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Unformatted text preview: BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE WebsiteCHAPTER8Periodic RelationshipsAmong the ElementsINTRODUCTIONMANYOF THE CHEMICAL PROPERTIES OF THE ELEMENTS CAN BE UN-DERSTOOD IN TERMS OF THEIR ELECTRON CONFIGURATIONS.8.1 DEVELOPMENT OF THE PERIODIC TABLEBECAUSE8.2 PERIODIC CLASSIFICATION OF THEELEMENTSELECTRONS FILL ATOMIC ORBITALS IN A FAIRLY REGULAR FASHION, IT ISNOT SURPRISING THAT ELEMENTS WITH SIMILAR ELECTRON CONFIGURA-8.3 PERIODIC VARIATION IN PHYSICALPROPERTIESTIONS, SUCH AS SODIUM AND POTASSIUM, BEHAVE SIMILARLY IN MANY8.4 IONIZATION ENERGYRESPECTS AND THAT, IN GENERAL, THE PROPERTIES OF THE ELEMENTS8.5 ELECTRON AFFINITYEXHIBIT OBSERVABLE TRENDS.8.6 VARIATION IN CHEMICAL PROPERTIES OFTHE REPRESENTATIVE ELEMENTSCHEMISTSIN THE NINETEENTH CENTURYRECOGNIZED PERIODIC TRENDS IN THE PHYSICAL AND CHEMICAL PROPERTIES OF ELEMENTS LONG BEFORE QUANTUM THEORY CAME ONTO THESCENE.ALTHOUGHTHESE CHEMISTS WERE NOT AWARE OF THE EXISTENCEOF ELECTRONS AND PROTONS, THEIR EFFORTS TO SYSTEMATIZE THE CHEMISTRY OF THE ELEMENTS WERE REMARKABLY SUCCESSFUL.THEIRMAINSOURCES OF INFORMATION WERE THE ATOMIC MASSES OF THE ELEMENTSAND OTHER KNOWN PHYSICAL AND CHEMICAL PROPERTIES.287BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website288PERIODIC RELATIONSHIPS AMONG THE ELEMENTS8.1DEVELOPMENT OF THE PERIODIC TABLEIn the nineteenth century, when chemists had only a vague idea of atoms and molecules and did not know of the existence of electrons and protons, they devised the periodic table using their knowledge of atomic masses. Accurate measurements of theatomic masses of many elements had already been made. Arranging elements according to their atomic masses in a periodic table seemed logical to those chemists, whofelt that chemical behavior should somehow be related to atomic mass.In 1864 the English chemist John Newlands noticed that when the elements werearranged in order of atomic mass, every eighth element had similar properties. Newlandsreferred to this peculiar relationship as the law of octaves. However, this law turnedout to be inadequate for elements beyond calcium, and Newlandss work was not accepted by the scientific community.In 1869 the Russian chemist Dmitri Mendeleev and the German chemist LotharMeyer independently proposed a much more extensive tabulation of the elements basedon the regular, periodic recurrence of properties. Table 8.1 shows an early version ofMendeleevs periodic table. Mendeleevs classification system was a great improvement over Newlandss for two reasons. First, it grouped the elements together more accurately, according to their properties. Equally important, it made possible the prediction of the properties of several elements that had not yet been discovered. For example,Mendeleev proposed the existence of an unknown element that he called eka-aluminumand predicted a number of its properties. (Eka is a Sanskrit word meaning first; thuseka-aluminum would be the first element under aluminum in the same group.) Whengallium was discovered four years later, its properties matched the predicted properties of eka-aluminum remarkably well:EKA-ALUMINUM (Ea)Atomic massMelting pointDensityFormula of oxideAppendix 1 explains the names andsymbols of the elements.GALLIUM (Ga)68 amuLow5.9 g/cm3Ea2O369.9 amu30.15C5.94 g/cm3Ga2O3Mendeleevs periodic table included 66 known elements. By 1900, some 30 more hadbeen added to the list, filling in some of the empty spaces. Figure 8.1 charts the discovery of the elements chronologically.Although this periodic table was a celebrated success, the early versions had someglaring inconsistencies. For example, the atomic mass of argon (39.95 amu) is greaterthan that of potassium (39.10 amu). If elements were arranged solely according to increasing atomic mass, argon would appear in the position occupied by potassium inour modern periodic table (see the inside front cover). But no chemist would place argon, an inert gas, in the same group as lithium and sodium, two very reactive metals.This and other discrepancies suggested that some fundamental property other thanJohn Alexander Reina Newlands (18381898). English chemist. Newlandss work was a step in the right direction in theclassification of the elements. Unfortunately, because of its shortcomings, he was subjected to much criticism, and evenridicule. At one meeting he was asked if he had ever examined the elements according to the order of their initial letters!Nevertheless, in 1887 Newlands was honored by the Royal Society of London for his contribution.Dmitri Ivanovich Mendeleev (18361907). Russian chemist. His work on the periodic classification of elements is regardedby many as the most significant achievement in chemistry in the nineteenth century.Julius Lothar Meyer (18301895). German chemist. In addition to his contribution to the periodic table, Meyer also discovered the chemical affinity of hemoglobin for oxygen.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website8.1TABLE 8.1REIHENDEVELOPMENT OF THE PERIODIC TABLE289The Periodic Table as Drawn by Mendeleev*GRUPPE IR2OH7GRUPPE IIROGRUPPE IIIR2O3GRUPPE IVRH4RO2GRUPPE VRH3R2O5GRUPPE VIRH2RO3GRUPPE VII GRUPPE VIIIRHR2O7RO4BeBCNOF12Li134K56(Cu 63)Zn 65 68 72As 75Se 78Rb 85Sr 87?Yt 88Zr 90Nb 94Mo 9678910(Ag 108) Cd 112In 113Sn 118Sb 122Te 125Cs 133Ba 137?Di 138 ?Ce 140 ()?Er 178 ?La 180 Ta 182W 1841112(Au9,4111214Na 23Mg 24 Al 27,3Si 2839Ca 40 44Ti 48V199)Hg200Ti204Pb 207Th 23116P 31S 3251Cr 52Bi208U240ClMn1935,555Fe56, Co 59,Ni 59, Cu 63.Br 80100Ru 104, Rh 104,Pd 106, Ag 108.J 127Os 195, Ir 197,Pt 198, Au 199.*Spaces are left for the unknown elements with atomic masses 44, 68, 72, and 100.FIGURE 8.1 A chronologicalchart of the discovery of the elements. To date, 112 elementshave been identified.120Synthetic elements100Mendeleevs firstperiodic tableNumber of elements8060Elements known prior to 1650:40Ag As Au C Cu FeHg Pb S Sb Sn20016501700175018001850Year discovered190019502000atomic mass must be the basis of periodicity. This property turned out to be associatedwith atomic number, a concept unknown to Mendeleev and his contemporaries.Using data from -scattering experiments (see Section 2.2), Rutherford estimatedthe number of positive charges in the nucleus of a few elements, but the significanceof these numbers was overlooked for several more years. In 1913 a young EnglishBackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website290PERIODIC RELATIONSHIPS AMONG THE ELEMENTSphysicist, Henry Moseley, discovered a correlation between what he called atomicnumber and the frequency of X rays generated by bombarding an element with highenergy electrons. Moseley noticed that the frequencies of X rays emitted from the elements could be correlated by the equationva(Zb)where v is the frequency of the emitted X rays and a and b are constants that are thesame for all the elements. Thus, from the square root of the measured frequency of theX rays emitted, we can determine the atomic number of the element.With a few exceptions, Moseley found that atomic number increases in the sameorder as atomic mass. For example, calcium is the twentieth element in order of increasing atomic mass, and it has an atomic number of 20. The discrepancies that hadpuzzled earlier scientists now made sense. The atomic number of argon is 18 and thatof potassium is 19, so potassium should follow argon in the periodic table.A modern periodic table usually shows the atomic number along with the elementsymbol. As you already know, the atomic number also indicates the number of electrons in the atoms of an element. Electron configurations of elements help to explainthe recurrence of physical and chemical properties. The importance and usefulness ofthe periodic table lie in the fact that we can use our understanding of the general properties and trends within a group or a period to predict with considerable accuracy theproperties of any element, even though that element may be unfamiliar to us.8.2PERIODIC CLASSIFICATION OF THE ELEMENTSFigure 8.2 shows the periodic table together with the outermost ground-state electronconfigurations of the elements. (The electron configurations of the elements are alsogiven in Table 7.3.) Starting with hydrogen, we see that subshells are filled in the order shown in Figure 7.23. According to the type of subshell being filled, the elementscan be divided into categories the representative elements, the noble gases, the transition elements (or transition metals), the lanthanides, and the actinides. The representative elements (also called main group elements) are the elements in Groups 1Athrough 7A, all of which have incompletely filled s or p subshells of the highest principal quantum number. With the exception of helium, the noble gases (the Group 8Aelements) all have a completely filled p subshell. (The electron configurations are 1s2for helium and ns2np6 for the other noble gases, where n is the principal quantum number for the outermost shell.)The transition metals are the elements in Groups 1B and 3B through 8B, whichhave incompletely filled d subshells, or readily produce cations with incompletely filledd subshells. (These metals are sometimes referred to as the d-block transition elements.)The nonsequential numbering of the transition metals in the periodic table (that is, 3B 8B, followed by 1B 2B) acknowledges a correspondence between the outer electronconfigurations of these elements and those of the representative elements. For example, scandium and gallium both have 3 outer electrons. However, because they are indifferent types of atomic orbitals, they are placed in different groups (3A and 3B). TheHenry Gwyn-Jeffreys Moseley (18871915). English physicist. Moseley discovered the relationship between X-ray spectra and atomic number. A lieutenant in the Royal Engineers, he was killed in action at the age of 28 during the British campaign in Gallipoli, Turkey.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website8.2PERIODIC CLASSIFICATION OF THE ELEMENTS11A291188A11H1s122A133A144A155A166A177A2He1s223Li2s14Be2s25B2s22p16C2s22p27N2s22p38O2s22p49F2s22p510Ne2s22p6311Na3s112Mg3s233B44B55B66B77B898B10111B122B13Al3s23p114Si3s23p215P3s23p316S3s23p417Cl3s23p518Ar3s23p6419K4s120Ca4s221Sc4s23d122Ti4s23d223V4s23d 324Cr4s13d525Mn4s23d526Fe4s23d627Co4s23d 728Ni4s23d 829Cu4s13d1030Zn4s23d1031Ga4s24p132Ge4s24p233As4s24p334Se4s24p435Br4s24p536Kr4s24p6537Rb5s138Sr5s239Y5s24d140Zr5s24d241Nb5s14d 442Mo5s14d543Tc5s24d544Ru5s14d 745Rh5s14d 846Pd4d1047Ag5s14d1048Cd5s24d1049In5s25p150Sn5s25p251Sb5s25p352Te5s25p453I5s25p554Xe5s25p6655Cs6s156Ba6s257La6s25d172Hf6s25d273Ta6s25d 374W6s25d475Re6s25d576Os6s25d677Ir6s25d 778Pt6s15d979Au6s15d1080Hg6s25d1081Tl6s26p182Pb6s26p283Bi6s26p384Po6s26p485At6s26p586Rn6s26p6787Fr7s188Ra7s289Ac7s26d1104Rf7s26d2105Ha7s26d 3106Sg7s26d4107Ns7s26d5108Hs7s26d6109Mt7s26d 71101111127s26d87s26d97s26d 1058Ce6s24f 15d159Pr6s24f 360Nd6s24f 461Pm6s24f 562Sm6s24f 663Eu6s24f 764Gd6s24f 75d165Tb6s24f 966Dy6s24f 1067Ho6s24f 1168Er6s24f 1269Tm6s24f 1370Yb6s24f 1471Lu6s24f 145d190Th7s26d291Pa7s25f 26d192U7s25f 36d193Np7s25f 46d194Pu7s25f 695Am7s25f 796Cm7s25f 76d197Bk7s25f 998Cf7s25f 1099Es7s25f 11100Fm7s25f 12101Md7s25f 13102No7s25f 14103Lr7s25f 146d1FIGURE 8.2 The groundstate electron configurations ofthe elements. For simplicity,only the configurations of theouter electrons are shown.TABLE 8.2 ElectronConfigurations of Group 1Aand Group 2A ElementsGROUP 1ALiNaKRbCsFrBackGROUP 2A1[He]2s[Ne]3s1[Ar]4s1[Kr]5s1[Xe]6s1[Rn]7s1ForwardBeMgCaSrBaRa[He]2s2[Ne]3s2[Ar]4s2[Kr]5s2[Xe]6s2[Rn]7s2Main Menumetals iron (Fe), cobalt (Co), and nickel (Ni) do not fit this classification and are allplaced in Group 8B. The Group 2B elements, Zn, Cd, and Hg, are neither representative elements nor transition metals. There is no special name for this group of metals.It should be noted that the designation of A and B groups is not universal. In Europethe practice is to use B for representative elements and A for transition metals, whichis just the opposite of the American convention. The International Union of Pure andApplied Chemistry (IUPAC) has recommended numbering the columns sequentiallywith Arabic numerals 1 through 18 (see Figure 8.2). The proposal has sparked muchcontroversy in the international chemistry community, and its merits and drawbackswill be deliberated for some time to come. In this text we will adhere to the Americandesignation.The lanthanides and actinides are sometimes called f-block transition elements because they have incompletely filled f subshells. Figure 8.3 distinguishes the groups ofelements discussed here.A clear pattern emerges when we examine the electron configurations of the elements in a particular group. The electron configurations for Groups 1A and 2A areshown in Table 8.2. All members of the Group 1A alkali metals have similar outer electron configurations; each has a noble gas core and an ns1 outer electron. Similarly, theGroup 2A alkaline earth metals have a noble gas core and an outer electron configuration of ns2. The outer electrons of an atom, which are the ones involved in chemicalbonding, are often called valence electrons. The similarity of the outer electron configurations (that is, they have the same number and type of valence electrons) is whatmakes the elements in the same group resemble one another in chemical behavior. ThisTOCStudy Guide TOCTextbook WebsiteMHHE Website292PERIODIC RELATIONSHIPS AMONG THE ELEMENTSRepresentativeelements1H34LiBe1112NaMg33BLanthanides133ATransitionmetals22AZincCadiumMercuryNoble