IKATAN IONIKIkatan ionik : ikatan yang terjadi antara ion positif dan ion negatif karena interaksi elektrostatikTerjadi apabila terdapat atom yang mempunyai energi ionisasi kecil dan bertemu dengan atom yang mempunyai afinitas elektron besarselisih EN sangat tinggi

Kekuatannya sangat tergantung pada ukuran ion dan muatan ionsemakin besar ukuran ion, ikatan ionik semakin lemah dan makin tinggi muatan ion, makin kuat ikatannya

Contoh :NaNa+ + e-Cl+ e-Cl-Na+ + Cl-Na+Cl-

MgMg2+ + 2e-O + 2e-O2-Mg2+ + O2-Mg2+O2-

AlAl3+ + 3e-O + 2e-O2-2 Al3+ + 3O2- (Al3+)2(O2-)3

The compounds with more 50% ionic character are normally considered to be ionic solids.

-Catatan : Apabila selisih ukuran antara kation dan anion cukup besar untuk memungkinkan terjadinya persinggungan anion-anion atau kation-kation, maka ikatan ionik antara ion akan diperlemah oleh tolakan persinggungan tersebut Jika kation relatif besar maka akan terjadi kontak antar kation Jika anion relatif besar maka akan terjadi kontak antar anion-++++++-----+(a)(b)

2. Kristal ionik yang mengandung anion kompleks cenderung untuk lebih mudah terurai apabila kation tersebut memiliki kemampuan yang tinggi untuk menarik elektron dari anion. Ukuran kation makin kecil, makin mudah mempolarisasi.MAM+A- M+A- kovalenkov polar ionik

Jika kemampuan kation untuk menarik elektron dari anion makin kuat maka ikatannya makin lemah.

Makin besar kemampuan kation untuk menarik elektron, akan meningkatkan karakter kovalen

Contoh : CaCO3 dan BaCO3r Ca2+ < Ba2+ sehingga Ca2+ memiliki kemampuan menarik elektron anion CO32- lebih besar daripada Ba2+, akibatnya polarisasi Ca2+ > Ba2+, sehingga CaCO3 lebih mudah terurai

Sifat-sifat senyawa ionik1. Pada keadaan padatan, senyawa mempunyai konduktivitas sangat rendah, sedangkan dalam fasa cair/larutan, mempunyai konduktivitas tinggi.Contoh :Secara teoritis, NaCl dalam padatan mengandung banyak ion sehingga mampu menghantarkan listrik yang baik, tetapi faktanya ion-ion terikat secara kuat sehingga tidak dapat bebas bergerak oleh karena itu konduktivitasnya mendekati 0 (sangat rendah).Jika padatan NaCl dilarutkan dalam air maka ion (+) dan (-) akan berdiri sendiri-sendiri sehingga dapat menghantarkan listrik dengan baik konduktivitas menjadi tinggi

2. Mempunyai titik lebur tinggiKarena ion (-) dari Cl tidak hanya berinteraksi dengan salah satu ion (+) dari Na tapi ion (-) dari Cl berikatan ke segala arah shg ikatan Na+ dengan Cl- menjadi semakin kuat shg energi untuk memutus ikatan Na+Cl- menjadi semakin besar t.l tinggiIkatan ionik ke segala arah sehingga memperkuat interaksi antar ion.

Contoh : titik lebur dari beberapa garam NaAnion: F- Cl- Br- I-Titik leleh: 1268 10731023 924+-+--++-+-+-+-+-++--+

3. Keras dan britleKeras : padatan tersusun secara kompak dari senyawa ionik yang sudah tidak dapat dimampatkan lagi. Kekerasan naik dengan menurunnya jarak antar ion dan naiknya muatan ion.britle : permukaannya mudah rontok (mripil)Contoh : LiFNaFM-X (pm) : 202231kekerasan : 3,33,2

4. Mudah larut dalam pelarut polar (pelarut yang mempunyai tetapan dielektrik tinggi, contoh : air) Tetapi tidak semua zat yang larut dalam pelarut polar adalah ionik

Lithium-Fluoride structure

ENERGI IKATAN DALAM PADATAN IONIKENERGI KISI

Adalah :~ energi yang diperlukan untuk mengubah 1 mol senyawa ionik padatan menjadi ion-ionnya dalam fasa gas~ energi yang dilepaskan apabila ion (+) dan (-) dalam keadaan gas berubah menjadi padatan.

