4U Matter Practice Test
SCH4U: Matter (Structure and Properties)
PRACTICE FOR THE UNIT TEST
You may use a clean copy of the Alchem periodic table, and a blank copy of the energy level diagram (given out when we learned that topic).
1. Write full and shorthandelectron configurations both atoms and ions, e.g., for Zn atoms and Zn ions, or for Se atoms and selenide ions. Be sure you know the different step to take when forming a cation … i.e., which electrons do you remove first? (It is in your text and we covered it in class.)
Zn: 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10[Ar] 4s2, 3d10
Zn2+: 1s2, 2s2, 2p6, 3s2, 3p6, 3d10[Ar] 3d10 (notice that electrons from the orbitals
with the highest principle number are lost)
Se: 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p4 [Ar] 4s2, 3d10, 4p4
Se2−: 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6[Ar] 4s2, 3d10, 4p6
2.Write the quantum numbers for some particular electron shown in an electron level diagram. For example, in the diagram below, one electron is in bold. What are its quantum numbers? Also, what atom does is represented by the diagram?
n = 3, l = 1, ml= −1, ms = + ½ (although the last one is arguable - which electron should be considered +½ and which should be - ½ ? It doesn’t really matter).
The atom represented is Fe just count the electrons!
s / p / d3d
4s
3p
3s
2p
2s
1s
3. How is the presence of a finite set of distinct spectral lines evidence for the quantized nature of electrons? In other words, when you examine the spectrum of hydrogen gas (or another gas), you only see a set of individual coloured lines. You do not see the full spectrum of all visible colours. How does this support the quantum theory of electrons?
See section 3.4
4. Classify bonds (as ionic, strongly polar covalent, weakly polar, or non-polar covalent) and support your answer.
i) Mg - Fionic: eneg. difference = 2.8, much more than 1.7
ii) C – Fstrongly polar covalent: eneg. diff. = 1.5, close to 1.7 but still less
iii) O – Fweakly polar covalent: eneg. diff. = 0.5, just slightly above 0.4
iv) Be - Fstrongly polar covalent! eneg. diff. = 2.5, so you’d expect ionic. However, this is one of those exceptions to the usual rules. In case you are interested, the reason is related to the surprisingly small diameter of Be. This gives it what is known as a high “effective nuclear charge”. This high charge means that with only 2 valence electrons in the 2s orbital, it can hold them more tightly than the electronegativity suggests. If you look at the atomic radius of Be on the periodic table, and compare it to all the other group I and II metals, you can see that it is quite a bit smaller. As a result, there is some covalent (sharing) character to the bond. However, it is certainly on the fringe of being ionic I think.
5. Name and draw the shape of several molecules or ions, using a Lewis structure and discussing VSEPR theory. An example is done in the first row.
Substance / Lewis diagram / Shape – draw and name it / VSEPR explanationoxygen (O2) / / O O
linear / Each O atom has 2 bonding pairs in a double bond. Count this as one pair. It also has 2 lone pairs. The total is 3 pairs. They are most spread out if trigonal planar. However only the two O atoms are considered, so the molecule is linear.
Carbonate ion, CO32− / / trigonal planar
/ C is at the centre of 3 regions of electron pairing. There are no lone pairs on the C to push the bonded atoms out of the plane.
P.S. The diagram I included here is not quite accurate; it shows 2 single bonds and 1 double bond. In fact all 3 bonds are equal “resonance” or “delocalized” bonds, such as we saw when learning about benzene.
Nitrogen triiodide, NI3 / / trigonal pyramidal
/ There are 3 bonding pairs, and 1 lone pair on the central N atom. These 4 pairs occupy four regions, and point towards the vertices of a tetrahedron. However, only the 3 atoms of I are considered, so the shape is not called tetrahedral.
Sulfur difluoride, SF2 / / V-shaped:
In this drawing, E represents electron pairs that are not bonded, A represents the S atom, and X represents F atoms. / There are 2 bonding pairs, and 2 lone pairs. As usual, with 4 pairs all together, the electron geometry is a tetrahedron. However, with only 2 atoms bonded to S, they occupy 2 of the corners of a tetrahedron, making a V-shape.
6. Contrast the bond polarity and overall polarity (if any) for different molecules. For example, try XeF2 and SnCl2. Include your evidence and reasoning, with diagrams and explanations.
The two XeF bonds may or may not be polar covalent - we don’t have the electronegativity of Xe to decide. (Now that Xe compounds have been made, Xe has been given an electronegativity of about 2.6, so when subtracted from F’s 4.0, the difference indicates each bond would be polar covalent. But it doesn’t matter - read on…)
The molecular geometry of XeF2 is linear - see appendix C3 for a diagram. So if the 2 bonds are polar, their dipoles would cancel each other.
In the case of SnCl2, there is a V-shape (see Appendix C3.) Also, the SnCl bonds are strongly polar (electronegativity difference is 1.2). Since they are not symmetrical, their dipoles do not cancel each other, and thus the molecule is polar overall.
I now wish I hadn’t chosen that example. It is true that the electronegativity difference between Sn and Cl is only 1.2, indicating a covalent character, and there are other findings that support a covalent type of bond. Still, we in gr.12 are so used to classifying metal-nonmetal bonds as ionic, that I should have given you a different example. I could have used SF2, as shown in the table above, for question #5.
7. Compare (similarities and differences) the three types of van der Waal’s forces, and arrange them from weakest to strongest. Then apply your knowledge to predict the type of intermolecular forces present, and the overall predict strength of these forces combined (strong, medium, weak), for two molecules, such as the two shown below.
See section 4.5, and the videos on the Engrade calendar. In brief, London forces are the strongest, especially for small molecules, dipole interactions are usually stronger, and hydrogen bonding is the strongest.
There are very weak London forces for 1-propanol, but hydrogen bonding will occur, so overall there will be relatively strong van der Waal’s forces. This allows it to stay as a liquid, whereas a similar sized molecule without H bonding (such as propane) would be a gas.
M.P. = -127, B.P. = 97
There will be no H bonding or dipole-dipole interactions for dodecane, but there will be significant London dispersion forces because it consists of so many atoms. So the overall intermolecular forces may be stronger than expected. Dodecane is liquid at room temperature, not gas, like molecules with very little intermolecular forces.
M.P. = -10, B.P. = 216.
Notice that it’s melting point is much higher than 1-propanol, so it is harder to melt. Its intermolecular forces are stronger. This shows that even though we say London forces are weak, as molecules get bigger, they become very significant.
1-propanol, H3C-CH2 – CH2OHdodecane, C12H26