Periodic Trends PDF

PERIODIC TRENDS Trends in Some Periodic Properties ⦿ The physical and chemical behavior of the elements is based on the electron configurations of their atoms. ⦿ e- configurations can be used to explain many of the repeating or “periodic” properties of the elements ⦿ Period...

PERIODIC TRENDS Trends in Some Periodic Properties ⦿ The physical and chemical behavior of the elements is based on the electron configurations of their atoms. ⦿ e- configurations can be used to explain many of the repeating or “periodic” properties of the elements ⦿ Periodic Law – The properties of elements are periodic functions of their atomic numbers. CONTENTS ⦿ Atomic Radii and Ionic Radii ⦿ Ionization Energy ⦿ Electron Affinity ⦿ Electronegativity Atomic Radii (AR) ⦿ Size in representative elements follow 2 rules ⚫ AR increases going down in a group ⚫ AR decreases going left to right in a period ⦿ Why? ⦿ Moving down in a group: ⚫ Electrons in a higher n and therefore larger ⦿ Moving across a period: ⚫ Z increases and electrons in same shell don't shield each other well so e- are attracted more strongly and are closer/smaller Atomic Radii Atomic Radii of Representative Elements Ionic Size ⦿ Ionic size increase down in a group. ⦿ Cations are smaller than the parent atom. ⦿ Anions are larger than the parent atom. ⦿ In an isoelectronic series, the most negative ion is largest, the most positive is smallest. ⦿ Atoms making more than one ion, most negative is the largest.  Figure 8.29 Ionic vs. atomic radius SW. Answer on size 4. Arrange each set of elements in terms of the following criteria in DECREASING size. A. Atomic Radii B. Ionic Radii 1. Ca, Mg, Sr 1. Ca2+, Sr2+, Mg2+ 2. K, Ga, Ca 2. K+, S2-, Cl-1 3. Rb, Br, Kr 3. Au+, Au3+ 4. Sr, Ca, Rb SAMPLE PROBLEM 1 Ranking Elements by Atomic Size PROBLEM: Using only the periodic table (not Figure 8.15), rank each set of main group elements in order of decreasing atomic size: (a) Ca, Mg, Sr (b) K, Ga, Ca (c) Br, Rb, Kr (d) Sr, Ca, Rb RECALL: Elements in the same group increase in size and you go down; elements decrease in size as you go across a period. SOLUTION: (a) Sr > Ca > Mg These elements are in Group 2A(2). (b) K > Ca > Ga These elements are in Period 4. (c) Rb > Br > Kr Rb has a higher energy level and is far to the left. Br is to the left of Kr. (d) Rb > Sr > Ca Ca is one energy level smaller than Rb and Sr. Rb is to the left of Sr. SAMPLE PROBLEM 2 Ranking Ions by Size PROBLEM: Rank each set of ions in order of decreasing size, and explain your ranking: (a) Ca2+, Sr2+, Mg2+ (b) K+, S2-, Cl - (c) Au+, Au3+ RECALL: Compare positions in the periodic table, formation of positive and negative ions and changes in size due to gain or loss of electrons. SOLUTION: (a) Sr2+ > Ca2+ > Mg2+ These are members of the same group (Group 2A) and therefore decrease in size going up the group. (b) S2- > Cl - > K+ The ions are isoelectronic (meaning they have the same no. of electrons); S2- has the smallest Z and therefore has the largest radius while K+ is a cation with a largest Z and therefore has the smallest radius. (c) Au+ > Au3+ The more positive the charge, the smaller the ion. Ionization energy ⦿ Energy required to remove an electron from an element (to make it positive or more positive) ⦿ Always a positive value ⦿ First ionization energy IE1– energy to remove 1st e- ⚫ Elements with small IE1 tend to form cations; and those with large IE1 form anions Ionization energy ⦿ Trends: ⚫ Generally as AR decreases, IE1 increases. ⚫ IE1 decreases going down in a group ⚫ IE1 increases from left to right, with some exceptions ⦿ Why? ⚫ Down a group – atom is larger so e- is farther ⚫ Across a period - smaller atomic radii (e- closer to nucleus hence nuclear pull is greater) ⚫ Exceptions: ○ Be (s2) to B (s2p1) because s shields p ○ N (s2p3) to O (s2p4) because repulsions in first paired electron  Figure 8.17 Periodicity of first ionization energy (IE1)  First ionization energies of the main-group elements Figure 8.18 SW. Answer on size 2. Arrange each set of elements in decreasing ionization energy (IE). Next to each item, give the reason for the arrangement. 