CHEM125 – CHAPTER 08 – REGULAR HOMEWORK

CHAPTER 08 REGULAR HOMEWORK

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Question 1

Give the number of unpaired electrons present in a ground-state atom of Fe. Hint given in feedback Answer:Correct

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 1

Question 2

Give the number of unpaired electrons present in a ground-state atom of S. Hint given in feedback Answer:Correct

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 2

Question 3

Give the number of unpaired electrons present in a ground-state atom of Fe2+. Answer:Correct

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 3

Question 4

Give the number of unpaired electrons present in a ground-state of the ion formed when one electron is ionized from an O atom to form the O+ ion Answer:Correct

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 4 Summarizing Atomic Radii for Main-Group Elements:

  • As we move down a column in the periodic table, the principal quantum number (n) of the electrons in the outermost principal energy level increases, resulting in larger orbitals and therefore larger atomic radii.
  • As we move to the right across a row in the periodic table, the effective nuclear charge experienced by the electrons in the outermost principal energy level increases, resulting in a stronger attraction between the outermost electrons and the nucleus, and smaller atomic radii.

Question 5

Give the number of unpaired electrons present in a ground-state atom of Se2- . Answer:Correct

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 4 Summarizing Atomic Radii for Main-Group Elements:

  • As we move down a column in the periodic table, the principal quantum number (n) of the electrons in the outermost principal energy level increases, resulting in larger orbitals and therefore larger atomic radii.
  • As we move to the right across a row in the periodic table, the effective nuclear charge experienced by the electrons in the outermost principal energy level increases, resulting in a stronger attraction between the outermost electrons and the nucleus, and smaller atomic radii.

Question 6

Using the abbreviated notation, the ground-state electron configuration of Y3+ is Select one:a. [Kr] Correctb. [Ar] 4s2 5d10 4p6 5s2 4d1c. [Kr] 5s2 4d4d. [Ar] 4s2 5d10 4p5 4d1

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 4 Summarizing Atomic Radii for Main-Group Elements:

  • As we move down a column in the periodic table, the principal quantum number (n) of the electrons in the outermost principal energy level increases, resulting in larger orbitals and therefore larger atomic radii.
  • As we move to the right across a row in the periodic table, the effective nuclear charge experienced by the electrons in the outermost principal energy level increases, resulting in a stronger attraction between the outermost electrons and the nucleus, and smaller atomic radii.

Y = 39 electrons Y3+ = 36 electrons —–> Kr (36 electrons)

Question 7

Using the abbreviated notation, the ground-state electron configuration of Br- is Select one:a. [Kr] Correctb. [Ar] 4s2 5d10 4p4c. [Kr] 5s24d10 5p5d. [Ar] 4s25d10 4p5 4d1

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 7 Br: 35 electrons Br – = 36 electrons —> Kr (36 electrons)

Question 8

Using the abbreviated notation, the ground-state electron configuration of Rh 2+ is Select one:a. [Kr] 5s2 4d7b. [Kr] 5s2 4d5c. [Kr] 4d7 Correctd. [Kr] 5s1 4d6

Explanation

Rh: 45 electrons Rh2+ : 43 electrons Rh 2+ : [Kr] 4d7

Question 9

Which electron configuration represents the most highly excited state of C? Select one:a. 1s22s22px12py1b. 1s22s22px2c. 1s22s12px12py12pz1d. 1s12s22px12py12pz1 Correct

Explanation

 Protons Electrons
 F 99  strongest attraction to nucleus (smallest)
 F -19  10 weaker attraction to nucleus (medium)
 O 2-  8 10 weakest attraction to nucleus (largest)

Question 10

Write the electron configuration of the copper atom and the 2+ cation of copper. Check the correct choices. Select one or more:a. Cu [Ar] 4s23d9b. Cu [Ar] 4s13d10 Correctc. Cu [Ar] 4s03d11d. Cu2+ [Ar] 4s23d9e. Cu2+ [Ar] 4s03d9 Correctf. Cu2+ [Ar] 4s23d7

