even answers 11.02-11.30.09

even answers 11.02-11.30.09 - Even-numbered answers for...

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Even-numbered answers for recitations Nov 2 – Nov 30 (7.8) Assuming eka- means one place below or under, eka-manganese on Table 7.2 is technetium,Tc. (7.12) (a) Electrostatic attraction for the nucleus lowers the energy of an electron, while electron– electron repulsions increase this energy. The concept of effective nuclear charge allows us to model this increase in the energy of an electron as a smaller net attraction to a nucleus with a smaller positive charge, Z eff . (b) In Be (or any element), the 1s electrons are not shielded by any core electrons, so they experience a much greater Z eff than the 2s electrons. (7.28) (a) As Z stays constant and the number of electrons increases, the electron–electron repulsions increase, the electrons spread apart, and the ions become larger. (b) Going down a column, the increasing average distance of the outer electrons from the nucleus causes the size of particles with like charge to increase. (c) Fe: [Ar]4s 2 3d 6 ; Fe 2+ : [Ar]3d 6 ; Fe 3+ : [Ar]3d 5 . The 4s valence electrons in Fe are on average farther from the nucleus than the 3d electrons, so Fe is larger than Fe 2+ . Since there are five 3d orbitals, in Fe 2+ , at least one orbital must contain a pair of electrons. Removing one electron to form Fe 3+ significantly reduces repulsion, increasing the nuclear charge experienced by each of the other d electrons and decreasing the size of the ion. (7.44) (a) Moving from F to I in group 7A, first ionization energies decrease, and atomic radii increase. The greater the atomic radius, the smaller the electrostatic attraction of an outer electron for the nucleus and the smaller the ionization energy of the element. (b)
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even answers 11.02-11.30.09 - Even-numbered answers for...

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