1. Kr, He, Ar 2. Sb, Te, Sn 3. K, Ca, Rb 4. I, Xe, Cs SAMPLE PROBLEM 3 Ranking Elements by First Ionization Energy PROBLEM: Using the periodic table only, rank the elements in each of the following sets in order of decreasing IE1: (a) Kr, He, Ar (b) Sb, Te, Sn (c) K, Ca, Rb (d) I, Xe, Cs RECALL: IE decreases down in a group; IE increases as you go across a period. SOLUTION: (a) He > Ar > Kr Group 8A(18) - IE decreases down a group. (b) Te > Sb > Sn Period 5 elements - IE increases across a period. (c) Ca > K > Rb Ca is to the right of K; Rb is below K. (d) Xe > I > Cs I is to the left of Xe; Cs is furtther to the left and down one period. Second and Succeeding Ionization Energies* ⦿ Second ionization energy IE2 – to remove 2nd e- ⚫ As the e- are removed, Z remains constant and remaining e- are harder to remove ⚫ Always larger than IE1 ⦿ A huge increase occurs in ionization energies when noble gas valence (or octet) is reached. ⦿ Result - we only involve outermost e- Figure 8.19 The first three ionization energies of beryllium (in MJ/mol) +1 4 Be 4 Be + 1e- IE1 +1 +2 4 Be 4 Be + 1e- IE2 +2 +3 4 Be 4 Be + 1e- IE3 Problem. Identify the 3rd Period element based on its ionization energies in Table 1 below. Then give its electron configuration. Table 1. Succeeding Ionization Energies (kJ / mol) IE1 1,012 IE2 1,903 IE3 2,910 IE4 4,956 IE5 6,278 IE6 22,230 SAMPLE PROBLEM 4 Identifying an Element from Successive Ionization Energies PROBLEM: Name the Period 3 element with the following ionization energies (in kJ/mol) and write its electron configuration: IE1 IE2 IE3 IE4 IE5 IE6 1012 1903 2910 4956 6278 22,230 HOW TO Look for a large increase in energy which indicates that all of SOLVE: the valence electrons have been removed. SOLUTION: The largest increase occurs after IE5, that is, after the 5th valence electron has been removed. Five electrons at Period 3 would mean that the valence configuration is 3s2 3p3 and the element must be phosphorous, P (Z = 15). The complete electron configuration is 1s2 2s2 2p6 3s2 3p3. Electron Affinity (EA) Ionization energies measures the energy changes associated with removing electrons from an atom to form positively charged ions. For example, The I1 energy of Cl gas, 1251 kJ/mol, is the energy change associated with the following process. Ionization Energy: Cl(g) Cl+(g) + e- ∆E = 1251 kJ/mol Electron Affinity: Cl(g) + e- Cl-(g) ∆E = -349 kJ/mol Electron Affinity The electron affinity of an element is the energy given off when a neutral atom in the gas phase gains an extra electron to form a negatively charged ion. A fluorine atom in the gas phase, for example, gives off energy when it gains an electron to form a fluoride ion. F(g) + e- F-(g) ∆Ho = -328.0 kJ/mol Electronegativity (EN) ⦿ Ability of an element to attract electrons when bonded with another atom ⦿ Arbitrary score given to elements; not really a measured property ⦿ Value was based on how elements share electrons when in a compound or molecule ⦿ Trends: ⚫ EN increases across a period; and ⚫ EN decreases along a group ⦿ Highest EN score is 4.0; assigned to F Electronegativity  Figure 8.21 Trends in three atomic properties