gases11AActinides5678910BCNOFNe131415161718AlSiPSClAr44B55B66B77B898B188A10111B122B144A155A166A177A2He192021222324252627282930313233343536KCaScTiVCrMnFeCoNiCuZnGaGeAsSeBrKr373839404142434445464748495051525354RbSrYZrNbMoTcRuRhPdAgCdInSnSbTeIXe555657727374757677787980818283848586CsBaLaHfTaWReOsIrPtAuHgTlPbBiPoAtRn110111112878889104105106107108109FrRaAcRfHaSgNsHsMtFIGURE 8.3 Classificationof the elements. Note thatthe Group 2B elements areoften classified as transitionmetals even though they donot exhibit the characteristics of the transition metals.5859606162636465666768697071CePrNdPmSmEuGdTbDyHoErTmYbLu90919293949596979899100101102103ThPaUNpPuAmCmBkCfEsFmMdNoLrobservation holds true for the other representative elements. Thus, for instance, thehalogens (the Group 7A elements) all have outer electron configurations of ns2np5, andthey have very similar properties. We must be careful, however, in predicting properties for Groups 3A through 7A. For example, the elements in Group 4A all have thesame outer electron configuration, ns2np2, but there is some variation in chemical properties among these elements: Carbon is a nonmetal, silicon and germanium are metalloids, and tin and lead are metals.As a group, the noble gases behave very similarly. With the exception of kryptonand xenon, the rest of these elements are totally inert chemically. The reason is thatthese elements all have completely filled outer ns and np subshells, a condition thatrepresents great stability. Although the outer electron configuration of the transitionmetals is not always the same within a group and there is no regular pattern in thechange of the electron configuration from one metal to the next in the same period, alltransition metals share many characteristics that set them apart from other elements.The reason is that these metals all have an incompletely filled d subshell. Likewise,the lanthanide (and the actinide) elements resemble one another because they have incompletely filled f subshells.EXAMPLE 8.1A neutral atom of a certain element has 15 electrons. Without consulting a periodictable, answer the following questions: (a) What is the ground-state electron configuration of the element? (b) How should the element be classified? (c) Are the atomsof this element diamagnetic or paramagnetic?BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website8.2PERIODIC CLASSIFICATION OF THE ELEMENTS293(a) Using the building-up principle and knowing the maximum capacityof s and p subshells, we can write the ground-state electron configuration of the element as 1s22s22p63s23p3.(b) Since the 3p subshell is not completely filled, this is a representative element.Based on the information given, we cannot say whether it is a metal, a nonmetal,or a metalloid.(c) According to Hunds rule, the three electrons in the 3p orbitals have parallelspins. Therefore, the atoms of this element are paramagnetic, with three unpairedspins. (Remember, we saw in Chapter 7 that any atom that contains an odd numberof electrons must be paramagnetic.)AnswerSimilar problem: 8.20.PRACTICE EXERCISEA neutral atom of a certain element has 20 electrons. (a) Write the ground-state electron configuration of the element, (b) classify the element, and (c) determine whetherthe atoms of this element are diamagnetic or paramagnetic.REPRESENTING FREE ELEMENTS IN CHEMICAL EQUATIONSHaving classified the elements according to their ground-state electron configurations,we can now look at the way chemists represent metals, metalloids, and nonmetals asfree elements in chemical equations. Because metals do not exist in discrete molecular units, we always use their empirical formulas in chemical equations. The empiricalformulas are the same as the symbols that represent the elements. For example, theempirical formula for iron is Fe, the same as the symbol for the element.For nonmetals there is no single rule. Carbon, for example, exists as an extensivethree-dimensional network of atoms, and so we use its empirical formula (C) to represent elemental carbon in chemical equations. But hydrogen, nitrogen, oxygen, andthe halogens exist as diatomic molecules, and so we use their molecular formulas (H2,N2, O2, F2, Cl2, Br2, I2) in equations. The stable form of phosphorus is molecular (P4),and so we use P4. For sulfur chemists often use the empirical formula (S) in chemicalequations, rather than S8 which is the stable form. Thus, instead of writing the equation for the combustion of sulfur asS8(s)Note that these two equations forthe combustion of sulfur have identical stoichiometry. This correspondence should not be surprising,since both equations describe thesame chemical system. In both casesa number of sulfur atoms react withtwice as many oxygen atoms.8O2(g) 88n 8SO2(g)we usually writeS(s)O2(g) 88n SO2(g)All the noble gases are monatomic species; thus we use their symbols: He, Ne, Ar, Kr,Xe, and Rn. The metalloids, like the metals, all have complex three-dimensional networks, and we represent them, too, with their empirical formulas, that is, their symbols: B, Si, Ge, and so on.ELECTRON CONFIGURATIONS OF CATIONS AND ANIONSBecause many ionic compounds are made up of monatomic anions and cations, it ishelpful to know how to write the electron configurations of these ionic species. Just asfor neutral atoms, we use the Pauli exclusion principle and Hunds rule in writing theground-state electron configurations of cations and anions. We will group the ions intwo categories for discussion.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website294PERIODIC RELATIONSHIPS AMONG THE ELEMENTSIons Derived from Representative ElementsIons formed from neutral atoms of most representative elements have the noble-gasouter-electron configuration of ns2np6. In the formation of a cation from the neutralatom of a representative element, one or more electrons are removed from the highestoccupied n shell. Following are the electron configurations of some neutral atoms andtheir corresponding cations:Na: [Ne]3s1Ca: [Ar]4s2Al: [Ne]3s23p1Na : [Ne]Ca2 : [Ar]Al3 : [Ne]Note that each ion has a stable noble gas configuration.In the formation of an anion, one or more electrons are added to the highest partially filled n shell. Consider the following examples:H:F:O:N:1s11s22s22p51s22s22p41s22s22p3HFO2N3::::1s2 or [He]1s22s22p6 or [Ne]1s22s22p6 or [Ne]1s22s22p6 or [Ne]All of these anions also have stable noble gas configurations. Notice that F , Na , andNe (and Al3 , O2 , and N3 ) have the same electron configuration. They are said tobe isoelectronic because they have the same number of electrons, and hence the sameground-state electron configuration. Thus H and He are also isoelectronic.Cations Derived from Transition MetalsBear in mind that the order of electron filling does not determine orpredict the order of electron removalfor transition metals.8.3In Section 7.10 we saw that in the first-row transition metals (Sc to Cu), the 4s orbitalis always filled before the 3d orbitals. Consider manganese, whose electron configuration is [Ar]4s23d 5. When the Mn2 ion is formed, we might expect the two electronsto be removed from the 3d orbitals to yield [Ar]4s23d 3. In fact, the electron configuration of Mn2 is [Ar]3d 5! The reason is that the electron-electron and electronnucleus interactions in a neutral atom can be quite different from those in its ion. Thus,whereas the 4s orbital is always filled before the 3d orbital in Mn, electrons are removed from the 4s orbital in forming Mn2 because the 3d orbital is more stable thanthe 4s orbital in transition metal ions. Therefore, when a cation is formed from an atomof a transition metal, electrons are always removed first from the ns orbital and thenfrom the (n 1)d orbitals.Keep in mind that most transition metals can form more than one cation and thatfrequently the cations are not isoelectronic with the preceding noble gases.PERIODIC VARIATION IN PHYSICAL PROPERTIESAs we have seen, the electron configurations of the elements show a periodic variationwith increasing atomic number. Consequently, there are also periodic variations in physical and chemical behavior. In this section and the next two, we will examine somephysical properties of elements that are in the same group or period and additionalproperties that influence the chemical behavior of the elements. First, lets look at theconcept of effective nuclear charge, which has a direct bearing on atomic size and onthe tendency for ionization.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website8.3PERIODIC VARIATION IN PHYSICAL PROPERTIES295EFFECTIVE NUCLEAR CHARGEIn Chapter 7, we discussed the shielding effect that electrons close to the nucleus haveon outer-shell electrons in many-electron atoms. The presence of shielding electronsreduces the electrostatic attraction between the positively charged protons in the nucleus and the outer electrons. Moreover, the repulsive forces between electrons in amany-electron atom further offset the attractive force exerted by the nucleus. The concept of effective nuclear charge allows us to account for the effects of shielding on periodic properties.Consider, for example, the helium atom, which has the ground-state electron configuration 1s2. Heliums two protons give the nucleus a charge of 2, but the full attractive force of this charge on the two 1s electrons is partially offset by electron-electron repulsion. Consequently we say that the 1s electrons shield each other from thenucleus. The effective nuclear charge (Zeff), which is the charge felt by an electron, isgiven byZeffZwhere Z is the actual nuclear charge (that is, the atomic number of the element) and(sigma) is called the shielding constant (also called the screening constant). The shielding constant is greater than zero but smaller than Z.One way to illustrate electron shielding is to consider the amounts of energy required to remove the two electrons from a helium atom. Measurements show that ittakes 2373 kJ of energy to remove the first electron from 1 mole of He atoms and5251 kJ of energy to remove the remaining electron from 1 mole of He ions. The reason it takes so much more energy to remove the second electron is that with only oneelectron present, there is no shielding, and the electron feels the full effect of the 2nuclear charge.For atoms with three or more electrons, the electrons in a given shell are shieldedby electrons in inner shells (that is, shells closer to the nucleus) but not by electronsin outer shells. Thus, in a neutral lithium atom, whose electron configuration is 1s22s1,the 2s electron is shielded by the two 1s electrons, but the 2s electron does not have ashielding effect on the 1s electrons. In addition, filled inner shells shield outer electrons more effectively than electrons in the same subshell shield each other.(a)ATOMIC RADIUS(b)FIGURE 8.4 (a) In metals suchas beryllium, the atomic radius isdefined as one-half the distancebetween the centers of two adjacent atoms. (b) For elements thatexist as diatomic molecules, suchas iodine, the radius of the atomis defined as one-half the distance between the centers of theatoms in the molecule.BackForwardMain MenuA number of physical properties, including density, melting point, and boiling point,are related to the sizes of atoms, but atomic size is difficult to define. As we saw inChapter 7, the electron density in an atom extends far beyond the nucleus, but we normally think of atomic size as the volume containing about 90 percent of the total electron density around the nucleus. When we must be even more specific, we define thesize of an atom in terms of its atomic radius, which is one-half the distance betweenthe two nuclei in two adjacent metal atoms.For atoms linked together to form an extensive three-dimensional network, atomicradius is simply one-half the distance between the nuclei in two neighboring atoms[Figure 8.4(a)]. For elements that exist as simple diatomic molecules, the atomic radius is one-half the distance between the nuclei of the two atoms in a particular molecule [Figure 8.4(b)].Figure 8.5 shows the atomic radii of many elements according to their positionsin the periodic table, and Figure 8.6 plots the atomic radii of these elements againstTOCStudy Guide TOCTextbook WebsiteMHHE Website296PERIODIC RELATIONSHIPS AMONG THE ELEMENTSFIGURE 8.5 Atomic radii (inpicometers) of representative elements according to their positionsin the periodic table. Note thatthere is no general agreement onthe size of atomic radii. We focus only on the trends in atomicradii, not on their precise values.Increasing atomic radius1A2A3A4A5A6A7A8AHHe3250BCNOFNe112989192737270NaMgAlSiPSClAr1861601431321281279998KCaGaGeAsSeBrKr227197135137139140114112RbSrInSnSbTeIXe248215166162159160133131CsBaTlPbBiPoAtRn265222171175170164142140Be152Increasing atomic radiusLitheir atomic numbers. Periodic trends are clearly evident. In studying the trends, bearin mind that the atomic radius is determined to a large extent by the strength of the attraction between the outer-shell electrons and the nucleus. The larger the effective nuclear charge, the stronger the hold of the nucleus on these electrons, and the smallerthe atomic radius. Consider the second-period elements from Li to F, for example.Moving from left to right, we find that the number of electrons in the inner shell (1s2)remains constant while the nuclear charge increases. The electrons that are added tocounterbalance the increasing nuclear charge are ineffective in shielding one another.Consequently, the effective nuclear charge increases steadily while the principal quantum number remains constant (n 2). For example, the outer 2s electron in lithium isshielded from the nucleus (which has 3 protons) by the two 1s electrons. As an approximation, we assume that the shielding effect of the two 1s electrons is to canceltwo positive charges in the nucleus. Thus the 2s electron only feels the attraction ofone proton in the nucleus; the effective nuclear charge is 1. In beryllium (1s22s2),each of the 2s electrons is shielded by the inner two 1s electrons, which cancel two ofthe four positive charges in the nucleus. Because the 2s electrons do not shield eachother as effectively, the net result is that the effective nuclear charge of each 2s electron is greater than 1. Thus as the effective nuclear charge increases, the atomic radius decreases steadily from lithium to fluorine.