## 5 Steps to a 5: AP Chemistry 2017 (2016)

**STEP **__4__

__4__

**Review the Knowledge You Need to Score High**

**CHAPTER 10**

**Spectroscopy, Light, and Electrons**

**IN THIS CHAPTER**

**Summary:** In developing the model of the atom, it was thought initially that all subatomic particles obeyed the laws of classical physics—that is, they were tiny bits of matter behaving like macroscopic pieces of matter. Later, however, it was discovered that this particle view of the atom could not explain many of the observations being made. About this time, the dual particle/wave model of matter began to gain favor. It was discovered that in many cases, especially when dealing with the behavior of electrons, describing some of their behavior in terms of waves explained the observations much better. Thus, the quantum mechanical model of the atom was born.

**Keywords and Equations**

Speed of light, *c* = 3.0 × 10^{8} ms^{−1}

Planck’s constant, *h* = 6.63 × 10^{−34} Js

**The Nature of Light**

Light is a part of the **electromagnetic spectrum** —radiant energy composed of gamma rays, X-rays, ultraviolet light, visible light, etc. __Figure 10.1__ shows the electromagnetic spectrum.

**Figure 10.1 The electromagnetic spectrum.**

The energy of the electromagnetic spectrum moves through space as waves that have three associated variables—frequency, wavelength, and amplitude. The **frequency,** *ν* , is the number of waves that pass a point per second. **Wavelength** (*λ* ) is the distance between two identical points on a wave. **Amplitude** is the height of the wave and is related to the intensity (or brightness, for visible light) of the wave. __Figure 10.2__ shows the wavelength and amplitude of a wave.

The energy associated with a certain frequency of light is related by the equation:

*E* = *hν* where *h* is Planck’s constant = 6.63 × 10^{−34} Js

In developing the quantum mechanical model of the atom, it was found that the electrons can have only certain distinct quantities of energy associated with them, and that in order for the atom to change its energy it has to absorb or emit a certain amount of energy. The energy that is emitted or absorbed is really the difference in the two energy states and can be calculated by:

Δ*E* = *hν*

All electromagnetic radiation travels at about the same speed in a vacuum, 3.0 × 10^{8} m/s. This constant is called the **speed of light ( c )** . The product of the frequency and the wavelength is the speed of light:

*c* = *νλ*

Let’s apply some of the relationships. What wavelength of radiation has photons of energy 7.83 × 10^{−19} J?

**Figure 10.2 Wavelength and amplitude of a wave.**

Answer:

Using the equations

Δ*E* = *hν* and *c* = *νλ*

we get

*ν* = Δ*E* /*h* and *λ* = *c* /*ν*

Inserting the appropriate values:

*n* = Δ*E* /*h* = 7.83 × 10^{−19} J/6.63 × 10^{−34} Js = 1.18 × 10^{15} s^{−1}

Then:

*λ* = *c* /*ν* = (3.0 × 10^{8} m/s)/(1.18 × 10^{15} s^{−1} ) = 2.5 × 10^{−7} m

This answer could have been calculated more quickly by combining the original two equations to give:

*λ* = *hc* /Δ*E*

**Wave Properties of Matter**

The concept that matter possesses both particle and wave properties was first postulated by de Broglie in 1925. He introduced the equation *λ* = *h* /*mv* , which indicates a mass (*m* ) moving with a certain velocity (*v* ) would have a specific wavelength (*λ* ) associated with it. (Note that this *v* is the velocity not *ν* the frequency.) If the mass is very large (a locomotive), the associated wavelength is insignificant. However, if the mass is very small (an electron), the wavelength is measurable. The denominator may be replaced with the momentum of the particle (*p* = *mv* ).

**Atomic Spectra**

Late in the 19th century, scientists discovered that when the vapor of an element was heated it gave off a **line spectrum** , a series of fine lines of colors, instead of a continuous spectrum like a rainbow. This was used in the developing quantum mechanical model as evidence that the energy of the electrons in an atom was **quantized** , that is, there could only be certain distinct energies (lines) associated with the atom. Niels Bohr developed the first modern atomic model for hydrogen using the concepts of quantized energies. The Bohr model postulated a **ground state** for the electrons in the atom, an energy state of lowest energy, and an **excited state** , an energy state of higher energy. In order for an electron to go from its ground state to an excited state, it must absorb a certain amount of energy. If the electron dropped back from that excited state to its ground state, that same amount of energy would be emitted. Bohr’s model also allowed scientists to develop a method of calculating the energy associated with a particular energy level for the electron in the hydrogen atom:

where *n* is the energy state. This equation can then be modified to calculate the energy difference between any two energy levels:

**Atomic Orbitals**

Bohr’s model worked well for hydrogen, the simplest atom, but didn’t work very well for any others. In the early 1900s, Schrödinger developed a more involved model and set of equations that better described atoms by using quantum mechanical concepts. His model introduced a mathematical description of the electron’s motion called a **wave function** or **atomic orbital** . Squaring the wave function (orbital) gives the volume of space in which the probability of finding the electron is high. This is commonly referred to as the **electron cloud** .

