321 Homework 1

2.10

Ground State _/__ _/__ _/__, maximum orbitals used, maximum unpaired electrons of same spin

Excited States _/\ _ _/__ ___; _/__ _\__ _/__, paired electrons (opposite spins

Impossible _//_ _/__ ___; electrons in the same orbital cannot have the same spin

Exchange Energy Maximized _/__ _/__ _/__

Coulombic Repulsion Maximized _/\_ _/__ ___, since electrons are in same orbital, r is smallest (same space)

 

 

2.11

Eex = S {N(N-1)}/2 K, N = no. of unpaired electrons.

For p, N = 3 -> 3(2)/2 = 3; for d, N = 5 -> 5(4)/2 = 10, a much larger number; thus the stabilization of a half-filled d shell is more pronounced.

 

 

2.14

5B 1s22s22p1 -> 1 unpaired electron

23V 1s22s22p63s23p64s23d3 -> 3 unpaired electrons

71Lu 1s22s22p63s23p64s23d104p65s24d105p66s25d14f14 -> 1 unpaired electron

 

2.15 ALL TERM SYMBOLS

B

1

 

 

/

0

 

/

 

-1

/

 

 

ML

-1

0

1

L = 1; unpaired electrons = 1; multiplicity = 2; 2P is only term symbol

Lu

2

 

 

 

 

/

1

 

 

 

/

 

0

 

 

/

 

 

-1

 

/

 

 

 

-2

/

 

 

 

 

ML

-2

-1

0

1

2

L = 2; unpaired electrons = 1; multiplicity = 2; 2D is only term symbol

V

2

 

 

/

 

/

/

/

 

/

/

1

 

/

 

/

 

 

/

/

 

/

0

/

 

 

/

/

/

 

/

/

/

-1

/

/

/

 

 

 

 

/

/

 

-2

/

/

/

/

/

/

/

 

 

 

ML

-3

-2

-1

-1

0

1

 

0

2

3

Upaired electrons = 3; multiplicity = 4; L = 3; 1; 4F and 4P

 

2

 

 

 

 

\

 

 

 

 

\

 

 

 

 

\

1

 

 

 

\

 

 

 

 

\

 

/

/

/

/\

/

0

 

 

\

 

 

/

/

/\

/

/

 

 

\

 

 

-1

/

/\

/

/

/

 

\

 

 

 

 

\

 

 

 

-2

/\

/

/

/

/

/\

/

/

/

/

/\

/

/

/

/

ML

-5

-4

-3

-2

-1

-4

-3

-2

-1

0

-3

-2

-1

0

1

2

/

/

/

/

/\

 

 

 

\

 

 

 

\

1

 

 

 

\

 

 

 

\

 

/

/

/\

/

0

 

 

\

 

 

/

/\

/

/

 

\

 

 

-1

 

\

 

 

 

/\

/

/

/

/\

/

/

/

-2

/\

/

/

/

/

 

 

 

 

 

 

 

 

ML

-2

-1

0

1

2

-2

-1

0

1

-1

0

1

2

2

/

/

/

/\

 

 

\

/

/

/\

/

/\

1

 

 

\

 

/

/\

/

 

\

 

/\

/

0

 

\

 

 

/\

/

/

/\

/

/

 

 

-1

/\

/

/

/

 

 

 

 

 

 

 

 

-2

 

 

 

 

 

 

 

 

 

 

 

 

ML

0

1

2

3

1

2

3

2

3

4

4

5

Unpaired electrons =, multiplicity = 2; L = 5, 4, 3, 2, 2, 1; 2H, 2G, 2F, 2x 2D; 2P

Microstates = 10! / (7! X 3!) = 10 x 3 x 4 = 120 microstates

Term Symbol

4P

4F

2H

2G

2F

2 x 2D

2P

Microstates

4 x 3

4 x 7

2 x 11

2 x 9

2 x 7

2 x 2 x 5

2 x 3

 

12

28

22

18

14

20

6

= 120 microstates

 

 

2.16 and 2.17

Ti = 1s2 2s2 2p6 3s2 3p6 4s2 3d2 -> Ti3+ = 1s2 2s2 2p6 3s2 3p6 3d1 because 4s (valence) electrons are lost first; [Ar] 3d1 ->

/

 

 

 

 

2

1

0

-1

-2

1 unpaired electron, multiplicity = 2, L = 2; Ground State Term Symbol = 2D

Mn = 1s2 2s2 2p6 3s2 3p6 4s2 3d5; - > Mn2+ 1s2 2s2 2p6 3s2 3p6 3d5 -> [Ar] 3d5

/

/

/

/

/

2

1

0

-1

-2

5 unpaired electrons, multiplicity = 6, L = 0; Ground State Term Symbol = 6S

Cu = 1s2 2s2 2p6 3s2 3p6 4s1 3d10; - > Cu2+ 1s2 2s2 2p6 3s2 3p6 3d9 -> [Ar] 3d9;

Not that Cu has an anomalous configuration but that it does not affect this ion.

/\

/\

/\

/\

/

2

1

0

-1

-2

1 unpaired electron, multiplicity = 2, L = 2; Ground State Term Symbol = 2D;

note hole formalism d9 = d1 hole

Pd = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10; - > Pd2+ 1s2 2s2 2p6 3s2 3p6 3d10 4p6 4d8 -> [Kr] 4d8; note that Pd has an anomalous configuration but that it does not affect this ion

/\

/\

/\

/

/

2

1

0

-1

-2

2 unpaired electrons, multiplicity = 3, L = 3; Ground State Term Symbol = 3F

Gd = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 5d1 4f7; - > Gd3+ [Xe] 4f7 ; not highest n electrons are removed first.

/

/

/

/

/

/

/

3

2

1

0

-1

-2

-3

7 unpaired electrons; multiplicity = 8, L = 0; Ground State Term Symbol = 8S

 

 

2.19

Co (1s2) (2s2,2p6) (3s2, 3p6) (3d7)(4s2); calculate effective nuclear charge for one of the 4s electrons

27 (at.no.) - [10 first and second shell (1.00) + 15 third shell (0.85) - 1 other 4s (0.35) ] = 26-23.1 = 3.9

Mn (1s2) (2s2,2p6) (3s2, 3p6) (3d5)(4s1); [anomalous configuration]; calculate for a 3d electron

25 - [18(1.00) + 4 (0.35)] = 25 - 19.4 = 5.6 [all electrons in lower n orbitals shield completely; higher n electrons don't shield at all.

 

2.20 First ionization energies

Li > Cs because of greater electrostatic attraction of electrons closer to positive nucleus

F > Br ditto

Cu > Sc because greater ENC from buildup across d elements attracts electrons more

Pt > Cu because greater ENC from filling f orbitals (even though more distant)

 

2.23 Atomic radii

Cs+ > K+ because electrons are in higher valence shell

La3+ > Lu3+ because La3+ has lower ENC

Br > Cl because electrons are in higher valence shell

Ca2+ > Zn2+ because Ca2+ has lower ENC

Fr > Cs because electrons are in higher valence shell

 

2.24 Electron affinity

Li > Cs because of smaller shell = greater Coulombic attractrion

F > Li because greater ENC attracts extra electron more strongly

F > Cs ditto greater ENC

Cl > F because "crowding" -> repulsion in 2nd shell leads to extra electron not being attracted as strongly

Cl > Br because of smaller shell = greater Coulombic attraction without "crowding" effect

S > O because of "crowding/repulsion in n = 2 shell

S > Se because S has smaller shell = greater Coulombic attraction without "crowding" effect