Evaluation of lattice energyExperimentallyBorn-Haber cycleTheoreticallyBorn-Mayer equationKapustinskii equation12is the standard molar enthalpy change accompanying the formation of a gas of ions from an ionic solidThe disruption of a lattice requires energy and it is, therefore, an endothermic processThis means that Lattice Enthalpies are always positive and the most stable crystal structure of a compound is the structure with the greatest lattice enthalpy under prevailing conditions

Dalam padatan/kristal ionik terdapat 2 jenis energi :1. Etarikan muatan berlawananEcoulomb2. Etolakan muatan sama Erepulsion

berlaku untuk 2 ion

Karena padatan terdiri dari banyak ion maka :z = muatan ionA = tetapan Madelung

Energi tolakan dari Born :B = tetapan Bornn = eksponen Born

Energi kisi = U

U = Ec + Er - + karena tandanya berlawanan maka dalam kurva U vs r akan di dapat harga U minimal pada r tertentu

At large inter nuclear distances (right side of graph) there is no overlap of electron clouds. This state is designated with PE = 0.As the nuclei approach each other (moving to the left on the graph), the atoms become stabilized due to sharing of their electron with the other atom. We see the PE decrease in this region.The minimum on the curve corresponds to the optimized balancing of attraction (nuclear-electron) and repulsion (nuclear-nuclear and electron-electron). The energy at the minimum is the bond strength which represents the amount that the molecule is stabilized over the free atoms. If you want to break the bond, you need to supply that much energy. The inter nuclear distance at this minimum is the bond length for that molecule.If the nuclei are brought any closer together, the nuclear-nuclear repulsion dominates and the PE for the molecule shoots up.At large inter nuclear distances (right side of graph) there is no overlap of electron clouds. This state is designated with PE = 0.

Pada U minimal U=U0 dan r=r0 sehingga :Pers. BORN-LANDE

n tergantung pada konfigurasi elektron ion yang bersangkutanHarga n untuk kristal ionik selalu > 1Tabel harga n

Konfigurasi e ionn2 (He)10 (Ne)18 (Ar)36 (Kr)54 (Xe)5791012

Contoh :

Na+Cl-11Na : 1s2 2s2 2p6 3s1Na+ : 1s2 2s2 2p6 e = 10Konfigurasi e Na+ = konf. e Ne sehingga n Na+ = 717Cl : 1s2 2s2 2p6 3s2 2p5Cl- : 1s2 2s2 2p6 3s2 2p6e = 18Konfigurasi e Cl- = konf. e Ar sehingga n Cl- = 9

Jadi, n untuk NaCl = (7+9)/2 = 8

The Madelung constant, A, depends on the relative distribution of cations and anions in the structureMENCARI HARGA A

for NaCl6Cl- at a distance r12 Na+ at r28 Cl- at r36 Na+ at 2rA = Madelung constant

Giving its dependence on large contribution coming from nearest neighbours,the Madelung constant increases with coordination number

Structure typeACoordination numbersZnS blende1.6384, 4NaCl1.7486, 6CsCl1.7638, 8

Tipe struktur kristal ionikNaClLiCl, KBr, RbI, AgCl, AgBr, BeO, MgO, CaO, SrO, BaOZnS (blende)CuCl, CuBr, CuI, AlP, SiC, BeS, CdS, HgS, ZnOZnS (wurtzite) AgI, ZnO, NH4F, AlNCsClCsBr, CsI, TlCl, TlBr

THEORETICAL EVALUATION OF THE LATTICE ENERGYBorn-Mayer Equation: it shows the dependency of the lattice enthalpy from the structure

The use of the Born-Land equation requires the knowledge of the structure of the compound (Madelung constant)If the structure of the compound is not known, or the Madelung constant is not available, the Kapustinskii equation can be used to calculate the lattice energyThe Kapunstinskii Equation

The enthalpy of a given chemical reaction is constant, regardless of the reaction happening in one step or many steps. Hesss LawEXPERIMENTAL EVALUATION OF THE LATTICE ENERGYBorn - Haber Cycle: a closed path of steps (reactions), which include the lattice formation of the compound MX from its constituent ionsWe need to define all the quantities included in a Born-Haber cycleThe standard enthalpy of decomposition of a compound into its elements in their reference states is the negative of its standard enthalpy of formation

Likewise the standard enthalpy of lattice formation from the gaseous ions is the negative of the lattice enthalpyFor a solid element, the standard enthalpy of atomisationis the standard enthalpy of sublimationFor a gaseous element, the standard enthalpy of atomisation is the standard enthalpy of dissociation

The standard enthalpy for the formation of ions from their neutral atoms is the enthalpy of ionisation (for the formation of cations) and the electron-gain enthalpy (for the formation of anions)The value of the lattice enthalpy can be calculated from the requirement that the sum of the enthalpy changes round a complete cycle is zero

Sublimation of K(s)+ 89Ionisation of K(g)+ 425Dissociation of Cl2(g)+ 244Electron gain by Cl(g)- 355Formation of KCl(s)- 438

The lattice energy is equal to -xOnly one chlorine atom from Cl2(g) is used to produce KCl and therefore half of the dissociation energy of Cl2 is used

Formation of an Ionic Solid1. Sublimation of the solid metal M(s) M(g) [endothermic]2. Ionization of the metal atoms M(g) M+(g) + e- [endothermic]3. Dissociation of the nonmetal 1/2X2(g) X(g) [endothermic]4. Formation of X ions in the gas phase: X(g) + e- X-(g) [exothermic]5. Formation of the solid MXM+(g) + X-(g) MX(s) [quite exothermic]

Sublimation of LiIonization of LiDissociation of F2Electron affinity of FFormation of solid

Why does the reaction stop at NaCl? Why doesn't it keep going to form NaCl2 or NaCl3?

