Covalent Bond (By Mutual Sharing of Electrons)
The covalent bond is formed when two atoms achieve stability by the sharing of an electron pair, each contributing one electron to the electron pair.
The arrangement of electrons in a covalent molecule is often shown by a Lewis structure in which only valency shells (outer shells) are depicted. For sake of clarity, the electrons on different atoms are denoted by dots and crosses.
Polarity of Bonds: A covalent bond is set up by sharing of electrons between two atoms. It is further classified as polar or non-polar depending upon the fact whether the electron pair is shared unequally between the atoms or shared equally. For example, the covalent bonds in H2 and Cl2 are called non-polar as the electron pair is equally shared between the two atoms.
In the case of hydrogen fluoride the bond is polar as the electron pair is unequally shared. Fluorine has a greater attraction for electrons or has higher electronegativity than hydrogen and the shared pair of electrons is nearer to the fluorine atom than hydrogen atom. The hydrogen end of the molecule, therefore, appears partial positive with respect to fluorine.
Bond polarities affect both physical and chemical properties of compounds containing polar bond. The polarity of a bond determines the kind of reaction that can take place at that bond and even affects the reactivity at nearby bonds. The polarity of bonds can lead to polarity of molecules and affect melting point, boiling point and solubility.
Characteristics of Covalent Compounds
Melting Point and Boiling Point: In covalent compounds, except those consisting of giant molecules, the molecules are less powerfully attracted to each other, as a result of which their melting points and boiling points are relatively low compared to ionic compounds, e.g.,
\(\underset{(Covalent\,\,compound)}{\mathop{SiC{{l}_{4}}\,\,b.p.=33K}}\,\,\,and\,\,\underset{(Covalent\,\,compound)}{\mathop{NaCl\,\,b.p.=1713\,K}}\)
Conductivity: Covalent substances (whether of the “molecular lattice” or “giant molecule” type) do not conduct electricity in the fused state since there are no free electrons or ions to carry the current. However, substances like graphite, which consists of separate layers, conduct electricity because the electrons have a passage in between the two flat layers.
Solubility: The characteristic solubility of covalent compounds in non-polar solvents such as benzene and carbon tetrachloride can be described to the similar covalent nature of the molecules of solute and solvent (i.e., like dissolves like). Covalent compounds in solution soluble more slowly as compared with the ionic compounds which soluble instantaneously in solution. The solubility of covalent compounds is, however, very much dependent upon the size of the molecule. Thus covalent substances having giant molecules are insoluble in virtually all solvents due to the big size of the molecule unit.
Fajan’s Rules
When two oppositely charged ions approach each other closely, the positively charged cation attracts the outermost electrons of the anion and repel its positively charged nucleus. This results in the distortion or polarization of the anion followed by some sharing of electrons between the two ions, i.e., the bond becomes partly covalent in character.
The magnitude of polarization depends upon following factors :
i) Charge on Either of the ions: As the charge on the cation increases, its tendency to polarize the anion increases. This brings more and more covalent nature in the electrovalent compound. Whereas with the increasing charge of anion, its ability to get polarized, by the cation, also increases.
For example, in the case of NaCl,\(MgC{{l}_{2}}\,\,and\,\,AlC{{l}_{3}}\) the polarization increases, thereby covalent character becomes more and more as the charge on the cation increases.
Similarly, lead forms two chlorides \(PbC{{l}_{2}}\,\,and\,\,PbC{{l}_{4}}\) having charges +2 and +4 respectively. \(PbC{{l}_{4}}\) shows covalent nature. Similarly among \(\text{NaCl},\text{ N}{{\text{a}}_{\text{2}}}\text{S},\text{ N}{{\text{a}}_{\text{3}}}\text{P},\)
the charge of the anions are increasing, therefore the increasing order of covalent character is NaCl < Na2S < Na3P.
