Practice Quiz -
solutions

1. Provide the
appropriate value for the missing quantum number:

n=1, l= 0, m_{l}= _______?, m_{s}=1/2

Since
l = 0, m_{l}
can only be zero as well.

2. How many
electrons can there be with m_{l} = 0 when n=5?

______________

Only 2 electrons are allowed in an orbital._{l} = 0. The answer is 10.

3. Write down the complete electronic
configurations for the following neutral atoms.

a)
chromium

This
was covered in class, and is also in the text.

b)
meitnerium
(This is a newly discovered element of Z=109 which appears in the periodic
table below iridium in group 9.)

The
best we can do is apply our Aufbau
rules – which may or may not apply in reality to these super-heavy elements.

Here
goes nothing:

1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^{10}6p^{6}7s^{2}5f^{14}6d^{7}

4. Pick the best answers for the following
multiple choice questions:

i) Hund’s rule
states that the lowest energy configuration for an atom is the one with

a)
maximum number of paired electrons

b)
lowest value of n

c)
minimum multiplicity

d)
maximum number of
unpaired electrons

ii)
The ability of a electron in an
orbital to shield another electron from the nuclear charge is:

a)
independent of the value of *l*

b)
depends only on n

c)
depends on m_{s} among other
things

d) depends on *l*

5. Calculate the number of electrons in 1.00 mL of water at 298 K.

Some general knowledge is assumed in
this question. The density of water is
about 1.00 g/mL at room temperature, so we’re talking
about one gram of water. In chemistry,
when we talk about the amount of stuff, we mean how many moles. So, find the number of moles of water here:

n = 1.00g / 18.02 g mol^{-1 } = 0.05549 mol

Now, each *molecule* of water has 10 electrons (equal to the sum of the atomic
numbers of the atoms). Thus the number
of moles of electrons is 0.05549 mol X 10 = 0.5549 mol

This could suffice for the answer: 0.5549 mol of electrons, since a mole is just a counting
number, like a dozen.

But, we could also say:

Number of e^{-} = 0.5549 mol X
6.022 x 10^{23} mol^{-1} = 3.3 x 10^{23}

6. The radius of an iridium atom is 135.5 pm. How long does it take for light (in a vacuum) to traverse the distance equal to a mole of iridium atoms placed right next to each other so as to form a straight line?

These atoms form a line this long: 2 X 135.5 x 10^{-12}
m X 6.022 x 10^{23} = 1.63 x 10^{14}m

*Think* about this – it’s a line billions of kilometers
long! Yet a mole of iridium, with a mass
of 192.2 g/mol and a density of 22.5 g/mL has a
volume of 8.5 mL or so. As an exercise, why not work out the volume
of a mole of iridium atoms, assuming they are packed this far apart.

7. Gallium consists of two naturally-occurring isotopes with masses of 68.926

and 70.925 amu. The average atomic mass for gallium is 69.72 amu.

Calculate the *fractional abundance*
for each isotope.

You
need to solve a system of two equations in two variables here.

Fractional
abundance is the decimal fraction of each isotope. For example, if you have half of the sample
being one isotope, its fractional abundance is 0.50.

Let *x* = the fractional abundance of the
first isotope and *y* = the second one.

Then
68.926*x* + 70.925*y* = 69.72

And,
since there are only two isotopes, *x*
+ *y _{ }* = 1

Substitute
*x* = 1 – *y* into the first equation and solve for *y*.

You
will find that *y* = 0.397.

8. a) How many orbitals are there in the n = 3 level of the hydrogen atom?

___________________. 9

b) Provide an/the appropriate value for the missing quantum number:

n = 6, l
= ____? , m* _{l}* = -3, m

c) How many electrons in the *first
shell* of an atom with Z >1 can have the quantum number m* _{l}* = 0 ?
________________ n = 1, so only one orbital – the
answer is 2

d) How many d-electrons are there
in a Pd atom? ____________ Pd has both 3d and 4d
electrons, so the answer is 18

e) A certain isotope E has a mass number equal to three times its atomic number and a neutron number that is equal to two times its atomic number. Identify isotope E.

Doggonit, I
should’ve saved this for the final exam!
Here’s a hint: it’s an isotope of a very common element.

9. Calculate the deBroglie wavelength of a baseball with a mass of 250 g moving at 30 m/s.

λ
= h / mv

Only 0.6 X 500 W = 300 W is
ending up as light at this wavelength.
Remember that a watt is a joule per second, so that this is 300 joules
per second.

How many of these photons are
there in 300 joules? Each photon has an
energy of

E = hn.

n = c / λ so E = hc / λ

Use this relation to calculate
the energy per photon.

E = hc
/ λ = (6.626 x 10^{-34 J} s )(3.0 ^{ }10^{8} m s^{-1}) /
(589.2 x 10^{-9} m) = 3.37 x 10^{-19} J per photon.

We need to find the number of photons in 300 J of this colour of light:

Number of photons = 300 J / 3.37
x 10^{-19} J per photon

Convert this answer to moles,
and you’re all done.