section 4a bonding and structure i (ionic bonding)

8/14/2019 Section 4A Bonding and Structure I (Ionic Bonding)

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Modern College AL Chemistry Notes (2009 10) Section 4A

Prepared by Mr. Chau Chi Keung, Richard Page 1

Name: ______________________________

Class: _______________

Class No.: ____________


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4.1 Ionic Bonding Revisited4.1.1. Formation of ionic bond

Ionic bonding: Interaction (electronic attraction) between oppositely charged ions by

transferring of electrons.Ions are formed as a result of the transfer of one or more electrons from the valence shell

of an atom of a metal to the valence shell of an atom of a non-metal.

Ions often, but not always, have noble gas electronic configurations. The lack of chemical

reactivity of the noble gases is an indication of the great stability of their electronic

structure; and ions with such electronic structure.

Dot and Cross diagram ( ) can be used to show the transfer of electrons from one

atom to another gives ions.

It should be noted that when drawing Dot and Cross diagrams, the dot and crossonly show the number of electrons present; they do not show the positions of the

electrons. The electrons are distributed in space as diffuse negative charged clouds.

Example 1: Formation of sodium oxide

Example 2: Formation of magnesium chloride

4.1.2. Nature of ionic bond



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Ionic bond is a result of balance between attraction and repulsion.

There are electrostatic attractions among oppositely charged particle and repulsion

among similarly charged particle.

The ions will come to equilibrium when the attractive and repulsive forces balance

one another and the ions will stay in positions in the crystal lattice.

The structure of ionic compound is called giant ionic structure or giant ionic lattice.

The particular arrangement of ions depends on their relative charges and sizes.

Ionic bonds are non-directional.

Ionic compounds have very high melting point (m.p.) and boiling point (b.p.).

Stronger ionic bond, higher m.p. and b.p.

Size of ions ; strength of ionic bond

Charge of ion ; strength of ionic bond

Consider the m.p. of the following two pairs of ionic compounds

KF (858 C) > KCl (711 C) (size: F < Cl )

MgO (2825 C) > Na 2O (1132 C) (charge: Mg 2+ > Na +)

4.2 Energetics of Formation of Ionic Compounds

4.2.1. Formation of cations from Group I and Group II elements

Group I and Group II elements tend to form cations by losing electrons to attain electronicconfigurations of noble gases.

The ease with which an atom loses an electron to form a positive ion is measured by its

ionization enthalpy.

For Group I elements (alkali metals)

Element Electronicconfiguration

1st I.E. / kJmol -1 Electronicconfiguration of M +

Lithium [He]2 s1 520 Li +: [He]

Sodium [Ne]3 s1

496 Na+

: [Ne]Potassium [Ar]4 s1 419 K +: [Ar]



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Rubidium [Kr]5 s1 403 Rb +: [Kr]Caesium [Xe]6 s1 376 Cs +: [Xe]

The 1 st I.E. of alkali metal atom is the lowest among the elements in the same period.

The ions have the stable noble gas electronic structure. This shows that alkali metals

form ions easily. They are the most reactive metals and powerful reducing agents.

Alkali metals do not form M 2+ ions. This is because of the very high 2 nd I.E. required

and the dipositive ions formed do not have the stable noble gas electronic structure.

For Group II elements (alkaline earth metals)Element Electronic

configuration1st I.E. / 2 nd I.E.

/kJmol 1Electronic

configuration of M 2+

Beryllium [He]2 s2 899 / 1757 Be 2+: [He]Magnesium [Ne]3 s2 738 / 1451 Mg 2+: [Ne]

Calcium [Ar]4 s2 590 / 1145 Ca 2+: [Ar]

Strontium [Kr]5 s2

549 / 1064 Sr 2+

: [Kr]Barium [Xe]6 s2 502 / 965 Ba 2+: [Xe]

The 2 nd I.E. is much greater than the 1 st I.E. ( Electron is being pulled away from a

positively charged ion have to overcome the attraction force)

4.2.2. Formation of anions from Group VI and Group VII elements

Group VI and Group VII elements tend to form anions by gaining electrons to attain

electronic configurations of noble gases.

