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Homework #1-6 on page 7

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We’re going to work on Questions #1-13 pg 4-6 In groups work on finishing these questions Tips for a good answer: Did you state what the trend is? (increase/ decrease) Did you explain WHY? Just saying “because it increases to the left” is not enough Did you use as many vocab words as possible? (see page 4 on your packet) These are similar to the test questions

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2 Elements in period three have 3 energy levels while those in period 2 have only two energy levels. The additional core electrons increase the distance between the valence electrons and the nucleus. This shielding causes a weaker attraction, so the electrons are farther away and the atom is larger.

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4 The elements in group 4A can become either +4 or -4 (even though the Ionic Radius Chart shows only +4.) If the chart showed the -4 ion of Carbon or Silicon, describe what you would expect it to look like. Explain why they would look this way? Carbon with a -4 charge would probably be the largest in the period. It would have the same number of electrons (10) as the anions but would have the lowest number of protons (6). The lowest effective nuclear charge means a lesser force of attraction and electrons aren’t pulled as tight, increasing the radius.

5 Sodium would be the largest while aluminum would be the smallest. Although all the ions have the same number of electrons (they are isoelectronic), there are more protons as you go left to right; the larger effective nuclear charge pulls the electrons in tighter, causing the radius to decrease. Na+ Mg2+ Al3+ 11 protons 12 protons 13 protons 10 electrons

6 There is a force of attraction because protons and electrons have an opposite charge ( and oppositely charged particles are attracted to each other). The force between two electrons is repulsive. As the distance between proton and electron increases, the force of attraction: decreases

7 The lowest first ionization energies all belong to metals in group 1A. This group contains the atoms with the lowest effective nuclear charge for their row/period. Therefore, the valence electrons feel less attraction to the nucleus, making the atom’s radius larger. (the largest in the period). Less energy is required to remove electrons that are farther from the nucleus.

8 What happens to the first ionization energies of the noble gases as you descend a group? Why does this happen? First ionization energy decreases as you descend a group. Additional energy levels filled with core electrons increase the amount of electron shielding. Since the valence electrons are farther from the nucleus less energy is required to remove them.

9 Atomic radii decrease across a period. As you go across the periodic table, the number of protons increases, increasing the effective nuclear charge, but the electrons are being added to the same energy level (no additional core electrons/electron shielding). Therefore, there is a stronger force of attraction between the valence electrons and the larger nucleus, so electrons are pulled in closer and radius decreases.

11 A sample contains an unknown element. The ionic radius of the unknown element is smaller than its atomic radius. cation Its electronegativity is between 0.70 and 1.00 Paulings. Group 1A and 2A It is extremely reactive and has an atomic mass of less than 35 g/mol. 1A What is this element? Li or Na

12. Answers will vary

If there is time Work on questions #4-10 on page 6 & 7

Homework #1-6 on page 7

10 A meteor crashed into the desert in Arizona. The scientists who discovered it performed tests to determine what elements were in the meteor. It has an electronegativity value greater than 3.00 Paulings. N, O, F, Cl Its 6th ionization energy is over 5 times larger than the 5th ionization energy. N, O, F The most abundant element has an ionic radius 1.95 times larger than its atomic radius. N What is this element? nitrogen

10 Examine the Successive Ionization Energy Chart. Why is the 3rd ionization energy for Beryllium so much higher than its 2nd ionization energy? When you remove the second electron, you form beryllium ion (Be+2). When you remove the third electron, you are trying to take an electron from Be2+ which has a full octet (a noble gas electron configuration). Because this electron arrangement is very stable, a very large amount of energy will be required.