chapter 4-dmc 101-arrangement of electron.pdf

Upload: arivalagan-revichandran

Post on 14-Apr-2018

215 views

Category:

Documents


0 download

TRANSCRIPT

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    1/33

    CHAPTER 4: THE ARRANGEMENT OF

    ELECTRONS IN ATOMS

    Reaction happens WHEN electronin atoms

    interact.

    Electromagnetic Radiation

    When electron interacts produces

    electromagnetic radiation

    Electromagnetic radiation carries energy Example:

    visible light, x-rays, radio waves

    Audi Majdan - DMC 101 - KLIUC

    Chemical

    Reaction

    Interaction of

    electrons in atoms=

    1

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    2/33

    Properties of electromagnetic radiation:

    (a)

    Move through vacuum at speed oflight, c = 3.00 x 108 m/s

    (b) Wave-like properties

    Properties Symbol Unit Pronunciation

    frequency (Hz) nuwavelength (m) lambda

    Audi Majdan - DMC 101 - KLIUC 2

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    3/33

    Relationship:

    = c

    c (speed of light) = 3.00 x 108 m/s

    Electron interaction depends on

    arrangement of electron in atoms.

    The arrangements of electronin atoms

    provides information on:(a) Number of electron

    (b) Position of electron

    (c) Energypossess by atom

    Audi Majdan - DMC 101 - KLIUC 3

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    4/33

    Quantum and Photons

    Electron orbiting the nucleus have it's

    own energy Energy can be released orabsorbed by the

    atoms during the interaction

    Quantum = a unit of energy released or

    absorbed by atoms

    A photon = a quantum of energy

    E = h or E = hc

    E = Energy of a quantum

    = frequencyh = Planck's constant

    = 6.63 x 10-34 Joule seconds (Js)

    Example:

    Calculate the smallest amount of energy (one

    quantum) that an object can absorb from yellow

    light with a wavelength of 589 nm.

    Audi Majdan - DMC 101 - KLIUC 4

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    5/33

    Solution

    1) Frequency:

    = c

    = c/

    = (3.00 x 108 m/s)

    (589 x 10-9 m)

    = 5.09 x 1014 s-1

    2) Energy:

    E = h

    E = (6.63 x 10-34 Js) x (5.09 x 1014 s-1)

    E = 3.37 x 10-19 J

    Exercise 1

    A laser emits light with a frequency of 4.69 x

    1014 s-1. What is the energy of one quantum ofthe laser?

    Electrons Arrangement in Atom

    Audi Majdan - DMC 101 - KLIUC 5

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    6/33

    Several systems are created to know the

    LOCATION ofELECTRONs in ATOM

    Bohr's Model of Atom

    Dalton's model of atomprovides the basisof atom structure

    Bohr's model of atom is an advanced

    model of atom structure

    Characteristics of Bohr's Model:

    (a) Electrons orbiting the nucleus(b) The electrons are restricted only to orbits

    of certain radius and energies

    Audi Majdan - DMC 101 - KLIUC

    ELECTRONS

    ARRANGEMENT

    Single electron atom Multiple electrons atom

    Bohrs Model Schrodingers Model

    6

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    7/33

    (c) An electron orbiting in an orbit has a

    fixed energy state and will not release or

    absorb energy

    The energy possess by the electron:

    RH = Rydberg constant

    = 2.18 x 10-18 J (for calculation of E)

    or

    = 1.097 x 107 m-1 (for calculation of )

    n = principle quantum number

    = orbits for the electron

    Energies always in negative value

    n=1 an electron in the first allowed orbit

    (closest to the nucleus)

    n=2 an electron in the next allowed orbit

    further from the nuclei n=3,4,5

    Audi Majdan - DMC 101 - KLIUC 7

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    8/33

    The energies of allowed orbits for the

    electrons:

