1. Atomic Bonding Problem Set

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ATOMIC BONDING | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.

PROBLEM 1: What type of bond exists predominantly in a sodium chloride crystal? A. π‘€π‘’π‘‘π‘Žπ‘™π‘™π‘–π‘ π΅π‘œπ‘›π‘‘ B. πΌπ‘œπ‘›π‘–π‘ π΅π‘œπ‘›π‘‘ C. πΆπ‘œπ‘£π‘Žπ‘™π‘’π‘›π‘‘ π΅π‘œπ‘›π‘‘ D. π‘ƒπ‘œπ‘™π‘Žπ‘Ÿ π΅π‘œπ‘›π‘‘

SOLUTION 1: The TOPIC OF PRIMARY BONDS can be referenced under the topic of ATOMIC BONDING on page 60 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. In a sodium chloride crystal, sodium and chlorine are present as sodium and chloride ions. These ions are arranged in a periodic pattern to form the crystal.



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The electrostatic attraction between the positive sodium ion and the negative chloride ion forms the basis of the ionic bond that exists in a sodium chloride crystal.

Therefore, the correct answer choice is B. π‘°π’π’π’Šπ’„ 𝑩𝒐𝒏𝒅 PROBLEM 2: Which of the following is the ground-state electron configuration of a copper atom? A. 1𝑠 ; 2𝑠 ; 2𝑝> 3𝑠 ; 3𝑝> 4𝑠A 3𝑑AB B. 1𝑠 ; 1𝑝> 2𝑠 ; 2𝑝> 3𝑠 ; 3𝑝> 3𝑑 C C. 1𝑠 ; 1𝑝; 2𝑠 ; 2𝑝> 3𝑠 ; 3𝑝> 4𝑠 ; 4𝑝> 4𝑑A D. 1𝑠 ; 2𝑠 ; 2𝑝> 3𝑠 ; 3𝑝> 4𝑠 ; 4𝑝D

SOLUTION 2: The TOPIC OF ELECTRON CONFIGURATION OF ELEMENTS is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize these concepts and understand their application independent of the NCEES Supplied Reference Handbook. Copper has an atomic number of 29, so we know a copper atom has 29 electrons. In answer choices (B) and (C), the β€œ1𝑝” levels do not exist. In answer choice (D), the β€œ3𝑑” level is a higher energy level than the β€œ4𝑠” level.

Therefore, the correct answer choice is B. πŸπ’”πŸ πŸπ’”πŸ πŸπ’‘πŸ” πŸ‘π’”πŸ πŸ‘π’‘πŸ” πŸ’π’”πŸ πŸ‘π’…πŸπŸŽ

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PROBLEM 3: Which of the following affects most of the electrical and thermal properties of materials? A. π‘‡β„Žπ‘’ π‘Žπ‘‘π‘œπ‘šπ‘–π‘ π‘€π‘’π‘–π‘”β„Žπ‘‘ 𝑒π‘₯π‘π‘Ÿπ‘’π‘ π‘ π‘’π‘‘ 𝑖𝑛 π‘”π‘Ÿπ‘Žπ‘šπ‘  π‘π‘’π‘Ÿ π‘”π‘Ÿπ‘Žπ‘š βˆ’ π‘Žπ‘‘π‘œπ‘š B. π‘‡β„Žπ‘’ π‘’π‘™π‘’π‘π‘‘π‘Ÿπ‘œπ‘›π‘ , π‘π‘Žπ‘Ÿπ‘‘π‘–π‘π‘’π‘™π‘Žπ‘Ÿπ‘™π‘¦ π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘›π‘π‘’ π‘’π‘™π‘’π‘π‘‘π‘Ÿπ‘œπ‘›π‘  C. π‘‡β„Žπ‘’ π‘šπ‘Žπ‘”π‘›π‘–π‘‘π‘’π‘‘π‘’ π‘œπ‘“ π‘’π‘™π‘’π‘π‘‘π‘Ÿπ‘–π‘π‘Žπ‘™ π‘β„Žπ‘Žπ‘Ÿπ‘”π‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘Ÿπ‘œπ‘‘π‘œπ‘› D. π‘‡β„Žπ‘’ π‘€π‘’π‘–π‘”β„Žπ‘‘ π‘œπ‘“ π‘‘β„Žπ‘’ π‘Žπ‘‘π‘œπ‘š

SOLUTION 3: The outermost electrons are responsible for determining most of the material’s properties.

Therefore, the correct answer choice is B. 𝑻𝒉𝒆 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏𝒔, π’‘π’‚π’“π’•π’Šπ’„π’–π’π’‚π’“π’π’š 𝒕𝒉𝒆 𝒗𝒂𝒍𝒆𝒏𝒄𝒆 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏𝒔 PROBLEM 4: What are valence electrons? A. π‘‡β„Žπ‘’ π‘œπ‘’π‘‘π‘’π‘Ÿ βˆ’ π‘ β„Žπ‘’π‘™π‘™ π‘’π‘™π‘’π‘π‘‘π‘Ÿπ‘œπ‘›π‘  B. πΈπ‘™π‘’π‘π‘‘π‘Ÿπ‘œπ‘›π‘  π‘€π‘–π‘‘β„Ž π‘Ž π‘π‘œπ‘ π‘–π‘‘π‘–π‘£π‘’ π‘β„Žπ‘Žπ‘Ÿπ‘”π‘’ C. π‘‡β„Žπ‘’ π‘’π‘™π‘’π‘π‘‘π‘Ÿπ‘œπ‘›π‘  π‘œπ‘“ π‘π‘œπ‘šπ‘π‘™π‘’π‘‘π‘’ π‘žπ‘’π‘Žπ‘›π‘‘π‘’π‘š π‘ β„Žπ‘’π‘™π‘™π‘  D. π‘‡β„Žπ‘’ 𝐾 βˆ’ π‘žπ‘’π‘Žπ‘›π‘‘π‘’π‘š π‘ β„Žπ‘’π‘™π‘™ π‘’π‘™π‘’π‘π‘‘π‘Ÿπ‘œπ‘›π‘ 



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SOLUTION 4: The valence electrons are the outermost electrons in the electron configuration of an atom.

