Module 4 Exam Quiz

Module 4vExam

• Question 1

  Write the subshell electron configuration (i.e.1s² 2s², etc.) for the Mn₂₅ atom.

  Correct Answer

  Mn₂₅ = 25 electrons = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁵

• Question 2

  Write the subshell electron configuration (i.e.1s² 2s², etc.) for the S₁₆ atom.

  Correct Answer

  S₁₆ = 16 electrons = 1s² 2s² 2p⁶ 3s² 3p⁴

• Question 3

  Write the subshell electron configuration (i.e.1s² 2s², etc.) for the Ca₂₀ atom and identify which are valence (outer shell) electrons and determine how many valence electrons there are.

  Correct Answer

  Ca₂₀ = 20 electrons = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² = 2 valence electrons

• Question 4

  Using up and down arrows, write the orbital diagram for the Fe₂₆ atom.

  Correct Answer

  Electron configuration of Fe

  

  Orbital diagram of Fe

  ↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓  ↑↓↑↓↑↓ ↑↓  ↑↓↑↑↑↑

  1s 2s  2p      3s     3p    4s    3d

• Question 5

  Using up and down arrows, write the orbital diagram for the Ti₂₂ atom and identify which are unpaired electrons and determine how many unpaired electrons there are.

  Correct Answer

  Electron configuration of Ti

  

  Orbital diagram

  ↑↓  ↑↓  ↑↓↑↓↑↓ ↑↓  ↑↓↑↓↑↓  ↑↓  ↑↑

  1s 2s     2p     3s    3p      4s  3d

  2 unpaired electrons

• Question 6

  Write the subshell electron configuration (i.e.1s² 2s², etc.) for the Fe₂₆ atom and then identify the last electron to fill and write the 4 quantum numbers (n, l, m_l and m_s) for this electron.

  Correct Answer

  Electron configuration of Fe

  

  Orbital diagram of Fe

  ↑↓ ↑↓  ↑↓↑↓↑↓ ↑↓  ↑↓↑↓↑↓  ↑↓  ↑↓↑↑↑↑

  1s 2s   2p      3s    3p      4s   3d

  Last electron to fill is in 3d orbital so n=3, and the subshell is "d", so l=2 because it has two nodes. The last electron is in the first of the subshell in the d-orbital so ml=-2 and it points downward and ms=-1/2. So:

  n=3

  l=2

  ml=-2

  ms= -1/2

• Question 7

  1.  Arrange the following elements in a vertical list from largest (top) to smallest (bottom) atomic size:    Cl, Br, I

  
  2.  Arrange the following elements in a vertical list from lowest (top) to highest (bottom) electronegativity:      S, P, Cl

  
  3.    Arrange the following elements in a vertical list from highest (top) to lowest (bottom) ionization energy: S, O, Se

  Correct Answer

  1.         I

              Br

              Cl

   

  2.         P

              S

              Cl

   

  3.         O

              S

              Se

• Question 8

  1.    List and explain which of the following is the smaller atom.

              C or N

   2.    List and explain which of the following atoms holds its valence electrons more tightly.

              Br or I

  Correct Answer

  1.    N is smaller than C since atomic size decreases as you go to the right in a period which means that N which is further to the right is smaller.

  2.    Br holds its valence electrons more tightly than I since electronegativity decreases as you go down a group which means that Br which is further up the group has the higher electronegativity and therefore the higher attraction for its valence electrons.

• Question 9

  On a piece of scratch paper, draw the orbital configuration of the Si₁₄ atom and use it to draw the Lewis structure for the Si₁₄ atom.  Then choose the correct Lewis structure for Si₁₄ from the options listed below.

  

  Correct Answer

  

  Valence electrons=4

  2 electrons are paired and 2 unpaired.

  Lewis structure of Si is C

• Question 10

  On a piece of scratch paper, draw the orbital configuration of the Ga₃₁ atom and use it to draw the Lewis structure for the Ga₃₁ atom.  Then choose the correct Lewis structure for Ga₃₁ from the options listed below.

   

  Correct Answer

  Valence electron of Ga is 

  

  There are 3 valence electrons 

  2 paired electrons and 1 lone electron

  Lewis structure: B