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PS_4_Solution[1] - 7 By Hs and Cx(neutrals are in...

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Chem 115 PS 4 Solution Spring ‘08 1. Hs must be in the 4 th column. 2. Gt is in the top row (strongest attraction for an electron [from another molecule] to form a covalent bond if the electron can enter shell closest to nucleus). Gt will be far to the right (high Z eff ), but not a noble gas, because the new electron would be extremely well shielded from the Gt nucleus. 3. Fu is in the 3 rd column, for reasons stated above for Gt. 4. Element #1. 5. Az is in the top row (strong attraction of an electron close to nucleus), last column (highest Z eff ). 6. The valence electrons of Jq are more mobile than in Gt, implying a lower Z eff or higher shell number; so Jq is to the left of or below Gt. Malleability implies Dw has fewer valence electrons than Jq, so Dw is to the left of Jq.
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Unformatted text preview: 7. By, Hs, and Cx (neutrals) are in consecutive order. Small size of the Cx + cation indicates that the Cx + cation has the largest nuclear charge. 8. Ev and Kp are in the same column, because their valence electrons are perfectly screened by all the core electrons. Ev must be in shell 3 and Kp in shell 1. 9. Removing two electrons gives a noble gas configuration. Noble gases have very high IE’s because of the high Z eff , which stems from an increased nuclear charge at the end of a row, but poor shielding by valence electrons. 10. Lo is in the last row, where electrons are most distant from the nucleus. These deductions lead to the following periodic table: Ir Kp Gt Az Dw Jq By Hs Ct Ev Fu Lo...
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