Therefore, the correct answer choice is A. 𝑻𝒉𝒆 𝒐𝒖𝒕𝒆𝒓 βˆ’ 𝒔𝒉𝒆𝒍𝒍 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏𝒔 PROBLEM 5: What is the strong bond between hydrogen atoms called? A. π‘‡β„Žπ‘’ π‘–π‘œπ‘›π‘–π‘ π‘π‘œπ‘›π‘‘ B. π‘‡β„Žπ‘’ π‘šπ‘’π‘‘π‘Žπ‘™π‘™π‘–π‘ π‘π‘œπ‘›π‘‘ C. πΌπ‘œπ‘›π‘–π‘ π‘Žπ‘›π‘‘ π‘šπ‘’π‘‘π‘Žπ‘™π‘™π‘–π‘ π‘π‘œπ‘›π‘‘π‘  D. π‘‡β„Žπ‘’ π‘π‘œπ‘£π‘Žπ‘™π‘’π‘›π‘‘ π‘π‘œπ‘›π‘‘

SOLUTION 5: The TOPIC OF PRIMARY BONDS can be referenced under the topic of ATOMIC BONDING on page 60 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Covalent bonds provide the strongest attractive forces between atoms.

Therefore, the correct answer choice is 𝐃. 𝑻𝒉𝒆 𝒄𝒐𝒗𝒂𝒍𝒆𝒏𝒕 𝒃𝒐𝒏𝒅𝒔



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PROBLEM 6: Which of the following elements would not be expected to form ionic compounds? A. Silicon B. Potassium C. Barium D. Silver

SOLUTION 6: The PERIODIC TABLE OF ELEMENTS can be referenced under the topic of CHEMISTRY on page 54 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. One way to predict whether a bond is ionic or covalent is to look how far apart the two atoms forming the bonds are in the periodic table. If one atom is of the far left (Group 1 or 2) and the other is on the far right (Group 5, 6, or 7), then the atoms will have large differences in electronegativity and will form an ionic bond. Most other pairs of atoms are close enough in electronegativity to form covalent bonds. For example, any two atoms in the main group elements (Groups 3, 4, 5, 6, 7), will usually form a covalent bond. Silicon does not form ionic compounds. Silicon forms covalent compounds.

Therefore, the correct answer choice is 𝑨. π‘Ίπ’Šπ’π’Šπ’„π’π’



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PROBLEM 7: What is the difference between π‘ˆ ;DC and π‘ˆ ;Dp ? A. π‘‡β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘Ÿπ‘œπ‘‘π‘œπ‘›π‘  B. π‘‡β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘›π‘’π‘’π‘‘π‘Ÿπ‘œπ‘›π‘  C. π‘‡β„Žπ‘’ π‘Žπ‘‘π‘œπ‘šπ‘–π‘ π‘›π‘’π‘šπ‘π‘’π‘Ÿπ‘  D. π‘‡β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘’π‘™π‘’π‘π‘‘π‘Ÿπ‘œπ‘›π‘ 

SOLUTION 7: The PERIODIC TABLE OF ELEMENTS can be referenced under the topic of CHEMISTRY on page 54 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. π‘ˆ ;DC and π‘ˆ ;Dp are isotopes and have the same atomic numbers. As they have the same atomic numbers, they have the number of protons and electrons. They differ in the number of neutrons.

Therefore, the correct answer choice is 𝑩. 𝑻𝒉𝒆 π’π’–π’Žπ’ƒπ’†π’“ 𝒐𝒇 𝒏𝒆𝒖𝒕𝒓𝒐𝒏𝒔 PROBLEM 8: An isotropic material is one with which of the following properties? A. π·π‘–π‘“π‘“π‘’π‘Ÿπ‘’π‘›π‘‘ π‘π‘Ÿπ‘œπ‘π‘’π‘Ÿπ‘‘π‘–π‘’π‘  𝑖𝑛 π‘Žπ‘™π‘™ π‘‘π‘–π‘Ÿπ‘’π‘π‘‘π‘–π‘œπ‘›π‘  B. π‘†π‘Žπ‘šπ‘’ π‘π‘Ÿπ‘œπ‘π‘’π‘Ÿπ‘‘π‘–π‘’π‘  𝑖𝑛 π‘Žπ‘™π‘™ π‘‘π‘–π‘Ÿπ‘’π‘π‘‘π‘–π‘œπ‘›π‘  C. π·π‘–π‘“π‘“π‘’π‘Ÿπ‘’π‘›π‘‘ π‘π‘Ÿπ‘œπ‘π‘’π‘Ÿπ‘‘π‘–π‘’π‘  π‘Žπ‘‘ π‘‘π‘–π‘“π‘“π‘’π‘Ÿπ‘’π‘›π‘‘ π‘™π‘œπ‘π‘Žπ‘‘π‘–π‘œπ‘›π‘  D. π‘†π‘Žπ‘šπ‘’ π‘π‘Ÿπ‘œπ‘π‘’π‘Ÿπ‘‘π‘–π‘’π‘  𝑖𝑛 π‘Žπ‘™π‘™ π‘™π‘œπ‘π‘Žπ‘‘π‘–π‘œπ‘›π‘ 

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SOLUTION 7:

An isotropic material is one with the same properties in all directions.

Therefore, the correct answer choice is B. The same properties in all directions.



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