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 4

Question 11

Arrange the elements lithium, carbon, and oxygen in order of increasing number of unpaired electrons. (There are two correct choices since 2 have the same number.) Select one or more:a. O, C, Lib. O, Li, Cc. C, Li, Od. C, O, Lie. Li, C, O Correctf. Li, O, C Correct

Explanation

Li: 1 unpaired electron C: 2 unpaired electrons O: 2 unpaired electrons

Question 12

In the reaction of lithium with oxygen, what expected compound is formed? If the compound is written LixOy, what is x+y? For example, water is H2O. In this case x =2, y = 1, and you write 3 in the space provided for the answer. Answer:Correct

Explanation

Li2O = 2+1 = 3

Question 13

Write the equation for the reaction, if any, of lithium with nitrogen. If there is a reaction forming LixNy, what is x+y? Answer:Correct

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 9

Question 14

Write the electron configuration of the alkali metal (M) that reacts with oxygen to yield an oxide, M2O, if 1.22 g of the metal reacts with 1.41 g of oxygen to form 2.63 g of the oxide. Select one:a. [He] 2s1 Correctb. [Ne] 3s1c. [Ar] 4s1d. [Kr] 5s1e. [Xe] 6s1

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 9 29.85 g – 16.0g (Oxygen) = 13.85g 13.85 / 2 (Metal) = 6.925g Li (Molar Mass) = 6.9

Question 15

In the reaction of iron and chlorine, what mass of iron is needed to prepare 22.97 g of the iron chloride product if, under the conditions of the reaction, the electron configuration of the iron cation in the product is 1s22s22p63s23p63d6? Answer:Correct

Explanation

Fe = +2 (loss 2 electrons) Final product = FeCl2 Fe + Cl2 -> FeCl2 Molecular mass of FeCl2 is 126.48 needed to prepare 22.97 g of the iron chloride product 22.97/ 126.48 = 0.1816 mol Mass of Iron (55.58 g/mol) x (0.1816 mol) = 10.1 g

Question 16

Using only a periodic table as a guide, arrange each of the following series of atoms in order of increasing size. Select one:a. Be, Li, Na Correctb. Be, Na, Lic. Na, Be, Lid. Na, Li, Bee. Li, Na, Bef. Li, Be, Na

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 4 Summarizing Atomic Radii for Main-Group Elements:

  • As we move down a column in the periodic table, the principal quantum number (n) of the electrons in the outermost principal energy level increases, resulting in larger orbitals and therefore larger atomic radii.
  • As we move to the right across a row in the periodic table, the effective nuclear charge experienced by the electrons in the outermost principal energy level increases, resulting in a stronger attraction between the outermost electrons and the nucleus, and smaller atomic radii.

Question 17

Using only a periodic table as a guide, arrange each of the following series of atoms in order of increasing size. Select one:a. F, N, P Correctb. F, P, Nc. P, F, Nd. P, N, Fe. N, P, Ff. N, F, P

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 12 Summarizing Atomic Radii for Main-Group Elements:

  • As we move down a column in the periodic table, the principal quantum number (n) of the electrons in the outermost principal energy level increases, resulting in larger orbitals and therefore larger atomic radii.
  • As we move to the right across a row in the periodic table, the effective nuclear charge experienced by the electrons in the outermost principal energy level increases, resulting in a stronger attraction between the outermost electrons and the nucleus, and smaller atomic radii.

Question 18

Using only a periodic table as a guide, arrange each of the following series of atoms in order of increasing size. Select one:a. O, I, Sn Correctb. O, Sn, Ic. Sn, I, Od. Sn, O, Ie. I, Sn, Of. I, O, Sn

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 12 Summarizing Atomic Radii for Main-Group Elements:

  • As we move down a column in the periodic table, the principal quantum number (n) of the electrons in the outermost principal energy level increases, resulting in larger orbitals and therefore larger atomic radii.
  • As we move to the right across a row in the periodic table, the effective nuclear charge experienced by the electrons in the outermost principal energy level increases, resulting in a stronger attraction between the outermost electrons and the nucleus, and smaller atomic radii.

