ELECTRON CONFIGURATION

Agenda • Electron Configuration (O/S) HW: Complete Questions on the Handouts

Bohr’s model

• Electrons orbit the nucleus in energy levels and are held there by electrostatic force of attraction ( attraction between positive nucleus and negatively charged electrons).

Bohr - Rutherford diagrams

• Putting all this together, we get B-R diagrams • To draw them you must know the # of protons, neutrons, and electrons. • Draw protons (p+), (n0) in circle (i.e. “nucleus”) • Draw electrons around in shells

He p+

2 2 n0

Li

Li shorthand

p+

3 4 n0

Draw BR diagrams for Ar, Ca, Sc.

3 p+ 4 n0

2e– 1e–

Quantum Mechanics Model Quantum Mechanics Model – modern description of the electron in atoms, derived from a mathematical equation (Schrodinger’s wave equation). Electrons within an atom can posses only discrete quantities of energy. Schrodinger’s equation describes the probability distribution of an electron. 90% probability of finding the electron

Orbital

Models of the Atom- Review

Dalton’s model Greek model (1803) (400 B.C.)

1803 John Dalton pictures atoms as tiny, indestructible particles, with no internal structure.

1800

Thomson’s plum-pudding model (1897)

Rutherford’s model (1909)

1897 J.J. Thomson, a British

1911 New Zealander

scientist, discovers the electron, leading to his "plum-pudding" model. He pictures electrons embedded in a sphere of positive electric charge.

Ernest Rutherford states that an atom has a dense, positively charged nucleus. Electrons move randomly in the space around the nucleus.

1805 ..................... 1895

1900

1905

1910

1904 Hantaro Nagaoka, a Japanese physicist, suggests that an atom has a central nucleus. Electrons move in orbits like the rings around Saturn.

1915

Bohr’s model (1913)

1926 Erwin Schrödinger

1913 In Niels Bohr's model, the electrons move in spherical orbits at fixed distances from the nucleus.

1920

1925

Charge-cloud model (present)

1930

develops mathematical equations to describe the motion of electrons in atoms. His work leads to the electron cloud model.

1935

1940

1945

1924 Frenchman Louis

1932 James

de Broglie proposes that moving particles like electrons have some properties of waves. Within a few years evidence is collected to support his idea.

Chadwick, a British physicist, confirms the existence of neutrons, which have no charge. Atomic nuclei contain neutrons and positively charged protons.

Heisenberg’s uncertainty principle We cannot know both the location and velocity of an electron (Heisenberg’s uncertainty principle), thus Schrodinger’s equation does not tell us the exact location of the electron, rather it describes the probability that an electron will be at a certain location in the atom. Today we say that the electrons are located in a region outside the nucleus called the electron cloud. An orbital is a region of space where there is a high probability of locating an electron.

Electron Cloud – Energy Levels The energy levels are analogous to the rungs of a ladder. The lowest rung of the ladder corresponds to the lowest energy level. A person can climb up or down a ladder by going from rung to rung. Similarly, an electron can jump from one energy level to another. A person on a ladder cannot stand between the rungs; similarly, the electrons in an atom cannot exist between energy levels.

ATOMIC SUBLEVELS • Bohr Rutherford atomic theory does not account for all the spectral lines present in Hydrogen. • It was proposed that energy levels were divided into sublevels. • The letters s, p, d and f were used to identify the sublevels. • Each sublevel orbital holds only 2 electrons, but the sublevels contain different numbers of orbitals: s - 1 orbital [maximum 2 e-] p - 3 orbitals [maximum 6 e-] d - 5 orbitals [maximum 10 e-] f - 7 orbitals [maximum 14 e-]

Shapes of s, p, and d-Orbitals s orbital 1 orbital p orbitals 3 orbitals

d orbitals 5 orbitals

Different energy levels contain only certain sublevels Energy level 1 - s sublevel only Energy level 2 - s and p sublevels only Energy level 3 - s, p and d sublevels only Energy level 4 - s, p, d and f sublevels only Electron configuration denotes the arrangement of the electrons in their energy levels, sublevels and orbitals

Size of the orbital in different energy levels n2 = # of orbitals in an energy level Total # of electrons = 2 n2

Energy level

Due to like charge repulsion, electrons of opposite spin direction can occur together in 1 orbital, maximum of 2 e- /orbital.

