part c: electron configuration\n12. how many electrons can each level hold? 1st =____ 2nd =____ 3rd…
part c: electron configuration\n12. how many electrons can each level hold? 1st =____ 2nd =____ 3rd =____\n13. what term is used for the electrons in the outermost shell or energy level?____\n14. scientists use two types of diagrams to show the electron - configuration for atoms. follow your teacher’s directions to complete the diagrams.\nsulfur\natomic # = 16\natomic mass = 32\nprotons = 16\nneutrons = 16\nelectron = 16\nbohr diagram\nshows all electrons\nlewis structure\nshows valence electrons\ns\n15. calculate the missing information and then draw the bohr diagram and lewis structure for each element.\nli\natomic # = 3\nmass # = 7\n# of p =____\n# of n =____\n# of e =____\nne\natomic # = 10\nmass # = 20\n# of p =____\n# of n =____\n# of e =____\nmg\natomic # = 12\nmass # = 24\n# of p =____\n# of n =____\n# of e =____\ncl\natomic # = 17\nmass # = 35\n# of p =____\n# of n =____\n# of e =____\nhe\natomic # = 2\nmass # = 4\n# of p =____\n# of n =____\n# of e =____\nsi\natomic # = 14\nmass # = 28\n# of p =____\n# of n =____\n# of e =____\n16. answer the questions below based on the elements in question #15.\n(1) which elements had a filled outermost shell?____\n(2) which element would be most likely to lose electrons in a chemical bond?____\n(3) which element would be most likely to gain electrons in a chemical bond?____\n(4) which elements are not likely to bond with other elements?____ why?____
Answer
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):
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of P = 10.
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of N=20 - 10 = 10.
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of E = 10.
Step5: Calculate sub - atomic particles for Mg
For Mg (Atomic # = 12):
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of P = 12.
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of N=24 - 12 = 12.
-
of E = 12.
Step6: Calculate sub - atomic particles for Cl
For Cl (Atomic # = 17):
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of P = 17.
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of N=35 - 17 = 18.
-
of E = 17.
Step7: Calculate sub - atomic particles for He
For He (Atomic # = 2):
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of P = 2.
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of N=4 - 2 = 2.
-
of E = 2.
Step8: Calculate sub - atomic particles for Si
For Si (Atomic # = 14):
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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:
- 2, 8, 8
- Valence electrons
- 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
- (1) He, Ne
- (2) Li
- (3) Cl
- (4) He, Ne; Because they have full valence shells.