Question 19

Using only a periodic table as a guide, arrange each of the following series of species in order of increasing size. Select one:a. O2-, F, F-b. O2-, F-, Fc. F-, F, O2-d. F-, O2-, Fe. F, O2-, F-f. F, F-, O2- Correct

Explanation

 Protons Electrons
 F 99  strongest attraction to nucleus (smallest)
 F -19  10 weaker attraction to nucleus (medium)
 O 2-  8 10 weakest attraction to nucleus (largest)

Question 20

Using only a periodic table as a guide, arrange each of the following series of species in order of increasing size. Select one:a. Al3+, Mg, Na Correctb. Al3+, Na, Mgc. Na, Al3+, Mgd. Na, Mg, Al3+e. Mg, Na, Al3+f. Mg, Al3+, Na

Explanation

 Protons Electrons
Al 3+1310  strongest attraction to nucleus (smallest)
Mg1212right across a row in the periodic table radius becomes smaller
Na1111moving left across the periodic table radius becomes larger

Question 21

Using only a periodic table as a guide, arrange each of the following series of species in order of increasing size. Select one:a. N, P, Si Correctb. N, Si, Pc. Si, N, Pd. Si, P, Ne. P, Si, Nf. P, N, Si

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 4

Question 22

Indicate which species in each pair has the higher ionization energy.Select one or more:a. Between K and I, I Correctb. Between K and I, Kc. Between Al and Al+, Ald. Between Al and Al+, Al+ Correcte. Between Ar and Cl-, Ar Correctf. Between Ar and Cl-, Cl-g. Between Mg and Ca, Mg Correcth. Between Mg and Ca, Ca

Explanation

Ionization Energy – required energy to remove an electron from orbital CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 9

  • First ionization energy decreases as we move down a column (or family) in the periodic table because electrons in the outermost principal level are increasingly farther away from the positively charged nucleus and are therefore held less tightly.
  • First ionization energy increases as we move to the right across a row (or period) in the periodic table because electrons in the outermost principal energy level generally experience a greater effective nuclear charge

Question 23

Using only a periodic table as a guide, arrange each of the following series of species in order of increasing first ionization energy. Select one:a. K, Se, S Correctb. K, S, Sec. S, K, Sed. S, Se, Ke. Se, K, Sf. Se, S, K

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 9

Question 24

Using only a periodic table as a guide, arrange each of the following series of species in order of increasing first ionization energy. Select one:a. Cr, As, O Correctb. Cr, O, Asc. O, Cr, Asd. O, As, Cre. As, Cr, Of. As, O, Cr

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 9

Question 25

Using only a periodic table as a guide, arrange each of the following series of species in order of increasing first ionization energy. Select one:a. O, O-, Fb. O, F, O-c. F, O, O-d. F, O-, Oe. O-, F, Of. O-, O, F Correct

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 9

Question 26

Which ground-state electron configuration represents the element with the largest size? Select one:a. 1s22s22p1 Correctb. 1s22s22p4c. 1s22s22p5d. 1s22s22p3

Explanation

The electron configuration with least amount of electrons has the largest size. As the number of electrons increases the attraction toward the nucleus increases causing the atom size to become smaller.

Question 27

Which ground-state electron configuration represents the element with the smallest ionization energy? Select one:a. 1s22s22p1 Correctb. 1s22s22p4c. 1s22s22p5d. 1s22s22p3

Explanation

The electron configuration with least amount of electrons has the largest size. Therefore it is requires the least amount of ionization energy to remove an electron because it is least closest to the nucleus.

Question 28

Which ground-state electron configuration represents the element with the most negative electron affinity? Select one:a. 1s22s22p1b. 1s22s22p4c. 1s22s22p5 Correctd. 1s22s22p3

Explanation

Electron affinity generally becomes more negative (adding an electron becomes more exothermic) as we move to the right across a period (row) in the periodic table.