The ease with which an atom gains an electron to form a negative ion is measure by its

electron affinity ( ).

The electron affinity for an element is the enthalpy change when 1 mole of electrons

is added to 1 mole of neutral atoms in the gaseous state to form 1 mole of gaseous

anions under standard conditions.

X(g) + e X (g) H = E.A.1 (in kJmol 1)

The 2 nd E.A. and 3 rd E.A. can be defined as:

X

(g) + e

X2

(g) H = E.A.2X2 (g) + e X 3 (g) H = E.A.3

The value of the electron affinity depends on the attraction between the incoming

electron and the nucleus and the shielding effect offered by the existing electrons.

For Group VII elements (halogens)


1st E.A. / kJmol 1 Electronicconfiguration of X

Fluorine [He]2 s22 p5 333 F : [Ne]

Chlorine [Ne]3 s2

3 p5

348 Cl

: [Ar]Bromine [Ar]3 d 104 s24 p5 324 Br : [Kr]



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Iodine [Kr]4 d 105 s25 p5 295 I : [Xe]

The halogens are expected to have the most negative 1 st E.A. because they complete

their outer energy level by gaining an electron to give the halide ion (X ) with stable

noble gas electronic configuration.

Going down a group, the atomic radii of the elements increase. The outermostelectrons become further away from the nucleus and so they would experience less

nuclear attraction.

Besides, as the number of occupied inner electron shells increase, the screening

effect becomes more significant. This also leads to a reduction of nuclear attraction.

The value for fluorine is anomalous ( ) because it includes an electron cloud

with high repulsion. The electron cloud of fluorine is very compact due to its small atomic size.

Requires higher electrostatic energy to force the electron into the region of outer

energy level.

For Group VI elements


1st E.A. / 2 nd E.A./kJmol 1

E.A.1 + E.A.2 /kJmol 1

Electronicconfiguration of X 2

Oxygen [He]2 s22 p4 141 / +791 +650 O 2 : [Ne]Sulphur [Ne] 3 s23 p4 200 / +649 +449 S 2 : [Ar]

Formation of X 2 ions requires the uptake of a second electron. This will involve the

second electron affinity of an element.

The enthalpy required for adding one mole of electrons to one mole of isolateduninegative ions in the gaseous state.

The 1 st E.A. of O and S are exothermic because the attraction between the incoming

electron and the nucleus is stronger than the repulsion between the incoming electron

and the existing electrons.

The 2 nd E.A. of O and S are endothermic because in the presence of the extra

electron, the extra electron charge offers a stronger repulsion with the incoming

electron. Also, the electron cloud of O and S will expand and the incoming electron

will occupy a position further from the nucleus which reduces the attraction with the

nucleus.

As a result, both formation of oxide and sulphide ions are endothermic. This

considerable expenditure of energy can be justified if the subsequent bonding in the

ionic compound causes an even greater release of energy.

4.2.3. Energetics of formation of ionic compounds (Part I) Lattice enthalpy ( ) and its determination

Ionization enthalpies and electron affinities give us information about the energy changesinvolved in the production of ions from neutral atoms, but this is only part of what



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happens when an ionic compound is formed.

The story is completed by knowledge of the lattice enthalpy of the compound.

