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Interpretation: The Lewis dot structure of the atoms has to be found by using their valence electrons present in the outermost shell.
Concept introduction: Lewis dot structures are also called electron dot structures. The diagrams are named after Gilbert N. Lewis, who described them in his 1916 article entitled The Atom and the Molecule.
A nucleus of an atom consists of neutrons and protons. Electrons are from closest to outermost shells around the nucleus. Proton and electron have +1 and -1 charge respectively. Neutron has no charge.
Number of protons is equal to number of electrons in the case of neutral atoms.
The number of electrons present in the outermost shell of an atom that participate in forming
If one valence electron is lost, an atom carries positive charge. If one valence electron is gained, an atom carries negative charge.
Each group present in the periodic table has specific common valence electrons. Therefore, the position of the atom in the periodic table should be known for drawing the Lewis dot structures.
To find: Lewis dot structure of a carbon atom that bears a positive charge and second-row element in which valence electrons resemble with positively charged carbon atom.
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Chapter 1 Solutions
Organic Chemistry Third Edition + Electronic Solutions Manual And Study Guide
- Complete combustion of a 0.6250 g sample of the unknown crystal with excess O2 produced 1.8546 g of CO2 and 0.5243 g of H2O. A separate analysis of a 0.8500 g sample of the blue crystal was found to produce 0.0465 g NH3. The molar mass of the substance was found to be about 310 g/mol. What is the molecular formula of the unknown crystal?arrow_forward4. C6H100 5 I peak 3 2 PPM Integration values: 1.79ppm (2), 4.43ppm (1.33) Ipeakarrow_forwardNonearrow_forward
- 3. Consider the compounds below and determine if they are aromatic, antiaromatic, or non-aromatic. In case of aromatic or anti-aromatic, please indicate number of I electrons in the respective systems. (Hint: 1. Not all lone pair electrons were explicitly drawn and you should be able to tell that the bonding electrons and lone pair electrons should reside in which hybridized atomic orbital 2. You should consider ring strain- flexibility and steric repulsion that facilitates adoption of aromaticity or avoidance of anti- aromaticity) H H N N: NH2 N Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic TT electrons Me H Me Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic πT electrons H HH…arrow_forwardA chemistry graduate student is studying the rate of this reaction: 2 HI (g) →H2(g) +12(g) She fills a reaction vessel with HI and measures its concentration as the reaction proceeds: time (minutes) [IH] 0 0.800M 1.0 0.301 M 2.0 0.185 M 3.0 0.134M 4.0 0.105 M Use this data to answer the following questions. Write the rate law for this reaction. rate = 0 Calculate the value of the rate constant k. k = Round your answer to 2 significant digits. Also be sure your answer has the correct unit symbol.arrow_forwardNonearrow_forward
- in which spectral range of EMR, atomic and ionic lines of metal liesarrow_forwardQ2: Label the following molecules as chiral or achiral, and label each stereocenter as R or S. CI CH3 CH3 NH2 C CH3 CH3 Br CH3 X &p Bra 'CH 3 "CH3 X Br CH3 Me - N OMe O DuckDuckarrow_forward1. For the four structures provided, Please answer the following questions in the table below. a. Please draw π molecular orbital diagram (use the polygon-and-circle method if appropriate) and fill electrons in each molecular orbital b. Please indicate the number of π electrons c. Please indicate if each molecule provided is anti-aromatic, aromatic, or non- aromatic TT MO diagram Number of π e- Aromaticity Evaluation (X choose one) Non-aromatic Aromatic Anti-aromatic || ||| + IVarrow_forward
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