Brief Answers to Critical Thinking Questions from Tutorial 1
Model 1: Bonding in Main Group Diatomic Molecules
1.
See below.
Molecule
Li2
Be2
B2
C2
N2
O2
F2
Ne2
Bond length / Å
2.67
-
1.59
1.24
1.09
1.21
1.42
-
Bond order
1
0
1
2
3
2
1
0
Paramagnetic?
✗
✗
✓
✗
✗
✓
✗
✗
Stable?
✓
✗
✓
✓
✓
✓
✓
✗
2.
Briefly:
• The molecules with zero bond orders are not stable so do not have bond lengths.
• Molecules with higher bond orders have shorter, stronger bonds.
• Because of the decrease in atomic size across the period, the bond length in O2 is shorter than
that in C2, the bond length of F2 is shorter than that in B2 which is itself shorter than that in Li2,
even though the bond orders are the same.
3.
2 (two)
4.
The decrease in bond order will lead to:
(a)
An increase in the bond length
(b)
A decrease in the vibrational frequency
(c)
A decrease in the dissociation energy
Model 2: Using Waves to Predict the Properties of
Particles
1.
See top diagram opposite.
2.
See top diagram opposite.
3.
See top diagram opposite.
4.
Shown by the blue line in top diagram opposite.
5.
See middle diagram opposite.
6.
See bottom diagram opposite.
Model 3: A Very Short Course in the Mathematics of Waves
!"
= Akcoskx
= -Ak2sinkx
1.
(a)
2.
It is zero when kx = 0, π, 2π, 3π, 4π, 5π …. (i.e. 0, 180o, 360o, 540o, 720o, 900o…..)
!"
(b)
!!!
!! !
Model 4: The Schrödinger Equation
1.
(a)
This function violates (iii)
If the wavefunction is infinite, its square is too. Such a function does not fit with the Born
interpretation. Function (a) is infinite over a range of values of x.
(b)
This function violates (ii).
The Born interpretation is that the square of the wavefunction is related to the probability. If the
function has more than one value at some position, the probability also has more than one value
which is absurd. Function (b) has multiple values where it loops back on itself.
(d)
This function violates (i).
This requirement stems from the wavefunction being a solution to a second-order differential
equation. For the second derivative to be defined, the function must be continuous.
Model 5: Solving the Schrödinger Equation
1.
When x = L, ψ(x) = AsinkL. This must equal zero. This occurs when kL = 0, π, 2π, 3π, 4π, 5π ....
In general, kL = nπ where n = 0, 1, 2, 3, 4, 5…. so
k = nπ / L
The general form of the wavefunctions is therefore:
ψ(x) = Asin(nπx/L)
where n = 1, 2, 3, 4, 5….
Note that that n cannot be zero since this would give ψ(x) = 0 everywhere.
2.
See below.
𝑛𝜋𝑥
𝑑𝜓 𝛿(𝐴sin 𝐿 )
𝐴𝑛𝜋
𝑛𝜋𝑥
=
= cos
𝑑𝑥
𝛿𝑥
𝐿
𝐿
𝑛𝜋𝑥
𝐴𝑛𝜋
𝑛𝜋𝑥
!
𝛿( 𝐿 cos 𝐿 )
𝑑 ! 𝜓 𝛿 (𝐴sin 𝐿 )
𝐴𝑛! 𝜋 !
𝑛𝜋𝑥
=
=
=
−
sin
𝑑𝑥 !
𝛿𝑥 !
𝛿𝑥
𝐿!
𝐿
Substituting this in to
!! !
!! ! !
−
!! ! ! ! !
!! ! ! !" !
!! ! ! !
!!
sin
= 𝐸𝜓, gives:
!"#
!
= +
!! !!
× Asin
!!!!
So,
E=
!! !!
!!!!
with n = 1, 2, 3, 4, 5 ...
!"#
!
= 𝐸𝜓