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  • #4866

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    Atomic Structure – Notes

    2.1 Isotopes

    (a) Isotopes are atoms of the same element which contain the same number of protons and electrons but different number of neutrons.

    (b) Isotopes have the same number of electrons and therefore the same chemical properties.

    (c) However, since they have different number of neutrons, they will have different masses and hence different physical properties such as melting point, density etc.

    (d) They are found in all naturally occurring elements except F, Na, Al, P, Mn, Co, As.

    Why do isotopes react similarly?
    This is because in a chemical reaction, it is the electrons that are transferred between different atoms; atoms either gain, lose or share electrons

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    #4907

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    Atomic Structure – Notes

    Approach of writing electronic configurations

    1. First, write down the electronic configurations for the neutral atom.

    2. Remember to add electrons to the 4s orbitals before the 3d orbitals.

    3. Remove or add the relevant number of electrons to the electronic configuration of the neutral atom to get that of the cation or anion
    respectively.

    4. Remember to remove electrons from the orbitals with the highest energy.Thus remove electrons from the 4s orbitals before the 3d orbitals.

    Some Definitions

    Isoelectronic: species containing the same number of electrons

    Isotopic: species containing the same number of protons

    Isotonic: species containing the same number of neutrons

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    #4931

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    IONISATION ENERGIES

    • Ionisation involves the removal of electron(s), forming a cation.

    • When the electrons in the atom occupy the lowest energy levels, the atom is said to be in its
    ground state.

    • When an electron is promoted to one of the higher energy levels, the atom is unstable and is said to be in an excited state.

    (a) M (g)—> M+ (g)+e– H = 1st I.E.

    First ionisation energy (1st I.E.) is the energy required to remove 1 mole of electrons from one mole of gaseous atoms in the ground state to form one mole of gaseous singly- charged cations.

    (b) M+ (g) —> M2+ (g)+e– H = 2nd I.E

    Second ionisation energy (2nd I.E.) is the energy required to remove 1 mole of electrons from one mole of gaseous singly-charged cations to form one mole of gaseous doubly- charged cations.

    • Ionisation energies affect the type of bond formed by the atom with other atoms. Elements with low ionisation energies will find it easy to lose an electron to form a cation, resulting in ionic bonds being formed.

    • Ionisation energies are positive values (i.e. endothermic) since energy is absorbed during ionisation to overcome the attraction between electron and nucleus.

    • The 2nd I.E. > 1st I.E. because more energy is required to remove an electron from a positive ion (compared to a neutral atom) due to greater electrostatic attraction between the positive ion and the valence electron.

    Factors influencing ionisation energies

    Ionisation energy (I.E.) of an atom is influenced mainly by two factors:

    Effective nuclear charge, Zeff
    • An electron in the atom faces two main electrostatic forces, namely the attractive force by the nucleus (nuclear charge) and the repulsive force by electrons closer than itself to the
    nucleus (shielding effect).

    • Effective nuclear charge is the combined effect of nuclear charge, Z and shielding effect, S, caused by inner electrons:

    Zeff = nuclear charge (Z)  shielding effect (S)

    • Higher Zeff ⇒ stronger forces of attraction between nucleus and valence electron,
    ⇒ higher ionisation energy

    (a) Size of (positive) nuclear charge, Z
    • indicates the electrostatic forces of attraction between the protons in the nucleus and the valence electrons
    • nuclear charge increases with an increase in proton number
    • stronger attraction between the positive nucleus and valence electrons
    • more energy is required to remove the valence electron

    (b) Shielding (or screening) effect of the inner electrons, S
    • shielding of the valence electrons from the electrostatic attraction of the positively charged nucleus by the inner electrons
    • electrons in the same subshell offer poor shielding for one another
    • shielding effect increases with an increase in the number of inner electrons
    • weaker attraction between the positive nucleus and valence electrons
    • less energy is required to remove the valence electrons

    Distance of the valence electron from the nucleus (i.e. the size/radius of the atom)
    • attraction of the positive nucleus for the valence electron decreases as distance of electron from nucleus increases
    • ionisation energy decreases as n increases

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    #4989

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    Trends in First Ionisation Energies

    (A) Across a period of the Periodic Table, the ionisation energies generally increase.