ii) Size of the cation: Polarisation of the anion increases as the size of the cation decreases i.e., the electrovalent compounds having smaller cations show moreof the covalent nature. For example, in the case of halides of alkaline earth metals, the covalent character decreases as we move down the group. Hence melting point increases
e.g. : BeCl2 < MgCl2 < CaCl2 < SrCl2 < BaCl2
iii) Size of anion: The larger the size of the anion, more easily it will be polarized by the cation i.e., as the size of the anion increases for a given cation, the covalent character increases. For example, in the case of halides of calcium, the covalent character increases from F– anion to I– anion i.e. \(\underset{\operatorname{cov}alent\,\,character}{\mathop{\xrightarrow{Ca{{F}_{2}}<CaC{{l}_{2}}<CaB{{r}_{2}}<Ca{{I}_{2}}}}} \)
Similarly, in case of trihalides of aluminium, the covalent character increases with increase in size of halide anion i.e.\(\underset{\begin{smallmatrix} \text{Covalent character increases as the} \\ \,\,\,\,\,\,\text{size of the halide ion increase} \end{smallmatrix}}{\mathop{\xrightarrow{Al{{F}_{3}}\,\,\,\,\,\,\,\,AlC{{l}_{3}}\,\,\,\,\,\,\,\,AlB{{r}_{3}}\,\,\,\,\,\,\,\,Al{{I}_{3}}}}}\)
iv) Nature of the cation: Cations with 18 electrons (s2p6d10) in outermost shell polarize an anion more strongly than cations of 8 electrons (s2p6) type. The d electrons of the 18 electron shell screen the nuclear charge of the cation less effectively than the s and p electrons of the 18-electron shell. Hence the 18-electron cations behave as if they had a greater charge. Copper (I) and Silver (I) halides are more covalent in nature compared with the corresponding sodium and potassium halides although charge on the ions is the same and the sizes of the corresponding ions are almost similar. This illustrates the effect of 18-electron configuration of Cu+ and Ag+ (4s2, p6, d10) ions.
Illustration: The decomposition temperature of Li2CO3 is less than that of Na2CO3. Explain.
Sol. As Li+ ion is smaller than Na+ ion, thus small cation (Li+) will favour more covalent character in Li2CO3 and hence it has lower decomposition temperature than that of Na2CO3.
Valence Bond Theory
This theory was proposed by Linus Pauling, who was awarded the Noble Prize for chemistry 1954. Atoms with unpaired electrons tend to combine with other atoms which also have unpaired electrons. In this way the unpaired electrons are paired up, and the atoms involved, all attain a stable electronic arrangement. This is usually a full shell of electrons (i.e. a noble gas configuration). Two electrons shared between two atoms constitute a bond. The number of bonds formed by an atom is usually the same as the number of unpaired electrons in the ground state, i.e. the lowest energy state. However, in some cases the atom may form more bonds than this. This occurs by excitation of the atom (i.e. providing it with energy) when electrons which were paired in the ground state are unpaired and promoted into suitable empty orbitals. This increases the number of unpaired electrons, and hence it increases number of bonds which can be formed.
A covalent bond results from the pairing of electrons (one from each bonding or combining atom). The spins of the two electrons must be opposite (antiparallel) because of the Pauli exclusion principle that no two electrons in one atom can have all four quantum numbers the same.
1. In HF, H has a singly occupied s-orbital that overlaps with a singly filled 2p orbital on F.
2. In H2O, the O atom has two singly filled 2p orbitals, each of which overlaps with a single occupied s-orbital of each from two H atoms.
3. In NH3, there are three singly occupied p orbitals on N which overlap with s orbital of each from three H atoms.
4. In CH4, the C atom in its ground state has the electronic configuration \(1{{s}^{2}},\,\,2{{s}^{2}},\,\,2p_{x}^{1},\,\,2p_{y}^{1}\) and only has two unpaired electrons, and so can form only two bonds. If the C atom is excited, then the 2s electrons may be unpaired, giving \(1{{s}^{2}},\,\,2{{s}^{2}},\,\,2p_{x}^{1},\,\,2p_{z}^{1}\). There are now four unpaired electrons which overlap with singly occupied s orbital each of four H atoms.
