Question

Why is the radius of \(\mathrm{Na}^{+}\) much smaller than the radius of Na? Why is the radius of \(\mathrm{Cl}^{-}\) much larger than the radius of \(\mathrm{Cl} ?\)

Short answer

Na⁺ is smaller than Na due to electron loss; Cl⁻ is larger than Cl due to electron gain.

Step-by-step solution

Understanding Atomic Structure of Na and Na+

Sodium (Na) has an atomic number of 11, meaning it has 11 protons and, when neutral, 11 electrons. When sodium loses an electron to become Na⁺, it has 10 electrons, leading to a full outer shell and reduced electron-electron repulsion. This results in a smaller atomic radius for Na⁺ compared to Na.

Exploring Electron Configuration in Na

In its neutral state, the outer electron shell of sodium (Na) is partially filled with one electron in the 3s orbital. The loss of this electron to form Na⁺ leads to a completely filled 2s² 2p⁶ shell, which is stable due to its noble gas configuration, significantly decreasing its radius.

Analyzing Atomic Structure of Cl and Cl−

Chlorine (Cl) has an atomic number of 17, which means it has 17 protons and, when neutral, 17 electrons. When chlorine gains an electron to become Cl⁻, it has 18 electrons. This increase in electron count results in increased electron-electron repulsion and a larger atomic radius for Cl⁻ compared to Cl.

Understanding Electron Configuration in Cl

A neutral chlorine atom has a p-orbital with 5 electrons in its outermost shell (3p⁵). The addition of an electron to form Cl⁻ fills the outer 3p subshell, resulting in increased electron repulsion and a larger radius due to the extra electron.

Key concepts

Ionic Size

When atoms form ions, their size can change significantly. Let's consider sodium (Na) and its ion

• Sodium, when neutral, has a certain size based on its electron cloud with 11 electrons.
• When it loses an electron to become \(\mathrm{Na}^{+}\), its electron count drops to 10.
• Fewer electrons result in less electron-electron repulsion, pulling electrons closer to the nucleus.
• Thus, \(\mathrm{Na}^{+}\) has a much smaller radius than neutral Na.

Now, consider chlorine (Cl) and its ion:

• Neutral chlorine has 17 electrons.
• It gains an electron to form \(\mathrm{Cl}^{-}\), giving it 18 electrons.
• This addition increases repulsion among electrons, expanding the electron cloud.
• The result? A much larger radius for \(\mathrm{Cl}^{-}\) compared to neutral Cl.

Understanding ionic size provides insight into how ions behave in different environments.

Electron Configuration

The arrangement of electrons, or electron configuration, drastically affects an atom's or ion's properties. For sodium and

• In its neutral state, Na has the configuration \(1s^2 \, 2s^2 \, 2p^6 \, 3s^1\).
• Losing one electron to become \(\mathrm{Na}^{+}\) leads to \(1s^2 \, 2s^2 \, 2p^6\), mimicking the stable noble gas configuration of neon.
• This stability is one reason why \(\mathrm{Na}^{+}\) is smaller; its electrons are held tightly by the nucleus.

For chlorine and

• Neutral Cl has an electron configuration of \(1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^5\).
• Gaining an electron results in \(3p^6\) for \(\mathrm{Cl}^{-}\), achieving a full outer shell.
• This extra electron fills the outer shell, increasing electron repulsion.
• Despite the added stability, the larger radius of \(\mathrm{Cl}^{-}\) is due to this increased repulsion.

Electron configuration is key to understanding changes in size and reactivity.

Electron-Electron Repulsion

Electrons naturally repel each other due to their negative charge. This concept has a significant impact on ionic size:

• In a neutral sodium atom, there's a balance of forces with 11 electrons.
• Removing one leads to \(\mathrm{Na}^{+}\), reducing electron-electron repulsion significantly.
• This makes it easier for remaining electrons to be attracted closer to the nucleus, reducing size.

For chlorine, notice:

• The addition of an 18th electron in \(\mathrm{Cl}^{-}\) increases repulsion within the electron cloud.
• This increased repulsion pushes electrons further from the nucleus.
• The result is an expanded size for \(\mathrm{Cl}^{-}\).

Recognizing how electron-electron repulsion influences atomic and ionic radii helps explain trends on the periodic table.