    Reason:
    • Nuclear charge increases as the proton number increases.
    • Successive members of the period has one electron added to the same outermost shell, thus, the increase in screening effect is negligible.
    • Effective nuclear charge increases.
    • More energy is required to remove the more tightly held electrons, hence ionisation energy generally increases.

    (B) Down a group of the Periodic Table, the ionisation energies generally decrease.

    Reason:
    • Nuclear charge increases as proton number increases.
    • Electrons are added to a higher principal quantum shell which is further away from nucleus.
    • Weaker electrostatic forces of attraction between the nucleus and valence electrons.
    • Less energy is required to remove the valence electrons, hence ionisation energy generally decreases

    (C)TRENDS IN THE ATOMIC RADIUS OF ELEMENTS

    Across a period,
    • nuclear charge increases
    • shielding effect is relatively constant (number of inner core electrons is the same across a period)
    • effective nuclear charge increases, resulting in electrons being pulled closer towards the nucleus
    • atomic radius generally decreases.

    Down a group,
    • nuclear charge increases
    • electrons are added to a higher principal quantum shall
    • valence electrons are increasingly further away from the nucleus
    • weaker electrostatic forces of attraction between valence electrons and nucleus
    • atomic radius generally increases.

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    #5133

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    Chemical Bonding

    IONIC OR ELECTROVALENT BONDING

    Formation of Ionic Bonds
    An ionic bond is the electrostatic force of attraction between two oppositely charged ions, which are formed by the transfer of electrons from one atom (usually metallic) to another
    (usually non-metallic).

    Metallic atoms usually have 1 to 3 valence electrons. By losing these valence electrons, they achieve a stable octet (noble gas) electronic configuration, forming positively-charged cations.

    Non-metallic atoms usually have 4 to 8 valence electrons. Atoms of elements from Group V to Group VII tend to accept electrons to form negatively-charged anions that have octet configurations.

    Ionic Bond Strength

    Ionic bonds are generally strong forces of attraction and a lot of energy is required to break such bonds.

    The strength of an ionic compound is indicated by its lattice energy. Lattice energy is defined as “the energy released when one mole of an ionic crystalline solid is formed from its constituent gaseous ions”. For example, lattice energy of NaCl is equal to the heat released for the equation: Na+(g) + Cl–(g) NaCl(s). The lattice energy is exothermic since it arises from the attraction between oppositely charged ions.

    Note:
    • Attraction between unlike charges results in stabilisation and energy is given out.
    • The greater the attraction, the greater will be the amount of energy given out.
    • Conversely, repulsion between like charges results in destabilisation and energy is taken in.

    In general, the greater the magnitude of the lattice energy, the stronger the ionic bond

    Structure of an Ionic Compound

    In an ionic compound, constituent ions are held in fixed positions in an orderly arrangement by strong ionic bonds.

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    #5446

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    Chemical Bonding

    Sigma bond and pi bond

    Depending on the way the orbitals overlap, the covalent bond formed is classified as either a sigma bond or a pi bond.

    A sigma bond is formed when two orbitals overlap head-on. The electron density of a sigma bond is concentrated between the nuclei of the two bonding atoms.

    3 Basic Examples:
    (a) s–s overlap
    (b) p–p overlap
    (c) s–p overlap

    A pi bond is formed when atomic orbitals overlap collaterally, i.e. side-on overlap

    A double bond consists of a sigma bond and a pi bond e.g. for oxygen O2

    A triple bond consists of a sigma bond and two pi bonds e.g. for nitrogen N2

    The sigma bond is stronger than a pi bond because head-on overlap of charge cloud in a sigma bond has a greater degree of overlap than the side-onoverlap in a pi bond.

    Remember:
    • A single covalent bond consists of one sigma bond.
    • A double covalent bond consists of one sigma bond and one pi  bond.
    • A triple covalent bond consists of one sigma bond and two pi  bonds.

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    #5572

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    2015 A Level Chemistry Schedule

    J1 H2 FRI 7.30pm – 9.30pm

    J1 H1 FRI 7.30pm – 9.30pm

    J1 H2 SAT 1.30pm – 3.30pm

    J1 H2 SUN 12pm – 2pm

    J2 H2 TUE 5.30pm – 7.30pm

    J2 H2 FRI 7.30pm – 9.30pm

    J2 H2 SAT 3.30pm – 5.30pm

    J2 H1 SAT 3.30pm – 5.30pm

    J2 H2 SUN 2pm – 4pm

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    #5679

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    Chemical Bonding

    Covalent Bond Strength
    The strength of a covalent bond is measured by its bond energy (also called bond enthalpy) which is defined as the average energy absorbed when one mole of a particular bond is broken in the gaseous state.