Schrödinger’s equation required the use of three **quantum numbers** to describe each electron within an atom, corresponding to the orbital size, shape, and orientation in space. It was also found that a quantum number concerning the spin of the electron was needed.

The first quantum number is the **principal quantum number ( n )** . It describes the energy (related to size) of the orbital and relative distance from the nucleus. The allowed (by the mathematics of the Schrödinger equation) values are positive integers (1, 2, 3, 4, etc.). The smaller the value of

*n*, the closer the orbital is to the nucleus. The number

*n*is sometimes called the atom’s

**shell**.

The second quantum number is the **angular momentum quantum number ( l )** . Its value is related to the principal quantum number and has allowed values of 0 up to (

*n*− 1). For example, if

*n*= 3, then the possible values of

*l*would be 0, 1, and 2 (3 − 1). This value of

*l*defines the shape of the orbital:

- If
*l*= 0, the orbital is called an s orbital and has a spherical shape with the nucleus at the center of the sphere. The greater the value of*n*, the larger the sphere. - If
*l*= 1, the orbital is called a p orbital and has two lobes of high electron density on either side of the nucleus. This makes for an hourglass or dumbbell shape. - If
*l*= 2, the orbital is a d orbital and can have variety of shapes. - If
*l*= 3, the orbital is an f orbital, with more complex shapes.

__Figure 10.3__ , on the next page, shows the shapes of the s, p, and d orbitals. These are sometimes called **sublevels** or **subshells** .

The third quantum number is the **magnetic quantum number ( m_{l} )** . It describes the orientation of the orbital around the nucleus. The possible values of

*m*

_{l}*depend on the value of the angular momentum quantum number,*

*l*. The allowed values for m

_{ l }are −

*l*through zero to +

*l*. For example, for

*l*= 2 the possible values of

*m*

_{l}*would be −2, −1, 0, +1, +2. This is why, for example, if*

*l*= 1 (a p orbital), then there are three p orbitals corresponding to m

_{ l}values of −1, 0, +1. This is also shown in

__Figure 10.3__.

The fourth quantum number, the **spin quantum number ( m_{s} **), indicates the direction the electron is spinning. There are only two possible values for

*m*

_{s}*, + ½ and −½.*

The quantum numbers for the six electrons in carbon would be:

Therefore, the electron configuration of carbon is 1s^{2} 2s^{2} 2p^{2} .

**Figure 10.3 The shapes of the s, p, and d atomic orbitals.**

**Photoelectron (Photoemission) Spectroscopy (PES)**

Photoelectron spectroscopy is one of a group of related techniques where high-energy photons remove an electron from an atom in a photoelectric effect process. The method relies on a measurement of the kinetic energy of the emitted electron. The kinetic energy is equal to the energy of the photon minus the binding energy of the electron. The binding energy is the energy holding the electron in the atom and can be rather difficult to measure.

X-ray photons can excite core electrons. For example, it is possible to focus on the 1s electrons of an oxygen atom. The binding energy is in part related to the effective nuclear charge experienced by the electron. In compounds, other atoms bonded to the atom of interest can influence the effective nuclear charge. Atoms donating electron density to the atom of interest decrease the effective nuclear charge, while electron-withdrawing atoms lead to an increase in the effective nuclear charge. An important factor in whether an atom donates or withdraws electron density is the relative electronegativity of the two atoms. This experimental method can be used to give information on which atoms are bonded to each other.

**Experiments**

No experimental questions related to this chapter have appeared on the AP exam in recent years.

**Common Mistakes to Avoid**

- Be sure not to confuse wavelength and frequency.
- The speed of light is 3.0 × 10
^{8}m/s. The exponent is positive. - The value of
*n*is never zero. - The values of
*l*and*m*_{l} - Do not confuse velocity (
*v*) and frequency (*ν*).