Lattice enthalpy (lattice energy): The enthalpy change when one mole of an ionic crystal

is formed from its constituent ions in the gaseous state under standard conditions . (

)

mA n+ (g) + nB m (g) A mBn(s) H = H lattice (in kJmol 1)

Take sodium chloride as an example, the expression should be:

Na +(g) + Cl (g) NaCl(s) H = H lattice [NaCl(s)]

Lattice enthalpy must be exothermic because it is associated with the formation of

electrostatic attraction between cations and anions (i.e. ionic bond). (

)

The value of lattice enthalpy is proportional to the strength of ionic bonds in an ioniccompound. The more exothermic (more negative) the lattice enthalpy, the more

stable the ionic compound formed. (

)

Mathematically, we have,

+

r r

Z Z E

Z+ and Z : Charge of cation and anion

r + and r : Internuclear distance = sum of radii of cation and anion

Lattice enthalpies cannot be determined directly and so they are calculated by using the

Born-Haber cycle ( ).Take sodium chloride as an example,

Analysis of the cycle:

H atom[Na(s)] /H atom[Cl 2(g)]: Standard enthalpy change of atomization (


Internuclear distance


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) The enthalpy change when 1 mole of gaseous atoms is formed from

its element in the standard state under standard conditions (

) For sodium: Na(s) Na(g) H atom [Na(s)] = +109 kJmol 1

For chlorine: 21

Cl 2(g) Cl(g) H atom [Cl 2(g)] = +121 kJmol 1

Always endothermic (energy is needed for vaporization and bond breaking)

H I.E. [Na(g)]: 1 st I.E. of sodium Formation of Na +(g)

H E.A.[Cl(g)]: 1 st E.A. of chlorine Formation of Cl (g)

H lattice [NaCl(s)]: Lattice enthalpy of sodium chloride Ionic bond formation

H f [NaCl(s)]: Standard enthalpy change of formation of sodium chloride

Formation of NaCl can be regarded as a combination of the above steps

H of all other steps can be determined experimentally, except L.E.

So far the whole process can be summarized by the following flow chart:

The values of these enthalpy terms are given as follows:

Process involved Enthalpy changeFormation of sodium chloride from itsconstituent elements

H f [NaCl(s)] = 411 kJmol 1

Na(s) is atomized to form Na(g) H atom [Na(s)] = +109 kJmol 1

Na(g) loses 1 electron to give Na +(g) (1 st I.E.) H I.E. [Na(g)] = +496 kJmol 1

Cl 2(g) is atomized to give Cl(g) H atom [Cl 2(g)] = +121 kJmol 1

Cl(g) gains 1 electron to give Cl (g) (1 st E.A.) H E.A. [Cl(g)] = 348 kJmol 1

Na +(g) and Cl (g) combine to give NaCl(s) H lattice [NaCl(s)]

By Hesss law, we have

Enthalpy change of Route 1 Enthalpy change of Route 2

H f[NaCl(s)]

= H atom [Na(s)] + H I.E.[Na(g)] + H atom [Cl 2(g)] + H E.A.[Cl(g)] + H lattice [NaCl(s)]

H lattice [NaCl(s)]

= H f [NaCl(s)] {H atom [Na(s)] + H I.E. [Na(g)] + H atom [Cl 2(g)] + H E.A. [Cl(g)]}

= 411 [109 + 496 + 121 + (348)]

= 789 kJmol 1

In general, the sequence of drawing a Born-Haber cycle is as follows:



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Besides Born-Haber cycle, we may also use an enthalpy level diagram to determine L.E.

of an ionic compound (Again, take NaCl as an example).



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Example 1(a) Draw a Born-Haber cycle for the formation of magnesium oxide.



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(b) With the following thermochemical data, calculate the lattice enthalpy of magnesium

oxide by using the Born-Haber cycle drawn in (a).

Given: H atom [Mg(s)] = +150 kJmol 1

H I.E. [Mg(g)] = +736 kJmol 1

H I.E. [Mg +(g)] = +1450 kJmol 1

H atom [O 2(g)] = +248 kJmol 1

H E.A. [O(g)] = 142 kJmol 1

H E.A. [O (g)] = +844 kJmol 1

H f [MgO(s)] = 602 kJmol 1

(Ans: 3888 kJmol 1) Example 2



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Calculate the lattice enthalpy of rubidium chloride (RbCl) using the following data:

