Question

CHAPTER 1-Practice Exercises (ORGANIC CHEM I) Dr. Pahlavan 1. Given the ground-state electron configuration for the following atoms or ions. С Cl 2. How many valence electrons the following elements have in their valence shell? C2 с Si 3. Write electron dot structures for each formula below. Show all bonding electrons as dashes and all non-bonding electrons as dots. a) CS в) СHCIS ) CCl4, tetrachloroethylene m) CHIN. methylamine o) CHaN b) НС: h) NH c) CHa i) HO d) CH j) ONO D CHIN k) CO(NH), Urea п) СН.ОН р) НСО 4. Loratadine- is a drug used to treat allergies, and marketed for its non-sedating properties C сн, b a What is the trgona IID molecular geometry about c? uTV) bond angle about d? G 90ut b planar V) formal charge of about ef II) electron domain geometry D hybridization for af sp VI) lone pairs about f lone pai s VII) total number of pi (n) bonds 5 Draw the correct Lewis dot structures (line bonds structures or Kekule structures) that are consistent with valence electron rules for a compound with the molecular formulas a) CaHO D NCCHICOCH:CH:CHO b) CHа c) CO в) CHCH :CН(ОН)СН:СО:н d) CaleO e) CHCH:CHСНОСН. 6. Using partial charges (8/8-), indicate the polarity (+) of the polar covalent bonds in the following: CI HI (as C-C H (d) (e) (b) (a)

Answer 1

1) Ground state electronic configurations

Atomic numbers are C = 6, N = 7, O= 8 , Cl =17

C = 6 = 1s² 2s² 2p²

C^2- = 8 = 1s² 2s² 2p⁴

N = 7 = 1s² 2s² 2p³

O = 8 = 1s² 2s² 2p⁴

O^2- = 10 = 1s² 2s² 2p⁴

Cl = 17 = 1s² 2s² 2p⁶ 3s² 3p⁵

Cl- = 18 = 1s² 2s² 2p⁶ 3s² 3p⁶

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2) valance electrons means electrons in their outermost shell (valance shell)

C: atomic number = 6

   C = 1s²2s² 2p²

Bold represents valance shell electrons.

Therefore,

valance electrons in C = 4

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Si: atomic number = 14

Si = 1s² 2s² 2p⁶3s² 3p²

Bold represents valance shell electrons.

Therefore,

valance electrons in Si = 4

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S: atomic number = 16

S = 1s² 2s² 2p⁶3s² 3p⁴

Bold represents valance shell electrons.

Therefore,

valance electrons in S = 6

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N: atomic number = 7

S = 1s²2s² 2p³

Bold represents valance shell electrons.

Therefore,

valance electrons in N = 5

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O: atomic number = 8

S = 1s²2s² 2p⁴

Bold represents valance shell electrons.

Therefore,

valance electrons in O = 6