chemistry form 6 sem 1 02
Post on 21-May-2017
224 Views
Preview:
TRANSCRIPT
4.1 Atomic Spectra.
� Spectrum ~ display of component of a beam of radiation.
� When white ray passes through a prism it forms rainbow colour.
� A continuous spectrum is spectrum composed of visible light of all
wavelength while a discontinuous spectrum which the line
spectrum represent the atomic emissions
� Productions of line spectrum depend on what types of element
used and the visible light emitted also depend on it.
� Spectrophotometer is the instrument that is used to split and
dispersed into individual wavelength.
� The study of line spectra can be used to
� Identify an element
� Determine the ionisation energy of an element
� Determine the arrangement of electrons in an atom of an
element.
The main components of a typical spectrophotometer
Monochromator
(wavelength selector)
disperses incoming
radiation into continuum of
component wavelengths
that are scanned or
individually selected.
Sample in compartment
absorbs characteristic
amount of each incoming
wavelength.
Computer
converts
signal into
displayed
data.
Source produces radiation in
region of interest. Must be
stable and reproducible. In
most cases, the source emits
many wavelengths.
Lenses/slits/collimaters
narrow and align beam.
Detector converts
transmitted radiation
into amplified electrical
signal.
7.3
4.2 The Atomic Emission Spectrum of Hydrogen
� In hydrogen, the atomic emission spectrum of hydrogen shows a
few distinct series of line. These are the
� Lyman series – in …………………………… region
� Balmer series – in …………………………. region.
� Paschen series – in ………………………… region
� The diagram below shows the example of the line emission
produced through the spectrophotometer.
ultraviolet
visible
infrared
4.2.1 Formation of atomic spectra.
� The absorption spectrum of a substance is a continuous spectrum of
radiation produced by examining the substance through
spectrophotometer.
� Hence, there is a absorption and emission process occur in the
formation of spectra where
Absorption spectroscopy Emission spectroscopy
↑ Occur when electron which is occupying at
lower energy level (ground level / state)
absorb light at a certain frequency.
↑ When enough energy is absorbed, electron is
able to ‘excite’ to a higher energy level.
↓ Occur when electron which is located at a high
energy level (after absorbed energy) released
the energy in the form of radiation
↓ When energy is released, electron falls to the
original ground state, emitting the radiation
and hence give the spectrum
n = 3
n = 2
n = 1
n = 3
n = 2
n = 1
� The following characteristics of any series should be noted :
� The spectrum consists of discrete lines, each having its own
discrete frequency.
� The wavelengths represented by broken lines are called spectrum
(plural : spectra) Each spectrum represents the wavelength (or
frequency) in a series. The frequency emitted can be related to
the spectral line by the Rydberg’s Equation
� The value of n1 and n2 for the atomic spectrum of hydrogen are
given in Table below
λ = wavelength (in m)
n1 = series of the spectra line
n2 = line where the emission of energy begin.
RH = Rydberg constant = 1.097 x 107 m-1
−=
2
2
2
1
111
nnRHλ
Series n1 n2
Lyman 1 2, 3, 4 …….
Balmer 2 3, 4, 5 …….
Paschen 3 4, 5, 6 …….
Example 1 : Calculate the wavelength of the first
line of Lyman series
Example 2 : Calculate the wavelength of the third
line of Balmer series
Example 3 : Calculate the wavelength of the forth
line of Paschen series
Example 4 : Calculate the wavelength of the last
line of Balmer series
In Lyman series, n1 = 1
First line in series, so n2 = 1 + 1 = 2
−=
2
2
2
1
111
nnRHλ
−×=
λ 22
7
2
1
1
110097.1
1
λ = 122 nm
In Balmer series, n1 = 2
First line in series, so n2 = 2 + 3 = 5
−=
2
2
2
1
111
nnRHλ
−×=
λ 22
7
5
1
2
110097.1
1
λ = 434 nm
In Passchen series, n1 = 3
First line in series, so n2 = 3 + 4 = 7
−=
2
2
2
1
111
nnRHλ
−×=
λ 22
7
7
1
3
110097.1
1
λ = 1005 nm
In Balmer series, n1 = 2
Last line in series, so n2 = 2 + ∞ = ∞
−=
2
2
2
1
111
nnRHλ
∞−×=
λ 22
7 1
2
110097.1
1
λ = 365 nm
Example 5 : Calculate the frequency of the second
line of Paschen series
Example 6 : Calculate the frequency of the last line
of Lyman series
m
sm100.3f;
)wavelength(
)lightofspeed(c f frequency, Since
18
λ
×=
λ=
−
• In 1900, Max Planck proposed the quantum theory. He mentioned that
radiant energy was emitted in discrete packets or “quantum” Each
quantum of energy emitted, E, is proportional to the frequency, f, of the
radiation
• Where energy, E = Planck constant (h) x frequency (f)
• h = 6.63 x 10-34 J s
)wavelength(
)lightofspeed(c)tstanconPlanck(hE,Energy
λ×=
In Passchen series, n1 = 3
Second line in series, n2 = 3 + 2 = 5
−=
2
2
2
1
111
nnRHλ
−×=
λ 22
7
5
1
3
110097.1
1
1 / λ = 7.80 x 105 m-1
f = c x (1 / λ) = (3.0 x 108)(7.80 x 105)
f = 2.34 x 1014 s-1
In Lyman series, n1 = 1
last line in series, so n2 = 1 + ∞ = ∞
−=
2
2
2
1
111
nnRHλ
∞−×=
λ 22
7 1
1
110097.1
1
1 / λ = 1.097 x 107 m-1
f =c x (1 / λ) =(3.0 x 108)(1.097 x 107)
f = 3.29 x 1015 s-1
Example 7 : Calculate the energy required to
excite one electron to the fourth line in
Passchen series
Example 8 : Calculate the energy required to
excite one electron to the last line in Balmer
series.