    The strength of a covalent bond is affected by the following factors:

    (a) The effectiveness of overlap of the orbitals
    In general, larger orbitals are more diffuse so that overlap is less effective in bonds formed by larger atoms compared to smaller atoms. For example the bond energies of chlorine to iodine are shown below.

    BE(Cl—Cl): 244 kJ mol–1 BE(Br—Br): 193 kJ mol–1 BE(I—I):151 kJ mol–1

    As the halogen increases in size, the valence orbital used in bonding is more diffuse so that the overlap of the orbitals is less effective from chlorine to iodine and bond energy decreases from chlorine to iodine.

    Stop and Think

    Question: Explain which bond is stronger, C—H or Si—H.
    Answer: C—H bond is stronger since C is smaller than Si so that valence
    orbital of C is less diffuse and overlap of its valence orbital with that
    of H is more effective.

    (b) The differences in electronegativities of the bonding atoms (bond polarity)

    Electronegativity of an element measures the relative tendency of its atom to
    attract the shared electron–pair in a covalent bond. Based on Pauling’s definition, fluorine, the most electronegative element, is given an arbitrary value of 4.0 and all values of the other elements are relative to it. The higher the value, the stronger the attraction.

    Electronegativity decrease down a group.
    Electronegativity increase­ across a period.

    A polar covalent bond results if the bonded atoms have different electronegativities.

    Partial charges (d+ and d–) arise on the two bonded atoms. The covalent bond is
    described as possessing some ionic character.

    For example in HF, F is more electronegative than H and hence attracts the
    bonding electrons more strongly. The electron density distribution of the H—F
    bond is asymmetrical and the H—F bond is polar, with F having a d– charge while
    H having a d+ charge:

    In addition to the existing covalent bond, there is now an increase in electrostatic attraction due to the two partial charges, which leads to increased bond strength.

    In general, the greater the difference in electronegativity, the more polar is the covalent bond (i.e. greater the bond polarity) and stronger the covalent bond.

    (c) Number of bonds between atoms (Single vs double vs triple bonds)

    For the same bonding atoms, an increase in the number of bonds increases the
    number of shared electrons between the two atoms i.e. there is increased
    electrostatic attraction between the bond pairs and the two nuclei, hence bond
    strength is increased.

    Hence the strength of triple bond > double bond > single bond.

    Bond Length
    (i) The covalent bond length is the distance between the nuclei of the two
    atoms in the bond.
    (ii) Generally, the stronger the covalent bond, the shorter is the bond length.
    (iii) The three factors of more effective overlap by less diffused orbitals, greater polarity of bond and increase in number of bonds between 2 atoms decrease bond length and increase bond strength.

    A molecule with strong covalent bonds generally has less tendency to undergo
    chemical change than does one with weak bonds. This is seen in nitrogen with its very large NºN bond energy (944 kJ mol–1). A very large amount of energy must be supplied to nitrogen to break the triple bond before nitrogen can react with other elements. Fluorine having a weak F–F bond (due to lone pair-lone pair repulsion in a very short bond) is very reactive.

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    #6771

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    Chemical Bonding

    Shapes of Molecules – Valence Shell Electron Pair Repulsion Theory (VSEPR)

    Covalent bonds are directional. Hence covalent molecules or ions with covalent bonds have definite shapes.

    VSEPR is used to predict the molecular geometry of a species. The key essential principle underlying the theory is that regions of electron density arrange themselves as far apart as possible so as to minimise electronic repulsion.

    Steps to deduce the molecular shape of a given species:

    1. Draw the dot–and–cross diagram of the required species. Remember that lone pairs must be included.

    2. Count the total number of regions of electron density around the central atom to determine the “electron pair geometry”

    (i) Each lone pair OR lone electron = ONE region of electron density

    (ii) Each single bond OR coordinate bond = ONE region of electron density

    (iii) Each double bond OR triple bond = ONE region of electron density

    3. Count the total number of lone pairs around the central atom.

    4. Deduce the molecular geometry / shape of the species using Table 1

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    #6802

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    Chemical Bonding

    Modification to the general shapes predicted by VSEPR

    Bond angles depend on repulsion between electron pairs. The closer the electron pairs are to the nucleus of the central atom, the greater is their repulsion power. Bond angles may thus deviate from the ideal as electron pairs present may have different repulsion power.

