inner electrons.-
For F- (Fluoride Ion):
- A neutral fluorine atom has 9 protons and 9 electrons.
- When it gains one electron to form F-, it now has 9 protons and 10 electrons.
- With more electrons than protons, the 9 protons exert a weaker pull on each of the 10 electrons. The effective nuclear charge experienced by the outermost electrons is relatively low.
- Additionally, the increased electron-electron repulsion among the 10 electrons in the electron cloud causes the cloud to expand, contributing to a larger ionic radius.
-
For Na+ (Sodium Ion):
- A neutral sodium atom has 11 protons and 11 electrons.
- When it loses one electron to form Na+, it now has 11 protons and 10 electrons.
- With more protons than electrons, the 11 protons exert a stronger pull on each of the 10 electrons. The effective nuclear charge is significantly higher compared to F-.
- This stronger nuclear attraction pulls the electron cloud closer to the nucleus, resulting in a smaller ionic radius.
Comparing F- and Na+: Isoelectronic Species
Both F- and Na+ are isoelectronic with Neon (Ne), meaning all three species have 10 electrons. However, their nuclear charges differ significantly:- F-: 9 protons
- Ne: 10 protons
- Na+: 11 protons
This difference in the number of protons, or nuclear charge, is the deciding factor for their sizes. The species with fewer protons pulling on the same number of electrons will have a larger size, and vice versa.
The table below summarizes the key differences contributing to their sizes:
| Property | F- (Fluoride Ion) | Na+ (Sodium Ion) |
|---|---|---|
| Number of Protons | 9 | 11 |
| Number of Electrons | 10 | 10 |
| Net Charge | -1 | +1 |
| Electron-to-Proton Ratio | Higher (10:9) | Lower (10:11) |
| Effective Nuclear Charge | Lower | Higher |
| Nuclear Attraction on Electrons | Weaker | Stronger |
| Ionic Radius | Larger | Smaller |
In conclusion, F- is larger than Na+ because it has fewer protons (9) attracting the same number of electrons (10), leading to a weaker effective nuclear charge and a more expanded electron cloud compared to Na+, which has more protons (11) attracting the same 10 electrons.
