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. If n = 5, l can be 0, 1, 2, 3 or 4. For every value of l, there is one occurance of m_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:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d⁷

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 10²³ mol^-1 = 3.3 x 10²³

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 10²³ = 1.63 x 10¹⁴m

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.

For the answer to the question, divide the distance by the speed of light to get the time.

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 68.926x + 70.925y = 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_s =  ½ .  5, 4 or 3 will work

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

10. A street lamp emits orange light of wavelength 589.2 nm. Assuming that 60% of the electrical power consumed is converted to light at this wavelength, calculate the number of moles of photons being emitted per second if the power consumption is 500 W.

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 )(3.0 10⁸ 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.