Problem 86 - Chemistry Matter and Change

Question

Apply Sodium forms a \(1+\) ion, while fluorine forms a \(1-\) ion. Write the electron conffiguration for each ion. Why don't these two elements form \(2+\) and \(2-\) ions, respectively?

Step-by-Step Solution

Answer

Na forms a \(1+\) ion and F forms a \(1-\) ion, achieving stable, noble gas configurations; \(2+\) or \(2-\) ions would be unstable.

Step 1: Identify Neutral Atoms

Sodium (Na) is element number 11, and Fluorine (F) is element number 9 on the periodic table. This helps us understand their electron configurations in their neutral forms.

Step 2: Write Electron Configurations for Neutral Atoms

For Sodium (Na), the electron configuration is \(1s^2 \, 2s^2 \, 2p^6 \, 3s^1\). For Fluorine (F), the electron configuration is \(1s^2 \, 2s^2 \, 2p^5\).

Step 3: Define Ion Configurations

Sodium forms a \(1+\) ion by losing one electron from its outermost shell. Fluorine forms a \(1-\) ion by gaining one electron.

Step 4: Write Electron Configurations for Ions

For \(Na^+\), the electron configuration becomes \(1s^2 \, 2s^2 \, 2p^6\) (same as Neon, \(Ne\)). For \(F^-\), the electron configuration becomes \(1s^2 \, 2s^2 \, 2p^6\) (same as Neon, \(Ne\)).

Step 5: Analyze Stability of Ion Configurations

Both \(Na^+\) and \(F^-\) have electron configurations corresponding to the noble gas Neon, which is highly stable due to a complete octet.

Step 6: Explain Why Other Ions Are Not Formed

Sodium does not form a \(2+\) ion because it would require losing another electron from a filled shell, requiring significantly more energy, thus highly unstable. Fluorine does not form a \(2-\) ion as it would require it to add two electrons beyond a complete shell, causing electron-electron repulsion and an unstable state.

Key Concepts

Electron Configuration

Electron configuration tells us how electrons are arranged around an atom. They distribute across "shells" and "orbitals". Think of orbitals like rooms in a house, where each room can hold a certain number of electrons.

• The first shell (closest to the nucleus) is filled first.
• Each orbital is filled in a specific order: starting from the 1s, then 2s, 2p, and so on.
• Electrons will fill the lowest energy orbitals first, following this order.

When writing the configuration, each shell and orbital is represented. For example, Sodium's neutral form is represented as: \(1s^2 \, 2s^2 \, 2p^6 \, 3s^1\). This configuration shows us that Sodium has one electron in its outermost "3s" shell. Fluorine's configuration, at \(1s^2 \, 2s^2 \, 2p^5\), shows it needs one more electron to complete its "2p" orbitals.

Sodium Ion

Sodium (Na), in its neutral state, has 11 electrons. However, when it becomes a sodium ion (\(Na^+\)), it loses one electron. This loss leads to a more stable electronic state: \(1s^2 \, 2s^2 \, 2p^6\). This configuration is the same as Neon, a noble gas, known for its stability.Why does Sodium prefer this state? Removing its lone outer electron from the "3s" shell is relatively easy, energy-wise. This loss lands sodium a full "2p" orbital, mimicking Neon’s stable structure. A full orbital or "shell" is highly favored because it leads to the least energy state.

Fluorine Ion

Fluorine (F), as a neutral atom, holds 9 electrons, leaving one space to fill its "2p" orbital. Upon gaining an electron, Fluorine becomes \(F^-\) and achieves the configuration \(1s^2 \, 2s^2 \, 2p^6\), equivalent to Neon.Fluorine is highly inclined to gain an electron for two reasons:

• Achieving a full outer shell, reaching stability.
• Becoming energetically more favorable, similar to Sodium.

The process of gaining an electron leaves Fluoride (the negative ion formed) with a configuration identical to a noble gas, eliciting the heightened stability due to a complete electron octet.

Stability of Noble Gas Configuration

Both Sodium and Fluorine ions aim to attain configurations similar to noble gases. Why? Because noble gas configurations are very stable, due to their complete outer shells.Such configurations are considered as possessing a "noble gas stability", characterized by:

• Lowest possible energy states.
• Full outer "shells", reducing reactive tendencies.

The stability extends further to influencing the behavior of atoms in reactions. Sodium achieves stability by losing an electron, while Fluorine does so by gaining one. This desirable state of relative inactivity underlies why Sodium does not form a \(2+\) ion, nor Fluorine a \(2-\) ion. Such formations would require disturbing the complete orbital achieved, making them energetically unfavorable.