Standard enthalpy change of formation of rubidium chloride = 431 kJmol 1

Standard enthalpy change of atomization of rubidium (Rb) metal = +82 kJmol 1

First ionization enthalpy of rubidium = +403 kJmol 1

Standard enthalpy change of dissociation of chlorine molecules = +242 kJmol 1

(H dissociation [Cl 2(g)])

Electron affinity of chlorine atoms = 348 kJmol 1

(Ans: 689 kJmol 1)



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Example 3Calculate the lattice enthalpy of calcium fluoride using the following data:

Standard enthalpy change of formation of calcium fluoride = 1203 kJmol 1

Standard enthalpy change of sublimation of calcium metal = +193 kJmol 1

First ionization enthalpy of calcium = +590 kJmol 1

Second ionization enthalpy of calcium = +1145 kJmol 1

Standard enthalpy change of dissociation of fluorine molecules = +158 kJmol 1

Electron affinity of fluorine atoms = 333 kJmol 1

(Ans: 2623 kJmol 1)



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4.2.4. Energetics of formation of ionic compounds (Part II)

Explaining the stoichiometry of ionic compounds

General rule: All substances in nature tend to keep their energy content as low as possible.

It would be expected that the compound which has the most negative standard enthalpy

change of formation would be the most stable.

For instance, energy calculation can be used to show whether MgCl, MgCl 2, or MgCl 3would most likely be the formula for magnesium chloride.

If it is assumed that MgCl has a sodium chloride lattice structure, and MgCl 3 has a

structure similar to AlCl 3, then a reasonable estimate of the lattice enthalpies for the

hypothetical crystals MgCl and MgCl 3 may be made.

Born-Haber cycles can then be constructed to estimate the standard enthalpy changes of

formation of these hypothetical crystals.

Enthalpy terms H / kJmol -1

H atom [Mg(s)] = Standard enthalpy change of atomization of magnesium +148

I.E.1 = 1 st I.E. of magnesium +738

I.E.2 = 2 nd I.E. of magnesium +1451

I.E.3 = 3 rd I.E. of magnesium +7733

H atom [Cl 2(g)] = Standard enthalpy change of atomization of chlorine +122

H E.A. [Cl(g)] = 1st

E.A. of chlorine 349* H lattice [MgCl(s)] = Lattice enthalpy of MgCl 753

H lattice [MgCl 2(s)] = Lattice enthalpy of MgCl 2 2526 H lattice [MgCl 3(s)] = Lattice enthalpy of MgCl 3 5440

*Note: Here the given lattice enthalpies are calculated based on the mathematical model.

Cycle 1 (MgCl)

H atom [Mg(s)]

Mg(s) + 21

Cl 2(g) Mg(g) + 21

Cl 2(g)

I.E.1 H atom [Cl2(g)]

H f [MgCl(s)]

Mg +(g) + Cl(g) + e

H E.A. [Cl(g)]

H lattice [MgCl(s)]MgCl(s) Mg +(g) + Cl (g)



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Cycle 2 (MgCl 2)

H atom [Mg(s)]

Mg(s) + Cl2(g) Mg(g) + Cl

2(g)

I.E.1 2 H atom [Cl 2(g)]

H f [MgCl 2(s)]

Mg +(g) + 2Cl(g) + e

I.E.2

Mg 2+(g) + 2Cl(g) + 2e

2 H E.A. [Cl(g)]

H lattice [MgCl 2(s)]

MgCl 2(s) Mg 2+(g) + 2Cl (g)

Cycle 3 (MgCl 3)

H atom [Mg(s)]

Mg(s) +2

3Cl 2(g) Mg(g) +

2

3Cl 2(g)

I.E.1 3 H atom [Cl 2(g)]

H f [MgCl 3(s)]

Mg +(g) + 3Cl(g) + e

I.E.2

Mg2+

(g) + 3Cl(g) + 2e

I.E.3

Mg 3+(g) + 3Cl(g) + 3e

3 H E.A. [Cl(g)] H lattice [MgCl 3(s)]

MgCl 3(s) Mg 3+(g) + 3Cl (g)

