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In general, atomic radius and ionization energy have opposite periodic trends. Why?

Short Answer

Expert verified
Atomic radius and ionization energy have opposite trends due to the balance of nuclear attraction and electron distance.

Step by step solution

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01

Understand Periodic Trends in Atomic Radius

The atomic radius is the measure of the size of an atom from its nucleus to the boundary of its surrounding cloud of electrons. As you move across a period (from left to right) on the periodic table, the atomic radius decreases. This is because additional protons increase the attractive force on the electron cloud, pulling the electrons closer to the nucleus and hence reducing the atomic size. Conversely, as you move down a group (from top to bottom), the atomic radius increases due to the addition of electron shells, which outweighs the increase in nuclear charge.
02

Understand Periodic Trends in Ionization Energy

Ionization energy is the energy required to remove an electron from an atom in its gaseous state. Across a period, ionization energy increases because the electrons are closer to the nucleus and more strongly attracted to it, making them harder to remove. Meanwhile, moving down a group decreases ionization energy, as the electrons are further from the nucleus and experience less nuclear attraction, making them easier to remove.
03

Relate Opposite Trends

The opposing trends arise because atomic radius reflects how closely electrons are held by the nucleus. A smaller atomic radius means that electrons are held more tightly by the nucleus, resulting in higher ionization energy. Conversely, a larger atomic radius means electrons are further away and less tightly held, resulting in lower ionization energy.
04

Conclude the Relationship

The relationship between these properties is fundamentally based on the balance of attraction between the protons in the nucleus and the electrons. As atomic radius decreases, electrons closer to the nucleus require more energy for removal (higher ionization energy). Conversely, increased atomic radius means electrons are further and less tightly bound, reducing ionization energy.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Atomic Radius
The atomic radius tells us how large an atom is, stretching from its central nucleus right to the outermost cloud of electrons. As you look across a period on the periodic table from left to right, the atomic radius tends to shrink. This happens because as you add more protons to the nucleus, their positive charge pulls the negative electrons closer. This stronger attraction tugs the electrons inward, reducing the overall size of the atom.

On the other hand, if you're moving down a group (from top to bottom), the atomic radius increases. This is due to additional electron shells being added, which outweigh the increased nuclear charge. Think of it as adding more layers to an onion—the nucleus's pull becomes less effective at reaching through all the layers, allowing electrons to sit further out.
Ionization Energy
Ionization energy is simply the energy you need to pluck an electron from an atom when it's in a gaseous state. As you journey across a period from left to right, the ionization energy ramps up. That is because electrons, now closer to the nucleus due to a shrinking atomic radius, become more tightly bound. The strong pull from the nucleus makes dislodging these electrons a more energy-demanding task.

Conversely, when you move down a group, ionization energy drops. Here, electrons sit farther from the nucleus due to additional electron layers or shells. Being further from the nucleus means that these electrons experience less attractive force, making them easier to remove. Less energy is needed to pop these distant electrons off the atom.
Periodic Table
The periodic table is essentially a map of all known elements, organized in a way that highlights repeating patterns, or periodic trends, in chemical behavior. The layout of the periodic table allows us to observe how atomic traits like atomic radius and ionization energy vary in a predictable way. Elements are arranged in rows called periods and columns called groups. Each period represents a cycle where elements start with a smaller atomic radius that gradually gets smaller until a new period begins. Simultaneously, ionization energy increases across a period.

By comparing elements within the same group, one can observe larger atomic radii and lower ionization energies as you move down. The presence of these trends makes the periodic table an invaluable tool for predicting and understanding chemical behavior and reactivity of the elements.
Atomic Size
Atomic size is a broad term often used interchangeably with atomic radius, but it could also consider other factors like the ionic or covalent radii. When discussing atomic size in the context of periodic trends, we are mostly interested in the atomic radius. The size of an atom plays a crucial role in its interactions with other atoms.

Since the atomic size shrinks across a period and grows down a group, this influences how atoms bond and form compounds. For instance, smaller atoms with high ionization energy tend to form strong covalent bonds, while larger atoms with low ionization energy are more likely to lose electrons and form ionic bonds. This interplay between size and bonding capabilities leads to the incredibly varied and complex world of chemistry.

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Most popular questions from this chapter

Write equations representing the following processes: (a) The electron affinity of \(S^{-}\) (b) The third ionization energy of titanium (c) The electron affinity of \(\mathrm{Mg}^{2+}\) (d) The ionization energy of \(\mathrm{O}^{2-}\)

Little is known of the chemistry of astatine, the last member of Group 7 A. Describe the physical characteristics that you would expect this halogen to have. Predict the products of the reaction between sodium astatide (NaAt) and sulfuric acid. (Hint: Sulfuric acid is an oxidizing agent.)

The ionization energies of sodium (in \(\mathrm{kJ} / \mathrm{mol}\) ), starting with the first and ending with the eleventh, are 496 , 4562,6910,9543,13,354,16,613,20,117,25,496 \(28,932,141,362,159,075 .\) Plot the log of ionization energy \((y\) axis \()\) versus the number of ionization \((x\) axis); for example, \(\log 496\) is plotted versus 1 (labeled \(I E_{1}\), the first ionization energy), \(\log 4562\) is plotted versus 2 (labeled \(I E_{2},\) the second ionization energy \(),\) and so on. (a) Label \(I E_{1}\) through \(I E_{11}\) with the electrons in orbitals such as \(1 s, 2 s, 2 p,\) and \(3 s\) (b) What can you deduce about electron shells from the breaks in the curve?

Which of the following species are isoelectronic with each other: \(\mathrm{C}, \mathrm{Cl}^{-}, \mathrm{Mn}^{2+}, \mathrm{B}^{-}, \mathrm{Ar}, \mathrm{Zn}, \mathrm{Fe}^{3+}, \mathrm{Ge}^{2+} ?\)

A \(\mathrm{M}^{2+}\) ion derived from a metal in the first transition metal series has four electrons in the \(3 d\) subshell. What element might \(\mathrm{M}\) be?

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