1.      A     

There are two ways to get this answer, the hard way and the easy way. To solve this problem the hard way, first multiply the percent abundance by the atomic mass of a given isotope and then add the products together. You’ll see that the hard way actually isn’t very hard:

30.00% = 40.00 amu: 30% is 310%, right? So if 10% of 40 is 4, then 34 = 30% of 40, which is equal to 12 amu.

50.00% = 41.00 amu, and 50% is 1 /2 of 41, which is 20.5 amu.

20.00% = 42.00 amu, and 20% = 210%, so 10% of 42 = 4.2, and 20% = 24.2, which is 8.4 amu.

Now add those three numbers together to get your answer: 12 + 20.5 + 8.4 = 40.9 amu. But the easy way to get the answer is to guesstimate: 50% of the element exists as the 41.00 amu isotope. Now, 30% of the remaining element is the 40.00 amu isotope, and only 20% of the element exists as the 42.00 isotope. Therefore you can estimate that the average of these three amounts should be less than 41 since there is more of the lighter isotope. The only answer choice that’s less than 41 is A.

2.      D     

The fourth principal energy level has four sublevels: s, p, d, and f. If the sublevel is completely filled, then s = 2 electrons, p = 6 electrons, d = 10 electrons, and f = 14 electrons; thus 2 + 6 + 10 + 14 = 32 total electrons for a full fourth principal energy level.

3.      B     

The set of quantum numbers given was n = 3, l = 1, ml = 0, ms = ±1/2. If n = 3, this means that it’s a third energy level electron; if l = 1, then it’s a p-sublevel electron; if ml = 0, then it’s in the middle position of the set of three p orbitals. The only tricky thing is that ms = + or - 1/2. This means it’s either a p2 or a p5 electron. However, if it were p5, then one of the answer choices would be argon (a noble gas), but it isn’t listed, so it must be the p2, which makes silicon the correct answer.

4.      A     

The configuration given is 1s22s22p63s23p63d44s2. The 3d4 is the important part—it means the element we desire is in the first row of the d-block elements and is the fourth element in that block, so it is Cr, or chromium.

5.      C     

The mass number and atomic number must be equal on both sides of the equation in order for the equation to be properly balanced. We are looking for a component that has a mass number of 4 and an atomic number of 2: helium. This is an example of alpha decay, and the answer is C, .

The complete equation is:

6.      B     

This problem is easily solved without a calculator, especially if you’ve been practicing your math skills. The half-life is given as 9.98 minutes, which is mighty close to 10 minutes. The total time is given as 60.0 minutes, so the sample undergoes six half-lives. Start with this mass and keep cutting it in half; each 10-minute half-life should be represented with an arrow, and you can even put numbers under each arrow if you want, in order to keep track., which is B.

7.      E     

This question asks about the second ionization energy. Remember that the second ionization energy of any element is always larger than its first ionization energy. The second ionization energy is significantly larger if the second electron comes from a completed sublevel or principal energy level. Na’s first electron removed is 3s1, while the second to be removed comes from 2p6. There is a huge increase in the amount of energy needed to remove that second electron because of the change in principal energy levels.

8.      T, T     

(Do not fill in CE, for “correct explanation.”) Remember this type of question? We told you about it in the first part of the book. Here we go: statement I is true: hydrogen has a lower IE than helium. The electron removed in each case is from the 1s sublevel, so Zeff becomes very important: H has a Zeff of 1, while He has a Zeff of 2. He attracts its electrons with more force; thus it requires more energy to remove them. Statement II is also true: The halogens are highly electronegative and form polar bonds with hydrogen. However, is statement II the correct explanation for why statement I is true? No, so you would not fill in the CE bubble.

9.      E     

Remember, atomic radii decrease moving across a period from left to right. This is because protons are added to the nucleus and these protons attract the electrons more strongly, pulling them in tighter and decreasing the atomic radius. Atomic radius increases moving down a group or family, however, because the value of n increases as you add another shell, the electrons spend more time away from the nucleus, dissipating the attraction between nuclear protons and themselves. Since fluorine and chlorine are both in the same group but chlorine is below fluorine, we know that chlorine is bigger than fluorine. Sulfur is to the left of both chlorine and fluorine, so it must be the smallest of all. Arranged in order of increasing size, they are S, F, Cl, or answer E.

10.      T, T     

(Do not fill in CE.) In general, the ionization energy increases with increasing Zeff, and the same is true for second ionization energies. B has a higher Zeff than does Be, and since the second ionization energy would be needed to remove protons from the same principal energy level, B’s higher Zeff means that its second ionization energy would be higher than that of Be.

11.      F, F     

(Do not fill in CE.) While oxygen has a lower first ionization energy than nitrogen due to the p4 anomaly, as you pass oxygen (going from left to right across the period), the trend of increasing ionization energy continues; the increased Zeff results in subsequent elements in the period (such as fluorine) having a higher ionization energy than oxygen. Oxygen does not have a higher Zeff than that of fluorine, so the second statement is also false, and you would not fill in the CE oval.

12.      C     

It simply requires the input of energy to remove an electron from any type of atom. This is because energy must be put into the system in order to overcome the favorable attraction between the electron and the positively charged atomic nucleus. A reaction that must consume energy in order to proceed is known as endothermic, while a reaction that gives off energy is exothermic; both of these reactions are endothermic, and the answer is C.

13.      C     

Draw each structure using the rules for drawing Lewis dot structures. Remember that resonance structures refer to two or more Lewis structures that are equally good descriptions of a molecule. So, you’re looking for structures in which you found yourself placing one double bond arbitrarily between a pair of atoms that, elsewhere in the molecule, share only a single pair of atoms. You should also be looking for a combination of single and multiple bonds occurring between the central atom and a member of C, N, O, P, or S.

14.      C     

SF6 has six bonding sites as seen by its Lewis structure, drawn below. It is octahedral. The four equatorial sites have F—S—F bond angles of 90˚, while the two axial sites have F—S—F bond angles of 180˚, so as you can see, choice C is the best answer.

15.      T, T, CE     

The first statement is true: not all molecules that contain a polar bond are themselves polar. And the second statement correctly explains why this is so: in order for a molecule to be polar, it must contain at least one polar bond or unshared pair of electrons that are not arranged symmetrically so as to cancel each other out. If they are arranged so that they do cancel each other out, the molecule will be nonpolar.

16.      T, T     

(Do not fill in CE.) While fluorine is indeed bigger than hydrogen, this is not the reason NH3 is more polar than NF3. The true reason for this is that the three N—F bonds are more polar than the N—H bonds: the difference between the electronegativities of N and H are greater than that between N and F, so in NH3, the nitrogen attracts electrons much more strongly than do the hydrogens, and a significant dipole moment is created.

17.      B     

An octahedral geometry is produced from six bonding sites, sp3d2 hybridization, and an expanded octet. You know it cannot be BeCl2, BF3, or CF4 since neither Be, B, nor C is from the third period or higher. SeF6 and PF5 are both possibilities. The Lewis structures for the remaining choices appear below:

You can see that PF5 has five bonding sites and is thus trigonal bipyramidal. SeF6 has six bonding sites and is octahedral.

18.      D     

All of the Lewis structures appear below. Focus only on the positions of the nuclei—you can’t see the lone pairs, but they determine the molecular shape by their repulsions.

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