From these three cycles, we have,



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H f [MgCl(s)] = 148 + 738 + 122 + (349) + (753) = 94 kJmol -1

H f [MgCl 2(s)] = 148 + 738 + 1451 + 2(122) + 2(349) + (2526) = 643 kJmol -1

H f [MgCl 3(s)] = 148 + 738 + 1451 + 7733 + 3(122) + 3(349) + (5440)

= +3949 kJmol -1

From the Born-Haber cycles, we can see that:

Formation of MgCl is just exothermic. The compound is energetically stable with

respect to its elements.

Formation of MgCl 2 is even more exothermic, and so the compound would be more

energetically stable with respect to its elements.

However, formation of MgCl 3 is highly endothermic. Thus MgCl 3 would be

extremely unstable with respect to its elements.

Therefore, it appears that the compound formed is the one whose formation involves the

greatest transfer of energy to the surroundings. (i.e. the most energetically stable one.)

Besides, the largest contributions in each cycle are made by the ionization enthalpies and

the lattice enthalpy, and that these two are always opposite in sign.

If it requires more energy to ionize the metal than it is returned as lattice enthalpy,

then the metal ion (and hence the ionic compound) will not be formed.

To sum up, energetics can help us to understand the stoichiometry of ionic compounds.

The calculation of the lattice energies and use of Born-Haber cycle can also be used

to explain why group I, II and III metal always form M +, M 2+ and M 3+ ion

respectively.

4.2.5. More about lattice enthalpy A. Factors affecting the lattice enthalpies

As mentioned previously, L.E. can be calculated based on mathematical model, we have,

+

r r

Z Z E

Z+ and Z : Charge of cation and anion

r + and r : Internuclear distance = sum of radii of cation andanion

Effect of ionic size

L.E. becomes less negative the size of ion increases ( electrostatic attraction )

Besides, small and similarity in size of ions result in more efficient packing and

more negative lattice energy

Effect of ionic charge

L.E. becomes more negative when the charge of ions increase ( electrostatic

attraction )

The charge density ( ) of the constituent ions


Internuclear distance


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Charge density of an ion is affected by both ionic size and ionic charge

For ions with the same charge, smaller ion, larger charge density

For ions with the similar size, greater charge, larger charge density

The higher the charge density of the ion, stronger will be the electrostatic attraction

and the lattice energy will be more negative.

B. Application of lattice enthalpiesLattice enthalpy is an important consideration in prediction of solubility of an ionic

compound.

When an ionic compound is dissolved, heat energy will be absorbed or evolved because:

Firstly, the crystal lattice must be broken down and this requires energy.

This energy will be numerically equal to the lattice enthalpy with the sign reversed

( Endothermic step).

The second step is the solvation ions: Interaction between the ions and the solventmolecules. Energy is evolved during solvation ( Exothermic step).

If water is the solvent the process of solvation is known as hydration and the energy

released is the hydration energy (H hyd) ( ).

The standard enthalpy change of solution (H soln ) is the enthalpy change when one mole of

a substance is dissolved in an infinite amount of solvent (usually water) so that further

dilution produces no detectable heat change under standard conditions.

To some extent, it reflects the solubility of a compound.

For a salt to be soluble in water, H soln usually has to be a negative or a small

positive value (Remember: G = H TS).

For ionic compounds, the value of H soln can be calculated by the following formula

H soln = H lattice + [H hyd (cation) +H hyd (anion)]

H soln = H hyd H lattice

Take NaCl as an example,

Firstly, the sodium chloride solid lattice is broken down to give its constituent ions

in the gaseous state.

NaCl(s) Na +(g) + Cl (g) H = H lattice = (776) = +776 kJmol 1

Then the resulting ions are hydrated (i.e. form interaction with water molecules)

Na +(g) + Cl (g) Na +(aq) + Cl (aq) H hyd = 772 kJmol 1

By Hesss Law, we have

H soln = H hyd H lattice = 772 (776) = +4 kJmol 1



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For salts which form crystals with water of crystallization (e.g. CuSO 4, MgSO 4, Na 2CO 3),

the value of H soln depends on whether the anhydrous or hydrated salt is used.

