- ground-state electronic configuration
- Hund's rule
- Pauli exclusion principle
- aufbau principle
Electron Configurations
![Atomic 1 0 4 Equals Atomic 1 0 4 Equals](https://upload.wikimedia.org/wikipedia/commons/5/5c/Atomic_orbitals_n123_m-eigenstates.png)
Rules for Assigning Electron Orbitals
Pauli Exclusion Principle
Hund's Rule
Occupation of Orbitals
The Aufbau Process
- B (Z=5) configuration: 1s2 2s2 2p1
- C (Z=6) configuration:1s2 2s2 2p2
- N (Z=7) configuration:1s2 2s2 2p3
- O (Z=8) configuration:1s2 2s2 2p4
- F (Z=9) configuration:1s2 2s2 2p5
- Ne (Z=10) configuration:1s2 2s2 2p6
The Number of Valence Electrons
Periodic table group | Valence electrons |
---|---|
Group 1: alkali metals | 1 |
Group 2: alkaline earth metals | 2 |
Groups 3-12: transition metals | 2* (The 4s shell is complete and cannot hold any more electrons) |
Group 13: boron group | 3 |
Group 14: carbon group | 4 |
Group 15: pnictogens | 5 |
Group 16: chalcogens | 6 |
Group 17: halogens | 7 |
Group 18: noble gases | 8** |
- Locate the nearest noble gas preceding phosphorus in the periodic table. Then subtract its number of electrons from those in phosphorus to obtain the number of valence electrons in phosphorus.
- Referring to Figure 1.3.1, draw an orbital diagram to represent those valence orbitals. Following Hund’s rule, place the valence electrons in the available orbitals, beginning with the orbital that is lowest in energy. Write the electron configuration from your orbital diagram.
- Ignore the inner orbitals (those that correspond to the electron configuration of the nearest noble gas) and write the valence electron configuration for phosphorus.
- Answer
- [Ne]3s23p5; 3s23p5
1s2 | row 1 | 2 electrons |
2s22p6 | row 2 | 8 electrons |
3s23p6 | row 3 | 8 electrons |
4s23d104p6 | row 4 | 18 electrons |
5s24d105p6 | row 5 | 18 electrons |
row 1–5 | 54 electrons |
Atomic 1 0 4 Equals Ounces
Summary
Exercises
Solutions
Questions
![Atomic 1 0 4 equals grams Atomic 1 0 4 equals grams](https://static.docsity.com/documents_pages/notas/2008/12/02/55917fcd80c9d14dc3bc8d1543877be6.png)
Atomic 1 0 4 Equals Equal
Solutions
Contributors and Attributions
- Dr. Dietmar Kennepohl FCIC (Professor of Chemistry, Athabasca University)
- Prof. Steven Farmer (Sonoma State University)
- William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
- Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
As you I just discussed in the Spectral Lines page, electrons fall to lower energy levels and give off light in the form of a spectrum. These spectral lines are actually specific amounts of energy for when an electron transitions to a lower energy level. If you assume the energy levels of an atom to be a staircase; if you roll a ball down the stairs the ball only has a few 'steps' that it can stop on. This is the same situation an electron is in. Electrons can only occupy specific energy levels in an atom. It most be on an energy level if it is in the atom. There is no in between. This is why you get lines and not a 'rainbow' of colors when electrons fall. Jahann Balmer in 1885 derived an equation to calculate the visible wavelengths that the hydrogen spectrum displayed. The lines that appear at 410nm, 434 nm, 486 nm, and 656 nm. These electrons are falling to the 2nd energy level from higher ones. This transition to the 2nd energy level is now referred to as the 'Balmer Series' of electron transitions. Johan Rydberg use Balmers work to derived an equation for all electron transitions in a hydrogen atom. Here is the equation: R= Rydberg Constant 1.0974x107m-1; λis the wavelength; nis equal to the energy level (initial and final) If we wanted to calculate energy we can adjust R by multipling by h (planks constant) and c (speed of light) Valhalla room 1 5 1 mac torrent crack download. Now we have Rydbergs equation to calculate energy. RE= -2.178 x 10-18J (it is negative because energy is being emitted) l = h c /E l= ( 6.626 x 10- 34J s) (3.0 x 108m/s)/E 1nm= 1 x 10-9m
Converting Wavelength to frequency c= 3.0 x 108m/s ;l= wavelength (m) ;v= frequency (s-1)
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