Lu 1.5 6 *Radii of group 18 elements is Van der W alls’ radii
4. Polarization and hydration energy
The cations, which are more polarizing, have high hydration energy. It is affected by radius as well as core electrons of the metal ions. One very clear example is very high hydration energy of Al+3.
Al3+ + Water Al+3(aq) ΔHHyd. = 4665KJmol–1 ΔHHYd(Ca+2) <ΔHHyd(Cd+2), as Cd+2 has more core electrons.
Lattice Energy
Ionic compounds are solids and have crystalline structure. Crystals contain alternate arrangement of ions i.e., + – + – etc., in all the three dimensions. It results into attractive and repulsive forces both in the lattice.
But calculations show that attractive force is always greater than the repulsive force. This situation brings stability to the structure.
Lattice energy is defined as: Energy required to separate one mole crystal into gaseous ions is called lattice energy.
NaCl(S) Na(+g) + Cl-(g) (E = + UO)
Or conversely, Energy releases when one mole crystal is produced from its gaseous ions is called lattice energy.
Na+(g) + Cl-(g) NaCl(S) + E (E = –UO)
Theoretically, lattice energy is given by Born – Lande equation. It is,
U
NMZ Z e
r n
o
o
= −
+. − 2 1 1
Where, N = Avogadro’s number
M = Madelung constant, it depends on crystal structure.
Z+ and Z– = Charges on cation and anion e = Charge on electron
ro = r+ + r_ [(r+ = radius of cation), (r_ = radius of anion)]
n = Born constant, it depends upon electron configuration of the ions.
Table 2.8
e – configuration Ions n
He Li+, Be2+, H– 5
Ne Na+, Mg2+, F–, O–2 7 Ar K+, Ca2+, Cl–, S–2 9 Kr Rb+, Sr2+, Br– 10
Xe Cs+, I– 12
For an ionic compound, Born constant (n) is average of the ions of the compound, e.g.,
NaCl, n = (7 + 9)/2 = 8
The value of Madelung constant (M) depends on crystal structure.
Table 2.9
Crystal type M NaClCsCl
ZnS
1.74756 1.76267 1.63806
In SI unit’s lattice energy equation is 8 10= = H
U Q
R ê
ơô
º
ẳằ
SH
Where = ˛ 0, is dielectric constant and it has the value (of a vacuum)
˛ 0 = 8.854 x 10–12 C2m–1J–1 Some conclusions from Born equation
(i) The contribution of the term (1 –1/n) is small to lattice energy.
(ii) UO 1/rO i.e., larger the ions lesser the lattice energy and vice versa.
(iii) UO Z+.Z– i.e., larger the product of the ionic charges greater the lattice energy.
rO(Å) Z+.Z– UO (KJmol–1)
LiF 2.01 1 x 1 = 1 –1004
MgO 2.10 2 x 2 = 4 –3933
(iv) Crystals having higher L.E. usually melt at high temperature.
Crystal L.E (KJmol–1)
(from Born eq.) mpt (oC)
LiCl – 825 614
NaCl – 764 800
KCl – 686 770
Melting point of LiCl, is low, inspite of higher L.E. It is due to covalent nature in LiCl.
(v) Lattice energy and water solubility
Solubility of ionic compounds depends on (a) L.E., and (b) Hydration energy. The conditions are (a) If LE – HE = ive, the substance is soluble.
(b) If LE – HE = ⊕ ive, the substance is insoluble.
Hydration energy: Ions in water do not exist isolated. They are surrounded by water molecules having weak attraction. It is called hydration.
1D
2 2
2 2 2 +
+ +
++
+ + +
+ +
&O + + + +
+ + 2
2
2 Figure 2.4
Table 2.10
Ion Charge
density Hydration energy (KJmol–1) Na+Mg2+
Al+3 F– Cl–
1.12.9 1/1.335.2 1/1.84
– 422.0 – 1954.0 – 4665.0 – 513.0 – 370.0 The energy released in this attractive
interaction is called hydrationenergy. This depends on charge/radius ratio of the ion i.e., charge density,
Charge density = Charge/radius
When charge density is high interaction with water is more and hydration energy is high.