Carbon in CH4 molecule uses its three p-orbitals \({{p}_{x}},\,\,{{p}_{y}}\,\,and\,\,{{p}_{z}},\) which are mutually at right angles to each other, and the s orbital is spherically symmetrical.
CH4 H – C – H = 1090 28’
A covalent bond is formed by the overlapping of atomic orbitals. Covalent bonds formed are of two types depending upon the way the orbitals overlap each other.
Sigma and Pi Bonds ( \(\sigma\) and \(\pi\) Bonds)
1. Sigma bond (\(\sigma\) bond): The bond formed by the overlapping of two half filled atomic orbitals along their axis is known as sigma bond. \(\sigma\) bond is a strong bond because overlapping in it takes place to large extent. The hybrid orbitals always from \(\sigma \) bond.
a) s – s overlapping
b) s – p overlapping
c) p – p overlapping
2. Pi bond \(\pi\) bond): The bond formed by the lateral overlapping of half filled atomic orbitals is known as pi bond. The sidewise overlapping takes place to less extent. Therefore, \(\pi\) bond formed is a weak bond. \(\pi\) bond overlapping takes place only at the sides of two lobes. A \(\pi\) bond is formed when a \(\sigma\) bond already exists between the combining atoms.
Example:
In A – B molecule the bond formed is Covalent bond.
Thus, all the single bonds are Covalent bonds. Double bond has one \(\sigma\) and one \(\pi\) bond. Triple bond has one \(\sigma\) and two \(\pi\) bonds.
Hybridisation
It is the mathematically fabricated concept that is introduced to explain the geometry or shapes of the covalent polyatomic molecules of ions containing covalent bonds.
It is a process of intermixing of atomic orbitals with small difference in energy or same energy and belong to the same atom, at the time of bonding so as to give another set of orbitals with equivalent shapes and energies.
Stereochemistry of the hybrid orbitals
\(s{{p}^{3}}\) Hybridisation: In ground state, the electronic configuration of carbon is \(\text{1}{{\text{s}}^{\text{2}}},\text{ 2}{{\text{s}}^{\text{2}}},\text{ 2}{{\text{p}}^{\text{2}}}\)
It is proposed that from 2s orbital, being quite near in energy to 2p orbitals, one electron may be promoted to the vacant 2pz orbital thus obtaining the excited atom. At this stage the carbon atom undoubtedly has four half-filled orbitals and can form four bonds. In the excited atom, all the four valence shell orbitals may mix up to give four identical \(s{{p}^{3}}\) hybrid orbitals. Each of these four \(s{{p}^{3}}\) orbital possesses one electron and overlaps with 1s orbital each of four H atoms thus forming four equivalent bonds in methane molecule. Due to the tetrahedral disposition of \(s{{p}^{3}}\) hybrid orbitals, the orbitals are inclined at an angle of 1090 28’. Thus all the H – C – H angles are equal to 1090 . 28’
$s{{p}^{2}}$ Hybridisation: When three out of the four valence obritals of carbon atom in excited state hybridize, we have three \(s{{p}^{2}}\) hybrid orbitals lying in a plane and inclined at an angle of 1200. If 2s and two 2p, orbitals of the excited carbon atom are hybridized, the new orbitals lie in the xy plane while the pure \(2{{p}_{z}}\) orbital lies at right angles to the hybridized orbitals with its two lobes above and below the plane of hybrid orbitals. Two such carbon atoms are involved in the formation of alkenes (compounds having double bonds). In the formation of ethene two carbon atoms (in \(s{{p}^{2}}\) hybridization state) form one sigma bond by ‘head-on’ overlap of two \(s{{p}^{2}}\) orbitals contributed one each by the two atoms. The remaining two sp2 orbitals of each carbon form \(\sigma \) bonds with each
s-orbital of each H atom. The unhybridized each 2p, orbital of the two carbon atoms undergo a side-wise overlap forming a \(\pi \) bond. Thus the carbon carbon double bond in ethene is made of one \(\sigma\) bond and one \(\pi\) bond. Since the energy of a \(\pi\) bond is less than that of a \(\sigma\) bond, the two bonds constituting the ethene molecule are not identical in strength. The molecule is a planar one.
