8. A
Recall that the superscript (i.e., the A in AC) refers to the mass number of an atom, which is equal to the number of protons plus the number of neutrons present in an element. (Sometimes a text will list the atomic number, Z, or total number of protons, under the mass number A.) According to the periodic table, carbon contains 6 protons; therefore, its atomic number (Z) = 6. An isotope contains the same number of protons and a different number of neutrons as the element. Carbon is most likely to have an atomic number of 12, for 6 protons and 6 neutrons. (C) and (D) are possible isotopes that would have more neutrons than does 12C. The 6C isotope is unlikely. It would mean that there were 6 protons and 0 neutrons, and it would probably collapse under the stress of the positive charge.
9. C
The Heisenberg uncertainty principle states that you cannot know the position and momentum of a particle simultaneously, which eliminates (A) and (D). Momentum depends on velocity (recall from Newtonian mechanics that p = mv), so in order to calculate momentum, velocity must be known. (C) explains this.
10. B
For the electron to gain energy, it must absorb photons to jump up to a higher energy level. This eliminates (A) and (D). Between (B) and (C), there is a bigger jump between n = 2 and n = 6 than there is between n = 3 and n = 4. Therefore, (B) represents the greatest energy gain.
11. A
The MCAT covers qualitative topics more often than quantitative topics in this unit. It is critical to be able to distinguish the fundamental principles that determine electron organization, which are usually known by the names of the scientists who discovered them. The Heisenberg uncertainty principle refers to the momentum and position of a single electron, and the Bohr model was an early attempt to describe the behavior of the single electron in a hydrogen atom. (D) is tempting, but (A) is more complete and therefore the correct answer. The element shown here, nitrogen, is often used to demonstrate Hund’s rule because it is the smallest element with a half-filled p subshell. Hund’s rule explains that electrons fill empty orbitals first, and in fact, the three p-electrons in this image each occupy a separate orbital. Hund’s rule is really a corollary of the Pauli exclusion principle, in that the Pauli exclusion principle suggests that each orbital contains two electrons of opposite spin. Additional electrons must fill new orbitals so the compound remains stable in its ground state.
12. A
The quickest way to solve this problem is to use the periodic table and find out how many protons are in Cs atoms; there are 55. Neutral Cs atoms would also have 55 electrons. A Cs cation is most likely to have a single positive charge because it has one unpaired s-electron. This translates to one fewer electron than the number or protons, or 54 electrons.
13. B
The easiest way to approach this problem is to set up a system of two algebraic equations, where x and y are the percentages of H (mass = 1 amu) and D (mass = 2 amu), respectively. Your setup should look like the following system:
x + y = 1 (proportion H (x) + proportion D ( y) in whole, x% + y% = 100%).
1x + 2y = 1.008 (the total atomic mass).
Substitute one variable for the other so the atomic mass is in terms of one variable (1–y = x), then solve for the other percentage ([1–y] + 2y = 1.008 simplifies to 0.008 = y, or 0.8% D). That plus 99.2 percent H makes 100 percent. Another way to examine this problem is if we had 50 percent of H and 50 percent of D, we could probably imagine an atomic mass of 1.5 amu. Therefore, if 50 percent D gave a mass of 1.5 amu, 80 percent D would yield an atomic mass of 1.8 amu. Eight percent would result in 1.08 amu, and 0.8 percent in 1.008 amu, which is the desired mass.
14. A
The terms in the answer choices refer to the magnetic spin of the two electrons. The quantum number ms represents this property as a measure of the electrons’ relative intrinsic angular momentum. These electrons’ spins are parallel, in that their spins are aligned in the same direction (ms = +½ for both species). (B), (C), and (D) suppose that ms = +½ for one electron and -½ for the other. (D) implies that the two electrons have opposite spins but lie within the same orbital.
15. D
When dealing with ions, you cannot directly approach electronic configurations based on the number of electrons they currently hold. First examine the neutral atom’s configuration, and then determine which electrons would have been removed.
Neutral Atom’s Configuration Ion’s Configuration Cr0: [Ar] 4s13d5—Mn0: [Ar] 4s23d5Mn+: [Ar] 4s13d5Fe0: [Ar] 4s23d6Fe2+: [Ar] 4s03d6
Due to the stability of half-filled d-orbitals, neutral chromium assumes the electron configuration of [Ar] 4s13d5. Mn must lose one electron from its initial configuration to become the Mn+ cation. That electron would come from the 4s-orbital, because the 3d-orbital would strongly prefer to remain half-full. This d-orbital’s desire to be half-full trumps the s-orbital’s desire to be completely full, because it is at a higher energy. Fe must lose two electrons to become Fe2+. They’ll both be lost from the same orbital, because the scientist can’t pick and choose which electrons to take! Moreover, the only way Fe2+ could hold the configuration in the question stem would be if one d-electron and one s-electron were lost together. Therefore, both answer choices (A) and (B) match the electron configuration in the question stem.
16. D
(D) is the only incorrect statement in the set. Electrons are assumed to be in motion in any energy level, even in the ground state. All of the other statements are true. An electron’s ground state describes its position at the lowest and most stable energy level, i.e., lowest n value. It must be an integer value in the ground state and in the excited state. This number gives a relative indication of the electron’s distance from the nucleus, so it must have the smallest radius of all the energy levels. The farther the electron moves from the nucleus, the greater energy it needs to overcome its attractive forces.
17. C
High pressure is unlikely to excite an electron out of the ground state unless it causes an extreme change in temperature, which would add enough energy to the system to promote the electron. There is not enough information in the problem to determine whether the stated high pressure would be high enough, as the answer is too generic. Irradiation has the same effect as temperature.