** Review Questions**

Use these questions to review the content of this chapter and practice for the AP Chemistry exam. First are 16 multiple-choice questions similar to what you will encounter in Section I of the AP Chemistry exam. Following those is a multipart free-response question like the ones in Section II of the exam. To make these questions an even more authentic practice for the actual exam, time yourself following the instructions provided.

**Multiple-Choice Questions**

Answer the following questions in 20 minutes. You may not use a calculator. You may use the periodic table and the equation sheet at the back of this book.

** 1 .** Which of the following represents the electron arrangement for the least reactive element?

** 2 .** Which of the following might refer to a transition element?

** 3 .** Which of the following electron arrangements refers to the most chemically reactive element?

** 4 .** Which of the following electron arrangements represents an atom in an excited state?

** 5 .** The ground-state configuration of Fe

^{2+}is which of the following?

(A) 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{5} 4s^{1}

(B) 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{6}

(C) 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{6} 4s^{2}

(D) 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{8} 4s^{2}

** 6 .** Which of the following contains only atoms that are diamagnetic in their ground state?

(A) Kr, Ca, and P

(B) Ne, Be, and Zn

(C) Ar, K, and Ba

(D) He, Sr, and C

** 7 .** Which of the following is the electron configuration of a halogen?

(A) 1s^{2} 1p^{6} 2s^{2} 2p^{3}

(B) 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{1}

(C) 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{3}

(D) 1s^{2} 2s^{2} 2p^{5}

** 8 .** Which of the following is a possible configuration for a transition metal atom?

(A) 1s^{2} 1p^{6} 2s^{2} 2p^{3}

(B) 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{1}

(C) 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{3}

(D) 1s^{2} 2s^{2} 2p^{5}

** 9 .** Which of the following electron configurations is not possible?

(A) 1s^{2} 1p^{6} 2s^{2} 2p^{3}

(B) 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{1}

(C) 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{3}

(D) 1s^{2} 2s^{2} 2p^{5}

** 10 .** This is a possible configuration of a transition metal ion.

(A) 1s^{2} 1p^{6} 2s^{2} 2p^{3}

(B) 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{1}

(C) 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{3}

(D) 1s^{2} 2s^{2} 2p^{5}

** 11 .** Which idea relates to the fact that the exact position of an electron is not known?

(A) Pauli exclusion principle

(B) electron shielding

(C) Hund’s rule

(D) Heisenberg uncertainty principle

** 12 .** Which of the following explains why oxygen atoms, in their ground state, are paramagnetic?

(A) Pauli exclusion principle

(B) electron shielding

(C) Hund’s rule

(D) Heisenberg uncertainty principle

** 13 .** An atomic orbital can hold no more than two electrons; this is a consequence of which of the following?

(A) Pauli exclusion principle

(B) electron shielding

(C) Hund’s rule

(D) Heisenberg uncertainty principle

** 14 .** Why does the 4s orbital fill before the 3d orbital starts to fill?

(A) Pauli exclusion principle

(B) electron shielding

(C) Hund’s rule

(D) Heisenberg uncertainty principle

** 15 .** Calcium reacts with element

*X*to form an ionic compound. If the ground-state electron configuration of

*X*is 1s

^{2}2s

^{2}2p

^{4}, what is the simplest formula for this compound?

(A) Ca*X*

(B) Ca*X* _{2}

(C) Ca_{4} *X* _{2}

(D) Ca_{2} *X* _{2}

** 16 .** Which of the following best explains the diffraction of electrons?

(A) Pauli exclusion principle

(B) Hund’s rule

(C) the wave properties of matter

(D) Heisenberg uncertainty principle

**Answers and Explanations for the Multiple-Choice Questions**

** 1 . D** —This configuration represents a noble gas (neon). The outer s and p orbitals are filled.

** 2 . C** —Transition elements have partially filled d orbitals. This configuration is for the metal zirconium, Zr.

** 3 . B** —The single electron in the s orbital indicates that this is the very reactive alkali metal lithium.

** 4 . A** —The 1s orbital is not filled. One indication of excited states is to have one or more inner orbitals unfilled.

** 5 . B** —The electron configuration for iron is 1s

^{2}2s

^{2}2p

^{6}3s

^{2}3p

^{6}3d

^{6}4s

^{2}. To produce an iron(II) ion, the two 4s electrons are removed first.

** 6 . B** —The elements that are normally diamagnetic are those in the same columns of the periodic table as Be, Zn, and He because all of the electrons are paired. All other columns are normally paramagnetic.

** 7 . D** —Halogens have a valence shell with s

^{2}p

^{5}.