Different types of hybridization depend upon the type of atomic orbitals, which are used for intermixing.
Note:
i) Orbitals participating in hybridization must have only small difference in their energies.
ii) Both half-filled and completely filled orbitals or without filled can get involved in hybridization.
iii) The number of hybrid orbitals is equal to the number of orbitals participating in hybridization.
iv) Hybrid orbital form more stronger bonds than pure atomic orbitals.
v) Same atom can assume different hybrid states under different situations.
vi) Hybrid orbitals form sigma bonds only.
vii) Hybriclined orbitals occupied by electrons according to Pauli’s and Hund’s rule.
viii)Hybridisation phenomenon is of orbitals only but not of electrons.
Method of predicting the Hybrid state of the central atom in covalent molecules and polyatomic ions:
The hybridization of the central atom in similar covalent molecule or polyatomic ion can be predicted by using the generalized formulae as described below:
Valence Shell Electron Pair Repulsion (Vsepr) Theory
In 1957 Gillespie and Nyhom gave this theory to predict and explain molecular shapes and bond angles more exactly. The theory was developed extensively by Gillespie as the Valence Shell Electron Pair Repulsion (VSEPR) theory. This may be summarized as:
Shapes of molecules based on VSEPR Theory
1.The shape of the molecule is determined by repulsions between electron pairs present in the valence shell.
2. A lone pair of electrons takes up more space round the central atom than a bond pair, since the lone pair is attracted to one nucleus whilst the bond pair is shared by two nuclei. It follows that repulsion between two lone pairs is greater than repulsion between a lone pair and a bond pair, which in turn is greater than the repulsion between two bond pairs. Thus the presence of lone pairs on the central atom causes slight distortion of the bond angles and also from the ideal shape. If the repulsion between a lone pair, and a bond pair is increased, it follows that the actual bond angles between the atoms must be decreased.The order of repulsion between lone pairs and bond pairs of electrons follows the order as:
Lone pair - lone pair > lone pair – bond pair > bond pair – bond pair.
3.The magnitude of repulsions between bond pairs of electrons depends on the electronegativity difference between the central atom and the other atoms.
4.Double bonds cause more repulsion than single bonds, and triple bonds cause more repulsion than a double bond.
Effect of Lone Pairs: Molecules with four electron pairs in their outer shell are based on a tetrahedron. In CH4 there are four bonding pairs of electrons in the outer shell of the C atom, and the structure is a regular tetrahedron with bond angle H – C – H of 109028’. In NH3, N atom has four electron pairs in the outer shell, made up of three bond pairs and one lone pair. Because of the lone pair, the bond angle H – N – H is reduced from the theoretical tetrahedral angle of 109028’ to 107028’. In H2O the oxygen atom has four electron pairs in the outer shell. The shape of the H2O molecule is based on a tetrahedron with two corners occupied by bond pairs and the other two corners occupied by lone pairs. The presence of two lone pairs reduces the bond angle further to 104027¢.
In a similar way, SF6 has six bond pairs in the outer shell and is a regular octahedron with bond angles of exactly 900. In BrF5, the Br also has six outer pairs of electrons, made up of five bond pairs and one lone pair. The lone pair reduces the bond angles to 84030¢. Whilst it might be expected that two lone pairs would distort the bond angles in an octahedral as in XeF4 but it is not so. Actual bond angle is 900, reason being that the lone pairs are trans to each other in the octahedron, and hence the atoms have a regular square planar arrangement.
