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Summarize the trend in metallic character as a function of position in the periodic table. Is it the same as the trend in atomic size? The trend in ionization energy?

Short Answer

Expert verified
Metallic character increases from right to left and top to bottom, matching atomic size trends but opposing ionization energy trends.

Step by step solution

01

Identify the Trend in Metallic Character

Metallic character increases as you move from right to left across a period and from top to bottom down a group in the periodic table.
02

Compare with Atomic Size Trend

Atomic size also increases from right to left across a period and from top to bottom down a group. Hence, the trend in metallic character aligns with the trend in atomic size.
03

Compare with Ionization Energy Trend

Ionization energy increases from left to right across a period and decreases from top to bottom down a group. This shows that the trend in ionization energy is opposite to the trend in metallic character.
04

Summarize the Comparisons

While both metallic character and atomic size increase moving from right to left and top to bottom, ionization energy does the opposite: it decreases in these directions.

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Key Concepts

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

Understanding Periodic Table Trends
The periodic table is not only a way to organize elements, but it also tells us about the properties and behavior of these elements. There are various trends observed in the periodic table that help us predict how elements will interact in different conditions.
One such trend is the **metallic character** of elements. Metallic character refers to how easily an element can lose electrons to form positive ions. Elements with high metallic character are usually shiny, good conductors of electricity and heat, malleable, and ductile.
The metallic character increases as you move from right to left across a period (horizontal rows) and from top to bottom down a group (vertical columns) in the periodic table. So, elements on the leftmost side, like sodium (Na), are more metallic compared to those on the right, like chlorine (Cl). Similarly, elements at the bottom of a group, like francium (Fr), are more metallic than those at the top, like lithium (Li).
Overall, understanding these trends helps us in predicting the nature of elements and their chemical behavior.
Exploring Atomic Size
Atomic size or atomic radius is the distance from the nucleus of an atom to the outermost shell of electrons.
Atomic size increases as you move from right to left across a period and from top to bottom down a group.
There are a few reasons behind these trends:
  • As you move down a group, each element has an additional electron shell compared to the element above it, making the atom larger.
  • As you move from right to left across a period, there are fewer protons in the nucleus, which means there is less positive charge to pull electrons closer to the nucleus, resulting in a larger atomic size.
So, in summary, elements on the left side and bottom of the periodic table have larger atomic sizes than those on the right side and at the top.
This trend in atomic size is very similar to the trend in metallic character, both of which increase from the right to the left side of a period and from the top to the bottom of a group.
Understanding Ionization Energy
Ionization energy is the energy required to remove an electron from an atom in the gas phase.
It provides insight into how tightly an atom holds onto its electrons. The trend in ionization energy is the opposite of the trends in atomic size and metallic character:
  • Ionization energy increases from left to right across a period.
  • Ionization energy decreases from top to bottom down a group.
This happens because as you move from left to right across a period, the number of protons in the nucleus increases. This increased positive charge pulls the electrons closer, making them harder to remove.
Conversely, as you move down a group, the outermost electron is further from the nucleus due to additional electron shells, and thus easier to remove.
High ionization energy means that an element does not lose its electrons easily, making it less metallic. Likewise, elements with low ionization energy lose electrons more readily, contributing to their higher metallic character.

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Most popular questions from this chapter

How many unpaired electrons are present in a ground-state atom from each of the following groups? (a) \(4 \mathrm{~A}(14)\) (b) \(7 \mathrm{~A}(17)\) (c) \(1 \mathrm{~A}(1)\) (d) \(6 \mathrm{~A}(16)\)

Before Mendeleev published his periodic table, German scientist Johann Döbereiner grouped elements with similar properties into "triads," in which the unknown properties of one member could be predicted by averaging known values of the properties of the others. To test this idea, predict the values of the following quantities: (a) The atomic mass of \(\mathrm{K}\) from the atomic masses of \(\mathrm{Na}\) and \(\mathrm{Rb}\) (b) The melting point of \(\mathrm{Br}_{2}\) from the melting points of \(\mathrm{Cl}_{2}\) \(\left(-101.0^{\circ} \mathrm{C}\right)\) and \(\mathrm{I}_{2}\left(113.6^{\circ} \mathrm{C}\right)\left(\right.\) actual value \(\left.=-7.2^{\circ} \mathrm{C}\right)\)

State the periodic law, and explain its relation to electron configuration. (Use Na and \(\mathrm{K}\) in your explanation.)

The \(\mathrm{EA}_{2}\) of an oxygen atom is positive, even though its \(\mathrm{EA}_{1}\) is negative. Why does this change of sign occur? Which other elements exhibit a positive \(\mathrm{EA}_{2}\) ? Explain.

Name the element described in each of the following: (a) Smallest atomic radius in Group \(6 \mathrm{~A}(16)\) (b) Largest atomic radius in Period 6 (c) Smallest metal in Period 3 (d) Highest IE \(_{1}\) in Group \(4 \mathrm{~A}(14)\) (e) Lowest IE \(_{1}\) in Period 5 (f) Most metallic in Group \(5 \mathrm{~A}(15)\) (g) Group \(3 \mathrm{~A}(13)\) element that forms the most basic oxide (h) Period 4 element with highest energy level filled (i) Condensed ground-state electron configuration of [Ne] \(3 s^{2} 3 p^{2}\) (j) Condensed ground-state electron configuration of \([\mathrm{Kr}] 5 s^{2} 4 d^{6}\) (k) Forms \(2+\) ion with electron configuration [Ar] \(3 d^{3}\) (1) Period 5 element that forms \(3+\) ion with pseudo-noble gas configuration (m) Period 4 transition element that forms \(3+\) diamagnetic ion (n) Period 4 transition element that forms \(2+\) ion with a half-filled \(d\) sublevel (o) Heaviest lanthanide (p) Period 3 element whose \(2-\) ion is isoelectronic with Ar (q) Alkaline earth metal whose cation is isoelectronic with \(\mathrm{Kr}\) (r) Group \(5 \mathrm{~A}(15)\) metalloid with the most acidic oxide

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