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Chapter 9: Problem 13

Which is (a) the smallest atom in group \(13 ;\) (b) the smallest of the following atoms: Te, In, Sr, Po, Sb? Why?

Short Answer

Expert verified
The smallest atom in group \(13\) is Boron (B) and the smallest atom among Te, In, Sr, Po, Sb is Tellurium (Te) due to atomic size trends on the periodic table.

Step by step solution

01

Identify Group 13 elements

First, identify all the elements in group 13. Group 13 of the periodic table consists of the elements Boron (B), Aluminum (Al), Gallium (Ga), Indium (In), Thallium (Tl), and element 113 (Ununtrium, Uut).
02

Find the smallest atom in Group 13

According to the atomic size trend, the atomic size increases as you move down a group because each subsequent element has an additional energy level. Therefore, Boron (B), being at the top of Group 13, is the smallest atom in this group.
03

Identify given elements

Identify the provided elements: Tellurium (Te), Indium (In), Strontium (Sr), Polonium (Po), and Antimony (Sb).
04

Compare atomic radii to determine the smallest atom

Using the trend of atomic size, the further to the right and down you go on the periodic table, the larger the atom. In this item selection, Tellurium (Te) is farthest to the right and has the smallest atomic radius.
05

Explain why certain elements have smaller atomic radii

The size of atoms (atomic radii) within a period or row on the periodic table decreases from left to right because of the increasing positive charge of the nucleus, which pulls the electrons closer to the center of the atom, causing the size to decrease.

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Most popular questions from this chapter

The fourth-period element with the largest atom is (a) \(\mathrm{K} ;\) (b) \(\mathrm{Br} ;\) (c) \(\mathrm{Pb} ;\) (d) \(\mathrm{Kr}\).

Refer only to the periodic table on the inside front cover, and arrange the following ionization energies in order of increasing value: \(I_{1}\) for \(\mathrm{F} ; I_{2}\) for \(\mathrm{Ba}\) \(I_{3}\) for \(\mathrm{Sc} ; I_{2}\) for \(\mathrm{Na} ; I_{3}\) for \(\mathrm{Mg}\). Explain the basis of any uncertainties.

All the isoelectronic species illustrated in the text had the electron configurations of noble gases. Can two ions be isoelectronic without having noble-gas electron configurations? Explain.

For each of the following pairs, indicate the atom that has the larger size: (a) Te or Br; (b) \(\mathrm{K}\) or \(\mathrm{Ca} ;\) (c) Ca or Cs; (d) \(\mathrm{N}\) or \(\mathrm{O} ;\) (e) \(\mathrm{O}\) or \(\mathrm{P} ;\) (f) \(\mathrm{Al}\) or \(\mathrm{Au}.\)

The element whose atoms have the electron configuration \([\mathrm{Kr}] 4 d^{10} 5 \mathrm{s}^{2} 5 p^{3}(\mathrm{a})\) is in group 13 of the periodic table; (b) bears a similarity to the element Bi; (c) is similar to the element \(\mathrm{Te} ;\) (d) is a transition element.
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