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PERKONGSIAN

NOTA CHEMISTRY

Chemistry
SK016

CHAPTER 1
MATTER

1.1 Atoms & Molecules

LEARNING OUTCOMES:

a)  Describe proton, electron and neutron in terms of the relative mass and relative charge

b)  Define proton number, Z, nucleon number, A and isotope.

c)  Write isotope notation.

LEARNING OUTCOMES:

a) Determine the oxidation number of an element in a chemical formula

b) Write and balance :

  i) chemical equation by inspection method

  ii) redox equation by ion-electron method

LEARNING OUTCOMES

c) Define limiting reactant and percentage yield.

d) Perform stoichiometric calculations using mole concept including limiting reactant and percentage yield.

Learning Outcomes:

Perform stoichiometric calculations using mole concept including limiting reactant and percentage yield.

Exercise

  A particular analytical chemistry procedure requires 0.0500 M K2CrO4. What volume of 0.250 M K2CrO4 must be diluted with water to prepare 100 mL of 0.0500 M K2CrO4?

  The laboratory procedure for preparing a solution by dilution is as follows:

  A pipette is used to withdraw a 20.0-mL sample of 0.250 M K2CrO4(aq).

  The pipetteful of 0.250 M K2CrO4 is discharged into a 100.0-mL volumetric flask.

  Following this, water is added to bring the level of the solution to the calibration mark etched on the neck of the flask. At this point the solution is 0.0500 M K2CrO4. 

Exercise

A 25.0-mL sample of HCl solution is titrated against Na2CO3 solution of 0.150 M. It requires 21.2 mL of Na2CO3 for complete neutralisation. Calculate the concentration of HCl solution.

SOALAN 

QUIZ 1 (MATTER – ATOM AND MOLECULES)

 

NAME:                                                                                           TUTORIAL:

Answer all the question.

         

        1.    Isotopes can be define as….                                                       [1 mark]

``

2. i. Given the nucleon number, proton number and charge in the following nuclides. Determine the neutron number and write the isotopic notation for each of the following species.

Elements	J	K	L	M
Nucleon Number	17		18	19
Proton Number	7	8		9
Neutron Number		9	10
Charge	0	2-	2+
Isotopic Notation				18       3+ M        9

                                                            Table 1.0

                                                                                                [8 marks]

            ii. From Table 1.0, which elements is isotopes?                                 [1 mark]

                                                                                                                       

TOTAL MARKS: 10

TOPIC 2

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CHAPTER 2

ATOMIC STRUCTURE

l2.1 Bohr’s Atomic Model

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l2.2 Quantum Mechanical Model

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l2.3 Electronic Configuration

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lLECTURE 1

2.1 Bohr’s Atomic Model

lBohr’s Atomic Model

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lAt the end of this topic students should  be able to:-

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a)Describe the Bohr’s atomic models.

b)Explain the existence of electron energy levels in an atom.

c)Calculate the energy of electron using :

l  En = - RH (1/n2) ,

l  RH = 2.18 x 10-18 J

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In 1913, a young Dutch physicist, Niels Bohr proposed a theory of atom that shook the scientific world.

The atomic model he described had electrons circling a central nucleus that contains positively charged protons.

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l10:61 (Yunus)

lKAMU TIDAK BERADA DALAM SUATU KEADAAN DAN TIDAK MEMBACA SUATU AYAT DA AL QURAN DAN KAMU TDAK MENGERJAKAN SESUATU PEKERJAAN , MELAINKAN KAMI MENJADI SAKSI ATAS MU DI WAKTU KAMU MELAKUKAN NYA. TIDAK LUPUT DARI PENGETAHUAN TUHANMU BIARPUN SEBESAR ZARAH (ATOM) DI BUMI ATAU PUN LANGIT. TIDAK ADA YANG LEBIH KECIL DAN TIDAK PULA YANG LEBIH BESAR DARI ITU, MELAINKAN (SEMUA TERCATAT) KITAB YANG NYATA (LAUH MAHFUZ)

l4:40 (AN NISAA)

lSESUNGGUHNYA ALLAH TIDAK MENGANIAYA SESEORANG WALAUPUN SEBESAR ZARAH, NESCAYA ALLAH AKAN MELIPAT GANDAKAN DAN MEMBERIKAN DARI SISINYA PAHALA YANG  BESAR.