The lattice energy would increase as the charge on the sodium atom increased from Na+ to Na2+ or Na3+. But to form an Na2+ ion, we have to remove a second electron from the sodium atom, and the second ionization energy of sodium (4562.4 kJ/mol) is almost 10 times as large as the first ionization energy. The increase in the lattice energy that would result from forming an Na2+ ion can't begin to compensate for the energy needed to break into the filled-shell configuration of the Na+ ion to remove a second electron. The reaction between sodium and chlorine therefore stops at NaCl.

LATTICE ENERGIES FOR ALKALI METALS HALIDES The bond between ions of opposite charge is strongest when the ions are small. The lattice energies for the alkali metal halides is therefore largest for LiF and smallest for CsI.

Ionic Sizes

Arrange the following ionic compounds in order of increasing lattice energy: LiCl, NaCl, MgCl2

NaCl < LiCl < MgCl2

LiCl < NaCl < MgCl2

MgCl2 < NaCl < LiCl

Correct Answer:NaCl < LiCl < MgCl2

LiCl < NaCl < MgCl2

MgCl2 < NaCl < LiClLattice energy depends on size of the ions and their charges. MgCl2 has the highest lattice energy because it has a more positive cation. LiCl is higher than NaCl because the interatomic distance between cation and ion is shorter.

F- Cl-Br-I-Li+1036853807757Na+923787747704K+821715682649Rb+785689660630Cs+740659631604

The ionic bond should also become stronger as the charge on the ions becomes larger.The lattice energies for salts of the OH- and O2- ions increase rapidly as the charge on the ion becomes larger.

OH- O2-Na+9002481Mg2+30063791Al3+562715,916

Kelarutan padatan ionik dalam pelarut polar dan non polarTitik leleh dan titik didih padatan ionikStabilitas padatan ionik

Whether a given ionic solid is soluble or insoluble in a polar (ionic) or non-polar solvent depends two factors viz, (a) lattice energy of the ionic solid, and (b) hydration energy of ionic solid.The lattice energy of a salt gives a rough indication of the solubility of the salt in water because it reflects the energy needed to separate the positive and negative ions in a salt.Higher is the magnitude of hydration energy , greater will be the tendency of the ionic solid to break into ions in water and then to get hydrated by H2O molecules and consequently the solubility of the solid will increase.

Apabila energi hidrsi lebih besar daripada energi kisi, padatan ionik akan larut dalam pelarut polar dan sebaliknya.Padatan ionik tidak dapat larut dalam pelarut nonpolar seperti benzena, CCl4, selama ion-ion padatan ionik tidak berinteraksi dengan molekul pelarut membentuk ion-ion terhidrat.

Ionic solids (or salts) contain positive and negative ions, which are held together by the strong force of attraction between particles with opposite charges. When one of these solids dissolves in water, the ions that form the solid are released into solution, where they become associated with the polar solvent molecules.

H2O NaCl(s) Na+(aq) + Cl-(aq)

We can generally assume that salts dissociate into their ions when they dissolve in water. Ionic compounds dissolve in water if the energy given off when the ions interact with water molecules compensates for the energy needed to break the ionic bonds in the solid and the energy required to separate the water molecules so that the ions can be inserted into solution.

Solubility is a result of an interaction between polar water molecules and the ions which make up a crystal. Two forces determine the extent to which solution will occur:

Force of attraction between H2O molecules and the ions of the solid This force tends to bring ions into solution. If this is the predominant factor, then the compound may be highly soluble in water.

Force of attraction between oppositely charged ions This force tends to keep the ions in the solid state. When it is a major factor, then water solubility may be very low.

Sodium and potassium salts are soluble in water because they have relatively small lattice energies. Magnesium and aluminum salts are often much less soluble because it takes more energy to separate the positive and negative ions in these salts. NaOH is very soluble in water (420 g/L), but Mg(OH)2 dissolves in water only to the extent of 0.009 g/L, and Al(OH)3 is essentially insoluble in water.

LiCl, LIBr, NaCl, NaBr larut dalam air karena energi hidrasinya lebih besar daripada energi kisinya. E hidrasi : LiCl = 883, LiBr = 854, NaCl= 775, NaBr = 741 E kisi : LiCl = 840,1 LiBr = 781,2 NaCl= 770,3 NaBr = 728,4KCl, KBr, KI tidak larut dalam air karena energi hidrasinya lebih kecil daripada energi kisinya. E hidrasi : KCl = 686, KBr = 657, KI= 619, E kisi : KCl = 701,2 KBr = 671,1 KI=632,2

The strength of the bond between the ions of opposite charge in an ionic compound depends on the charges on the ions and the distance between the centers of the ions when they pack to form a crystal.An estimate of the strength of the bonds in an ionic compound can be obtained by measuring the lattice energy of the compound.Greater is magnitude of lattice energy of the ionic solid, greater is the stability of the ionic solid.

Accompanying : menyertaiDisruption : gangguanPrevailing : umum*K = 1.21 MJ mol-11A=100pm**