Problem set 2 Crib 2011

Problem set 2 Crib 2011 - CHM 24100 HW #2 Crib Spring 2011...

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Unformatted text preview: CHM 24100 HW #2 Crib Spring 2011 1.24 e) V: [Ar] 4s23d3 so V5+: [Ar] f) Mo: [Kr] 5s14d5 so Mo4+: [Kr] 4d2 1.28 Lanthanide contraction results due to the imperfect screening by the 4f electrons. Dramatic increase in Z between Nb and Ta. 11.5 ii A) 2CsOH(aq) + H2 B) CsO2(s) C) Cs2O + O2 D) CsNH2 + H2 (solvated e- as an intermediate) 12.1 Be has large polarizing power and high charge density due to its small radius. Descending group II increases the size of the atom, enhances electropositive character, and results in higher ionic characteristics in their compounds for larger group II ions. See also II below. 12.3 A) Ba(OH)2(aq) + H2(g) B) BaCO3 C) BaC2 (carbide) D) BaCl2 + 2H2O I. See last page II. B: [He] 2s2 2p1 Al: [Ne] 3s2 3p1 Ga: [Ar] 4s23d104p1 B > Al 2p vs. 3p orbital Ga > Al appearance of 3d shell explains Ga high ionization energy (imperfect screening) 18 extra protons in nucleus III. Entropy of hydration follows the trend Na+>Mg2+>Al3+ because the second solvation shell can carry more water molecules as the charge on the ion increases. These extra water molecules cause for a higher ordering and thus a more negative entropy. IV. 2Na(s) + xss O2(g) + heat Na2O2(s) Mg3N2(s) + 6H2O(l) 3Mg(OH)2 + 2NH3 2K(s) + 2H2O(l) 2KOH + H2 V. Sr2+ Li1+ K1+ I. ...
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