Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Arrange the following groups of atoms in order of increasing size. a. \(\mathrm{Te}, \mathrm{S}, \mathrm{Se}\) b. \(\mathrm{K}, \mathrm{Br}, \mathrm{Ni}\) c. \(\mathrm{Ba}, \mathrm{Si}, \mathrm{F}\)

Short Answer

Expert verified
Arrange the groups of atoms in order of increasing size as follows: a. \(\mathrm{S}, \mathrm{Se}, \mathrm{Te}\) b. \(\mathrm{Br}, \mathrm{Ni}, \mathrm{K}\) c. \(\mathrm{F}, \mathrm{Si}, \mathrm{Ba}\)

Step by step solution

01

a. Arrange \(\mathrm{Te}, \mathrm{S}, \mathrm{Se}\) in order of increasing size

We can find these elements in the same group (Group 16) of the periodic table. Since atomic size increases from top to bottom in a group, we can arrange them in the order of their position in the periodic table: S < Se < Te So the order of increasing size for these atoms is \(\mathrm{S}, \mathrm{Se}, \mathrm{Te}\).
02

b. Arrange \(\mathrm{K}, \mathrm{Br}, \mathrm{Ni}\) in order of increasing size

These elements belong to different groups but they are in the same period (Period 4). Since atomic size decreases from left to right in a period, we can arrange them in the order of their position in the periodic table: K > Ni > Br So the order of increasing size for these atoms is \(\mathrm{Br}, \mathrm{Ni}, \mathrm{K}\).
03

c. Arrange \(\mathrm{Ba}, \mathrm{Si}, \mathrm{F}\) in order of increasing size

These elements are neither in the same group nor in the same period. Therefore, we have to use both the group and period trends to determine their order. Ba is in Group 2 and Period 6. Si is in Group 14 and Period 3. F is in Group 17 and Period 2. Comparing Ba and Si: Ba has more energy levels than Si since it is below Si in the periodic table. Thus, Ba has a larger atomic size than Si. Comparing Si and F: Si is left of F in the periodic table within Period 3. Thus, Si has a larger atomic size than F. From the above comparisons, the order of increasing size for these atoms is \(\mathrm{F}, \mathrm{Si}, \mathrm{Ba}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Periodic Table Group Trends
When studying the periodic table, it's important to understand how atoms behave within the same group or column. A group consists of elements that have similar chemical properties because they have the same number of valence electrons.
However, one key trend to remember is that atomic size tends to increase as you move down a group in the periodic table.
  • Each row (or period) down a group indicates a higher principal energy level or shell. This means additional shells are added, making the atomic radius larger.
  • This results in atoms being larger at the bottom of the group than those at the top.
For example, in Group 16 with elements like sulfur (S), selenium (Se), and tellurium (Te), Te will be the largest because it is at the bottom of the group, making it occupy more space. This is why atomic size increases from S to Se to Te.
Periodic Table Period Trends
In contrast to group trends, period trends involve comparing elements across a period or row on the periodic table. When moving from left to right across a period, the atomic radius usually decreases.
This can be explained by the fact that elements across a period share the same number of shells, but more protons are added to the nucleus without adding new energy levels.
  • The additional protons increase the nuclear charge, pulling electrons closer to the nucleus.
  • This tighter pull decreases the size of the electron cloud, thereby reducing the atomic radius.
Let's take the elements potassium (K), nickel (Ni), and bromine (Br) from Period 4. As we move from K to Ni to Br, the atomic size decreases because the atoms become more tightly bound by the increased nuclear charge without adding new outer orbitals.
Atomic Radius
The atomic radius is a key concept for understanding atomic size. It is essentially the distance from the center of the nucleus to the outermost shell of an electron in an atom. Atomic radius can vary based on several factors:
  • Number of energy levels: More levels mean a larger radius.
  • Effective nuclear charge: Greater charge results in a smaller radius as electrons are pulled closer.
  • Shrinkage across a period: With an increase in protons but not energy levels, radius shrinks.
For instance, when comparing elements in the same group, the added energy level leads to larger atomic radii. Conversely, across a period, although the energy levels remain the same, the atomic radius decreases due to a stronger effective nuclear charge.
Elements Comparison
Comparing atomic sizes of elements is about understanding both their position in a group and across a period. Some elements fall neither in the same group nor the same period, and such cases require careful analysis. Let's look at barium (Ba), silicon (Si), and fluorine (F) as examples.
  • Barium is further down in Group 2 and hence has a larger atomic radius than Si and F due to more energy levels.
  • Silicon and fluorine are both in different periods; Si is left of F in Period 3, making it larger because the atomic size decreases across a period.
In such comparisons, it helps to consider the number of energy levels first (group trend) and then think about the effective nuclear charge (period trend) for an informed decision on atomic size. These principles help predict the order of atomic sizes across diverse elements.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The following numbers are the ratios of second ionization energy to first ionization energy: Na: \(\quad 9.2\) P: \(1.8\) Mg: \(\quad 2.0\) \(\mathrm{S}: \quad 2.3\) Al: \(\quad 3.1\) \(\mathrm{Cl}: \quad 1.8\) Si: \(\quad 2.0\) Ar: \(1.8\) Explain these relative numbers.

Three elements have the electron configurations \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}\), \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{4}\), and \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2}\). The first ionization en- ergies of these elements (not in the same order) are \(0.590,0.999\), and \(0.738 \mathrm{MJ} / \mathrm{mol}\). The atomic radii are 104,160 , and \(197 \mathrm{pm}\). Identify the three elements, and match the appropriate values of ionization energy and atomic radius to each configuration. Complete the following table with the correct information.

A certain microwave oven delivers 750 . watts \((\mathrm{J} / \mathrm{s})\) of power to a coffee cup containing \(50.0 \mathrm{~g}\) water at \(25.0^{\circ} \mathrm{C}\). If the wavelength of microwaves in the oven is \(9.75 \mathrm{~cm}\), how long does it take, and how many photons must be absorbed, to make the water boil? The specific heat capacity of water is \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\), and assume only the water absorbs the energy of the microwaves.

The bright yellow light emitted by a sodium vapor lamp consists of two emission lines at \(589.0\) and \(589.6 \mathrm{~nm}\). What are the frequency and the energy of a photon of light at each of these wavelengths? What are the energies in \(\mathrm{kJ} / \mathrm{mol}\) ?

Write equations corresponding to the following. a. the fourth ionization energy of Se b. the electron affinity of \(\mathrm{S}^{-}\) c. the electron affinity of \(\mathrm{Fe}^{3+}\) d. the ionization energy of \(\mathrm{Mg}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free