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website8.3FIGURE 8.6 Plot of atomicradii (in picometers) of elementsagainst their atomic numbers.297PERIODIC VARIATION IN PHYSICAL PROPERTIES300CsRb250KAtomic radius (pm)200NaLiPo150IBr100ClF5001020304050Atomic number60708090Within a group of elements we find that atomic radius increases with increasingatomic number. For the alkali metals in Group 1A, the outermost electron resides inthe ns orbital. Since orbital size increases with the increasing principal quantum number n, the size of the metal atoms increases from Li to Cs. We can apply the same reasoning to the elements in other groups.EXAMPLE 8.2Referring to a periodic table, arrange the following atoms in order of increasing radius: P, Si, N.Note that N and P are in the same group (Group 5A) and that N is aboveP. Therefore, the radius of N is smaller than that of P (atomic radius increases aswe go down a group). Both Si and P are in the third period, and Si is to the left ofP. Therefore, the radius of P is smaller than that of Si (atomic radius decreases aswe move from left to right across a period). Thus the order of increasing radius isN P Si.AnswerSimilar problems: 8.37, 8.38.PRACTICE EXERCISEArrange the following atoms in order of decreasing radius: C, Li, Be.IONIC RADIUSIonic radius is the radius of a cation or an anion. It can be measured by X-ray diffraction (see Chapter 11). Ionic radius affects the physical and chemical properties ofBackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website298PERIODIC RELATIONSHIPS AMONG THE ELEMENTSFIGURE 8.7 Comparison ofatomic radii with ionic radii.(a) Alkali metals and alkali metalcations. (b) Halogens and halideions.300300CsRb250250KI200NaCl LiCs+150K+Rb+Radius (pm)Radius (pm)200Br 100150FIBr100ClNa+50010FLi+5020 30 40 50Atomic number(a)6001020 30 40 50Atomic number(b)60an ionic compound. For example, the three-dimensional structure of an ionic compounddepends on the relative sizes of its cations and anions.When a neutral atom is converted to an ion, we expect a change in size. If theatom forms an anion, its size (or radius) increases, since the nuclear charge remainsthe same but the repulsion resulting from the additional electron(s) enlarges the domain of the electron cloud. On the other hand, removing one or more electrons froman atom reduces electron-electron repulsion but the nuclear charge remains the same,so the electron cloud shrinks, and the cation is smaller than the atom. Figure 8.7 showsthe changes in size that result when alkali metals are converted to cations and halogens are converted to anions; Figure 8.8 shows the changes in size that occur when alithium atom reacts with a fluorine atom to form a LiF unit.Figure 8.9 shows the radii of ions derived from the familiar elements, arrangedaccording to elements positions in the periodic table. We can see parallel trends between atomic radii and ionic radii. For example, from top to bottom both the atomicradius and the ionic radius increase within a group. For ions derived from elements indifferent groups, a size comparison is meaningful only if the ions are isoelectronic. Ifwe examine isoelectronic ions, we find that cations are smaller than anions. For example, Na is smaller than F . Both ions have the same number of electrons, but NaFIGURE 8.8 Changes in thesizes of Li and F when they reactto form LiF.+LiBackForwardMain MenuFTOCLi +Study Guide TOCFTextbook WebsiteMHHE Website8.3Li+PERIODIC VARIATION IN PHYSICAL PROPERTIES299Be2+N365K+Ca2+13399140136S2Cl184Mg2+95171181Se2Br195Te2I31Na+F19860O2Al3+Fe3+Ti4+V5+Sc3+Rb+148Fe2+Cr3+50Ni2+Mn2+Co2+Zn2+Ga3+Cu+816859 6480Sr2+607772699662Sb5+In3+ Sn4+Ag+12611374Cd2+978171Pb4+Cs+Ba2+169135Au+Hg2+Tl3+622212161371109584FIGURE 8.9 The radii (in picometers) of ions of familiar elements arranged according to the elements positions in the periodic table.(Z 11) has more protons than F (Z 9). The larger effective nuclear charge of Naresults in a smaller radius.Focusing on isoelectronic cations, we see that the radii of tripositive ions (ionsthat bear three positive charges) are smaller than those of dipositive ions (ions that beartwo positive charges), which in turn are smaller than unipositive ions (ions that bearone positive charge). This trend is nicely illustrated by the sizes of three isoelectronicions in the third period: Al3 , Mg2 , and Na (see Figure 8.9). The Al3 ion has thesame number of electrons as Mg2 , but it has one more proton. Thus the electron cloudin Al3 is pulled inward more than that in Mg2 . The smaller radius of Mg2 compared with that of Na can be similarly explained. Turning to isoelectronic anions, wefind that the radius increases as we go from ions with uninegative charge ( ) to thosewith dinegative charge (2 ), and so on. Thus the oxide ion is larger than the fluorideion because oxygen has one fewer proton than fluorine; the electron cloud is spreadout more in O2 .EXAMPLE 8.3For each of the following pairs, indicate which one of the two species is larger:(a) N3 or F ; (b) Mg2 or Ca2 ; (c) Fe2 or Fe3 .(a) N3 and F are isoelectronic anions. Since N3 has only seven protons and F has nine, N3 is larger.(b) Both Mg and Ca belong to Group 2A (the alkaline metals). The Ca2 ion islarger than Mg2 because Cas valence electrons are in a larger shell (n 4) thanare Mgs (n 3).AnswerBackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website300PERIODIC RELATIONSHIPS AMONG THE ELEMENTSFIGURE 8.10 The third-periodelements. The photograph of argon, which is a colorless, odorless gas, shows the color emittedby the gas from a dischargetube.Magnesium (Mg)Sodium (Na)Silicon (Si)Aluminum (Al)Phosphorous (P)Chlorine (Cl2)Sulfur (S)Similar problems: 8.43, 8.45.Argon (Ar)(c) Both ions have the same nuclear charge, but Fe2 has one more electron andhence greater electron-electron repulsion. The radius of Fe2 is larger.PRACTICE EXERCISESelect the smaller ion in each of the following pairs: (a) K , Li ; (b) Au , Au3 ;(c) P3 , N3 .VARIATION OF PHYSICAL PROPERTIES ACROSS A PERIOD AND WITHIN A GROUPFrom left to right across a period there is a transition from metals to metalloids to nonmetals. Consider the third-period elements from sodium to argon (Figure 8.10). Sodium,the first element in the third period, is a very reactive metal, whereas chlorine, the sec-BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website8.4IONIZATION ENERGY301ond-to-last element of that period, is a very reactive nonmetal. In between, the elements show a gradual transition from metallic properties to nonmetallic properties.Sodium, magnesium, and aluminum all have extensive three-dimensional atomic networks, which are held together by forces characteristic of the metallic state. Silicon isa metalloid; it has a giant three-dimensional structure in which the Si atoms are heldtogether very strongly. Starting with phosphorus, the elements exist in simple, discretemolecular units (P4, S8, Cl2, and Ar) which have low melting points and boiling points.Within a periodic group the physical properties vary more predictably, especiallyif the elements are in the same physical state. For example, the melting points of argon and xenon are 189.2C and 111.9C, respectively. We can estimate the melting point of the intermediate element krypton by taking the average of these two values as follows:melting point of Kr[( 189.2C)( 111.9C)]2150.6CThis value is quite close to the actual melting point of 156.6C.The Chemistry in Action essay on p. 302 illustrates one interesting application ofperiodic group properties.8.4IONIZATION ENERGYNot only is there a correlation between electron configuration and physical properties,but a close correlation also exists between electron configuration (a microscopic property) and chemical behavior (a macroscopic property). As we will see throughout thisbook, the chemical properties of any atom are determined by the configuration of theatoms valence electrons. The stability of these outermost electrons is reflected directlyin the atoms ionization energies. Ionization energy is the minimum energy (in kJ/mol)required to remove an electron from a gaseous atom in its ground state. In other words,ionization energy is the amount of energy in kilojoules needed to strip one mole ofelectrons from one mole of gaseous atoms. Gaseous atoms are specified in this definition because an atom in the gas phase is virtually uninfluenced by its neighbors andso there are no intermolecular forces (that is, forces between molecules) to take intoaccount when measuring ionization energy.The magnitude of ionization energy is a measure of how tightly the electron isheld in the atom. The higher the ionization energy, the more difficult it is to removethe electron. For a many-electron atom, the amount of energy required to remove thefirst electron from the atom in its ground state,energyX(g) 88n X (g)e(8.1)is called the first ionization energy (I1). In equation (8.1), X represents an atom of anyelement, e is an electron, and g denotes the gaseous state. The second ionization energy (I2) and the third ionization energy (I3) are shown in the following equations:energyenergyX (g) 88n X2 (g)23X (g) 88n X (g)esecond ionizationethird ionizationThe pattern continues for the removal of subsequent electrons.When an electron is removed from a neutral atom, the repulsion among the remaining electrons decreases. Since the nuclear charge remains constant, more energyBackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website302PERIODIC RELATIONSHIPS AMONG THE ELEMENTSBackThe Third Liquid Element?Of the 112 known elements, 11 are gases under atmospheric conditions. Six of these are the Group 8Aelements (the noble gases He, Ne, Ar, Kr, Xe, andRn), and the other five are hydrogen (H2), nitrogen(N2), oxygen (O2), fluorine (F2), and chlorine (Cl2).Curiously, only two elements are liquids at 25C: mercury (Hg) and bromine (Br2).We do not know the properties of all the knownelements because some of them have never been prepared in quantities large enough for investigation. Inthese cases we must rely on periodic trends to predicttheir properties. What are the chances, then, of discovering a third liquid element?Let us look at francium (Fr), the last member ofGroup 1A, to see if it might be a liquid at 25C. Allof franciums isotopes are radioactive. The most stable isotope is francium-223, which has a half-life of21 minutes. (Half-life is the time it takes for one-halfof the nuclei in any given amount of a radioactive sub-stance to disintegrate.) This short half-life means thatonly very small traces of francium could possibly exist on Earth. And although it is feasible to preparefrancium in the laboratory, no weighable quantity ofthe element has been prepared or isolated. Thus weknow very little about franciums physical and chemical properties. Yet we can use the group periodictrends to predict some of those properties.Take franciums melting point as an example. Theplot shows how the melting points of the alkali metalsvary with atomic number. From lithium to sodium, themelting point drops 81.4; from sodium to potassium,34.6; from potassium to rubidium, 24; from rubidium to cesium, 11. On the basis of this trend, we canpredict that the change from cesium to francium wouldbe about 5. If so, the melting point of francium wouldbe 23C, which would make it a liquid under atmospheric conditions.180Li150Melting point (C)Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in ActionChemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry12090NaK60RbCs30406080100Atomic numberA plot of the melting points of the alkali metals versustheir atomic numbers. By extrapolation, the melting pointof francium should be 23C.0False-colored image of francium-210 atoms created bybombarding gold with oxygen and trapped by laserbeams. The central spot is 1 mm in diameter and consists of about 10,000 atoms. The yellow light is the fluorescence of the Fr atoms induced by laser. The otherbright areas come from laser light scattered off the glasssurfaces.ForwardMain MenuTOCFrStudy Guide TOC20Textbook WebsiteMHHE Website8.4TABLE 8.3IONIZATION ENERGY303The Ionization Energies (kJ/mol) of the First 20 ElementsZELEMENTFIRSTSECONDTHIRDFOURTHFIFTHSIXTH1234567891011121314151617181920HHeLiBeBCNOFNeNaMgAlSiPSClArKCa1312237352089980110861400131416802080495.9738.1577.9786.31012999.512511521418.7589.5525173001757243023502860339033703950456014501820158019042250229726663052114511815148503660462045805300605061206900773027503230291033603820390044104900210052500062207500747084009370954010500116004360496046605160577059006500328203800094001100011000122001340013600148001600062406990654072408000810047261530001300015200150001660018000184002000021000850093008800960011000is needed to remove another electron from the positively charged ion. Thus, ionizationenergies always increase in the following order:I1I2I3...Table 8.3 lists the ionization energies of the first 20 elements. Ionization is always anendothermic process. By convention, energy absorbed by atoms (or ions) in the ionization process has a positive value. Thus ionization energies are all positive quantities. Figure 8.11 shows the variation of the first ionization energy with atomic number. The plot clearly exhibits the periodicity in the stability of the most loosely heldelectron. Note that, apart from small irregularities, the first ionization energies of elements in a period increase with increasing atomic number. This trend is due to the increase in effective nuclear charge from left to right (as in the case of atomic radii variation). A larger effective nuclear charge means a more tightly held outer electron, andhence a higher first ionization energy. A notable feature of Figure 8.11 is the peaks,which correspond to the noble gases. The high ionization energies of the noble gases,stemming from their stable ground-state electron configurations, account for the factthat most of them are chemically unreactive. In fact, helium (1s2) has the highest firstionization energy of all the elements.At the bottom of the graph in Figure 8.11 are the Group 1A elements (the alkalimetals) which have the lowest first ionization energies. Each of these metals has onevalence electron (the outermost electron configuration is ns1) which is effectivelyshielded by the completely filled inner shells. Consequently, it is energetically easy toremove an electron from the atom of an alkali metal to form a unipositive ion (Li ,Na , K , . . .). Significantly, the electron configurations of these cations are isoelectronic with those noble gases just preceding them in the periodic table.