Chem 481. Chapter 1. Atomic Stucture And Periodic Table Worksheet With Answers Page 5
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1
2
3
4
5
6
7
8
9
10shell, which we would expect to be further from the nucleus, it experiences so much additional nuclear charge that it is
actually slightly harder to remove a Ra electron than a Ba electron.
21.
Account for the trends, element by element, across period 3 in (a) ionization energy, (b) enthalpy of electron attachment,
and (c) electronegativity.
st
element
electron configuration
1
IP (eV)
Electron-Gain enthalpy
Electronegativity (χ)
1
Na
[Ne]3s
5.14
-0.548
0.93
2
Mg
[Ne]3s
7.64
+0.4
1.31
2
1
Al
[Ne]3s
3p
5.98
-0.441
1.61
2
2
Si
[Ne]3s
3p
8.15
-1.385
1.90
2
3
P
[Ne]3s
3p
11.0
-0.747
2.19
2
4
S
[Ne]3s
3p
10.36
-2.077
2.58
2
5
Cl
[Ne]3s
3p
13.10
-3.617
3.16
2
6
Ar
[Ne]3s
3p
15.76
+1.0
n/a
All the magnitudes increase from left to right across the period, since each trend reflects the increasing Z* as electrons are
being added to orbitals of the same shell. For IP, this reflects the difficulty of removing an electron. For EG enthalpy, this
reflects energy released when an additional electron is attached to the neutral atom, which is more likely when Z* is larger.
Electronegativity also grows (remember it is defined as the average of the previous two categories in Mulliken's definition!)
There are some reversals: the IP of Al is smaller than that of Mg. This is because the electron removed from the former
comes from a new subshell, 2p, which will be easier to do than removing a filled s subshell for Mg. Also, S is easier than P,
because phosphorus has a half-filled p subshell, which also has additional stability. The E-G enthalpies of Mg and Ar are
anomalous. Mg is full (3s), so that the additional electron would have to go into the 3p level. There is no driving force for
this procedure, and it is in fact endothermic. Similary, Ar needs to attract an additional electron into the next shell, and this
is also not favourable. The electronegativity grows much more gently and regularly than the IP, as expected for a property
which is the average of IP and E-G.
22.
Unlike the electronegativities of the other main groups, those of Groups 13 and 14 pass through a minimum on
descending the group. To what can you attribute this trend?
χ for groups 13 and 14 is lowest for Al and Si. This can be attributed to the Scandide contraction, i.e. to the filling of the
first d level between Al-Si and Ga -Ge. Thus Ga and Ge have anomalously high electronegativities.
23.
What are the names, symbols, and roles of the four quantum numbers associated with the hydrogen atom?
•
n = principal quantum number - controls the energy of the electrons in a hydrogen atom
•
l = orbital angular momentum q.n. - dictates the shape of the orbital
•
m
= orbital magnetic q.n. - determines the orientation of the orbital in 3-D space
l
•
m
= electron spin q.n. - determines the spin of the electron, which may be up or down
s
24.
Sketch the orbitals of atomic hydrogen for the first three quantum shells. In which of these orbitals are the signs
associated with the wave functions arranged symmetrically?
For sketches of the orbitals, see the lecture notes on p.23. Note that the s and d orbitals are symmetric.
25.
What orbitals do the following quantum number combinations describe? Are they even permissible solutions of the
Schrödinger equation? (i)
n = 2
l = 1
m
= -1 (ii)
n = 1
l = 0
m
= -4
l
l
(i) n = 2
l = 1
m
= -1 a 2p orbital (allowed)
l
(ii) n = 1
l = 0
m
= -4 a 1s orbital, not permissible with a value of m
= -4
l
l
26.
Which of the following orbitals are allowed for the hydrogen atom by quantum mechanics? 5p; 5h; 6d
5p is permissible, since for n =5, l can be up to 4, which includes l = 1
5h is not permissible, since this requires l = 6
6d is permissible, since for n =6, l can be up to 5, which includes l = 2
27. Consider the following radial probability d ensity-distribution plot and respond to the associated questions.
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