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Chapter 2: Problem 100

From the following list of elements $-\mathrm{Mg}, \mathrm{Li}, \mathrm{T}, \mathrm{Pb}, \mathrm{Se}, \mathrm{Cl}, \mathrm{Xe}, \mathrm{Si},$ C-pick the one that best fits each description. Use each element only once: (a) an alkali metal, \((\mathbf{b})\) an alkaline earth metal, \((\mathbf{c})\) a noble gas, (d) a halogen, \((\mathbf{e})\) a metalloidin group \(14,(\mathbf{f})\) a nonmetal listed in group \(14,(\mathbf{g})\) a metal that forms a \(3+\) ion, \((\mathbf{h})\) a nonmetal that forms a 2 - ion, (i) an element that is used as radiation shielding.

Short Answer

Expert verified
(a) \(Li\) (b) \(Mg\) (c) \(Xe\) (d) \(Cl\) (e) \(Si\) (f) C (g) \(T\) (h) \(Se\) (i) \(Pb\)

Step by step solution

01

An alkali metal

To find an alkali metal, we look for elements in Group 1 of the periodic table. The only element in the given list that is in Group 1 is \(Li\), so \(Li\) is an alkali metal.
02

An alkaline earth metal

To find an alkaline earth metal, we look for elements in Group 2 of the periodic table. The only element in the given list that is in Group 2 is \(Mg\), so \(Mg\) is an alkaline earth metal.
03

A noble gas

To find a noble gas, we look for elements in Group 18 of the periodic table. The only element in the given list that is in Group 18 is \(Xe\), so \(Xe\) is a noble gas.
04

A halogen

To find a halogen, we look for elements in Group 17 of the periodic table. The only element in the given list that is in Group 17 is \(Cl\), so \(Cl\) is a halogen.
05

A metalloid in group 14

To find a metalloid in Group 14, we look for elements in Group 14 that have properties between metals and nonmetals. The only element in the given list that is in Group 14 and is a metalloid is \(Si\), so \(Si\) is a metalloid in group 14.
06

A nonmetal listed in group 14

To find a nonmetal listed in Group 14, we look for elements that have properties of nonmetals and are in Group 14. The only element in the given list that is in Group 14 and is a nonmetal is C, so C is a nonmetal listed in group 14.
07

A metal that forms 3+ ion

To find a metal that forms a 3+ ion, we can look for elements that lose 3 electrons when they form ions. In the given list, \(T\) represents the element Thallium, which can form a 3+ ion as \(T^{3+}\), so \(T\) is a metal that forms a 3+ ion.
08

A nonmetal that forms a 2- ion

To find a nonmetal that forms a 2- ion, we can look for elements that gain 2 electrons when they form ions. In the given list, \(Se\) is a nonmetal that forms a 2- ion as \(Se^{2-}\), so \(Se\) is a nonmetal that forms a 2 - ion.
09

An element that is used as radiation shielding

Among the given elements, the one that is used as radiation shielding is \(Pb\). \(Pb\) is a heavy metal that can absorb and block high-energy radiation such as gamma rays and X-rays, making it a suitable material for radiation shielding.

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Most popular questions from this chapter

Selenium, an element required nutritionally in trace quantities, forms compounds analogous to sulfur. Name the following ions: (a) \(\mathrm{SeO}_{4}^{2-}\), (b) \(\mathrm{Se}^{2-}\), (c) \(\mathrm{HSe}^{-}\), (d) \(\mathrm{HSeO}_{3}^{-}\).

Discovering which of the three subatomic particles proved to be the most difficult-the proton, neutron, or electron? Why?

One way in which Earth's evolution as a planet can be understood is by measuring the amounts of certain isotopes in rocks. One quantity recently measured is the ratio of \({ }^{129} \mathrm{Xe}\) to \({ }^{130}\) Xe in some minerals. In what way do these two isotopes differ from one another? In what respects are they the same?

Iron has three major isotopes: \({ }^{54} \mathrm{Fe}\) (atomic mass = $53.9396 \mathrm{u} ;\( abundance \)5.85 \%),{ }^{56} \mathrm{Fe}(\( atomic mass \)=$ \(55.9349 \mathrm{u} ;\) abundance \(91.75 \%\), and \({ }^{57} \mathrm{Fe} \quad(\) atomic mass \(=56.9354 \mathrm{u} ;\) abundance \(2.12 \%) .\) Calculate the atomic weight (average atomic mass) of iron.

(a) What is the mass in \(\mathrm{u}\) of a carbon- 12 atom? (b) Why is the atomic weight of carbon reported as 12.011 in the table of elements and the periodic table in the front inside cover of this text?
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