** 8 . B** —Transition metals have partially filled d orbitals (d

^{1−10}), along with an s

^{1}or s

^{2}.

** 9 . A** —The 1p orbital does not exist.

** 10 . C** —The outer s electrons are not present in most transition metal ions; however, d electrons may be present. C could be V

^{2+}, Cr

^{3+}, or Mn

^{4+}(among other choices).

** 11 . D** —This is part of the Heisenberg uncertainty principle.

** 12 . C** —The four electrons in the oxygen 2p orbitals are arranged with one pair and two unpaired electrons with spins parallel. This makes the oxygen atom paramagnetic. This arrangement is due to Hund’s rule.

** 13 . A** —The Pauli exclusion principle restricts the number of electrons that can occupy a single orbital.

** 14 . B** —The d orbitals are shielded more efficiently than the s orbitals. Thus, the less shielded d orbitals do not fill as readily as s orbitals with similar energy.

** 15 . A** —Calcium will form a +2 ion (Ca

^{2+}), and

*X*will need to gain two electrons to fill its outer shell and become a −2 ion (

*X*

^{2−}). The simplest formula for a compound containing a +2 ion and a −2 ion would be Ca

*X*. The other answers involve different charges or a formula that has not been simplified.

** 16 . C** —Diffraction is a wave phenomenon.

**Free-Response Question**

You have 15 minutes to answer the following question. You may use a calculator and the tables in the back of the book.

**Question 1**

(a) The bond energy of fluorine is 159 kJ mol^{−1} .

- Determine the energy, in J, of a photon of light needed to break an F−F bond.
- Determine the frequency of this photon in s
^{−1}.

iii. Determine the wavelength of this photon in nanometers.

(b) Barium imparts a characteristic green color to a flame. The wavelength of this light is 551 nm. Determine the energy involved in kJ/mol.

**Answer and Explanation for the Free-Response Question**

(a) If you do not remember them, several of the equations given at the beginning of the exam are necessary. In addition, the values of Planck’s constant, Avogadro’s number, and the speed of light are necessary. These constants are also given on the exam.

- This is a simple conversion problem:

Give yourself 1 point if you got this answer.

- This part requires the equation Δ
*E*=*hν*(this equation is given on the equation page of the AP exam).

Give yourself 1 point for this answer. If you got the wrong answer in the preceding part, but used it correctly here (in place of the 2.64 × 10^{−19} J), you still get 1 point.

iii. (a) The equation *c* = *νλ* is needed. (This equation is given on the equation page of the AP exam.)

Again, give yourself 1 point for the correct answer. If you correctly used a wrong answer from the preceding part, you still get 1 point.

(b) This can be done as a one-step or a two-step problem. The AP test booklet gives you the equations to solve this directly as a two-step problem. This method will be followed here. The two equations may be combined to produce an equation that will allow you to do the problem in one step.

Using *c* = *λν* :

Using Δ*E* = *hν* :

Give yourself 1 point for each of these answers. If you did the problem as a one-step problem, give yourself 2 points if you got the final answer correct or 1 point if you left out any of the conversions.

The total for this question is 5 points, minus 1 point if any answer does not have the correct number of significant figures.

** Rapid Review**

- Know the regions of the electromagnetic spectrum.
- The frequency,
*ν*, is defined as the number of waves that pass a point per second. - The wavelength,
*λ*, is the distance between two identical points on a wave. - The energy of light is related to the frequency by
*E*=*hν*. - The product of the frequency and wavelength of light is the speed of light:
*c*=*νλ*. - An orbital or wave function is a quantum mechanical, mathematical description of the electron.
- If all electrons in an atom are in their lowest possible energy level, then the atom is said to be in its ground state.
- If any electrons in an atom are in a higher energy state, then the atom is said to be in an excited state.
- The energy of an atom is quantized, existing in only certain distinct energy states.
- Quantum numbers are numbers used in Schrödinger’s equation to describe the orbital size, shape, and orientation in space, and the spin of an electron.
- The principal quantum number,
*n*, describes the size of the orbital. It must be a positive integer. It is sometimes referred to as the atom’s shell. - The angular momentum quantum number,
*l*, defines the shape of the electron cloud. If*l*= 0, it is an s orbital; if*l*= 1, it is a p orbital; if*l*= 2, it is a d orbital; if*l*= 3, it is an f orbital, etc. - The magnetic quantum number,
*m*_{l}*l*through 0 to +*l*. - The spin quantum number,
*m*_{s} - Be able to write the quantum numbers associated with the first 20 electrons.

****