Chemistry Homework #4
Answer Key
 

Problems from Kotz & Treichel
 

11)

Mg: [Ne]3s2        ­¯       ­¯       ­¯ ­¯­¯       ­¯
                            1s         2s             2p              3s
 

S: [Ne]3s23p4     ­¯       ­¯       ­¯ ­¯­¯       ­¯       ­¯ ­ ­
                           1s         2s             2p              3s               3p

In each case the number of valence electrons equals the column number of the element on the Periodic Table.
 
 

23)

Na+                     ­¯        ­¯        ­¯ ­¯­¯
                            1s         2s             2p

Al3+                     ­¯        ­¯        ­¯ ­¯­¯
                            1s         2s             2p

Ge2+                    ­¯       ­¯        ­¯ ­¯ ­¯        ­¯       ­¯ ­¯ ­¯       ­¯    ­¯ ­¯­¯ ­¯ ­¯
                            1s         2s                 2p              3s             3p                4s                3d

F-                        ­¯        ­¯        ­¯ ­¯­¯
                            1s         2s             2p
 

41 a) n = 2, l = 1: 6 electrons (in the 2p orbitals)

    b) n = 3: 18 electrons (in the 3s, 3p and 3d orbitals)

    c) n = 3, l= 2: 10 electrons (in the 3d orbitals)

    d) n = 4, l= 1, ml = -1, ms = -½: 1 electrons (in a single 4p orbital)

    e) n = 5, l= 0, ml = 1: none, ml cannot be greater than l
 
 

43    a) upper value for l is n-1. An allowed set would be: n = 2, l= 1, ml = 0, ms = +½
        b) ms must be +½ or -½. An allowed set would be: n = 2, l= 1, ml = -1, ms = +½
        c) This is an allowed set of quantum numbers.

  49    a) Cs has the larger radius (it has a higher occupied principal quantum level)

        b) O2- has the larger radius (it has more electrons with the same number of protons; electron-electron repulsion will result in a larger radius)

        c) As has the larger radius (it has a smaller nuclear charge and greater relative electron-electron repulsion)

  53) The order of ionization energy is: K < Ca < Si < As
 
 

57 a) The order of increasing radius is: F < O < S

    b) Sulfur has the largest ionization energy

    c) The order of increasing radius is: Ne < F- < O2- < N3-

    d) The order of ionization energy is: Sr > Ba > Cs
 
 

67 a) A is a metal (calcium)

    b) B is a non-metal (bromine)

    c) Bromine will have the larger ionization energy

    d) Bromine will have the smaller atomic radius
 
 
 

75 a) Vanadium

    b) Vanadium is in group 5B and in the fourth period.

    c) Vanadium is a transition metal!

    d) Vanadium is paramagnetic, having 3 unpaired electrons.

    e) For the five valence electrons, the quantum numbers are:
            4s:   n = 4, l= 0, ml = 0, ms = +½
                   n = 4, l= 0, ml = 0, ms = -½

            3d:   n = 3, l= 2, ml = 2, ms = +½
                    n = 3, l= 2, ml = 1, ms = +½
                    n = 3, l= 2, ml = 0, ms = +½

            Note that in the 3d orbitals, other values of ml would also be valid.
 

79)   First ionization energy for potassium: K ® K+ + e- ;        [Ar]4s1 ® [Ar]
        Second ionization energy for potassium: K+® K2+ + e- ;  [Ar] ® [Ne]3s23p5

The second ionization energy is much larger than the first because the second electron removed is from the much more stable 3p orbital. The first electron removed is from the higher energy 4s and is easier to separate remove from the atom.
 
    85) Group 5A elements typically have much lower electron affinities relative to 4A elements because the 4A elements have a valence electron configuration of ns2np2. The p orbitals are less than half-full and an additional electron can occupy a previously vacant orbital. The 5A elements have an electron configuration of ns2np3. Here each p orbital is occupied by a single electron. Any additional electron would need to spin pair with an electron that is already in a p orbital. This is energetically unfavorable.   Additional Problems
 
 

1) The number of unpaired electrons in each atom is:

        carbon: 2 unpaired electrons
        fluorine: 1 unpaired electron
        neon: 0 unpaired electrons
 
 

2) The atoms with the higher electron affinity are:

a) Cu > Zn: an additional electron in copper would occupy the 4s orbital, which is only partially full. An additional electron in zinc would occupy the much higher energy 4p orbital.

b) K > Ca: an additional electron in potassium would occupy the 4s orbital, which is only partially full. An additional electron in zinc would occupy the higher energy 3d orbital.

c) Cl > S: both additional electrons would occupy 3p orbitals; chlorine, because of its greater nuclear charge has a greater attraction for an additional electron.

d) H > Li: additional electron in hydrogen would occupy the 1s orbital; the additional electron in lithium would occupy the higher energy 2s.

e) Ge > As: both electrons would occupy 4p orbitals, but the additional electron of germanium would be the only electron in a 4p, while that for arsenic would be forced to spin pair with an existing 4p electron. This would cause significantly greater electron-electron repulsion.