Question

part c: electron configuration

1. how many electrons can each level hold? 1st =__ 2nd = 3rd =__
2. what term is used for the electrons in the outermost shell or energy level?____
3. scientists use two types of diagrams to show the electron - configuration for atoms. follow your teacher’s directions to complete the diagrams.

sulfur
atomic # = 16
atomic mass = 32
protons = 16
neutrons = 16
electron = 16
bohr diagram
shows all electrons
lewis structure
shows valence electrons
s

1. calculate the missing information and then draw the bohr diagram and lewis structure for each element.

li
atomic # = 3
mass # = 7

of p =____

of n =____

of e =____

ne
atomic # = 10
mass # = 20

of p =____

of n =____

of e =____

mg
atomic # = 12
mass # = 24

of p =____

of n =____

of e =____

cl
atomic # = 17
mass # = 35

of p =____

of n =____

of e =____

he
atomic # = 2
mass # = 4

of p =____

of n =____

of e =____

si
atomic # = 14
mass # = 28

of p =____

of n =____

of e =____

1. answer the questions below based on the elements in question #15.

(1) which elements had a filled outermost shell?____
(2) which element would be most likely to lose electrons in a chemical bond?____
(3) which element would be most likely to gain electrons in a chemical bond?____
(4) which elements are not likely to bond with other elements?__ why?__

Explanation:

Step1: Recall electron - shell capacity rules

The first energy level can hold 2 electrons, the second can hold 8 electrons, and the third can hold 8 electrons (in simple atomic - structure models for these elements). So, 1st = 2, 2nd = 8, 3rd = 8.

Step2: Define valence electrons

The term used for the electrons in the outermost shell or energy level is valence electrons.

Step3: Calculate sub - atomic particles for Li

For Li (Atomic # = 3):

• The number of protons (# of P) is equal to the atomic number, so # of P = 3.
• The number of neutrons (# of N)=Mass # - Atomic # = 7 - 3 = 4.
• In a neutral atom, the number of electrons (# of E) is equal to the number of protons, so # of E = 3.

Step4: Calculate sub - atomic particles for Ne

For Ne (Atomic # = 10):

• # of P = 10.
• # of N=20 - 10 = 10.
• # of E = 10.

Step5: Calculate sub - atomic particles for Mg

For Mg (Atomic # = 12):

• # of P = 12.
• # of N=24 - 12 = 12.
• # of E = 12.

Step6: Calculate sub - atomic particles for Cl

For Cl (Atomic # = 17):

• # of P = 17.
• # of N=35 - 17 = 18.
• # of E = 17.

Step7: Calculate sub - atomic particles for He

For He (Atomic # = 2):

• # of P = 2.
• # of N=4 - 2 = 2.
• # of E = 2.

Step8: Calculate sub - atomic particles for Si

For Si (Atomic # = 14):

• # of P = 14.
• # of N=28 - 14 = 14.
• # of E = 14.

Step9: Determine elements with filled outermost shell

He has 2 electrons filling its first shell and Ne has 8 electrons filling its second shell. So, He and Ne have filled outermost shells.

Step10: Determine element likely to lose electrons

Li has 1 valence electron and is most likely to lose electrons in a chemical bond to achieve a stable electron configuration.

Step11: Determine element likely to gain electrons

Cl has 7 valence electrons and is most likely to gain 1 electron to achieve a stable octet.

Step12: Determine non - bonding elements

He and Ne are noble gases. They have full valence shells, so they are not likely to bond with other elements because they are already in a stable electron configuration.

Answer:

1. 2, 8, 8
2. Valence electrons

15.

• Li: # of P = 3, # of N = 4, # of E = 3
• Ne: # of P = 10, # of N = 10, # of E = 10
• Mg: # of P = 12, # of N = 12, # of E = 12
• Cl: # of P = 17, # of N = 18, # of E = 17
• He: # of P = 2, # of N = 2, # of E = 2
• Si: # of P = 14, # of N = 14, # of E = 14

16.

• (1) He, Ne
• (2) Li
• (3) Cl
• (4) He, Ne; Because they have full valence shells.