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2.  The energy of an electron in a hydrogen atom is quantised, that is, the electron has only a fixed set of allowed orbits, called stationary states.

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lThe energy of an electron in its level is given by:

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       RH (Rydberg constant)  or A = 2.18´10-18J.

       n (principal quantum number) = 1, 2, 3   ….¥    (integer)

Note:

ln identifies the orbit of electron

lEnergy is zero if electron is located infinitely far from nucleus

lEnergy associated with forces of attraction are taken to be negative (thus, negative sign)

lLECTURE 2

At the end of this topic students should  be able to:-

d) Describe the formation of line spectrum of hydrogen atom

e) Calculate the energy change of an electron during transition.

  DE = RH (1/n12 - 1/n22) ,

  where  RH = 2.18 x 10-18 J

f)  Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition

  DE = hn    where n= c/λ

lEmission Spectra

lContinuous Spectrum

  A spectrum consists all wavelength components (containing an unbroken sequence of frequencies) of the visible portion of the electromagnetic spectrum are present.

  It is produced by incandescent solids, liquids, and compressed gases.

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lWhen white light from incandescent lamp is passed through a slit then  a prism, it separates into a spectrum.

lThe white light spread out into a rainbow of colours produces a continuous spectrum.

lThe spectrum is continuous in that all wavelengths are presents and each colour merges into the next without a break.

lLine Spectrum (atomic spectrum)

A spectrum consists of discontinuous & discrete lines produced by excited atoms and ions as the electrons fall back to a lower energy level. The radiation emitted is only at a specific wavelength or frequency. It means each line corresponds to a specific wavelength or frequency.

lLine spectrum are composed of only a few wavelengths giving a series of discrete line separated by blank areas

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lExercise: Complete the following table

lExercise

The following diagram depicts the line spectrum of hydrogen atom. Line A is the first line of the Lyman series.

lExercise

Describe the transitions of electrons that lead to the lines W, and Y, respectively.

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lHomework

Calculate En for n = 1, 2, 3, and 4. Make a one-dimensional graph showing energy, at different values of n, increasing vertically. On this graph, indicate by vertical arrows transitions that lead to lines in

a)  Lyman series

b)  Paschen series

lSignificance of Atomic Spectra

lIn Lyman series, the frequency of the convergence of spectral lines can be used to find the ionisation energy of hydrogen atom:

                IE = hn¥

lThe frequency of the first line of the Lyman series > the frequency of the first line of the Balmer series.

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lExercise

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lRadiant energy emitted when the electron moves from higher-energy state to lower-energy state is given by the difference in energy between energy levels:

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lThe amount of energy released by the electron is called a photon of energy.

lA photon of energy is emitted in the form of radiation with appropriate frequency and wavelength.

                                   where;

                                 h (Planck’s constant) =6.63 ´ 10-34 J s

                                 n  = frequency

              

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lExercises:

1)Calculate the energy of an electron in the second energy level of a hydrogen atom. (-5.448 x 10-19 J)

2)Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom.

3)    Calculate the energy change (J), that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom.

       (answer: 1.55 x 10-19J)

4)Calculate the frequency and wavelength (nm) of the radiation emitted in question 3.

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lLECTURE 3

At the end of this topic students should  be able to:-

g)  Perform calculations involving the Rydberg equation for Lyman, Balmer, Paschen, Brackett and Pfund series .

  1/ λ = RH (1/n12 - 1/n22) ,

  where  RH = 1.097 x 107 m-1 and n1<n2

h) Calculate the ionisation energy of     hydrogen atom from Lyman series.

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i)State the weaknesses of Bohr’s atomic model.

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j)State the dual nature of electron    using the Broglie’s postulate and     Hesseinberg’s uncertainty principle.

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lWavelength emitted by the transition of electron between two energy levels is calculated using Rydberg equation:

lExample

  Calculate the wavelength, in nanometers of the spectrum of hydrogen corresponding to ni = 2 and nf = 4 in the Rydberg equation.

lExample

  Use the Rydberg equation to calculate the wavelength of the spectral line of hydrogen atom that would result when an electron drops from the fourth orbit to the second orbit, then identified the series the line would be found.

