Videos
(a)
Interpretation:
The element with its full and condensed electron configuration and the number of inner electrons is to be determined from the given partial orbital diagram.
Concept introduction:
The electronic configuration tells about the distribution of electrons in various atomic orbitals.
The full electronic configuration of an atom tells about the distribution of electrons in its various atomic orbital.
The condensed electronic configuration is a way to write the electronic configuration where the inner shell configurations are compressed to the nearest noble gas configuration and only the valence shell configuration is written in the expanded form.
The partial orbital diagram is the one that shows the distribution of valence electrons only.
The inner electrons are present in the inner orbitals of the atom and the valence electrons are present in the outermost shell of the atom.
The number of inner electrons is calculated by subtracting the valence electrons from the total number of electrons
(b)
Interpretation:
The element with its full and condensed electron configuration and the number of inner electrons is to be determined from the given partial orbital diagram.
Concept introduction:
The electronic configuration tells about the distribution of electrons in various atomic orbitals.
The full electronic configuration of an atom tells about the distribution of electrons in its various atomic orbital.
The condensed electronic configuration is a way to write the electronic configuration where the inner shell configurations are compressed to the nearest noble gas configuration and only the valence shell configuration is written in the expanded form.
The partial orbital diagram is the one that shows the distribution of valence electrons only.
The inner electrons are present in the inner orbitals of the atom and the valence electrons are present in the outermost shell of the atom.
The number of inner electrons is calculated by subtracting the valence electrons from the total number of electrons
(c)
Interpretation:
The element with its full and condensed electron configuration and the number of inner electrons is to be determined from the given partial orbital diagram.
Concept introduction:
The electronic configuration tells about the distribution of electrons in various atomic orbitals.
The full electronic configuration of an atom tells about the distribution of electrons in its various atomic orbital.
The condensed electronic configuration is a way to write the electronic configuration where the inner shell configurations are compressed to the nearest noble gas configuration and only the valence shell configuration is written in the expanded form.
The partial orbital diagram is the one that shows the distribution of valence electrons only.
The inner electrons are present in the inner orbitals of the atom and the valence electrons are present in the outermost shell of the atom.
The number of inner electrons is calculated by subtracting the valence electrons from the total number of electrons
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Chapter 8 Solutions
MCGRAW: CHEMISTRY THE MOLECULAR NATURE
- 3. Provide the missing compounds or reagents. 1. H,NNH КОН 4 EN MN. 1. HBUCK = 8 хно Panely prowseful kanti-chuprccant fad, winddively, can lead to the crading of deduc din-willed, tica, The that chemooices in redimi Грин. " like (for alongan Ridovi MN نيا . 2. Cl -BuO 1. NUH 2.A A -BuOK THE CF,00,H Ex 5)arrow_forward2. Write a complete mechanism for the reaction shown below. NaOCH LOCH₁ O₂N NO2 CH₂OH, 20 °C O₂N NO2arrow_forward4. Propose a synthesis of the target molecules from the respective starting materials. a) b) LUCH C Br OHarrow_forward
- The 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_forward1. 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_forward
- The 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_forward3. 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_forward
- Question 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_forwardIn 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_forward
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