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE WebsitePERIODIC RELATIONSHIPS AMONG THE ELEMENTSFIGURE 8.11 Variation of thefirst ionization energy withatomic number. Note that the noble gases have high ionizationenergies, whereas the alkali metals and alkaline earth metalshave low ionization energies.2500HeNeFirst ionization energy (kJ/mol)3042000ArKr1500XeHRn1000500Li0Na10RbK2030Cs4050Atomic number (Z )60708090The group 2A elements (the alkaline earth metals) have higher first ionization energies than the alkali metals do. The alkaline earth metals have two valence electrons(the outermost electron configuration is ns2). Because these two s electrons do notshield each other well, the effective nuclear charge for an alkaline earth metal atom islarger than that for the preceding alkali metal. Most alkaline earth compounds containdipositive ions (Mg2 , Ca2 , Sr2 , Ba2 ). The Be2 ion is isoelectronic with Li andwith He, Mg2 is isoelectronic with Na and with Ne, and so on.As Figure 8.11 shows, metals have relatively low ionization energies compared tononmetals. The ionization energies of the metalloids generally fall between those ofmetals and nonmetals. The difference in ionization energies suggests why metals always form cations and nonmetals form anions in ionic compounds. (The only important nonmetallic cation is the ammonium ion, NH4 .) For a given group, ionization energy decreases with increasing atomic number (that is, as we move down the group).Elements in the same group have similar outer electron configurations. However, asthe principal quantum number n increases, so does the average distance of a valenceelectron from the nucleus. A greater separation between the electron and the nucleusmeans a weaker attraction, so that it becomes increasingly easier to remove the firstelectron as we go from element to element down a group. Thus the metallic characterof the elements within a group increases from top to bottom. This trend is particularlynoticeable for elements in Groups 3A to 7A. For example, in Group 4A, carbon is anonmetal, silicon and germanium are metalloids, and tin and lead are metals.Although the general trend in the periodic table is for first ionization energies toincrease from left to right, some irregularities do exist. The first exception occurs between Group 2A and 3A elements in the same period (for example, between Be andB and between Mg and Al).The Group 3A elements have lower first ionization energies than 2A elements because they all have a single electron in the outermost p subshell (ns2np1), which is well shielded by the inner electrons and the ns2 electrons.Therefore, less energy is needed to remove a single p electron than to remove a paireds electron from the same principal energy level. The second irregularity occurs betweenBackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website8.5ELECTRON AFFINITY305Groups 5A and 6A (for example, between N and O and between P and S). In the Group5A elements (ns2np3) the p electrons are in three separate orbitals according to Hundsrule. In Group 6A (ns2np4) the additional electron must be paired with one of the threep electrons. The proximity of two electrons in the same orbital results in greater electrostatic repulsion, which makes it easier to ionize an atom of the Group 6A element,even though the nuclear charge has increased by one unit. Thus the ionization energiesfor Group 6A elements are lower than those for Group 5A elements in the same period.The following example compares the ionization energies of some elements.EXAMPLE 8.4(a) Which atom should have a smaller first ionization energy: oxygen or sulfur?(b) Which atom should have a higher second ionization energy: lithium or beryllium?(a) Oxygen and sulfur are members of Group 6A. They have the same valence electron configuration (ns2np4), but the 3p electron in sulfur is farther fromthe nucleus and experiences less nuclear attraction than the 2p electron in oxygen.Thus, following the general rule that the ionization energy of elements decreases aswe move down a periodic group, we predict that sulfur should have a smaller firstionization energy. Table 8.3 confirms our reasoning.(b) The electron configurations of Li and Be are 1s22s1 and 1s22s2, respectively. Forthe second ionization process we writeAnswerLi (g) 88n Li2 (g)1s21s1Be (g) 88n Be2 (g)1s22s11s2Similar problem: 8.53.Since 1s electrons shield 2s electrons much more effectively than they shield eachother, we predict that it should be much easier to remove a 2s electron from Bethan to remove a 1s electron from Li . This is consistent with the data in Table 8.3.PRACTICE EXERCISE(a) Which of the following atoms should have a larger first ionization energy: N orP? (b) Which of the following atoms should have a smaller second ionization energy: Na or Mg?8.5ELECTRON AFFINITYAnother property that greatly influences the chemical behavior of atoms is their ability to accept one or more electrons. This property is called electron affinity, which isthe energy change that occurs when an electron is accepted by an atom in the gaseousstate to form an anion. The equation isX(g)e 88n X (g)(8.2)The sign of the electron affinity is opposite to the one we use for ionization energy. Aswe saw in Section 8.5, a positive ionization energy means that energy must be suppliedto remove an electron. A positive electron affinity, on the other hand, signifies that en-BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website306PERIODIC RELATIONSHIPS AMONG THE ELEMENTSTABLE 8.4 Electron Affinities (kJ/mol) of Some RepresentativeElements and the Noble Gases*1AH73Li60Na53K48Rb47Cs452ABe0Mg0Ca2.4Sr4.7Ba143AB27Al44Ga29In29Tl304A5AC122Si134Ge118Sn121Pb110N0P72As77Sb101Bi1106A7AF328Cl349Br325I295At?O141S200Se195Te190Po?8AHe0Ne0Ar0Kr0Xe0Rn0*The electron affinities of the noble gases, Be, and Mg have not been determined experimentally,but are believed to be close to zero or negative.ergy is liberated when an electron is added to an atom. To clarify this apparent paradox, let us consider the process in which a gaseous fluorine atom accepts an electron:F(g)e 88n F (g)H328 kJ/molThe sign of the enthalpy change indicates that this is an exothermic process; however,the electron affinity of fluorine is assigned a value of 328 kJ/mol. Thus we can thinkof electron affinity as the energy that must be supplied to remove an electron from anegative ion. For the removal of an electron from a fluoride ion, we haveF (g) 88n F(g)eH328 kJ/molTwo features of electron affinity to remember are: (1) The electron affinity of an element is equal to the enthalpy change that accompanies the ionization process of its anion, and (2) a large positive electron affinity means that the negative ion is very stable(that is, the atom has a great tendency to accept an electron), just as a high ionizationenergy of an atom means that the atom is very stable.Experimentally, electron affinity is determined by removing the additional electronfrom an anion. In contrast to ionization energies, however, electron affinities are difficult to measure because the anions of many elements are unstable. Table 8.4 shows theelectron affinities of some representative elements and the noble gases, and Figure 8.12plots the electron affinities of the first 20 elements versus atomic number. The overalltrend is an increase in the tendency to accept electrons (electron affinity values becomemore positive) from left to right across a period. The electron affinities of metals aregenerally lower than those of nonmetals. The values vary little within a given group.The halogens (Group 7A) have the highest electron affinity values. This is not surprising when we realize that by accepting an electron, each halogen atom assumes the stable electron configuration of the noble gas immediately to its right. For example, theelectron configuration of F is 1s22s22p6, or [Ne]; for Cl it is [Ne]3s23p6 or [Ar]; andso on. Calculations show that the noble gases all have electron affinities of less thanzero. Thus the anions of these gases, if formed, would be inherently unstable.The electron affinity of oxygen has a positive value (141 kJ/mol), which meansthat the processBackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website8.5FIGURE 8.12 A plot of electron affinity against atomic number for the first 20 elements.307ELECTRON AFFINITY400ClFBrIElectron affinity (kJ/mol)300200SiSnGeC100LiNa010K20Rb30Atomic number (Z )O(g)Cs4050e 88n O (g)60H141 kJis favorable (exothermic). On the other hand, the electron affinity of the Ohighly negative ( 780 kJ/mol), which means the processO (g)e 88n O2 (g)Hion is780 kJis endothermic even though the O2 ion is isoelectronic with the noble gas Ne. Thisprocess is unfavorable in the gas phase because the resulting increase in electron-electron repulsion outweighs the stability gained by achieving a noble gas configuration.However, note that O2 is common in ionic compounds (for example, Li2O and MgO);in solids, the O2 ion is stabilized by the neighboring cations. We will study the stability of ionic compounds in the next chapter.The following example shows why the alkaline earth metals do not have a greattendency to accept electrons.EXAMPLE 8.5Why are the electron affinities of the alkaline earth metals, shown in Table 8.4, either negative or small positive values?AnswerThe valence configuration of the alkaline earth metals is ns2. For theprocessM(g)e 88n M (g)where M denotes a member of the Group 2A family, the extra electron must enterthe np subshell, which is effectively shielded by the two ns electrons (the np elec-BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website308PERIODIC RELATIONSHIPS AMONG THE ELEMENTSSimilar problem: 8.61.trons are farther away from the nucleus than the ns electrons) and the inner electrons. Consequently, alkaline earth metals have little tendency to pick up an extraelectron.PRACTICE EXERCISEIs it likely that Ar will form the anion Ar ?8.6VARIATION IN CHEMICAL PROPERTIES OF THE REPRESENTATIVE ELEMENTSIonization energy and electron affinity help chemists understand the types of reactionsthat elements undergo and the nature of the elements compounds. On a conceptuallevel these two measures are related in a simple way: Ionization energy indexes the attraction of an atom for its own electrons, whereas electron affinity expresses the attraction of an atom for an additional electron from some other source. Together theygive us insight into the general attraction of an atom for electrons. With these conceptswe can survey the chemical behavior of the elements systematically, paying particularattention to the relationship between chemical properties and electron configuration.We have seen that the metallic character of the elements decreases from left toright across a period and increases from top to bottom within a group. On the basis ofthese trends and the knowledge that metals usually have low ionization energies whilenonmetals usually have high electron affinities, we can frequently predict the outcomeof a reaction involving some of these elements.GENERAL TRENDS IN CHEMICAL PROPERTIES1A2A3A4ALiBeBCNaMgAlSiFIGURE 8.13 Diagonal relationships in the periodic table.BackForwardBefore we study the elements in individual groups, let us look at some overall trends.We have said that elements in the same group resemble one another in chemical behavior because they have similar outer electron configurations. This statement, althoughcorrect in the general sense, must be applied with caution. Chemists have long knownthat the first member of each group (the element in the second period from lithium tofluorine) differs from the rest of the members of the same group. Lithium, for example, exhibits many, but not all, of the properties characteristic of the alkali metals.Similarly, beryllium is a somewhat atypical member of Group 2A, and so on. The difference can be attributed to the unusually small size of the first element in each group(see Figure 8.5).Another trend in the chemical behavior of the representative elements is the diagonal relationship. Diagonal relationships are similarities between pairs of elementsin different groups