Question 29

Arrange the elements lithium, carbon, and oxygen in order of increasing size. Select one:a. O, C, Li Correctb. O, Li, Cc. C, Li, Od. C, O, Lie. Li, C, Of. Li, O, C

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 4

Question 30

Arrange the elements lithium, carbon, and oxygen in order of increasing first ionization energy. Select one:a. O, C, Lib. O, Li, Cc. C, Li, Od. C, O, Lie. Li, C, O Correctf. Li, O, C

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 9

Question 31

Arrange the elements lithium, carbon, and oxygen in order of increasing second ionization energy. Select one:a. O, C, Lib. O, Li, Cc. C, Li, Od. C, O, Li Correcte. Li, C, Of. Li, O, C

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 9 count the ionization energy by the number the elements prior to the select element.

Question 32

Write the equation for the reaction, if any, of lithium with water. What products are formed? Select one or more:a. No reactionb. H2 gas Correctc. O2 gasd. Li+(aq) Correcte. OH-(aq) Correctf. H+(aq)

Explanation

2Li + H2O = Li2O + H2 = Li + OH + H

Question 33

Which of the following reactions produce H2 gas? Select one or more:a. Na(s) + cold water Correctb. NaH + cold water Correctc. Na2O + cold waterd. Mg + cold watere. Ni + cold water

Explanation

CHEM125 - CHAPTER 08 - REGULAR HOMEWORK 22

Question 34

Write the electron configuration of the alkali metal (M) that reacts with oxygen to yield an oxide, M2O, if 1.22 g of the metal reacts with 1.41 g of oxygen to form 2.63 g of the oxide. Select one:a. [He] 2s1 Correctb. [Ne] 3s1c. [Ar] 4s1d. [Kr] 5s1e. [Xe] 6s1

Explanation

Hint, determine the molecular weight of the metal.

Question 35

In the reaction of iron and chlorine, what mass of iron is needed to prepare 3.85 g of the iron chloride product if, under the conditions of the reaction, the electron configuration of the iron cation in the product is 1s22s22p63s23p63d6? Answer:Correct

Explanation

Procedure: (1) Determine what the iron chloride is. (2) Go from grams iron chloride to moles of iron chloride to moles Fe to grams Fe.

Question 36

Which of the following reactions produce H2 gas?Select one or more: a. Na(s) + cold water Correctb. NaH + cold water Correctc. Na2O + cold waterd. Mg + cold watere. Ni + cold water

Explanation

Explained Above

Question 37

Which of the following reactions produce a basic solution (the hydroxide ion)?Select one or more: a. Na(s) + cold water Correctb. NaH + cold water Correctc. Na2O + cold water Correctd. Mg + cold watere. Ni + cold water

Explanation

A basic solutions contains OH ions

  •  Na(s) + cold water all contain OH ions in final product
  •  NaH + cold water
  •  Na2O + cold water

Question 38

Which of the following reactions produces iron, the metal?Select one: a. FeCl3(aq) + Zn(s) Correctb. FeCl3(aq) + Cu(s)c. FeCl3(aq) + Ca(NO3)2(aq)d. FeCl3(aq) + Br2(l)e. FeCl3(aq) + F2(g)

Explanation

FeCl3 + Zn → ZnCl2 + Fe

Question 39

Which of the following reactions produces chlorine gas?Select one: a. FeCl3(aq) + Zn(s)b. FeCl3(aq) + Cu(s)c. FeCl3(aq) + Ca(NO3)2(aq)d. FeCl3(aq) + Br2(l)e. FeCl3(aq) + F2(g) Correct

Explanation

2FeCl3 + 3F2 = 2FeF3 + 3Cl2

Question 40

Match oxides with correct property.

CaOAnswer 1 Choose…AmphotericBasicAcidic Correct
K2OAnswer 2 Choose…AmphotericBasicAcidic Correct
CO2Answer 3 Choose…AmphotericBasicAcidic Correct
N2O5Answer 4 Choose…AmphotericBasicAcidic Correct
Al2O3Answer 5 Choose…AmphotericBasicAcidic Correct

Explanation