In Passchen series, n1 = 3
Fourth line in series, n2 = 3 + 4 = 7
−=
2
2
2
1
111
nnRHλ
−×=
λ 22
7
7
1
3
110097.1
1
1 / λ = 9.95 x 105 m-1
f = c x (1 / λ) = (3.0 x 108)(9.95 x 105)
f = 2.99 x 1014 s-1
E = h f
E = 6.63 x 10-34 x 2.99 x 1014
E = 1.98 x 10-19 J / e
In Balmer series, n1 = 2
Last line in series, n2 = 2 + ∞ = ∞
−=
2
2
2
1
111
nnRHλ
∞−×=
λ 22
7 1
2
110097.1
1
1 / λ = 2.74 x 106 m-1
f = c x (1 / λ) = (3.0 x 108)(2.74 x 106)
f = 8.23 x 1014 s-1
E = h f
E = 6.63 x 10-34 x 8.23 x 1014
E = 5.45 x 10-19 J / e
4.3 Electronic Energy Levels
� Hydrogen emission spectrums consist of several series of discrete
lines in different limited wavelength.
� This implies that hydrogen atoms do not emit light at of all possible
energies but only emit light with certain amount of energy.
� Bohr made these assumptions to explain line in spectrum.
� Electron move in orbit around nucleus (like planet and sun)
� Energy of electron is quantised (only have certain energy). Electron
closer to nucleus has lower energy and vice versa.
� Electron of atom will fill up the lowest energy level first. But when
energy is given to the electron, it moves from a lower level to a
higher level. This process is called excitation.
� If the energy loss from the excited level to original level, electron
will emit electromagnetic radiation. This is what caused the
emission of spectral lines. The convergence of spectral line show
the difference between successive energy level become ↓ with
increasing distance of the energy levels from the nucleus
� Electron are not randomly distributed but are arranged
in a series of shells or orbits which are situated at
various distance from the molecules corresponding to
differ E-level.
� Though Bohr theory proves its unsatisfactory when
come to spectra of more complicated atoms. Thus
Rydberg equation can only used for atoms or ions
contain only 1 electron.
� For atom or ion with many electrons, the analysis of
atomic spectra become more difficult.
Wavelength increase Frequency increase
4.4 Calculate the Ionisation energy of Hydrogen from its
Line Spectrum
� When energy level of the hydrogen spectrum increased, there’s a
possibility where it reaches the convergence limit of the energy
level where n =
� When this occur, electron are not able to return to ………………..,
and eventually the hydrogen atom is …………….., according to the
following equation
� Equation :
� For this to occur, a certain amount of energy is required. The
energy required to remove one electron is called
……………………………… Here, a positive charged ion is formed.
� There are 2 methods of calculating ionisation energy of hydrogen
through the line spectrum
ground level ionised
H (g) � H+ (g) + e-
ionisation energy
Using Rydberg and Planck equations
� Supposed that electron at energy level n1 = 1 absorbed energy and it is
excited. When reaches the convergence limit, n2 = , the wavelength can
be calculated
� Using Rydberg equation n1 = n2 =
� For each electron, the energy absorbed can be determined using the
Planck equation
� Since this much energy is required to excite one electron, so for 1 mole of
electron, the energy required
� Hence, the ionisation energy of hydrogen is …………. kJ mol-1
−=
2
2
2
1
111
nnRHλ
α−=
λ 22
7 1
1
110x097.1
1
∞
1 ∞710x097.1
1=
λ
E = h f E = h c 1/λE = (6.63 x 10-34)(3.0 x 108)(1.097 x 107)
E = 2.182 x 10-18 J / e
Since in 1 mol contain 6.02 x 1023 e, so
Ionisation energy, ∆H = E x NA
= 2.182 x 10-18 J / e x 6.02 x 1023
= 1.313 x 106 J / mol+ 1313
4.5 Atomic Orbital
� Heinsberg uncertainty principle = position & momentum of an
electron cannot be known with great precision but probability of
finding an electron in a certain position can be calculated.