    The lower the energy (more negative), the

    more stable the atom

    Audi Majdan - DMC 101 - KLIUC 8

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    9/33

    Ground state = electron is in the lowest

    energy state (n=1) the most stable state

    Electrons can jump from one orbit to

    another

    Excited state = electron is in other state

    than n=1 (n 2)

    When n = infinity

    Electron completely separated from the

    atom

    E = (-2.18 x 10-18 J) x (1/)

    = 0

    Reference or zero energy state = the state

    in which the electron is separated from the

    nucleus

    Audi Majdan - DMC 101 - KLIUC 9

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    10/33

    Electron moves to higher state (higher n)

    = Energyabsorbed

    Electron moves to lower state (lower n)

    = Energyemitted

    The difference in energy levels between the

    final and initial orbits:

    E = Ef - Ei

    or

    ni = initial orbit

    nf = final orbit

    E=+ve when energy absorbed:nf> ni = electron moves to higher n

    Audi Majdan - DMC 101 - KLIUC 10

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    11/33

    E=-vewhen energy emitted:ni > nf = electron falls to lower n

    Example:

    Calculate the energy of light that corresponds to

    the transition of the electron from the n=4 to the

    n=2 state of the hydrogen atom. Is the lightabsorbed or emitted by the atom?

    Solution

    ni = 4

    nf = 2

    RH = 2.18 x 10-18 J (for calculation of E)

    E = (2.18 x 10-18 J) x ((1/16)-(1/4))= -4.09 x 10-19 J

    (light is emitted)

    Audi Majdan - DMC 101 - KLIUC 11

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    12/33

    Exercise 2

    Determine the energy of light that corresponds

    to the transition of electron from n=3 to n=5

    state for hydrogen atom. Identify whether the

    energy is absorbed or released.

    Bohr's Model for Hydrogen atom

    Line Spectra

    Line spectra = spectrum of coloured line

    with different and separated wavelengths

    when electron fall (move to lower orbit)

    Example:

    A rainbow represents the spectrum of

    wavelengths of light contained in the light

    emitted by the sun

    Forhydrogen atom: 5 series ofline spectra

    observed depending on the type of electrontransition

    Audi Majdan - DMC 101 - KLIUC 12

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    13/33

    Audi Majdan - DMC 101 - KLIUC 13

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    14/33

    Series nf ni Spectrum

    Region

    Lymann 1 2,3,4.. Ultra violet

    Balmer 2 3,4,5.. Visible light

    Paschen 3 4,5,6.. Infrared

    Brackett 4 5,6,7.. Infrared

    Pfund 5 6,7,8.. Infrared

    To determine wavelengths of the spectrum:

    =

    22

    111

    fi

    H

    nnR

    Example:

    Calculate the wavelength of energy released

    during a transition from the n=2 to n=4.

    =

    22

    111

    fi

    H

    nnR

    RH = 1.097 x 107 m-1 (for calculation of )

    =

    22

    7

    4

    1

    2

    110097.1

    1

    ( )1875.010097.11

    7

    =

    Audi Majdan - DMC 101 - KLIUC 14

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    15/33

    16

    1006.21

    = m

    m710862.4 =

    Exercise 3

    Calculate the wavelength of the hydrogen

    emission line that corresponds to the transition

    of the electron from the n = 3 to the n =1 state.

    Schrodinger's Model of Atom

    Difference between Bohrs Model and

    Schrodingers Model

    Properties Bohr Schrodinger

    Type of atom Single electron

    atom

    Multiple

    electrons atom

    Audi Majdan - DMC 101 - KLIUC 15

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    16/33

    Electrons

    Location

    In a defined

    orbit

    Probability

    electron

    distribution

    only

    Notation

    system

    One quantum

    number:

    n

    Four quantum

    numbers:

    n, l, ml, ms

    Probability distribution of an electron in the

    ground state in a hydrogen atom:

    Audi Majdan - DMC 101 - KLIUC 16

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    17/33

    Uncertainty Principle: it is impossible to

    simultaneously know both the position

    and momentum of an object as small asan electron

    Schrodingers Notation System

    The Bohrs model: one quantum number (n)to describe an orbit

    The Schrdinger model: four quantum

    numbers:

    Audi Majdan - DMC 101 - KLIUC 17

    More probability

    to find an

    electron

    Less probability to

    find an electron

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    18/33

    (a) Principal quantum number (n),

    (b) Orbital quantum number (l),

    (c) Magnetic quantum number (ml)

    (d) Spin quantum number (ms)

    to describe an orbital.