(vi) Lattice energy and core electrons
Lattice energy also depends on core electrons. It is because ions having more core electrons are more polarizing. Polarization effect adds to lattice energy. Thus, CdCl2 has high lattice energy over CaCl2. It is because Cd+2 has more core electrons (18) than Ca+2 (8 electrons). Both these ions have almost equal ionic radii (rCa+2 ~ rCd+2 ~ 0.9Å).
General properties of ionic compounds
(a) All ionic compounds are solids, hard and brittle.
(b) Ionic solids have crystal structure.
(c) Ions in the solid are arranged alternately. e.g., + – + – + – + – + – +
– + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + –
(d) Ionic compounds have high melting point (due to lattice structure)
(e) Ionic compounds are generally soluble in water. The solution contains ions. Therefore, solution of ionic compounds is electrolyte.
(f) Ionic compounds in fused state are also conductor. It is due to the fact that melt contains free ions.
(g) Ionic compounds in solid state are insulators as ions are held at fixed points in the lattice structure and can not move on application of pot. difference.
Covalent bonding
A covalent bond is formed by the sharing of one (or more) pairs of electrons with opposite spin between two atoms. It gives stable electron configuration (either 2 – electron, duct or 8 – electron i.e., octet) to each atom. The electron pairs remain somewhere between the two nuclei of the combining atoms. It is, therefore,
attracted by both the nuclei simultaneously resulting into bonding. Thus, the bond in H2 results from the sharing of hydrogen electrons between the two atoms H. + H. H H (H – H).
Due to sharing of electrons both H atoms achieve electron configuration of He. Two or three pairs may also be shared, e.g., in N2,
11 11
Both N atoms achieve octet configuration due to sharing of three pairs of electrons between them. The above presentation of a molecule is called Lewis structure (or Dot structure). [G. N. Lewis was the first to put forward an electronic interpretation of the chemical bond].
Electrons of atoms are present in atomic orbitals (s, p, d etc.). Thus sharing of electrons involves interaction of atomic orbitals. Orbital interaction in chemical terms is called overlap of orbitals. Therefore, a covalent bond is formed by overlap of orbitals. As no more then two electrons can occupy an orbital, only half-filled atomic orbitals can take part in overlap. The larger is the overlap stronger is the bond. The model described above is Valence Bond Model for covalent bonding.
σ and π bonds
Orbitals can overlap in two distinct ways, (i) At an axis (or linearly or head on) and, (ii) Side-ways (or laterally)
The above situations give different type of covalent bonds.
σ -bond
When two orbitals overlap at an axis, the bond formed is called ‘σ – bond’. The axis taken is internuclear axis (or bond axis) i.e., the axis on which both the nuclei lie.
S S
S±S ı ,QWHUQXFOHDUD[LV
V V
,QWHUQXFOHDUD[LV V±V ı
V S
,QWHUQXFOHDUD[LV V±S ı
V S
Figure 2.5
The degree of overlap is large, so, a σ – bond is stronger. [Hybrid orbitals also can form σ – bond].
A σ – bond orbital is symmetrical around the bond axis. In such bonding rotation of one atom relative to the other about the bond axis does not change the extent of overlap. Therefore, free rotation is possible about a σ – bond.
π - bond
When two orbitals overlap side by side (or sideways or laterally) a π bond is formed. A π bond may involve two p orbitals (or d and p orbitals or d and d orbitals).
(p – p) π bond
When two p orbitals centered at two atoms overlap side by side a (p – p)π bond is formed. For most effective overlap the orbitals should lie perpendicular to the bond axis.
%RQGD[LV
S\ S\
Figure 2.6
A π – bond orbital is divided in two parts by the bond axis i.e., it is not symmetrical about the bond axis. It has a modal plane in the plane of the molecule. Free rotation about a π bond is not possible because it would break the π – bond.
As degree of overlap is small a π – bond is weaker than a σ – bond. The second period elements (B, C, N, O, F) form very strong (p – p)π bonds. Molecules like N2, O2, CO2, NO2, BF3 etc. all have strong (p – p) π bonds. The (p – p)π bonding is not effective in third period elements (Si, P, S, Cl etc.). It is because of the fact that
(i) 3p orbitals are large and,
Figure 2.7 (ii) The elements have more of core electrons (8 electrons).
