Chemistry 105 - Problem Set 09 – Electron Configurations PDF

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Problem Set 09 – Electron Configurations Chem 105 1. (a) 3.86 Explain how the electron configurations of the group 2 elements are linked to their location in the periodic table. The electron configuration of [core]ns2 that all the group 2 elements have places them all in the same column in the s block of the periodic table. (b) 3.87 How do we know from examining the periodic table that the 4s orbital is filled before the 3d orbitals? As we start from argon core electrons, we move to potassium and calcium, which are located in the s block of the periodic table. It is not until Sc, Ti, V, etc., that we begin to fill electrons in to the 3d subshell. (c) 3.88 Why do so many transition metals form ions with a 2+ charge? In all of the transition metals, the s shells (4s, 5s, 6s, 7s) are filled. Because the s shells of the transition metal have a higher principle quantum number (ie [Kr]4d25s2 for Zr), the electrons in the outer s shells are at a slightly higher energy than the corresponding d shells. Additionally, the s shells carry 2 electrons like the group 2 metals in the periodic chart and as a result of these facts, the transition metal s shells are more likely to be reduced to the 2+ charge. 2. (a) Write out the full electron configuration and the condensed electron configuration for the following atoms or ions: Atomic number 15: 1s22s22p63s23p3 or [Ne] 3s23p3 Atomic number 18: 1s22s22p63s23p6 or [Ar] Atomic number 19: 1s22s22p63s23p64s1 or [Ar] 4s1 Atomic number 22: 1s22s22p63s23p64s23d2 or [Ar] 4s23d2 Atomic number 24: 1s22s22p63s23p64s13d5 or [Ar] 4s13d5 Atomic number 29: 1s22s22p63s23p64s13d10 or [Ar] 4s13d10 Fe2+: 1s22s22p63s23p63d6 or [Ar]3d6 Cu+: 1s22s22p63s23p63d10 or [Ar]3d10 Br –: 1s22s22p63s23p64s23d104p6 or [Ar] 4s23d104p6

3. What ions are the first 4 atoms listed in #2 most likely to form, if any? Atomic number 15 = Phosphorous = 3- anion (gains 3 electrons to fill the 3p shell) Atomic number 18 = Argon = no ion likely (noble gas) Atomic number 19 = Potassium = 1+ cation (will lose its one 4s electron) Atomic number 22 = Titanium =2+ or 4+ cation (often loses all 4 of its valence electrons) 4. Which of the following configurations depicts an excited oxygen atom (not in its ground state)? a. 1s22s22p1 b. 1s22s22p2 c. [He]2s22p4 d. 1s22s22p23s2 e. 1s22s22p4 5. (a) How many valence electrons are in cobalt? What ion(s) is cobalt most likely to form? 9, cations (b) How many valence electrons are in iodine? What ion is iodine likely to form? 7, anions 6. Determine whether the following atoms or ions are paramagnetic or diamagnetic. a) Na+ Diamagnetic f) Li Paramagnetic b) Ca Diamagnetic g) Mg2+ Diamagnetic

c) F– Diamagnetic d) I Paragmagnetic e) Zn2+ Diamagnetic

h) Br– Diamagnetic i) Ag+ Diamagnetic j) Cl Paramagnetic

7. Draw an orbital diagram showing all filled or partially-filled valence subshells for each of the following atoms: 2p a. Oxygen Oxygen has 6 valence electrons, arranged as follows: 2s

b. Chlorine has 7 valence electrons:

c. Carbon has 4 valence electrons:

d. Phosphorous has 5 valence electrons:

8. Something is amiss with each of the following electron configurations. Assuming that the number of electrons present is correct, write what the ground state electron configuration should be and name the element to which it corresponds (assume they are atoms, not ions). a) 1s22s23s1 3s1 is an excited electron, ground state is 1s22s22p1 b) [Ne]2s22p3 [Ne] = 1s22s22p6 so the value of n for 2s22p3 is incorrect. They should be 3s23p3 (i.e., Pauli exclusion principle prevents having 4 electrons in 2s and 9 electrons in 2p). Or, [Ne] could be replaced with [He] to allow a valence of 2s22p3. c) [Ne]3s23d5 3d5 electrons would be excited. The 3p (and 4s) subshells are lower in energy than 3d, this should be [Ne]3s23p5....