The anhydrous form usually gives out heat on dissolving whilst the hydrated form

may absorb heat on dissolving.

This is because ions of the hydrated salt are hydrated by the water of crystallization

before the salt dissolves.

4.3 Ionic Crystals

The structure of an ionic crystal can be determined by a technique called X-ray diffraction.

The model of an ionic crystal is a lattice array: a regular three-dimensional arrangement of

cations surrounded by anions and each anion by cations.This arrangement gives the maximum attractive force between oppositely charged ions

while minimizing repulsion between like ions.

A unit cell ( ) is the smallest basic part of the crystal lattice. It can be repeatedly

stacked together at various directions to fill the space completely, and to reproduce the

whole crystal structure of the compound. (

)

4.3.1. Sodium chloride latticeFigure (a) shows a space-filling model of the structure of sodium chloride. Although it

only shows a few Na + and Cl ions, there will be millions and millions of ions in even the

smallest visible crystals of sodium chloride.

The positions of Na + and Cl ions in the crystal lattice of sodium chloride are emphasized

in figure (b).



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Note: Counting ions in a unit cell

Ion inside the unit cell = 1

Ion lies on each face = 21

Ion lies on each edge =4

1

Ion at each corner =8

1

For a face-centred cubic lattice like NaCl, you may refer to the figure below:

In the crystal lattice of NaCl, ions are placed at each corner and each of the 6 faces.

Each positive sodium ion is surrounded by 6 Cl ions and each negative chloride ion is

surrounded by 6 Na + ions.

The ions are arranged in a face-centred-cubic ( ) pattern. The structure of NaCl

is said to have 6:6 coordination because the Na + ions have a coordination number of 6

and the Cl ions also have a coordination number of 6.

Coordination number ( ): The number of adjacent atoms or ions which are in

direct contact with a certain atom or ion.


(a) (b)

Question: For NaCl lattice shown on P.17, whatare the numbers of Na + and Cl ions in a unit cell?


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Example 4 The figure below shows a unit cell of NaCl:

X-ray analysis shows that a side of this unit cell has a length of 5.641 10 8cm. Given the

density of sodium chloride is 2.165 gcm 3. Calculate a value for the Avogadro constant, L.

(Formula mass of NaCl = 58.44)

Density = NaClmole1of VolumemassformulaRelative

Volume of 1 mole NaCl =4

)10641.5(L

38 ( Each unit cell has 4 units of NaCl)

165.244.58

4)10641.5(L

38

=

L = 6.015 10 23

4.3.2. Caesium chloride (CsCl) lattice ( )Figure (c) shows a space-filling model of caesium chloride. The Cs + ion is similar in size

to the Cl ion. The general shape of the lattice is simple cubic ( ). The

positions of Cs + and Cl ions in the caesium chloride lattice are shown in figure (d).


NaCl: A face-centred cubic lattice


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In the crystal lattice of caesium chloride, ions are placed at each corner and in the centre.

There are 8 Cl ions around each Cs + ion, so the coordination number of Cs + ions in thestructure is 8.

In the same way, the coordination number of Cl ions is also 8.

The structure of caesium chloride is therefore said to have 8:8 coordination .

4.3.3. Calcium fluoride (CaF 2) lattice

The figure below shows the unit cell of calcium fluoride


(c)

(d)

Question: For CsCl, what are the numbers of Cs + and Cl ions in a unit cell?


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In the crystal lattice of CaF 2, the calcium ions are arranged in a face-centred-cubic pattern

while the fluoride ions are arranged in a simple cubic pattern.

Each Ca 2+ ion is surrounded by 8 F ions and each F ion is surrounded tetrahedrally by 4

Ca 2+ ions.