and periods of the periodic table. Specifically, the first three members of the second period (Li, Be, and B) exhibit many similarities to those elementslocated diagonally below them in the periodic table (Figure 8.13). The reason for thisphenomenon is the closeness of the charge densities of their cations. (Charge densityis the charge of an ion divided by its volume.) Elements with comparable charge densities react similarly with anions and therefore form the same type of compounds. Thusthe chemistry of lithium resembles that of magnesium in some ways; the same holdsfor beryllium and aluminum and for boron and silicon. Each of these pairs is said toexhibit a diagonal relationship. We will see a number of examples of this relationshiplater.Main MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website8.6VARIATION IN CHEMICAL PROPERTIES OF THE REPRESENTATIVE ELEMENTS309Bear in mind that a comparison of the properties of elements in the same groupis most valid if we are dealing with elements of the same type with respect to theirmetallic character. This guideline applies to the elements in Groups 1A and 2A, whichare all metals, and to the elements in Groups 7A and 8A, which are all nonmetals. InGroups 3A through 6A, where the elements change either from nonmetals to metalsor from nonmetals to metalloids, it is natural to expect greater variation in chemicalproperties even though the members of the same group have similar outer electron configurations.Now let us take a closer look at the chemical properties of the representative elements and the noble gases. (We will consider the chemistry of the transition metalsin Chapter 22.)Hydrogen (1s1)There is no totally suitable position for hydrogen in the periodic table. Traditionallyhydrogen is shown in Group 1A, but it really could be a class by itself. Like the alkalimetals, it has a single s valence electron and forms a unipositive ion (H ), which ishydrated in solution. On the other hand, hydrogen also forms the hydride ion (H ) inionic compounds such as NaH and CaH2. In this respect, hydrogen resembles the halogens, all of which form uninegative ions (F , Cl , Br , and I ) in ionic compounds.Ionic hydrides react with water to produce hydrogen gas and the corresponding metalhydroxides:2NaH(s)2H2O(l) 88n 2NaOH(aq)CaH2(s)2H2O(l) 88n Ca(OH)2(s)H2(g)2H2(g)Of course, the most important compound of hydrogen is water, which forms when hydrogen burns in air:2H2(g)LiNaKGroup 1A Elements (ns1, nO2(g) 88n 2H2O(l)2)Figure 8.14 shows the Group 1A elements, the alkali metals. All of these elements havelow ionization energies and therefore a great tendency to lose the single valence electron. In fact, in the vast majority of their compounds they are unipositive ions. Thesemetals are so reactive that they are never found in the pure state in nature. They reactwith water to produce hydrogen gas and the corresponding metal hydroxide:Rb2M(s)Cs2H2O(l) 88n 2MOH(aq)H2(g)where M denotes an alkali metal. When exposed to air, they gradually lose their shinyappearance as they combine with oxygen gas to form oxides. Lithium forms lithiumoxide (containing the O2 ion):4Li(s)O2(g) 88n 2Li2O(s)The other alkali metals all form oxides and peroxides (containing the O2 ion). For2example,2Na(s)O2(g) 88n Na2O2(s)Potassium, rubidium, and cesium also form superoxides (containing the O2 ion):K(s)BackForwardMain MenuTOCO2(g) 88n KO2(s)Study Guide TOCTextbook WebsiteMHHE Website310PERIODIC RELATIONSHIPS AMONG THE ELEMENTSFIGURE 8.14 The Group 1Aelements: the alkali metals.Francium (not shown) is radioactive.Lithium (Li)Rubidium (Rb)Potassium (K)Sodium (Na)Cesium (Cs)The reason that different types of oxides are formed when alkali metals react with oxygen has to do with the stability of the oxides in the solid state. Since these oxides areall ionic compounds, their stability depends on how strongly the cations and anions attract one another. Lithium tends to form predominantly lithium oxide because this compound is more stable than lithium peroxide. The formation of other alkali metal oxidescan be explained similarly.BeMgCaSrBaGroup 2A Elements (ns2, n2)Figure 8.15 shows the Group 2A elements. As a group, the alkaline earth metals aresomewhat less reactive than the alkali metals. Both the first and the second ionizationenergies decrease from beryllium to barium. Thus the tendency is to form M2 ions(where M denotes an alkaline earth metal atom), and hence the metallic character increases from top to bottom. Most beryllium compounds (BeH2 and beryllium halides,such as BeCl2) and some magnesium compounds (MgH2, for example) are molecularrather than ionic in nature.The reactivities of alkaline earth metals with water vary quite markedly. Berylliumdoes not react with water; magnesium reacts slowly with steam; calcium, strontium,and barium are reactive enough to attack cold water:Ba(s)2H2O(l) 88n Ba(OH)2(aq)H2(g)The reactivities of the alkaline earth metals toward oxygen also increase from Be toBa. Beryllium and magnesium form oxides (BeO and MgO) only at elevated temperatures, whereas CaO, SrO, and BaO form at room temperature.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website8.6VARIATION IN CHEMICAL PROPERTIES OF THE REPRESENTATIVE ELEMENTSBeryllium (Be)Magnesium (Mg)Strontium (Sr)Barium (Ba)FIGURE 8.15 The Group 2Aelements: the alkaline earthmetals.311Calcium (Ca)Radium (Ra)Magnesium reacts with acids in aqueous solution, liberating hydrogen gas:Mg(s)2H (aq) 88n Mg2 (aq)H2(g)Calcium, strontium, and barium also react with aqueous acid solutions to produce hydrogen gas. However, because these metals also attack water, two different reactionswill occur simultaneously.The chemical properties of calcium and strontium provide an interesting exampleof periodic group similarity. Strontium-90, a radioactive isotope, is a major product ofan atomic bomb explosion. If an atomic bomb is exploded in the atmosphere, the strontium-90 formed will eventually settle on land and water, and it will reach our bodiesvia a relatively short food chain. For example, if cows eat contaminated grass and drinkcontaminated water, they will pass along strontium-90 in their milk. Because calciumand strontium are chemically similar, Sr2 ions can replace Ca2 ions in our bones.Constant exposure of the body to the high-energy radiation emitted by the strontium90 isotopes can lead to anemia, leukemia, and other chronic illnesses.BAlGaGroup 3A Elements (ns2np1, nThe first member of Group 3A, boron, is a metalloid; the rest are metals (Figure 8.16).Boron does not form binary ionic compounds and is unreactive toward oxygen gas andwater. The next element, aluminum, readily forms aluminum oxide when exposed toair:4Al(s)InTlBackForwardMain Menu2)3O2(g) 88n 2Al2O3(s)Aluminum that has a protective coating of aluminum oxide is less reactive than elemental aluminum. Aluminum forms only tripositive ions. It reacts with hydrochloricacid as follows:TOCStudy Guide TOCTextbook WebsiteMHHE Website312PERIODIC RELATIONSHIPS AMONG THE ELEMENTSFIGURE 8.16elements.The Group 3ABoron (B)Aluminum (Al)Gallium (Ga)Indium (In)2Al(s)6H (aq) 88n 2Al3 (aq)3H2(g)The other Group 3A metallic elements form both unipositive and tripositive ions.Moving down the group, we find that the unipositive ion becomes more stable than thetripositive ion.The metallic elements in Group 3A also form many molecular compounds. Forexample, aluminum reacts with hydrogen to form AlH3, which resembles BeH2 in itsproperties. (Here is an example of the diagonal relationship.) Thus, from left to rightacross the periodic table, we are seeing a gradual shift from metallic to nonmetalliccharacter in the representative elements.CSiGeGroup 4A Elements (ns2np2, nThe first member of group 4A, carbon, is a nonmetal, and the next two members, silicon and germanium, are metalloids (Figure 8.17). These elements do not form ioniccompounds. The metallic elements of this group, tin and lead, do not react with water,but they do react with acids (hydrochloric acid, for example) to liberate hydrogen gas:Sn(s)SnPb(s)PbBackForward2)2H (aq) 88n Sn2 (aq)22H (aq) 88n Pb (aq)H2(g)H2(g)The Group 4A elements form compounds in both the 2 and 4 oxidation states.For carbon and silicon, the 4 oxidation state is the more stable one. For example, CO2is more stable than CO, and SiO2 is a stable compound, but SiO does not exist undernormal conditions. As we move down the group, however, the trend in stability isreversed. In tin compounds the 4 oxidation state is only slightly more stable than the2 oxidation state. In lead compounds the 2 oxidation state is unquestionably themore stable one. The outer electron configuration of lead is 6s26p2, and lead tends toMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website8.6VARIATION IN CHEMICAL PROPERTIES OF THE REPRESENTATIVE ELEMENTSCarbon (graphite)Carbon (diamond)Germanium (Ge)FIGURE 8.17 The Group 4Aelements.NPAsSbBiSilicon (Si)Tin (Sn)Lead (Pb)lose only the 6p electrons (to form Pb2 ) rather than both the 6p and 6s electrons (toform Pb4 ).Group 5A Elements (ns2np3, nP4O10(s)SSeTePoBackForwardMain Menu2)In Group 5A, nitrogen and phosphorus are nonmetals, arsenic and antimony are metalloids, and bismuth is a metal (Figure 8.18). Thus we expect a greater variation inproperties within the group.Elemental nitrogen is a diatomic gas (N2). It forms a number of oxides (NO, N2O,NO2, N2O4, and N2O5), of which only N2O5 is a solid; the others are gases. Nitrogenhas a tendency to accept three electrons to form the nitride ion, N3 (thus achieving theelectron configuration 2s22p6, which is isoelectronic with neon). Most metallic nitrides(Li3N and Mg3N2, for example) are ionic compounds. Phosphorus exists as P4 molecules. It forms two solid oxides with the formulas P4O6 and P4O10. The importantoxoacids HNO3 and H3PO4 are formed when the following oxides react with water:N2O5(s)O313H2O(l) 88n 2HNO3(aq)6H2O(l) 88n 4H3PO4(aq)Arsenic, antimony, and bismuth have extensive three-dimensional structures. Bismuthis a far less reactive metal than those in the preceding groups.Group 6A Elements (ns2np4, n2)The first three members of Group 6A (oxygen, sulfur, and selenium) are nonmetals,and the last two (tellurium and polonium) are metalloids (Figure 8.19). Oxygen is adiatomic gas; elemental sulfur and selenium have the molecular formulas S8 and Se8,respectively; tellurium and polonium have more extensive three-dimensional structures.TOCStudy Guide TOCTextbook WebsiteMHHE Website314PERIODIC RELATIONSHIPS AMONG THE ELEMENTSFIGURE 8.18 The Group 5Aelements. Molecular nitrogen is acolorless, odorless gas.ineblaailAvotNonrsietVxTee-Nitrogen (N2)Arsenic (As)Antimony (Sb)Sulfur (S8)Selenium (Se8)White and red phosphorous (P)Bismuth (Bi)Tellurium (Te)FIGURE 8.19 The Group 6A elements sulfur, selenium, and tellurium. Molecular oxygen is acolorless, odorless gas. Polonium (not shown) is radioactive.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website8.6VARIATION IN CHEMICAL PROPERTIES OF THE REPRESENTATIVE ELEMENTS315FIGURE 8.20 The Group 7Aelements chlorine, bromine, andiodine. Fluorine is a greenishyellow gas that attacks ordinary glassware. Astatine isradioactive.(Polonium is a radioactive element that is difficult to study in the laboratory.) Oxygenhas a tendency to accept two electrons to form the oxide ion (O2 ) in many ionic compounds. Sulfur, selenium, and tellurium also form dinegative anions (S2 , Se2 , andTe2 ). The elements in this group (especially oxygen) form a large number of molecular compounds with nonmetals. The important compounds of sulfur are SO2, SO3,and H2S. Sulfuric acid is formed when sulfur trioxide reacts with water:SO3(g)FClBrGroup 7A Elements (ns2np5, nH2O(l ) 88n H2SO4(aq)2)All the halogens are nonmetals with the general formula X2, where X denotes a halogen element (Figure 8.20). Because of their great reactivity, the halogens are neverfound in the elemental form in nature. (The last member of Group 7A, astatine, is aradioactive element. Little is known about its properties.) Fluorine is so reactive thatit attacks water to generate oxygen:I2F2(g)At2H2O(l) 88n 4HF(aq)O2(g)Actually the reaction between molecular fluorine and water is quite complex; the products formed depend on reaction conditions. The reaction shown above is one of several possible changes.The halogens have high ionization energies and high electron affinities. Anionsderived from the halogens (F , Cl , Br , and I ) are called halides. They are isoelectronic with the noble gases immediately to their right in the periodic table. For example, F is isoelectronic with Ne, Cl with Ar, and so on. The vast majority of thealkali metal halides and alkaline earth metal halides are ionic compounds. The halogens also form many molecular compounds among themselves (such as ICl and BrF3)and with nonmetallic elements in other groups (such as NF3, PCl5, and SF6). The halogens react with hydrogen to form hydrogen halides:H2(g)X2(g) 88n 2HX(g)When this reaction involves fluorine, it is explosive, but it becomes less and less violent as we substitute chlorine, bromine, and iodine. The hydrogen halides dissolve inwater to form hydrohalic acids. Hydrofluoric acid (HF) is a weak acid (that is, it is aBackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website316PERIODIC RELATIONSHIPS AMONG THE ELEMENTSHelium (He)Neon (Ne)FIGURE 8.21 All noble gasesare colorless and odorless. Thesepictures show the colors emittedby the gases from a dischargetube.HeNeArKrXeArgon (Ar)Krypton (Kr)Xenon (Xe)weak electrolyte), but the other hydrohalic acids (HCl, HBr, and HI) are all strong acids(strong electrolytes).Group 8A Elements (ns2np6, n2)All noble gases exist as monatomic species (Figure 8.21). Their atoms have completelyfilled outer ns and np subshells, which give them great stability. (Helium is 1s2.) TheGroup 8A ionization energies are among the highest of all elements, and these gaseshave no tendency to accept extra electrons. For years these elements were called inertgases, and rightly so. Until 1963 no one had been able to prepare a compound containing any of these elements. The British chemist Neil Bartlett shattered chemistslong-held views of these elements when he exposed xenon to platinum hexafluoride,a strong oxidizing agent, and brought about the following reaction (Figure 8.22):Xe(g)PtF6(g) 88n XePtF6(s)Since then, a number of xenon compounds (XeF4, XeO3, XeO4, XeOF4) and a fewkrypton compounds (KrF2, for example) have been prepared (Figure 8.23). Despite theimmense interest in the chemistry of the noble gases, however, their compounds do nothave any commercial applications, and they are not involved in natural biologicalprocesses. No compounds of helium, neon, and argon are known.COMPARISON OF GROUP 1A AND GROUP 1B ELEMENTSWhen we compare the Group 1A elements (alkali metals) and the Group 1B elements(copper, silver, and gold), we arrive at an interesting conclusion. Although the metalsin these two groups have similar outer electron configurations, with one electron in theoutermost s orbital, their chemical properties are quite different.Neil Bartlett (1932 ). English chemist. Bartletts work is mainly in the preparation and study of compounds with unusual oxidation states and in solid-state chemistry.