    (a) The principle quantum number (n)

    n = shell (the energy level/orbits) in the

    atom

    n = 1, 2, 3, ...

    Example: n = 1 (electron is at 1st shell)

    As n increases, the electron density is

    further away from the nucleus

    As n increases, the electron has a higher

    energy and is less tightly bound to the

    nucleus

    (b) The azimuthal quantum number (l)

    Audi Majdan - DMC 101 - KLIUC 18

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    19/33

    l= subshell in a shell

    The value ofldepends on n

    For every n, there are n value of l

    l= (n-1)

    lis referred to by a letter:

    s = 0, p =1, d =2, f =3

    l 0 1 2 3 4

    notation s p d f g

    Example:

    n = 1 l= (n-1)

    = 1-1= 0

    (n = 1: one subshelllwith value of 0)

    Notation: 1s

    n = 2 l= 2-1

    Audi Majdan - DMC 101 - KLIUC 19

    n = 1 l= 0

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    20/33

    = 1

    (n = 2: two subshelllwith value of 0 and 1)

    Notation: 2s (n = 2 and l= 0 )

    Notation: 2p (n = 2 and l= 1)

    n lSubshell

    notation

    1 0 1s2

    0 2s

    1 2p

    3

    0 3s

    1 3p

    2 3d

    40 4s1 4p

    2 4d

    3 4f

    (c) The magnetic quantum number (ml)

    ml = orbitals in a subshell Orbital is the place where electrons located

    ml = all values between 'l', 0, -'l'

    Audi Majdan - DMC 101 - KLIUC 20

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    21/33

    nValues

    ofl

    Subshell

    Designat

    ion

    Values

    ofml

    Number of

    Orbitals in

    Subshell

    Total

    number

    of orbitals

    in shell

    1 0 1s 0 1 1

    2 0 2s 0 14

    1 2p 1,0,-1 3

    3 0 3s 0 1

    91 3p -1,0,1 3

    2 3d-2,-

    1,0,1,25

    4 0 4s 0 1

    16

    1 4p -1,0,1 3

    2 4d-2,-

    1,0,1,25

    3 4f

    -3,-2,-

    1,0,1,2,

    3

    7

    Audi Majdan - DMC 101 - KLIUC 21

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    22/33

    (d) The electron spin quantum number (ms)

    ms = direction of spin of the electron

    ms = + or -

    Explanation on The Quantum Number

    Electron shell = consisting ofsubshell

    Subshell = consisting oforbitals

    Orbitals = consisting ofelectrons

    Audi Majdan - DMC 101 - KLIUC 22

    +

    -

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    23/33

    Example:

    All orbitals that have principle quantum

    number of 3 (n=3) is said to be in third shell.

    The third electron shell ('n'=3) consists of the

    3s, 3p and 3dsubshells

    The 3s subshellcontains 1 orbital,

    the 3p subshellcontains 3 orbitals

    the 3d subshellcontains 5 orbitals

    The third electron shellis comprised of

    nine different orbitals, but all orbital with

    the same energy (true for hydrogen only)

    Each subshell is divided into orbitals:

    Subshell Number of orbitals

    s 1

    p 3

    d 5

    f 7

    Audi Majdan - DMC 101 - KLIUC 23

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    24/33

    Energies of all hydrogen electron orbitals

    until n=3:

    At room temperatures all hydrogen atoms

    are in their ground states

    The electron may be promoted to an excitedstate by the absorption of a photon

    Example:

    a) Predict the number of subshells in the

    fourth shell, n=4.

    b) Give the label for each of the subshells

    in (a)

    c) How many orbitals are there in each of

    the subshells in (c)?

    Audi Majdan - DMC 101 - KLIUC 24

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    25/33

    Answer:

    a) n = 4 l= n-1

    = 3

    Therefore: 0, 1, 2, 3

    4 subshells

    b) s = 0, p = 1, d = 2, f = 3

    Therefore: 4s, 4p, 4d, 4f

    c) 4s l= 0, value forml ( from -l to l)

    = 0 one 4s orbital

    4p three 4p orbitals;

    4d five 4d orbitals;

    4f seven 4f orbitals

    Exercise 4

    a) What is the designation for the subshell

    with n =5 and l=1?

    b)How many orbitals are in this subshell?

    c) Indicate the values ofml for each ofthese orbitals.

    Orbitals in Multiple-Electron Atoms

    Audi Majdan - DMC 101 - KLIUC 25

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    26/33

    The hydrogen atom only 1 electron.

    All subshells ofhydrogen atoms with the

    same principle quantum number (n) have thesame energetic level

    In multiple electrons atom: electron-

    electron repulsion influences the energy

    levels of the orbitals. Example:

    Energy state 2s orbital < 2p orbital in a

    many-electron atom

    Electron spin and the Pauli Exclusion

    Principle

    Audi Majdan - DMC 101 - KLIUC 26

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    27/33

    Pauli exclusion principle = No two

    electrons in an atom can have the same

    set of four quantum numbers (n, l, mland

    ms)

    For any given orbital has fixed values of:

    n

    l

    ml

    ms quantum number is use to differentiate

    between 2 electron in an orbital (+ and

    - )

    Only two electrons at most can occupy the

    same orbital, and they have opposite valuesfor magnetic spin

    Example:

    2 electron in one orbital can have quantum

    number of

    (1, 0, 0, +) & (1, 0, 0, -)

    Electron Configurations

    Audi Majdan - DMC 101 - KLIUC 27

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    28/33

    Electron configuration = the way electrons

    are distributed among the orbitals

    Orbitals are occupied in order of

    increasing energy, with no more than two

    electrons per orbital

    Example

    Lithium

    1. Lithium has 3 electrons.2. Start : Place two electrons in the 1s,

    lowest energy orbital. These two electrons

    have opposite ms (+ & -).

    3. Place the third electron in the next

    highest energy level orbital = the 2s orbital:

    Audi Majdan - DMC 101 - KLIUC 28

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    29/33

    The arrows indicate the value of the

    magnetic spin (ms) quantum number

    (up for + and down for - )

    Diagram representation:

    Audi Majdan - DMC 101 - KLIUC 29

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    30/33

    1s2 2s1

    Audi Majdan - DMC 101 - KLIUC 30

    Li

    1s 2s

    Orbital

    Diagram

    Electronic

    Configuration

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    31/33

    Writing electronic configurations: Period 1, 2,

    3

    All 2p orbitals are equal energy any 2p

    orbital can be filled

    Carbon has 6 electrons. 3 possible

    arrangements can be made in 2p orbital.

    Which one is true?

    Audi Majdan - DMC 101 - KLIUC 31

    (a) (b) (c)

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    32/33

    Hund's rule:

    For degenerate orbitals, the lowest energy is

    attained when the number of electrons with thesame spin is maximized

    The second 2p electron in Carbon is placed

    in another 2p orbital, but with the same

    spinas the first 2p electron

    Core electrons: the electrons in the stable

    (Noble gas) configuration

    Valence electrons: the electrons in the outer

    shell (beyond the stable core)

    Audi Majdan - DMC 101 - KLIUC 32

  • 7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf

    33/33

    Exercise 5

    1. Draw the orbital diagram

    representation for the electron

    configuration of oxygen, atomic

    number 8.

    2. Write the electron configuration of

    phosphorus, element 15.