The structure of calcium fluoride is said to have8:4 coordination

because the Ca 2+ ions

have a coordination number of 8 and the F ions have a coordination number of 4.

Example 5The following is a unit cell of an oxide of titanium

(a) What are the coordination numbers of titanium and oxygen respectively?

(b) State the number of titanium ions and oxide ions per unit cell. Explain your answer.

(c) Write the formula of the compound.

Note: The unit cell also gives us hints about the stoichiometry of an ionic compound.

No. of ions in a unit cell

Coordination number

In general, there are two factors governing the packing of ions in a giant ionic lattice:

Close packing consideration Each ion tends to have the highest number of neighbors of opposite charge, as

this would increase the lattice stability

Relative size of ions



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If the cation has a larger size than the anion, then more anions can be packed

around the cation. On the other hand, if the cation is much smaller than the anion, then before the

anions come close together, the anions already repel each other.

(Correct packing for NaCl) (Incorrect packing of CsCl)

In figure (a), the Na + ion is small enough to fit into the empty space between the 4 larger Cl ions.

Since the Cs + ion is much larger, there would be a large wastage of space

if CsCl still packs in the NaCl type lattice in (b).

The arrangement of the ions in an ionic crystal is determined by the ionic radii ratio (r +/r ):Type of structure Coordination Ionic radii

ratio (r +/r )Examples

Simple cubic 8:8 >0.73 CsCl, CsBr, CsI, NH 4Cl and NH 4Br Face centered cubic 6:6 0.41 0.73 Halides of Li +, K +, Na +, Rb + (e.g.

KCl, NaBr); Oxides and sulphides of Ca 2+, Mg 2+ and Ba 2+; Silver halides(AgF, AgCl, AgBr and AgI)

4.4 Ionic Radii ( )

4.4.1. Some general discussionSimilar to atomic radius, the size of an ion is mainly governed by the strength of the

effective nuclear charge ( ).

In general, ionic radii increase down a group because of increased number of occupied

electron shells. ( )

As the inner electrons can effectively screen the outermost electron and number of

inner electrons increase, the outermost electrons are less attracted by the nucleus.

Going across a period, both cationic and anionic radius decrease. (

)

As the number of protons increase, the effective nuclear charge increases.

The electrostatic attraction between outer electrons and positively charged nucleusincreases. As a result, the electrons are pulled closer towards the nucleus, causing a



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reduction in ionic radius.

For cations, the size of a cation is always smaller than the size of its parent atom as

electron is removed. ( )

The proton to electron ratio increases. The nuclear charge will have a greater effect

on the fewer remaining electrons, which makes the electron cloud contracts. (

)

Besides, for a metal cation, it has 1 less occupied electron shell than its parent atom.

( )

For anions, the size of an anion is always bigger than the size of its parent atom as electron

is added. ( )

The additional electron enters the highest energy level without an increase in nuclear

charge. This causes greater repulsion among the electrons (repulsion between theexisting electrons and the newly added electron). (

)

Since the number of protons becomes less than that of electrons, the nuclear

attraction becomes less effective and so there is an expansion of the electron cloud.

(

)

4.4.2. Isoelectronic species ( / )For example, O 2 ion, F ion, Na + ion and Mg 2+ ion are isoelectronic because they contain

the same number of electron and have the same electronic configuration: 1 s22 s22 p6.

Constant number of electrons

Number of proton increase across this isoelectronic series

Nuclear charge

Attraction between the nucleus and the electrons

Ionic radius decrease across the series.

For a series of isoelectronic anions (e.g. N 3 , O 2 and F ), the ionic size increases as the

number of negative charge increases.There will be repulsion between electrons of a negative ion which causes an increase

in size and the more electrons in an anion the larger it is.

The variation is ionic size of different series of isoelectronic species is shown below:



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Comparison between atomic and ionic radii in nanometers (nm) is shown below:



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section 4a bonding and structure i (ionic bonding)

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