    Factors affecting the bond angles are:
    (i) Number of lone pairs present

    lone pair – lone pair > lone pair – bond pair > bond pair – bond pair
    repulsion repulsion repulsion

    A lone pair exerts greater repulsion than a bond pair as it is closer to the
    nucleus as it is attracted by only one positive nucleus compared to bond–pair
    electrons which are attracted by two nuclei.

    (ii) Electronegativity of central/ side atom

    For species with the same central atom and the same number of lone pairs of
    electrons on the central atom, the bond angles may differ depending on whether the atoms attached to the central atom are more or less electronegative than the
    central atom.

    Atoms which are more electronegative than the central atom tend to draw electron
    density of the bond pair towards itself. The bond pair electrons are thus further from the nucleus and exert less repulsion.

    On the other hand, atoms which are less electronegative than the central atom has less attraction for the bond pair of electrons so that electron density is drawn towards the central atom. The bond pair electrons are thus closer to the nucleus and exert greater repulsion resulting in bond angle being bigger than that mentioned above.

    (iii) Size of atoms attached

    For species with the same central atom and the same number of lone pairs of
    electrons on the central atom, the bond angles may differ depending on the size of the atoms attached to the central atom.

    For side atoms which are very much bigger than the central atom, the bond angle is bigger than the bond angle in the species with smaller side atoms. That is because the electron clouds on the side atoms may exert a repulsion which prevents the bonds from coming too close together.

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    #6858

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    Testimonial

    This tuition centre has friendly, dedicated and professional tutors. One of them is Mr Ong. He is a patient and jovial teacher whose chemistry between him and his students is excellent. Needless to say, his chemistry is also ‘up there’. As a result, my chemistry grades leapt from E (even) in the Prelims to A in the A-Levels! Even though my chemistry grades in JC were mediocre at best even up to the point of Prelims, Mr Ong kept faith in me and encouraged and motivated me to work harder. In the days leading up to the Chemistry A-Levels he was always present in the Whatsapp to answer my questions to the best of his ability and efficiency

    Mo FuJie TJC

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    #7093

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    JC1 Promo Exams and JC2 Prelim Exams Preparatory Classes

    for Chemistry

    Open for Registration Now!

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    #7516

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    Hi All A-Level/JC/H2/H1 Chemistry Students

    Attached 2015 A Level H2 Chemistry Paper 1 answers.
    Please contact Mr Ong @9863 9633 uf your answers is different from mine.

    1 B 11 D 21 B 31 D
    2 B 12 C 22 D 32 A
    3 C 13 D 23 B 33 C
    4 C 14 B 24 A 34 A
    5 B 15 D 25 D 35 B

    6 D 16 D 26 D 36 A
    7 B 17 A 27 D 37 C
    8 C 18 C 28 C 38 C
    9 A 19 B 29 B 39 A
    10 B 20 A 30 A 40 B

    #7586

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    Chemistry Tuition Center Singapore

    Question

    A 2.85 g sample of haematite iron ore, Fe O , was dissolved in hydrochloric acid and the solution diluted to 250 cm3 in a standard flask. A 25.0 cm3 of this solution was completely reduced with excess tin (II) chloride to form a solution of iron(II) ions.

    After the remaining tin(II) ions were removed with a suitable reagent, the solution of iron(II) ions was titrated against an acidified solution of 0.020 mol dm-3 potassium dichromate(VI) and required 26.40 cm3 for complete oxidation back to iron(III) ions.

    (i) Give the balanced equation for the reaction between iron(II) and dichromate(VI) ions.

    (ii) Calculate the percentage of iron(III) oxide, Fe2O3, in the ore.

    Answer

    (i) Cr2O72–(aq) + 14H+(aq) + 6Fe2+(aq) —> 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)

    (ii) 88.7% – Contact Mr Ong @ 98639633 if you need the full work solution

    Please contact Angie @ 96790479 or Mr Ong @ 98639633 if you need A level/JC Chemistry Tuition

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