AP Chemistry : Atomic Structure and Properties

Study concepts, example questions & explanations for AP Chemistry

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Example Questions

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Example Question #1 : Composition Of Mixtures

Which of the following mixtures is not a homogeneous mixture:

Possible Answers:

Coffee

A table salt solution

Wine

Tea

Oil and vinegar

Correct answer:

Oil and vinegar

Explanation:

A homogeneous mixture is a uniform mixture of two or more components that appears uniform throughout.  The oil and vinegar mixture would not appear uniform, and thus is a heterogeneous mixture.

Example Question #1 : Atomic Structure And Properties

Which of the following is a heterogeneous mixture.

Possible Answers:

All of the above

Oatmeal chocolate chip cookie

Orange juice with no pulp

Motor oil

Gasoline

Correct answer:

Oatmeal chocolate chip cookie

Explanation:

A heterogeneous mixture is a mixture of two or more components that do not appears uniform throughout.  The oatmeal chocolate chip cookie mixture would not appear uniform, and thus is a heterogeneous mixture.

Example Question #1 : Composition Of Mixtures

Which of the following are mixtures

Possible Answers:

Top soil

All of the above

Distilled water

b and c

Reeses Peanut butter cup

Correct answer:

b and c

Explanation:

A mixture contains two or more different substances.  Distilled water only contains liquid water, and is not a mixture.

Example Question #3 : Atomic Structure And Properties

What differentiates a homogeneous mixture from a pure substance?

Possible Answers:

A pure substance can be separated into elements by physical means.

A pure substance is always a compound while mixtures can be elements or compounds.

A mixture has a defined composition.

The composition of a pure substance is fixed.

A homogeneous mixture is uniform throughout.

Correct answer:

The composition of a pure substance is fixed.

Explanation:

A heterogeneous mixture is a mixture of two or more components that do not appears uniform throughout.  A pure substance contains only one component that has a fixed composition.

Example Question #1 : Composition Of Mixtures

A slice of watermelon would best be described as?

Possible Answers:

None of the above

A homogeneous mixture of compounds

A heterogeneous mixture of compounds

A heterogeneous mixture of elements and compounds

A pure compound

Correct answer:

A heterogeneous mixture of elements and compounds

Explanation:

A heterogeneous mixture is a mixture of two or more components that do not appears uniform throughout, but contains components that are elemental (single atom type) and compounds (contain two or more different elements).

Example Question #1 : Photoelectron Spectroscopy

 01

 

Using the PES spectra above, what element is illustrated?

Possible Answers:

Li

N

Na

Ne

C

Correct answer:

N

Explanation:

The peak at ~ 40 MJ binding energy corresponds to the 1s orbital.  Since there are peaks at lower binding energy, it is implied that this orbital is filled (by 2 electrons).  There are two peaks at lower binding energy corresponding to the 2s and 2p orbitals, respectively.  The 2s peak has the same intensity as the 1s peak indicating it has the same number of electrons (2).  The 2p peak is 3/2 as tall as the 2s peak and indicates that indicates it has 3 electrons in the orbital.  The electron configuration of 1s22s22p3 makes the element N.

Example Question #2 : Spectroscopy And Electromagnetic Spectrum

02

Using the PES spectra above, what answer explains the differences in the position and intensity of the 3s peaks between Na and Mg?

Possible Answers:

Mg is larger in size making the electrons in the 3s orbital farther from the nucleus and thus easier to remove

Mg experiences a larger effective nuclear charge and has more electrons in the 3s orbital.

Na experiences a larger effective nuclear charge on the 3s electrons and are thus harder to remove.

None of the above

Na experiences a greater shielding effect from the 2s and 1s electrons making the 3s electrons easier to remove.

Correct answer:

Mg experiences a larger effective nuclear charge and has more electrons in the 3s orbital.

Explanation:

Mg has a Z=12 while Na is Z=11.  The increased number of protons in Mg gives rise to a greater effective nuclear charge.  This greater effective nuclear charge binds the electrons in their respective orbitals more tightly than an element with a lower effective nuclear charge.  The peak height of Mg is 2x that of Na for the 3s orbital because it has twice the electrons present.

Example Question #1 : Photoelectron Spectroscopy

03

Using the spectra above, answer the following:  Why is the oxygen 1s electrons further to the right than the nitrogen 1s orbital?

Possible Answers:

The effective nuclear charge on oxygen is greater than that of nitrogen.

Nitrogen has 3 p-electrons and are thus unpaired leading to lower electron-electron repulsions.

There is a greater degree of electron shielding occurring in the oxygen 2p orbital.

Oxygen has more electrons and thus a greater amount of electron-electron repulsion present in the atom.

Nitrogen is larger than oxygen thus making the 1s electrons easier to remove.

Correct answer:

The effective nuclear charge on oxygen is greater than that of nitrogen.

Explanation:

Oxygen has a Z=8 vs nitrogen’s Z=7. This greater effective nuclear charge binds the electrons in their respective orbitals more tightly than an element with a lower effective nuclear charge.

Example Question #2 : Atomic Structure And Properties

Using the spectra above, answer the following:  What is the electron configuration of the element shown above?

Possible Answers:

Correct answer:

Explanation:

The binding energy at ~ 53 Mj represents the 1s orbital.  With other peaks present, this 1s orbital is fully occupied by 2 electrons.  The peaks at lower energy would correspond to the 2s and 2p, respectively.  Since the 2p peak is twice the intensity of the 2s or 1s peak, it has twice the electrons (4).  This gives the electron configuration 1s22s22p4.

Example Question #5 : Spectroscopy And Electromagnetic Spectrum

05

 

Using the spectra above, answer the following:  What is the wavelength required, in m, to remove a valence electron from the element shown above?

Possible Answers:

None of the above

Correct answer:

Explanation:

Step 1.  Convert the Binding Energy from MJ/mol to J

 

  x    x  

 

Step 2:

 

 

 

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