� Orbital - the space within which there’s a ………………… probability of
finding an ……………………. The electron in the orbital is describe as
occupying 3-D space around the nucleus. The nucleus is described as
being surrounded by
Orbital – boundary within which there
is 95% (high) chances to find an
electron
high
electron
Type of
orbitalShape of orbital
Number of
electron
filled
S
P
2
6
Type of
orbitalShape of orbital
Number of
electron
filled
f 7 orbitals
10
14
� Filling of orbitals in shell
� Electrons filled in the shell of an atom through occupation of
orbitals in each shell
� The formula used to calculate the number of orbitals that can
occupy each shell is
No of orbitals in shell = n2 ; (n = the number of shell)
Shell,
n
No. of
orbitalsType of orbitals Orbital in shell
1
2
3
4
n2 = 12 = 1 s 1s
n2 = 22 = 4 s , p 2s 2p
n2 = 32 = 9 s , p , d 3s 3p 3d
n2 = 42 = 16 s , p , d , f 4s 4p 4d 4f
4.6 Electronic configuration
� Electronic configuration of an element is the expression on how
the electrons inside the element filled in its orbitals accordingly
� There are 3 rules for assigning electrons to the orbitals of an
atom in the ground state.
1 . ………………………………………………………………….
2 . ………………………………………………………………………….
3 . ………………………………………………………………………….
1.The Aufbau Principle ~ e- occupy orbitals in the order of the
energy levels of the orbitals. Orbital with the lowest energy are
always occupied first.
� The arrangement of energy level from the lowest to the highest
is shown in the diagram below
Aufbau Principle
Pauli’s Exclusion Principle
Hund’s Rule
� Energy level : 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p
2. The Pauli exclusion principle ~ 2 electron may occupy the same orbital but
these 2 electrons must have opposite spin.
Same spin (cannot occur) Opposite spin
(Pauli exclusion principle)
3. The Hund’s Rule ~ when electrons are placed in a set of orbitals with equal
or degenerated energies, the electrons must occupy them singly with parallel
spins before they occupy the orbital in pair In other words, an atom tend to
have as many unpaired electrons as possible. Example : when filling in 3
electrons in p orbital
ElementNo of
e- Orbital diagramElectronic
configuration
Hydrogen
H____
1s
Helium
He____
1s
Lithium
Li ____ ____
1s 2s
Beryllium
Be ____ ____
1s 2s
Boron
B____ ____ ____ ____ ____
1s 2s 2 p
Carbon
C____ ____ ____ ____ ____
1s 2s 2 p
Nitrogen
N____ ____ ____ ____ ____
1s 2s 2 p
1 1s1
2 1s2
3 1s22s1
4 1s22s2
5 1s22s22p1
6 1s22s22p2
7 1s22s22p3
ElementNo of
e- Orbital diagramElectronic
configuration
Oxygen
O____ ____ ____ ____ ____
1s 2s 2 p
Fluorine
F____ ____ ____ ____ ____
1s 2s 2 p
Neon
Ne____ ____ ____ ____ ____
1s 2s 2 p
Sodium
Na____ ____ ____ ____ ____ ____
1s 2s 2 p 3s
Magnesium
Mg____ ____ ____ ____ ____ ____
1s 2s 2 p 3s
Aluminium
Al____ ____ ____ ____ ____ ____ ___ ___ ___
1s 2s 2 p 3s 3 p
Silicon
Si____ ____ ____ ____ ____ ____ ____ ____ ____
1s 2s 2 p 3s 3 p
8 1s22s22p4
9 1s22s22p5
10 1s22s22p6
11 1s22s22p63s1
12 1s22s22p63s2
13 1s22s22p63s23p1
14 1s22s22p63s23p2
ElementNo of
e- Orbital diagram Electronic configuration
Phospho-
rous, P____ ____ ____ ____ ____ ____ ____ ____ ____
1s 2s 2 p 3s 3 p
Sulphur
S____ ____ ____ ____ ____ ____ ____ ____ ____
1s 2s 2 p 3s 3 p
Chlorine
Cl____ ____ ____ ____ ____ ____ ____ ____ ____
1s 2s 2 p 3s 3 p
Argon
Ar____ ____ ____ ____ ____ ____ ____ ____ ____
1s 2s 2 p 3s 3 p
Potassium
K____ ____ ____ ____ ____ ____ ____ ____ ____ ___
1s 2s 2 p 3s 3 p 4s
Calcium
Ca____ ____ ____ ____ ____ ____ ____ ____ ____ ___
1s 2s 2 p 3s 3 p 4s
15 1s22s22p63s23p3
16 1s22s22p63s23p4
17 1s22s22p63s23p5
18 1s22s22p63s23p6
19 1s22s22p63s23p64s1
20 1s22s22p63s23p64s2
1s22s22p63s23p63d14s2
1s22s22p63s23p63d24s2
1s22s22p63s23p63d34s2
1s22s22p63s23p63d54s1
1s22s22p63s23p63d54s2
1s22s22p63s23p63d64s2
1s22s22p63s23p63d74s2
1s22s22p63s23p63d84s2
1s22s22p63s23p63d104s1
1s22s22p63s23p63d104s2
� According to Hund’s rule : ………………………………………………………..…
………………………………………
� For the case of chromium, it has the valence electronic as
………………………………... instead of …………………………….