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  Calculate the wavelengths of the fourth line in  the Balmer series of hydrogen.

l Different values of RH and its usage

1.RH = 1.097 ´ 107 m-1

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  Calculate the energy liberated when an electron  from the fifth energy level falls to the second energy level in the hydrogen atom.

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Calculate what is;

i   ) Wavelength

ii  ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

     in Lyman series   

 Wave number = 1/wavelength

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lDefinition : Ionization energy is the minimum energy required to remove one mole of electron from one mole of gaseous atom/ion.

      M (g) ® M+ (g) + e       DH = +ve

lThe hydrogen atom is said to be ionised when electron is removed from its ground state (n = 1) to n = ¥.

lAt n = ¥, the potential energy of electron is zero, here the nucleus attractive force has no effect on the electron  (electron is free from nucleus).

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ln1 = 1, n2 = ∞

          ∆E  = RH (1/n12 – 1/n22)

                 = 2.18 X 10 -18 (1/12 – 1/ ∞ 2)

                 = 2.18 X 10 -18 (1 – 0)

        = 2.18 X 10 -18 J

   Ionisation energy

                 = 2.18 X 10 -18x 6.02 X 1023J mol-1                                               =1.312 x 106 J mol-1

                 = 1312 kJ mol-1

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      10.97  10.66    10.52       10.27            9.74                     8.22

wave number (x106 m-1)

  The Lyman series of the spectrum of hydrogen is shown above. Calculate the ionisation energy of hydrogen from the spectrum.

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ΔE   = hc/λ

  =h x c / λ   = h x c x wave no.

  = 6.626 x 10-34 J s  x 3 x 108 m s-1 x 10.97x 106 m-1  = 218.06x 10-20 J

  = 2.18 x 10-18J

Ionisation energy

         = 2.18 X 10 -18x 6.02 X 1023 J mol-1

         =1.312 x 106 J mol-1

         = 1312 kJ mol-1

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Compute the ionisation energy of hydrogen atom in kJ mol-1. 

lThe weakness of Bohr’s Theory

1.His theory could not be extended to predict the energy levels and spectra of atoms and ions with more than one electron. It only can explain the hydrogen spectrum or ions contain one electron eg He+, Li2+.

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2.Electrons are restricted to orbit the nucleus at certain fixed distances

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3.It cannot explain for the dual nature of electron

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4.It cannot explain for the extra lines formed in the hydrogen spectrum.

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Davisson & Germer observed the diffraction of electrons when a beam of electrons was directed at a nickel crystal. Diffraction patterns produced by scattering electrons from crystals are very similar to those produced by scattering X-rays from crystals. This experiment demonstrated that electrons do indeed possess wavelike properties.

Thus, can the ‘position’ of a wave be specified???

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lde Broglie’s Postulate

In 1924 Louis de Broglie proposed that not only light but all matter has a dual nature and possesses both wave and corpuscular properties. De Broglie deduced that the particle and wave properties are related by the expression:

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lHeisenberg’s Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain.

lLECTURE 4

l2.2  QUANTUM MECHANICAL MODEL

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l    At the end of this topic students should be able to:-

a)Define the term orbital.

b)State the four quantum numbers of an electron in an orbital (n, l,m,s).

l  i)    principal quantum number, n

l    ii)  angular momentum quantum number, l

l    iii) magnetic quantum number, m

l    iv) electron spin quantum number, s.

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lAtomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron.

lQuantum Numbers

Each of the electrons in an atom is described and characterised by a set of four quantum numbers, namely

a)  principal quantum number, n

b)  angular momentum quantum number, l

c)  magnetic quantum number, m

d)  electron spin quantum number, s.

lPrincipal Quantum Number, n

lThe value of n determines the energy of an orbital and thereby the energy of the electron in that particular orbital.

lThe principal quantum number may have only integral values: n =1, 2, 3, …, ¥.

lAngular Momentum Quantum Number, l

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lMagnetic Quantum Number, m

lThe direction or orientation of the magnetic field is determined by the value of m.

lPossible values of m depend on the value of l. For a given l, m can be : -l, …, 0, …, +l

  (-l £ m £ +l)

  If l = 0, m can only be 0    Þ one orbital in s-subshell

    If l = 1, m can be -1, 0, +1 Þ  three orbitals in p-subshell

  If l = 2, m can be -2, -1, 0, +1, +2 Þ five orbitals in d-subshell

lMagnetic Quantum Number, m

The number of m values indicates the number of orbitals in a subshell with a particular l value. 