(d – p)π bond
A π bond may also involve d and p orbitals.
Such bonding is important in the compounds containing third (or higher) period elements (Si, P, S, Cl etc.). These elements have vacant 3d orbitals. These vacant d – orbital form (d – p)π bonding when Si, P, S etc., are bonded with N, O, F which have lone pair electrons in their p – orbitals. Molecules like SO2, SO3, SiF4, N(SiH3)3 etc. all have effective (d – p)π bonds.
Comparison of σ and π bond Table 2.11
Properties σ-bond π-bond
1. Nature of overlap Linear overlap of orbitals Side by side overlap of orbitals.
2. Extent of overlap Large Small
3. Bond strength Strong bond Relatively weak bond 4. Symmetry σ – bond orbital symmetrical
around the bond axis. π – bond orbital not symmetrical about the bond axis.
5. Nodal plane No nodal plane There is nodal plane which lies in the plane of the molecule [(p – p)π bond].
6. Rotational
property Free rotation about σ bond
axis is possible No free rotation about π – bond axis is possible.
7. Ploarizability Least polarizable Most polarizable.
Formation of molecules having σ bonds only
(a) H2: It is formed by overlap of 1s orbitals of H atoms.
V V
+ +
++ ++
Figure 2.8
(b) HF: It is formed by overlap of 1s orbital of H with half-filled 2pz orbital of F.
+) +)
+ ) V S[ S\S]
)
Figure 2.9
(c) F2: It is formed by overlap of half-filled 2pz orbitals centered on two F atoms.
) )
)) ))
Figure 2.10
(d) H2O: It is formed by overlap of two half-filled 2p orbitals centred on O – atom with 1s orbital of both the H – atom.
2 2
+
+
2 + V S[S\S] +
Figure 2.11
(e) NH3: It is formed by overlap of three half-filled 2p orbitals centred on N – atom with 1s orbital of H – atoms.
1 +
1 1 + 1
+
+ +
V S[S\S] +
Figure 2.12
Molecules having σ and π bond both
(a) O2 molecule: The O – atom has two half-filed 2p orbitals.
2
VS[S\S]
The 2pz orbitals overlap axially to form a σ bond (z–axis is taken as bond axis). The 2py orbitals, which are perpendicular to the bond axis, overlap side by side to form a π bond.
=D[LV
S] S]
22ı ʌ
Figure 2.13
Thus O2 has double bond between O – atoms, in which one is a σ bond and the other is a π bond.
[The above structure of O2 shows all electrons paired i.e., a diamagnetic molecule. But O2 is found paramagnetic i.e., has unpaired electrons. It can be explained by molecular orbital model for covalent bonding not by valence bond model.]
(b) N2 molecule: The N atom has three half-filled 2p orbitals.
1 VS[S\S]
The 2pz orbitals overlap axially to form a σ–bond (Z–axis, bond axis). The remaining p–orbitals, 2px and 2py which are perpendicular on the bond axis overlap side ways to form two π bonds (2px–2px and 2py–2py).
=D[LV 11
Figure 2.14
Thus N2 has a triple bond between N atoms, in which one is a σ – bond and the others are π–bonds (i.e., two π–bonds).
Electronegativity [χ (chi)]
Electronegativity is the property of a bonded atom. it is defined as: “ the tendency of an atom in a molecule to attract bond pair electrons towards itself is called electronegativity”. Elements differ in their electronegativity.
Therefore, a covalent bond between two dissimilar atoms is always polar, e.g., HCl. In this molecule Cl is
more electronegative (χCl = 3.0) than H (χH = 2.1). So, bond pair electrons are inclined towards Cl, making it partially negative and H partially positive,
As polarity is partial, δ is always less than unity (i.e., δ < 1). On the other hand a covalent bond between two similar atoms is always non – polar because both the atoms have the same χ, e.g., H2, N2, O2, F2, Cl2 etc.
Hence covalent bonds can be of two types:
(a) Polar covalent bond: It is always formed between two different non-metals, Example, the bonds O – H, N – H, F – H, S – O, N – O are polar covalent.