� This is due to the extra stability achieved due to the equal /
symmetrical distribution of charge around the atom within the half-filled
(p3 and d5) orbital and full-filled orbital (p6 and d10)
� What happened before and after the electron filling in the 4s orbital can
be explained by the diagram below
Position of 3 d and 4 s orbital before
electron filling in
Position of 3d and 4s orbital after
electron filling in
____ ____ ____ ____ ____
3 d
____
4s
____
4s
____ ____ ____ ____ ____
3 d
Electron filling in orbitals at the same
energy level with single electron before pairing
3d54s1 3d44s2
� When this occurs, the 3rd electron is expected to be filled in 3d orbital
(Aufbau principle). Until the 6th electron is filling in, a degeneration of
orbital occur where now the energy level between 3d and 4s orbital are
the same, thus one of the electron from the 4s “jumped” to 3d orbital.
When this happened, an extra stability occur at the half-filled 3 dorbital has achieved. When the 7th electron is filled, it will be placed in
4s orbital
Position of 3 d and 4 s orbital before
the
6th electron filling in
Position of 3d and 4s orbital after the
6th electron filling in
____
4s
____ ____ ____ ____ ____
3 d
____
4s
____ ____ ____ ____ ____
3 d
� Same thing happened when comes to the case of copper. An
extra stability is achieved when the orbital in 3 d is full-filled
with electrons.
Position of 3 d and 4 s orbital before
the
11th electron filling in
Position of 3d and 4s orbital after the
11th electron filling in
____
4s
____ ____ ____ ____ ____
3 d
____
4s
____ ____ ____ ____ ____
3 d
� 3Li : ……………………………. Li+ : ………………………………
� 5B : …………………………….. B3+ : ……………………………….
� 7N : …………………………….. N3- : ……………………………….
� 9F : ……………………………… F- : ……………………………….
� 12Mg : …………………………. Mg2+ : …………………………….
� 14Si : ………………………….. Si4+ : ……………………………….
� 16S : ……………………………. S2- : ………………………………
� 19K : …………………………………… K+ : …………………………………….
� 21Sc : ………………………………….. Sc2+ : ………………………………….
� 23V : ………….………………………. V2+ : …………………………………..
� 28Ni : ………………………………….. Ni+ : …………………………………..
� 30Zn : …………………………………. Zn2+ : …………………………………..
� 35Br : …………………………………. Br- : ……………………..............
1s22s1 1s2
1s22s22p1 1s2
1s22s22p3 1s22s22p6
1s22s22p5 1s22s22p6
1s22s22p63s2 1s22s22p6
1s22s22p63s23p2 1s22s22p6
1s22s22p63s23p4 1s22s22p63s23p6
1s22s22p63s23p64s1 1s22s22p63s23p6
1s22s22p63s23p63d14s2 1s22s22p63s23p63d1
1s22s22p63s23p63d34s2 1s22s22p63s23p63d3
1s22s22p63s23p63d84s2 1s22s22p63s23p63d84s1
1s22s22p63s23p63d104s2 1s22s22p63s23p63d10
1s22s22p63s23p63d104s24p5 1s22s22p63s23p63d104s24p6
Block of Periodic
TableGroup of elements
s-block Group 1 and 2 (Those with valence electron ns1 and ns2)
p-blockGroup 13 , 14 , 15 , 16 , 17 , 18
(Those with valence electrons ns2 np1, 2, 3, 4, 5, 6)
d-blockGroup 3 – 12
(Transition element with valence electron (n–1)d1-10 ns2 )
f-block Lanthanides and Actinides Series
•From the Periodic Table, elements can be classified accordingly to the block.
� Based on the valence electron, the Group of an element can be
determined.
Valence electron Group
ns1 1
ns2 2
ns2np1 13
ns2np2 14
ns2np3 15
ns2np4 16
ns2np5 17
ns2np6 18
top related