The values of n = 2 and l = 1 indicate that we have a 2p-subshell, and in this subshell we have three 2p-orbitals (because there are three values of m, given by -1, 0 and +1)

lElectron Spin Quantum Number, s

lThe value of s determines the direction of spinning motions of an electron (either clockwise or counter clockwise) which is spinning on its own axes, as Earth does.

lThe electron spin quantum number has a value of 

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lExercise: Complete the following table

lExercise

State whether or not each of the following symbols is an acceptable designation for an atomic orbital. Explain what is wrong with the unacceptable symbols.

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LECTURE 5

At the end of this topic students should be able to:-

  c) Sketch the shapes of s,p and d orbitals   with the correct orientations.

lShape of Atomic Orbitals

a)s orbitals

üSpherical shape with the nucleus at the centre.

üThe probability of finding electrons at the distance r from the nucleus is the same from all direction.

üWhen l = 0

üAs n increases s orbital

ü gets larger

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b) p orbitals

l When l = 1

ldumbbell shaped

l three p-orbitals px, py, and pz.

lcorrespond m of -1, 0, and +1.

lAs n increases, the p-orbitals get larger.

lAll p-orbitals have a node at the nucleus.

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lLECTURE 6 & 7

l2.3 ELECTRONIC CONFIGURATION

lElectronic Configuration

l    At the end of this topic students should

l    be able to:-

a)State and use Aufbau principle, Hund’s rule and Pauli exclusion principle

b)Apply rules in (a) to fill electrons into atomic orbitals.

c)Write the electronic configuration of an atom and monostomic ions by using spdf notation .

ld)    Explain the anomalous electronic configurations of chromium and copper.

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lRepresenting Electronic Configuration

Method 1: Orbital diagram

lRules for Assigning Electrons to Orbitals

lRelative Energy Level of Atomic Orbitals

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ii)  Pauli Exclusion Principle

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iii)  Hund’s Rule

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lHund’s rule states that when electrons are added to the orbital of equivalent energy (for degenerate orbital), each orbital are filled singly with electron of the parallel spins (same) first before it is paired.

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Indicate which of the following orbital diagrams are acceptable or unacceptable for an atom in ground state. Explain what mistakes have been made in each and draw the correct orbital diagram:

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Draw ‘electrons-in-boxes’ diagram of the electronic configuration of titanium, Ti (Z = 22). Also, write the ground-state electronic configurations for Ti and Ti2+ ion.

lIMPORTANT!

lIn an empty atom, the 4s orbital has a lower energy compared to that of the 3d orbital. That is why electrons fill the 4s orbital first before filling the 3d orbital.

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lHowever, once electron/s is/are added to the 3d orbital, the 4s electrons are repelled to a higher energy level. The 3d orbitals now have lower energy than 4s.

lPoints to remember

lThe electronic configuration of atom or monatomic ion at ground state

  Þ Distribution of electrons obeys Aufbau principle, Pauli exclusion principle and Hund’s rule

lEach atomic orbital can only accommodate a maximum of 2 electrons

lAtomic orbital is a 3-D region in space around the nucleus where there is a high probability of finding an electron.

lAssigning electrons to subshells

  s-orbital Þ a max of 2 electrons (ns2)

  p-orbitals Þ a max of 6 electrons (np6)

  d-orbitals Þ a max of 10 electrons (nd10)

lThe Anomalous Electronic Configurations of Cr and Cu

lCr and Cu have electron configurations which are inconsistent with the Aufbau principle. The anomalous are explained on the basis that a filled or half-filled orbital is more stable.

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Ø

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24Cr :  18[Ar]

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The actual orbital notation

24Cr :  18[Ar]

Half filled orbital is more stable

  (possesses an extra, added stability)

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Copper predicted orbital notation

ØCu : [Ar]

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The actual orbital notation

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ØCu : [Ar]

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Write the ground-state electronic configuration and explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

lWriting Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle.

Example:

i. Cl-

ii.  O2-

lWriting Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example:

i. Mg2+

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ii. K+

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iii. Fe2+

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win - win