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Chapter 9: Problem 4

The following melting points are in degrees Celsius. Show that melting point is a periodic property of these elements: $\mathrm{Al}, 660 ; \mathrm{Ar},-189 ; \mathrm{Be}, 1278 ; \mathrm{B}, 2300 ; \mathrm{C}$ $3350 ; \mathrm{Cl},-101 ; \mathrm{F},-220 ; \mathrm{Li}, 179 ; \mathrm{Mg}, 651 ; \mathrm{Ne},-249 ; \mathrm{N}$ $-210 ; \mathrm{O},-218 ; \mathrm{P}, 590 ; \mathrm{Si}, 1410 ; \mathrm{Na}, 98 ; \mathrm{S}, 119.$

Short Answer

Expert verified
The melting point of the given elements is a periodic property. Evidence for this includes the general increase and then decrease in melting points as you move from left to right across each period in the Periodic Table. The trends can be explained by differences in atomic structure and electron configuration.

Step by step solution

01

Sorting the elements

First, sort the elements according to their position in the Periodic Table. Here is how they will look arranged from left to right, top to bottom: Li, Be, B, C, N, O, F, Ne, Na, Mg, Al, Si, P, S, Cl, Ar.
02

Organizing melting points

Next, list the melting points of the elements in the same order as Step 1: 179, 1278, 2300, 3350, -210, -218, -220, -249, 98, 651, 660, 1410, 590, 119, -101, -189.
03

Identifying trends

Now examine this list for patterns. As you move across each period (row) from left to right, the melting points generally increase, then decrease. This trend repeats in the second period. This is evidence that melting point is a periodic property of these elements.
04

Explaining the trends

The trend in melting points can be explained by atomic structure. Elements on the left of the period (metals) have loosely held valence electrons which require less energy (lower temperature) to break free, hence lower melting points. At the right end are the noble gases with full electron shells, with low melting points due to weak interatomic forces. Between these two are elements with more tightly bound electrons, hence higher melting points.

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Most popular questions from this chapter

Concerning the incomplete seventh period of the periodic table, what should be the atomic number of the element (a) for which the filling of the \(6 d\) subshell is completed; (b) that should most closely resemble bismuth; (c) that should be a noble gas?

Indicate the smallest and the largest species (atom or ion) in the following group: Al atom, F atom, As atom, \(\mathrm{Cs}^{+}\) ion, \(\mathrm{I}^{-}\) ion, \(\mathrm{N}\) atom.

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.

In multielectron atoms many of the periodic trends can be explained in terms of \(Z_{\text {eff }}\) Consider the following statements and discuss whether or not the statement is true or false. (a) Electrons in a \(p\) orbital are more effective than electrons in the \(s\) orbitals in shielding other electrons from the nuclear charge. (b) \(\mathrm{Z}_{\text {eff }}\) for an electron in an \(s\) orbital is lower than that for an electron in a \(p\) orbital in the same shell. (c) \(Z_{\text {eff }}\) is usually less than \(Z.\) (d) Electrons in orbitals having \(\ell=1\) penetrate better than those with \(\ell=2.\) (e) \(\mathrm{Z}_{\text {eff }}\) for the orbitals of the elements \(\mathrm{Na}(3 s)\) \(\mathrm{Mg}(3 s), \mathrm{Al}(3 p), \mathrm{P}(3 p),\) and \(\mathrm{S}(3 p)\) are in the order \(Z_{\text {eff }}(\mathrm{Na})<\mathrm{Z}_{\text {eff }}(\mathrm{Mg})>\mathrm{Z}_{\text {eff }}(\mathrm{Al})<\mathrm{Z}_{\text {eff }}(\mathrm{P})>\mathrm{Z}_{\text {eff }}(\mathrm{S}).\)

Explain why the several periods in the periodic table do not all have the same number of members.
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