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website8.6VARIATION IN CHEMICAL PROPERTIES OF THE REPRESENTATIVE ELEMENTS317FIGURE 8.22 (a) Xenon gas(colorless) and PtF6 (red gas) separated from each other. (b) Whenthe two gases are allowed tomix, a yellow-orange solid compound is formed.(a)FIGURE 8.23 Crystals ofxenon tetrafluoride (XeF4).(b)The first ionization energies of Cu, Ag, and Au are 745 kJ/mol, 731 kJ/mol, and890 kJ/mol, respectively. Since these values are considerably larger than those of thealkali metals (see Table 8.3), the Group 1B elements are much less reactive. The higherionization energies of the Group 1B elements result from incomplete shielding of thenucleus by the inner d electrons (compared with the more effective shielding of thecompletely filled noble gas cores). Consequently the outer s electrons of these elementsare more strongly attracted by the nucleus. In fact, copper, silver, and gold are so unreactive that they are usually found in the uncombined state in nature. The inertnessand rarity of these metals make them valuable in the manufacture of coins and in jewelry. For this reason, these metals are also called coinage metals. The difference inchemical properties between the Group 2A elements (the alkaline earth metals) and theGroup 2B metals (zinc, cadmium, and mercury) can be explained in a similar way.PROPERTIES OF OXIDES ACROSS A PERIODOne way to compare the properties of the representative elements across a period is toexamine the properties of a series of similar compounds. Since oxygen combines withalmost all elements, we will compare the properties of oxides of the third-period elements to see how metals differ from metalloids and nonmetals. Some elements in thethird period (P, S, and Cl) form several types of oxides, but for simplicity we will consider only those oxides in which the elements have the highest oxidation number. Table8.5 lists a few general characteristics of these oxides. We observed earlier that oxygenSome Properties of Oxides of the Third-Period ElementsNa2OMgOAl2O3SiO2P4O10SO3Cl2O7m8888m888888ABLE 8.5m8888888Type of compoundStructureMelting point (C)Boiling point (C)Acid-base natureMain MenuTOC28003600Basic20452980AmphotericStudy Guide TOCmolecular units16.891.544.882m8888888 Acidic16102230580?Textbook Websitem8888888Forward1275?Basicm88Backm8888888 Ionicm888888 Molecularm88 Extensive three-dimensionalm888 DiscreteMHHE Website318PERIODIC RELATIONSHIPS AMONG THE ELEMENTShas a tendency to form the oxide ion. This tendency is greatly favored when oxygencombines with metals that have low ionization energies, namely, those in Groups 1Aand 2A, plus aluminum. Thus Na2O, MgO, and Al2O3 are ionic compounds, as indicated by their high melting points and boiling points. They have extensive threedimensional structures in which each cation is surrounded by a specific number of anions, and vice versa. As the ionization energies of the elements increase from left toright, so does the molecular nature of the oxides that are formed. Silicon is a metalloid; its oxide (SiO2) also has a huge three-dimensional network, although no ions arepresent. The oxides of phosphorus, sulfur, and chlorine are molecular compounds composed of small discrete units. The weak attractions among these molecules result inrelatively low melting points and boiling points.Most oxides can be classified as acidic or basic depending on whether they produce acids or bases when dissolved in water or react as acids or bases in certainprocesses. Some oxides are amphoteric, which means that they display both acidic andbasic properties. The first two oxides of the third period, Na2O and MgO, are basicoxides. For example, Na2O reacts with water to form the base sodium hydroxide:Na2O(s)H2O(l) 88n 2NaOH(aq)Magnesium oxide is quite insoluble; it does not react with water to any appreciable extent. However, it does react with acids in a manner that resembles an acid-base reaction:MgO(s)2HCl(aq) 88n MgCl2(aq)H2O(l)Note that the products of this reaction are a salt (MgCl2) and water, the usual productsof an acid-base neutralization.Aluminum oxide is even less soluble than magnesium oxide; it too does not react with water. However, it shows basic properties by reacting with acids:Al2O3(s)6HCl(aq) 88n 2AlCl3(aq)3H2O(l)It also exhibits acidic properties by reacting with bases:Al2O3(s)Note that this acid-base neutralization produces a salt but no water.2NaOH(aq)3H2O(l) 88n 2NaAl(OH)4(aq)Thus Al2O3 is classified as an amphoteric oxide because it has properties of both acidsand bases. Other amphoteric oxides are ZnO, BeO, and Bi2O3.Silicon dioxide is insoluble and does not react with water. It has acidic properties, however, because it reacts with very concentrated bases:SiO2(s)2NaOH(aq) 88n Na2SiO3(aq)H2O(l)For this reason, concentrated aqueous, strong bases such as NaOH(aq) should not bestored in Pyrex glassware, which is made of SiO2.The remaining third-period oxides are acidic. They react with water to form phosphoric acid (H3PO4), sulfuric acid (H2SO4), and perchloric acid (HClO4):P4O10(s)SO3(g)Cl2O7(l )6H2O(l) 88n 4H3PO4(aq)H2O(l) 88n H2SO4(aq)H2O(l) 88n 2HClO4(aq)This brief examination of oxides of the third-period elements shows that as themetallic character of the elements decreases from left to right across the period, theiroxides change from basic to amphoteric to acidic. Metallic oxides are usually basic,and most oxides of nonmetals are acidic. The intermediate properties of the oxides (asBackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website8.6VARIATION IN CHEMICAL PROPERTIES OF THE REPRESENTATIVE ELEMENTS319shown by the amphoteric oxides) are exhibited by elements whose positions are intermediate within the period. Note also that since the metallic character of the elementsincreases from top to bottom within a group of representative elements, we would expect oxides of elements with higher atomic numbers to be more basic than the lighterelements. This is indeed the case.EXAMPLE 8.6Classify the following oxides as acidic, basic, or amphoteric: (a) Rb2O, (b) BeO,(c) As2O5.(a) Since rubidium is an alkali metal, we would expect Rb2O to be a basic oxide. This is indeed true, as shown by rubidium oxides reaction with water toform rubidium hydroxide:AnswerRb2O(s)H2O(l) 88n 2RbOH(aq)(b) Beryllium is an alkaline earth metal. However, because it is the first member ofGroup 2A, we expect that it may differ somewhat from the other members of thegroup. Furthermore, beryllium and aluminum exhibit a diagonal relationship, so thatBeO may resemble Al2O3 in properties. It turns out that BeO, like Al2O3, is an amphoteric oxide, as shown by its reactions with acids and bases:BeO(s)2H (aq)3H2O(l) 88n Be(H2O)2 (aq)4BeO(s)2OH (aq)H2O(l) 88n Be(OH)2 (aq)4(c) Since arsenic is a nonmetal, we expect As2O5 to be an acidic oxide. This prediction is correct, as shown by the formation of arsenic acid when As2O5 reacts withwater:As2O5(s)Similar problem: 8.70.3H2O(l) 88n 2H3AsO4(aq)PRACTICE EXERCISEClassify the following oxides as acidic, basic, or amphoteric: (a) ZnO, (b) P4O10,(c) CaO.Chemistry in Action Chemistry in ActionChemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action ChemistryBackDiscovery of the Noble GasesIn the late 1800s John William Strutt, Third Baron ofRayleigh, who was a professor of physics at theCavendish Laboratory in Cambridge, England, accurately determined the atomic masses of a number ofelements, but he obtained a puzzling result with nitrogen. One of his methods of preparing nitrogen wasby the thermal decomposition of ammonia:2NH3(g) 88n N2(g)3H2(g)it oxygen, carbon dioxide, and water vapor.Invariably, the nitrogen from air was a little denser (byabout 0.5%) than the nitrogen from ammonia.Lord Rayleighs work caught the attention of SirWilliam Ramsay, a professor of chemistry at theUniversity College, London. In 1898 Ramsay passednitrogen, which he had obtained from air byRayleighs procedure, over red-hot magnesium to convert it to magnesium nitride:Another method was to start with air and remove fromForwardMain MenuTOCStudy Guide TOC3Mg(s)N2(g) 88n Mg3N2(s)Textbook WebsiteMHHE Website320PERIODIC RELATIONSHIPS AMONG THE ELEMENTSChemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action ChemistryAfter all of the nitrogen had reacted with magnesium,Ramsey was left with an unknown gas that would notcombine with anything.With the help of Sir William Crookes, the inventor of the discharge tube, Ramsay and Lord Rayleighfound that the emission spectrum of the gas did notmatch any of the known elements. The gas was a newelement! They determined its atomic mass to be39.95 amu and called it argon, which means thelazy one in Greek.Once argon had been discovered, other noblegases were quickly identified. Also in 1898 Ramsayisolated helium from uranium ores (see Chemistry inAction essay on p. 255). From the atomic masses ofhelium and argon, their lack of chemical reactivity,and what was then known about the periodic table,Ramsay was convinced that there were other unreactive gases and that they were all members of one periodic group. He and his student Morris Travers setout to find the unknown gases. They used a refrigeration machine to first produce liquid air. Applying atechnique called fractional distillation, they then allowed the liquid air to warm up gradually and collected components that boiled off at different temperatures. In this manner they analyzed and identifiedthree new elements neon, krypton, and xenon inonly three months. Three new elements in three monthsis a record that may never be broken!The discovery of the noble gases helped to complete the periodic table. Their atomic masses suggested that these elements should be placed to theright of the halogens. The apparent discrepancy withthe position of argon was resolved by Moseley, as discussed in the chapter.Finally, the last member of the noble gases, radon,was discovered by the German chemist Frederick DornSUMMARY OF FACTSAND CONCEPTSBackForwardSir William Ramsay (18521916).in 1900. A radioactive element and the heaviest elemental gas known, radons discovery not only completed the Group 8A elements, but also advanced ourunderstanding about the nature of radioactive decayand transmutation of elements.Lord Rayleigh and Ramsay both won Nobel Prizesin 1904 for the discovery of argon. Lord Rayleigh received the prize in physics and Ramsays award wasin chemistry.1. Nineteenth-century chemists developed the periodic table by arranging elements in the increasing order of their atomic masses. Discrepancies in early versions of the periodic tablewere resolved by arranging the elements in order of their atomic numbers.2. Electron configuration determines the properties of an element. The modern periodic tableclassifies the elements according to their atomic numbers, and thus also by their electronconfigurations. The configuration of the valence electrons directly affects the properties ofthe atoms of the representative elements.3. Periodic variations in the physical properties of the elements reflect differences in atomicstructure. The metallic character of elements decreases across a period from metals throughMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE WebsiteQUESTIONS AND PROBLEMS321the metalloids to nonmetals and increases from top to bottom within a particular group ofrepresentative elements.4. Atomic radius varies periodically with the arrangement of the elements in the periodictable. It decreases from left to right and increases from top to bottom.5. Ionization energy is a measure of the tendency of an atom to resist the loss of an electron.The higher the ionization energy, the stronger the attraction between the nucleus and an electron. Electron affinity is a measure of the tendency of an atom to gain an electron. The morepositive the electron affinity, the greater the tendency for the atom to gain an electron. Metalsusually have low ionization energies, and nonmetals usually have high electron affinities.6. Noble gases are very stable because their outer ns and np subshells are completely filled. Themetals among the representative elements (in Groups 1A, 2A, and 3A) tend to lose electronsuntil their cations become isoelectronic with the noble gases that precede them in the periodictable. The nonmetals in Groups 5A, 6A, and 7A tend to accept electrons until their anionsbecome isoelectronic with the noble gases that follow them in the periodic table.KEY WORDSAmphoteric oxide, p. 318Atomic radius, p. 295Diagonal relationship, p. 308Electron affinity, p. 305Ionic radius, p. 297Ionization energy, p. 301Isoelectronic, p. 294Valence electrons, p. 291Noble gases, p. 290Representative elements, p. 290QUESTIONS AND PROBLEMSDEVELOPMENT OF THE PERIODIC TABLEReview Questions8.1 Briefly describe the significance of Mendeleevs periodic table.8.2 What is Moseleys contribution to the modern periodic table?8.3 Describe the general layout of a modern periodictable.8.4 What is the most important relationship among elements in the same group in the periodic table?PERIODIC CLASSIFICATION OF THE ELEMENTSReview Questions8.5 Which of the following elements are metals, nonmetals, or metalloids? As, Xe, Fe, Li, B, Cl, Ba, P, I,Si.8.6 Compare the physical and chemical properties of metals and nonmetals.8.7 Draw a rough sketch of a periodic table (no detailsare required). Indicate regions where metals, nonmetals, and metalloids are located.8.8 What is a representative element? Give names andsymbols of four representative elements.8.9 Without referring to a periodic table, write the nameand give the symbol for an element in each of the following groups: 1A, 2A, 3A, 4A, 5A, 6A, 7A, 8A,transition metals.BackForwardMain MenuTOC8.10 Indicate whether the following elements exist asatomic species, molecular species, or extensive threedimensional structures in their most stable states at25C and 1 atm and write the molecular or empiricalformula for each one: phosphorus, iodine, magnesium, neon, arsenic, sulfur, boron, selenium, and oxygen.8.11 You are given a dark shiny solid and asked to determine whether it is iodine or a metallic element.Suggest a nondestructive test that would allow you toarrive at the correct answer.8.12 What are valence electrons? For representative elements, the number of valence electrons of an elementis equal to its group number. Show that this is truefor the following elements: Al, Sr, K, Br, P, S, C.8.13 Write the outer electron configurations for the (a) alkali metals, (b) alkaline earth metals, (c) halogens,(d) noble gases.8.14 Use the first-row transition metals (Sc to Cu) as anexample to illustrate the characteristics of the electron configurations of transition metals.8.15 How does the electron configuration of ions derivedfrom representative elements give them stability?8.16 What do we mean when we say that two ions or anatom and an ion are isoelectronic?8.17 What is wrong with the statement The atoms of element X are isoelectronic with the atoms of elementY?Study Guide TOCTextbook WebsiteMHHE Website322PERIODIC RELATIONSHIPS AMONG THE ELEMENTS8.18 Give three examples of first-row transition metal (Scto Cu) ions whose electron configurations are represented by the argon core.8.29Problems8.19 In the periodic table, the element hydrogen is sometimes grouped with the alkali metals (as in this book)and sometimes with the halogens. Explain why hydrogen can resemble the Group 1A and the Group 7Aelements.8.20 A neutral atom of a certain element has 17 electrons.Without consulting a periodic table, (a) write theground-state electron configuration of the element,(b) classify the element, (c) determine whether theatoms of this element are diamagnetic or paramagnetic.8.21 Group the following electron configurations in pairsthat would represent similar chemical properties oftheir atoms:(a) 1s22s22p63s2(b) 1s22s22p3(c) 1s22s22p63s23p64s23d104p6(d) 1s22s2(e) 1s22s22p6(f) 1s22s22p63s23p38.22 Group the following electron configurations in pairsthat would represent similar chemical properties oftheir atoms:(a) 1s22s22p5(b) 1s22s1(c) 1s22s22p6(d) 1s22s22p63s23p5(e) 1s22s22p63s23p64s1(f) 1s22s22p63s23p64s23d104p68.23 Without referring to a periodic table, write the electron configuration of elements with the followingatomic numbers: (a) 9, (b) 20, (c) 26, (d) 33. Classifythe elements.8.24 Specify the group of the periodic table in which eachof the following elements is found: (a) [Ne]3s1,(b) [Ne]3s23p3, (c) [Ne]3s23p6, (d) [Ar]4s23d8.8.25 A M2 ion derived from a metal in the first transition metal series has four electrons in the 3d subshell.What element might M be?8.26 A metal ion with a net 3 charge has five electronsin the 3d subshell. Identify the metal.8.27 Write the ground-state electron configurations of thefollowing ions: (a) Li , (b) H , (c) N3 , (d) F ,(e) S2 , (f) Al3 , (g) Se2 , (h) Br , (i) Rb , (j) Sr2 ,(k) Sn2 , (l) Te2 , (m) Ba2 , (n) Pb2 , (o) In3 ,(p) Tl , (q) Tl3 .8.28 Write the ground-state electron configurations of thefollowing ions, which play important roles in bio-BackForwardMain MenuTOC8.308.318.32chemical processes in our bodies: (a) Na , (b) Mg2 ,(c) Cl , (d) K , (e) Ca2 , (f) Fe2 , (g) Cu2 ,(h) Zn2 .Write the ground-state electron configurations of thefollowing transition metal ions: (a) Sc3 , (b) Ti4 ,(c) V5 , (d) Cr3 , (e) Mn2 , (f) Fe2 , (g) Fe3 ,(h) Co2 , (i) Ni2 , (j) Cu , (k) Cu2 , (l) Ag ,(m) Au , (n) Au3 , (o) Pt2 .Name the ions with 3 charges that have the following electron configurations: (a) [Ar]3d3, (b) [Ar],(c) [Kr]4d6, (d) [Xe]4f145d6.Which of the following species are isoelectronic witheach other? C, Cl , Mn2 , B , Ar, Zn, Fe3 , Ge2 .Group the species that are isoelectronic: Be2 , F ,Fe2 , N3 , He, S2 , Co3 , Ar.PERIODIC VARIATION IN PHYSICAL PROPERTIESReview Questions8.33 Define atomic radius. Does the size of an atom havea precise meaning?8.34 How does atomic radius change (a) from left to rightacross a period and (b) from top to bottom in a group?8.35 Define ionic radius. How does the size of an atomchange when it is converted to (a) an anion and (b) acation?8.36 Explain why, for isoelectronic ions, the anions arelarger than the cations.Problems8.37 On the basis of their positions in the periodic table,select the atom with the larger atomic radius in eachof the following pairs: (a) Na, Cs; (b) Be, Ba;(c) N, Sb; (d) F, Br; (e) Ne, Xe.8.38 Arrange the following atoms in order of decreasingatomic radius: Na, Al, P, Cl, Mg.8.39 Which is the largest atom in Group 4A?8.40 Which is the smallest atom in Group 7A?8.41 Why is the radius of the lithium atom considerablylarger than the radius of the hydrogen atom?8.42 Use the second period of the periodic table as an example to show that the size of atoms decreases as wemove from left to right. Explain the trend.8.43 Indicate which one of the two species in each of thefollowing pairs is smaller: (a) Cl or Cl ; (b) Na orNa ; (c) O2 or S2 ; (d) Mg2 or Al3 ; (e) Au orAu3 .8.44 List the following ions in order of increasing ionicradius: N3 , Na , F , Mg2 , O2 .8.45 Explain which of the following cations is larger, andwhy: Cu or Cu2 .8.46 Explain which of the following anions is larger, andwhy: Se2 or Te2 .Study Guide TOCTextbook WebsiteMHHE WebsiteQUESTIONS AND PROBLEMS8.47 Give the physical states (gas, liquid, or solid) of therepresentative elements in the fourth period (K, Ca,Ga, Ge, As, Se, Br) at 1 atm and 25C.8.48 The boiling points of neon and krypton are 245.9Cand 152.9C, respectively. Using these data, estimate the boiling point of argon.IONIZATION ENERGYReview Questions8.49 Define ionization energy. Ionization energy measurements are usually made when atoms are in thegaseous state. Why? Why is the second ionization energy always greater than the first ionization energyfor any element?8.50 Sketch the outline of the periodic table and showgroup and period trends in the first ionization energyof the elements. What types of elements have thehighest ionization energies and what types the lowest ionization energies?Problems8.51 Use the third period of the periodic table as an example to illustrate the change in first ionization energies of the elements as we move from left to right.Explain the trend.8.52 In general, ionization energy increases from left toright across a given period. Aluminum, however, hasa lower ionization energy than magnesium. Explain.8.53 The first and second ionization energies of K are419 kJ/mol and 3052 kJ/mol, and those of Ca are590 kJ/mol and 1145 kJ/mol, respectively. Comparetheir values and comment on the differences.8.54 Two atoms have the electron configurations 1s22s22p6and 1s22s22p63s1. The first ionization energy of oneis 2080 kJ/mol, and that of the other is 496 kJ/mol.Match each ionization energy with one of the givenelectron configurations. Justify your choice.8.55 A hydrogenlike ion is an ion containing only one electron. The energies of the electron in a hydrogenlikeion are given byEn(2.181018J)Z 21n2where n is the principal quantum number and Z is theatomic number of the element. Calculate the ionization energy (in kJ/mol) of the He ion.8.56 Plasma is a state of matter consisting of positivegaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons andtherefore would exist as Hg80 . Use the equation inProblem 8.55 to calculate the energy required for thelast ionization step, that is,BackForwardMain MenuTOCHg79 (g) 88n Hg80 (g)323eELECTRON AFFINITYReview Questions8.57 (a) Define electron affinity. (b) Electron affinity measurements are made with gaseous atoms. Why?(c) Ionization energy is always a positive quantity,whereas electron affinity may be either positive ornegative. Explain.8.58 Explain the trends in electron affinity from aluminumto chlorine (see Table 8.4).Problems8.59 Arrange the elements in each of the following groupsin increasing order of the most positive electron affinity: (a) Li, Na, K, (b) F, Cl, Br, I.8.60 Specify which of the following elements you wouldexpect to have the greatest electron affinity: He, K,Co, S, Cl.8.61 Considering their electron affinities, do you think itis possible for the alkali metals to form an anion likeM , where M represents an alkali metal?8.62 Explain why alkali metals have a greater affinity forelectrons than alkaline earth metals.VARIATION IN CHEMICAL PROPERTIES OF THEREPRESENTATIVE ELEMENTSReview Questions8.63 What is meant by the diagonal relationship? Nametwo pairs of elements that show this relationship.8.64 Which elements are more likely to form acidic oxides? Basic oxides? Amphoteric oxides?Problems8.65 Use the alkali metals and alkaline earth metals as examples to show how we can predict the chemicalproperties of elements simply from their electron configurations.8.66 Based on your knowledge of the chemistry of the alkali metals, predict some of the chemical propertiesof francium, the last member of the group.8.67 As a group, the noble bases are very stable chemically (only Kr and Xe are known to form compounds).Why?8.68 Why are Group 1B elements more stable than Group1A elements even though they seem to have the sameouter electron configuration, ns1, where n is the principal quantum number of the outermost shell?8.69 How do the chemical properties of oxides changefrom left to right across a period? From top to bottom within a particular group?Study Guide TOCTextbook WebsiteMHHE Website324PERIODIC RELATIONSHIPS AMONG THE ELEMENTS8.70 Write balanced equations for the reactions betweeneach of the following oxides and water: (a) Li2O,(b) CaO, (c) SO3.8.71 Write formulas for and name the binary hydrogencompounds of the second-period elements (Li to F).Describe how the physical and chemical propertiesof these compounds change from left to right acrossthe period.8.72 Which oxide is more basic, MgO or BaO? Why?8.73 State whether each of the following properties of therepresentative elements generally increases or decreases (a) from left to right across a period and(b) from top to bottom within a group: metallic character, atomic size, ionization energy, acidity of oxides.8.74 With reference to the periodic table, name (a) a halogen element in the fourth period, (b) an element similar to phosphorus in chemical properties, (c) the mostreactive metal in the fifth period, (d) an element thathas an atomic number smaller than 20 and is similarto strontium.8.75 Write equations representing the following processes:(a) The electron affinity of S .(b) The third ionization energy of titanium.(c) The electron affinity of Mg2 .(d) The ionization energy of O2 .8.76 Arrange the following isoelectronic species in orderof (a) increasing ionic radius and (b) increasing ionization energy: O2 , F , Na , Mg2 .8.77 Write the empirical (or molecular) formulas of compounds that the elements in the third period (sodiumto chlorine) should form with (a) molecular oxygenand (b) molecular chlorine. In each case indicatewhether you would expect the compound to be ionicor molecular in character.8.78 Element M is a shiny and highly reactive metal (melting point 63C), and element X is a highly reactivenonmetal (melting point 7.2C). They react to forma compound with the empirical formula MX, a colorless, brittle white solid that melts at 734C. Whendissolved in water or when in the molten state, thesubstance conducts electricity. When chlorine gas isbubbled through an aqueous solution containing MX,a reddish-brown liquid appears and Cl ions areformed. From these observations, identify M and X.