(b) Non-polar covalent bond: It is present in homonuclear diatomic molecules like, H2, F2, Cl2, O2, N2, Na2, B2 etc.
Electronegativity of some elements Li
1.0
Be 1.5
B 2.0
H 2.1
C 2.5
N 3.0
O 3.5
F 4.0
Br 2.8 Na
0.9
Mg 1.2
Al 1.5
Si 1.8
P 2.1
S 2.5
Cl 3.0
I 2.5
Electronegativity is a relative property of an element and not an absolute measure of the attraction for electrons. Degree of polarity in a bond depends on electronegativity difference. A rough estimate is:
(a) If Δχ = 1.8, bond is 50% ionic and 50% covalent.
(b) If Δχ < 1.8, bond is less ionic and more covalent.
(c) If Δχ > 1.8 bond is more ionic and less covalent.
The per cent ionic nature can roughly be calculated by the equation,
% Ionic nature = 16 (χA – χB) + 3.5(χA – χB)2
(The polar nature is best-calculated using dipole moment data).
Some important points Table 2.12
Ion χ Ion χ
Fe+2 Fe+3 1.83
1.96 Sn+2
Sn+4 1.8 1.96 (i) Electronegativity is relative property.
(ii) Electronegativity of an element change with its oxidation state (It is because attraction for electrons increases with increase in oxidation state).
Table 2.13
Hybrid Csp3 < Csp2 < Csp S – character 25% 33.3% 50%
χ 2.5 2.7 3.3
Electronegativity also changes with hybrid nature, e.g., an sp3 carbon is less electronegative than sp2 carbon, which in turn is less electronegative than sp carbon. It is due to change in s–characters of the hybrids,
Coordinate covalent bond
A covalent bond in which both the shared electrons are provided by only one atom is called a coordinate covalent bond. It is also called electron donor–acceptor bond. It is denoted by an arrow extending from donor towards acceptor. ( )
(OHFWURQGRQRUV1++2&O352HWF (OHFWURQDFFHSWRUV+&X)H%)HWF
Let us consider H+ and NH3. The H+ has vacant orbital and N of NH3 has lone pair. The H+ ion is, thus, an acceptor and NH3 is a donor. They can combine due to formation of a coordinate bond.
+1+ >+1 +@{>1+@
$PPRQLXPLRQ
The other example is formation of addition compound of BF3 and NH3. BF3 has vacant orbital (B is sp2 in BF3) and NH3 has a lone pair (N is sp3 in NH3 and lone pair is in sp3 orbital). These orbitals overlap resulting into the formation of addition compound.
) % )
) 1 +
++ )) %
) 1
+ ++
Both BF3 and NH3 molecules get distorted due to the formation of addition compound. Due to this combination hybrid nature of B changes from sp2 to sp3.
A coordinate bond has directional character because orbitals are involved in its formation.
Transition metals (or ions) forms a vast number of compounds due to formation of coordinate bond.
Such compounds are known as coordination compounds. For example, [Cu(NH3)4]2+, [Co(NH3)6]3+, [Fe(CN)6]4– etc.
+1
1+ 1+
1+
&X
)H 1&
±
&1±
&1±
&1±
&1±
&1±
±
Figure 2.15
Electronegativity and Dipole moment
A covalent bond between two unlike atoms is always a polar covalent bond. It is due to the difference in the electronegativities of the two atoms. Example, HCl (χH = 2.1 and χCl = 3.0). Thus HCl is represented as
d+H Cl- d- i.e., polar centers in the molecule.
A polar bond has bond dipole moment, à. It is given by, à = q x r
where r = bond length, q = charge on either end.
The bond dipole moment (à) is a vector quantity and it is represented as, (i.e., arrow extending from positive to the negative end)
H –––– Cl
Unit of à
(i) The CGS unit of bond moment is debye, D (in the honour of P. Debye who contributed much towards understanding of polar molecules).
1 D = 10−18 esu.cm
(ii) SI Unit: The SI unit is coloumb metre (cm).
1 D = 3.33 x 10−3o cm
For a heteronuclear diatomic molecule the bond moment is the molecular dipolemoment. But for a polyatomic molecule the molecular dipolemoment is the resultant of the bond moments. It is calculated using vector method.
Molecular dipole moment gives various informations: