Concept explainers
(a)
Interpretation:
The maximum number of electrons which can hold by shell No. 1 should be determined.
Concept Introduction:
In a planetary model of an atom, negative charged electrons are arranged around the positive charged electron in a series of shells which is like orbits.
The electrons present in the outermost energy level or shell is known as valence electrons. These electrons are available for bonding and outermost shell is known as valence shell.
(b)
Interpretation:
The maximum number of electrons which can hold by shell No. 2 should be determined.
Concept Introduction:
In a planetary model of an atom, negative charged electrons are arranged around the positive charged electron in a series of shells which is like orbits.
The electrons present in the outermost energy level or shell is known as valence electrons. These electrons are available for bonding and outermost shell is known as valence shell.
(c)
Interpretation:
The way should be described by which answers to a) and b) are contained in periodic table.
Concept Introduction:
The distribution of electrons in atom into orbitals is said to be electronic configuration. The electronic configuration for every element present in the periodic table is unique or different.
Periodic Table contains periods and groups. There are 18 groups and 7 periods in the periodic table. The vertical columns are known as groups and horizontal rows are known as periods. The numbering of periods is done as 1 to 7 from top to bottom and groups are named as 1A, 2A, 3B to 8B, 1B, 2B, 3A to 8A from left to right where A represents representative elements and B represents
In periodic table, different blocks within the periodic table correspond to the s, p, d and f sublevels. Thus, on the basis of increasing atomic number in the periodic table, electronic configuration of atoms can be written.
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- 4. Propose a synthesis of the target molecules from the respective starting materials. a) b) LUCH C Br OHarrow_forwardThe following mechanism for the gas phase reaction of H2 and ICI that is consistent with the observed rate law is: step 1 step 2 slow: H2(g) +ICI(g) → HCl(g) + HI(g) fast: ICI(g) + HI(g) → HCl(g) + |2(g) (1) What is the equation for the overall reaction? Use the smallest integer coefficients possible. If a box is not needed, leave it blank. + → + (2) Which species acts as a catalyst? Enter formula. If none, leave box blank: (3) Which species acts as a reaction intermediate? Enter formula. If none, leave box blank: (4) Complete the rate law for the overall reaction that is consistent with this mechanism. (Use the form k[A][B]"..., where '1' is understood (so don't write it) for m, n etc.) Rate =arrow_forwardPlease correct answer and don't use hand rating and don't use Ai solutionarrow_forward
- 1. For each of the following statements, indicate whether they are true of false. ⚫ the terms primary, secondary and tertiary have different meanings when applied to amines than they do when applied to alcohols. • a tertiary amine is one that is bonded to a tertiary carbon atom (one with three C atoms bonded to it). • simple five-membered heteroaromatic compounds (e.g. pyrrole) are typically more electron rich than benzene. ⚫ simple six-membered heteroaromatic compounds (e.g. pyridine) are typically more electron rich than benzene. • pyrrole is very weakly basic because protonation anywhere on the ring disrupts the aromaticity. • thiophene is more reactive than benzene toward electrophilic aromatic substitution. • pyridine is more reactive than nitrobenzene toward electrophilic aromatic substitution. • the lone pair on the nitrogen atom of pyridine is part of the pi system.arrow_forwardThe following reactions are NOT ordered in the way in which they occur. Reaction 1 PhO-OPh Reaction 2 Ph-O -CH₂ heat 2 *OPh Pho -CH2 Reaction 3 Ph-O ⚫OPh + -CH₂ Reaction 4 Pho Pho + H₂C OPh + CHOPh H₂C -CH₂ Reactions 1 and 3 Reaction 2 O Reaction 3 ○ Reactions 3 and 4 ○ Reactions 1 and 2 Reaction 4 ○ Reaction 1arrow_forwardSelect all possible products from the following reaction: NaOH H₂O a) b) ОН HO O HO HO e) ОН f) O HO g) h) + OHarrow_forward
- 3. Draw diagrams to represent the conjugation in these molecules. Draw two types of diagram: a. Show curly arrows linking at least two different ways of representing the molecule b. Indicate with dotted lines and partial charges (where necessary) the partial double bond (and charge) distribution H₂N* H₂N -NH2arrow_forwardQuestion 2 of 25 point Question Attempt 3 of Ulimited Draw the structure for 3-chloro-4-ethylheptane. Part 2 of 3 Click and drag to start drawing a structure. Draw the structure for 1-chloro-4-ethyl-3-lodooctane. Click and drag to start drawing a structure. X G X B c Part 3 of 30 Draw the structure for (R)-2-chlorobutane. Include the stereochemistry at all stereogenic centers. Check Click and drag to start drawing a structure. G X A 。 MacBook Pro G P Save For Later Submit Assignment Privacyarrow_forwardPlease correct answer and don't used hand raitingarrow_forward
- In a silicon and aluminum alloy, with 12.6% silicon, what are the approximate percentages of the phases present in the constituent that is formed at the end of solidification? Temperature (°C) 1500 1000 L B+L 1415- α+L 577' 500 1.65 12.6 99.83 α+B B 0 Al 20 40 60 Weight percent silicon 80 Siarrow_forwardPlease correct answer and don't used hand raitingarrow_forwardPlease correct answer and don't used hand raitingarrow_forward
Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning
World of Chemistry, 3rd editionChemistryISBN:9781133109655Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCostePublisher:Brooks / Cole / Cengage Learning