(You may need to consult a handbook of chemistryfor the melting-point values.)8.79 Match each of the elements on the right with its description on the left:(a) A dark-red liquidCalcium (Ca)(b) A colorless gas that burnsGold (Au)in oxygen gasHydrogen (H2)ForwardMain Menu8.818.82ADDITIONAL PROBLEMSBack8.80TOC8.838.848.858.868.878.888.898.908.918.928.93(c) A reactive metal that attacksArgon (Ar)waterBromine (Br2)(d) A shiny metal that is used injewelry(e) A totally inert gasArrange the following species in isoelectronic pairs:O , Ar, S2 , Ne, Zn, Cs , N3 , As3 , N, Xe.In which of the following are the species written indecreasing order by size of radius? (a) Be, Mg, Ba,(b) N3 , O2 , F , (c) Tl3 , Tl2 , Tl .Which of the following properties show a clear periodic variation? (a) first ionization energy, (b) molarmass of the elements, (c) number of isotopes of anelement, (d) atomic radius.When carbon dioxide is bubbled through a clear calcium hydroxide solution, the solution appears milky.Write an equation for the reaction and explain howthis reaction illustrates that CO2 is an acidic oxide.You are given four substances: a fuming red liquid,a dark metallic-looking solid, a pale-yellow gas, anda yellow-green gas that attacks glass. You are told thatthese substances are the first four members of Group7A, the halogens. Name each one.For each pair of elements listed below, give threeproperties that show their chemical similarity:(a) sodium and potassium and (b) chlorine andbromine.Name the element that forms compounds, under appropriate conditions, with every other element in theperiodic table except He, Ne, and Ar.Explain why the first electron affinity of sulfur is200 kJ/mol but the second electron affinity is649 kJ/mol.The H ion and the He atom have two 1s electronseach. Which of the two species is larger? Explain.Predict the products of the following oxides with water: Na2O, BaO, CO2, N2O5, P4O10, SO3. Write anequation for each of the reactions. Specify whetherthe oxides are acidic, basic, or amphoteric.Write the formulas and names of the oxides of thesecond-period elements (Li to N). Identify the oxidesas acidic, basic, or amphoteric.State whether each of the following elements is a gas,a liquid, or a solid under atmospheric conditions. Alsostate whether it exists in the elemental form as atoms,as molecules, or as a three-dimensional network: Mg,Cl, Si, Kr, O, I, Hg, Br.What factors account for the unique nature of hydrogen?The air in a manned spacecraft or submarine needsto be purified of exhaled carbon dioxide. Write equations for the reactions between carbon dioxide and(a) lithium oxide (Li2O), (b) sodium peroxide(Na2O2), and (c) potassium superoxide (KO2).Study Guide TOCTextbook WebsiteMHHE WebsiteQUESTIONS AND PROBLEMS8.94 The formula for calculating the energies of an electron in a hydrogenlike ion is given in Problem 8.55.This equation cannot be applied to many-electronatoms. One way to modify it for the more complex), where Z is theatoms is to replace Z with (Zis a positive dimensionlessatomic number andquantity called the shielding constant. Consider thehelium atom as an example. The physical significanceof is that it represents the extent of shielding thatthe two 1s electrons exert on each other. Thus the) is appropriately called the effectivequantity (Znuclear charge. Calculate the value of if the firstionization energy of helium is 3.94 10 18 J peratom. (Ignore the minus sign in the given equation inyour calculation.)8.95 Why do noble gases have negative electron affinityvalues?8.96 The atomic radius of K is 216 pm and that of K is133 pm. Calculate the percent decrease in volume thatoccurs when K(g) is converted to K (g). [The volume of a sphere is ( 4 ) r3, where r is the radius of the3sphere.]8.97 The atomic radius of F is 72 pm and that of F is136 pm. Calculate the percent increase in volume thatoccurs when F(g) is converted to F (g). (See Problem8.96 for the volume of a sphere.)8.98 A technique called photoelectron spectroscopy isused to measure the ionization energy of atoms. Asample is irradiated with UV light, and electrons areejected from the valence shell. The kinetic energiesof the ejected electrons are measured. Since the energy of the UV photon and the kinetic energy of theejected electron are known, we can writehIE12mu2where is the frequency of the UV light, and m andu are the mass and velocity of the electron, respectively. In one experiment the kinetic energy of theejected electron from potassium is found to be 5.3410 19 J using a UV source of wavelength 162 nm.Calculate the ionization energy of potassium. Howcan you be sure that this ionization energy corresponds to the electron in the valence shell (that is, themost loosely held electron)?8.99 Referring to the Chemistry in Action essay on p. 319,answer the following questions. (a) Why did it takeso long to discover the first noble gas (argon) onEarth? (b) Once argon had been discovered, why didit take relatively little time to discover the rest of thenoble gases? (c) Why was helium not isolated by thefractional distillation of liquid air?8.100 The energy needed for the following process is1.96 104 kJ/mol:BackForwardMain MenuTOCLi(g) 88n Li3 (g)3253eIf the first ionization energy of lithium is 520 kJ/mol,calculate the second ionization energy of lithium, thatis, the energy required for the processLi (g) 88n Li2 (g)e(Hint: You need the equation in Problem 8.55.)8.101 An element X reacts with hydrogen gas at 200C toform compound Y. When Y is heated to a higher temperature, it decomposes to the element X and hydrogen gas in the ratio of 559 mL of H2 (measured atSTP) for 1.00 g of X reacted. X also combines withchlorine to form a compound Z, which contains 63.89percent by mass of chlorine. Deduce the identity ofX.8.102 A student is given samples of three elements, X, Y,and Z, which could be an alkali metal, a member ofGroup 4A, and a member of Group 5A. She makesthe following observations: Element X has a metallic luster and conducts electricity. It reacts slowlywith hydrochloric acid to produce hydrogen gas.Element Y is a light-yellow solid that does not conduct electricity. Element Z has a metallic luster andconducts electricity. When exposed to air, it slowlyforms a white powder. A solution of the white powder in water is basic. What can you conclude aboutthe elements from these observations?8.103 Using the following boiling-point data and the procedure in the Chemistry in Action essay on p. 302,estimate the boiling point of francium:metalLiNaK RbCsboiling point (C) 1347 882.9 774 688 678.48.104 What is the electron affinity of the Na ion?8.105 The ionization energies of sodium (in kJ/mol), starting with the first and ending with the eleventh, are495.9, 4560, 6900, 9540, 13400, 16600, 20120,25490, 28930, 141360, 170000. Plot the log of ionization energy (y-axis) versus the number of ionization (x-axis); for example, log 495.9 is plotted versus1 (labeled I1, the first ionization energy), log 4560 isplotted versus 2 (labeled I2, the second ionization energy), and so on. (a) Label I1 through I11 with theelectrons in orbitals such as 1s, 2s, 2p, and 3s.(b) What can you deduce about electron shells fromthe breaks in the curve?8.106 Experimentally, the electron affinity of an elementcan be determined by using a laser light to ionize theanion of the element in the gas phase:Study Guide TOCX (g)h 88n X(g)Textbook WebsiteeMHHE Website326PERIODIC RELATIONSHIPS AMONG THE ELEMENTS8.1078.1088.1098.1108.1118.112Referring to Table 8.4, calculate the photon wavelength (in nanometers) corresponding to the electronaffinity for chlorine. In what region of the electromagnetic spectrum does this wavelength fall?Explain, in terms of their electron configurations,why Fe2 is more easily oxidized to Fe3 than Mn2to Mn3 .The standard enthalpy of atomization of an elementis the energy required to convert one mole of an element in its most stable form at 25C to one mole ofmonatomic gas. Given that the standard enthalpy ofatomization for sodium is 108.4 kJ/mol, calculate theenergy in kilojoules required to convert one mole ofsodium metal at 25C to one mole of gaseous Naions.Write the formulas and names of the hydrides of thefollowing second-period elements: Li, C, N, O, F.Predict their reactions with water.Based on knowledge of the electronic configurationof titanium, state which of the following compoundsof titanium is unlikely to exist: K3TiF6, K2Ti2O5,TiCl3, K2TiO4, K2TiF6.Name an element in Group 1A or Group 2A that isan important constituent of each of the following substances: (a) remedy for acid indigestion, (b) coolantin nuclear reactors, (c) Epsom salt, (d) baking powder, (e) gunpowder, (f) a light alloy, (g) fertilizer thatalso neutralizes acid rain, (h) cement, and (i) grit foricy roads. You may need to ask your instructor aboutsome of the items.In halogen displacement reactions a halogen elementcan be generated by oxidizing its anions with a halogen element that lies above it in the periodic table.This means that there is no way to prepare elemental fluorine, since it is the first member of Group 7A.Indeed, for years the only way to prepare elementalfluorine was to oxidize F ions by electrolytic means.Then, in 1986, a chemist reported that by reactingpotassium hexafluoromanganate(IV) (K2MnF6) withantimony pentafluoride (SbF5) at 150C, he had generated elemental fluorine. Balance the followingequation representing the reaction:K2MnF6SbF5 88n KSbF6MnF3F28.113 Write a balanced equation for the preparation of(a) molecular oxygen, (b) ammonia, (c) carbon dioxide, (d) molecular hydrogen, (e) calcium oxide.Indicate the physical state of the reactants and products in each equation.8.114 Write chemical formulas for oxides of nitrogen withthe following oxidation numbers: 1, 2, 3, 4,5. (Hint: There are two oxides of nitrogen with 4oxidation number.)BackForwardMain MenuTOC8.115 Most transition metal ions are colored. For example,a solution of CuSO4 is blue. How would you showthat the blue color is due to the hydrated Cu2 ions2and not the SO4 ions?8.116 In general, atomic radius and ionization energy haveopposite periodic trends. Why?8.117 Explain why the electron affinity of nitrogen is approximately zero, while the elements on either side,carbon and oxygen, have substantial positive electronaffinities.8.118 Consider the halogens chlorine, bromine, and iodine.The melting point and boiling point of chlorine are101.0C and 34.6C while those of iodine are113.5C and 184.4C, respectively. Thus chlorine isa gas and iodine is a solid under room conditions.Estimate the melting point and boiling point ofbromine. Compare your values with those from ahandbook of chemistry.8.119 While it is possible to determine the second, third,and higher ionization energies of an element, thesame cannot usually be done with the electron affinities of an element. Explain.8.120 The only confirmed compound of radon is radon fluoride, RnF. One reason that it is difficult to study thechemistry of radon is that all isotopes of radon areradioactive so it is dangerous to handle the substance.Can you suggest another reason why there are so fewknown radon compounds? (Hint: Radioactive decaysare exothermic processes.)8.121 Little is known of the chemistry of astatine, the lastmember of Group 7A. Describe the physical characteristics that you would expect this halogen to have.Predict the products of the reaction between sodiumastatide (NaAt) and sulfuric acid. (Hint: Sulfuric acidis an oxidizing agent.)8.122 As discussed in the chapter, the atomic mass of argon is greater than that of potassium. This observation created a problem in the early development ofthe periodic table because it meant that argon shouldbe placed after potassium. (a) How was this difficultyresolved? (b) From the following data, calculate theaverage atomic masses of argon and potassium: Ar36 (35.9675 amu; 0.337%), Ar-38 (37.9627 amu;0.063%), Ar-40 (39.9624 amu; 99.60%); K-39(38.9637 amu; 93.258%), K-40 (39.9640 amu;0.0117%), K-41 (40.9618 amu; 6.730%).8.123 Calculate the maximum wavelength of light (innanometers) required to ionize a single sodium atom.8.124 Predict the atomic number and ground-state electronconfiguration of the next member of the alkali metals after francium.8.125 Why do elements that have high ionization energiesalso have more positive electron affinities? WhichStudy Guide TOCTextbook WebsiteMHHE WebsiteQUESTIONS AND PROBLEMSgroup of elements would be an exception to this generalization?8.126 The first four ionization energies of an element areapproximately 738 kJ/mol, 1450 kJ/mol, 7.7103 kJ/mol, and 1.1 104 kJ/mol. To which periodicgroup does this element belong? Why?8.127 Some chemists think that helium should properly becalled helon. Why? What does the ending in helium(-ium) suggest?8.128 (a) The formula of the simplest hydrocarbon is CH4(methane). Predict the formulas of the simplest compounds formed between hydrogen and the followingelements: silicon, germanium, tin, and lead.(b) Sodium hydride (NaH) is an ionic compound.Would you expect rubidium hydride (RbH) to bemore or less ionic than NaH? (c) Predict the reactionbetween radium (Ra) and water. (d) When exposedto air, aluminum forms a tenacious oxide (Al2O3)coating that protects the metal from corrosion. Whichmetal in Group 2A would you expect to exhibit similar properties? Why?BackForwardMain MenuTOC3278.129 Match each of the elements on the right with its description on the left:(a) A pale yellow gasNitrogen (N2)that reacts with water.Boron (B)(b) A soft metal that reactsAluminum (Al)with water to produceFluorine (F2)hydrogen.Sodium (Na)(c) A metalloid that is hardand has a high meltingpoint.(d) A colorless, odorlessgas.(e) A metal that is more reactive thaniron, but does not corrodein air.Answers to Practice Exercises: 8.1 (a) 1s22s22p63s23p64s2,(b) it is a representative element, (c) diamagnetic. 8.2 Li BeC. 8.3 (a) Li , (b) Au3 , (c) N3 . 8.4 (a) N, (b) Mg. 8.5 No.8.6 (a) amphoteric, (b) acidic, (c) basic.Study Guide TOCTextbook WebsiteMHHE Website...
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