1s

2s

3s

ELECTRON CONFIGURATION • An electron configuration is a written representation of the arrangement of electrons in an atom. 1st rule - electrons occupy orbital’s that require the least amount of energy for the electron to stay there.

• You notice, for example, that the 4s sublevel requires less energy than the 3d sublevel; therefore, the 4s orbital is filled with electrons before any electrons enter the 3d orbital!!!! So just follow the above chart or the periodic table orbital blocks and you can’t go wrong!!!!)

Examples: energy level

# of e- in orbital

Hydrogen: 1s1 orbital/sublevel Helium: 1s2 Lithium: 1s22s1 Carbon: 1s22s22p2 You try to write the notation for Iron

Electron Configurations Orbital Filling Element

1s

2s

2px 2py 2pz

3s

Electron Configuration

H

1s1

He

1s2

C

NOT CORRECT 1s22s1 Violates Hund’s Rule 1s22s22p2

N

1s22s22p3

O

1s22s22p4

F

1s22s22p5

Ne

1s22s22p6

Na

1s22s22p63s1

Li

Fe = 1s1 2s22p63s23p64s23d6

1s

e-

ee-

e-

e-

eee-

e-

e-

e-

e-

ee-

2px 2py 2pz

2s

e-

+26

e-

e-

ee-

e-

ee-

e-

ee-

e-

3s

26 electrons. Iron has ___

3px 3py 3pz

4s

3d

3d

3d

3d

3d

Short Hand or Noble Gas Notation: Short Hand or Noble Gas Notation: Use the noble gases that have complete inner energy levels and an outer energy level with complete s and p orbital’s. Use the noble gas that just precedes the element you are working with. Ex: Boron is 1s22s22p1 The noble gas preceding Boron is He, so the short way is [He]2s22p1

Ex. Sulfur is ls22s22p63s23p4

Short way: [Ne]3s23p4

Practice Problems: 1.Write electron configuration for the following atom: iodine 1s22s22p63s23p6 4s23d104p65s24d105p5

2.Write shorthand electron configuration for the following: I [Kr] 5s24d105p5

Electron configurations for Ions First, determine if the element will lose or gain electrons. Secondly, what number of electrons will be gained or lost? It is recommended that you write the electron configuration for the atom and then determine what will happen.

Cations For cations (positive ions) – look at the element and decide how many electrons will be lost when it ionizes and keep that in mind when writing the E. C. The last number in the E. C. will now be LESS than what is written on your periodic table. Ex. Write the electron configuration for stable magnesium ion: 1s22s22p63s2 is for the atom Mg is a metal and will lose its valence (outer) electrons so the E.C. for Mg2+ is 1s22s22p6

Anions For anions (negative ions) – look at the element and decide how many electrons that element will GAIN when it ionizes. The last number in the E. C. will be MORE than what is written on the periodic table. Ex. Write the electron configuration for stable sulphide ion: 1s22s22p63s23p4 + 2 is for the atom S is a non-metal and will gain 2 electrons to become isoelectronic with a noble gas. so the E.C. for S2- is 1s22s22p63s23p6

Irregular Electron Configurations Sometimes the electron configuration is NOT what we would predict it to be. Sometimes electrons are moved because (l) it will result in greater stability for that atom or (2) for some unknown reason? It is very important to define “stable” here. STABLE means: • all (equal energy) orbital’s are FULL • all orbital’s are half-full • all orbital’s are totally empty.

Isoelectronicity Two or more entities (atoms, ions) are described as being isoelectronic with each other if they have the same number of electrons or the same electron configuration. Ex: State an neutral entity that is isoelectronic with Cl-. Cl-

1s22s22p63s23p6

Ar

1s22s22p63s23p6

Ex: State a charged entity that is isoelectronic with Cl-. S2-

1s22s22p63s23p6

Ca2+

1s22s22p63s23p6 4s2

1s

2 4

(n-1) d

5 6 7

( n-2) 4f 5f