Molecules with five pairs of electrons are all based on a trigonal bipyramid. Lone pairs distort the structures as before. The lone pairs always occupy the equatorial positions (in an triangle), rather than the axial positions (up and down). Thus in \(I_{3}^{-}\) ion, the central ionidine atom has five electron pairs in the outer shell, made of two bond pairs and three lone pairs. The lone pairs occupy all three equitorial positions and the three atoms occupy the top, middle, and bottom positions in the trigonal bipyramid, thus giving a linear arrangement with a bond angle of exactly 1800.
Effect of Electronegativity: NF3 and NH3 both have structures based on a tetrahedron with one corner occupied by a lone pair. The high electronegativity of F push the bonding electrons further away from N than in NH3. Hence the lone pair in NF3 causes a greater distortion from tetrahedral and gives a F – N – F bond angle of 102030¢, compared with 1070 . 48¢ in NH3. The same effect is found in H2O (bond angle 104027¢) and F2O (bond angle 1020).
Some examples using the VSEPR Theory
Phosphorus pentachloride PCl5 :
Gaseous PCl5 is covalent. (The electronic structure P is \(1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{3}}\)). All five outer electrons are used to form bonds to the five Cl atoms. In the PCl5 molecule the valence shell of the P atom contains five electron pairs: hence the structure is a trigonal bipyramid. There are no lone pairs, so the structure is not distorted. However, a trigonal bipyramid is not a completely regular structure, since some bond angels are 900 and others 1200. Symmetrical structures are usually more stable than asymmetrical ones.
Note: Thus \(\text{PC}{{\text{l}}_{\text{5}}}\) is highly reactive, and in the solid state it splits into \({{[PC{{l}_{4}}]}^{+}}\,\,and\,\,{{[PC{{l}_{6}}]}^{-}}\) ions, which have tetrahedral and octahedral structures respectively.
Chlorine trifluoride \(Cl{{F}_{3}}\): The chlorine atom is at the centre of the molecule and determines its shape. The electronic configuration of Cl is \(1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{5}}\). Three electrons form bonds to F, and four electrons do not take part in bonding. Thus in \(\text{Cl}{{\text{F}}_{\text{3}}},\) the Cl atom has five electron pairs in the outer shell, hence the structure is a trigonal bipyramid. There are three bond pairs and two lone pairs.
It was noted previously that a trigonal bipyramid is not a regular shape since the bond angles are not all the same. It therefore follows that all the corners are not equivalent. Lone pair occupy two of the corners, and F atoms occupy the other three corners. Three different arrangements are theoretically possible, as shown in figure below.
The most stable structure will be the one of lowest energy, that is the one with the minimum repulsion between the five orbitals. The great repulsion occurs between two lone pairs. Lone pair bond pair repulsions are next strongest, and bond pair-bond pair repulsions are weakest. Groups at 900 repel each other strongly, whilst groups 1200 apart repel each other much less.
Structure I is the most symmetrical, but has six 900 repulsions between lone pairs and atoms. Structure II has one 900 repulsion between two lone pairs, plus three 900 repulsions between lone pairs and atoms. These factors indicate that structure III is the most probable. The observed bond angles are 80040’, which is close to the theoretical 900. This confirms that the correct structure is III, and the slight distortion from 900 is caused by the presence of the two lone pairs.
As a general rule, if lone pairs occur in a trigonal bipyramid they will be located in the equatorial position (round the middle) rather than the axial positions (top and bottom), since this arrangement minimizes repulsive forces.
Sulphur hexafluoride \(S{{F}_{6}}\): The electronic structure of S is \(1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}\) All six of the outer electrons are used to form bonds with the fluorine atoms. Thus in \(\text{S}{{\text{F}}_{\text{6}}}\) the sulphur has six electron pairs in the outer shell: hence the structure is octahedral. There are no lone pairs, so the structure is completely regular with bond angles of 900
Illustration :
Ether and water have same hybridization at oxygen. What bond angle would you expect for them.
Sol. In \({{H}_{2}}O\) bond angle is less than 109028’ due to lone pair and bond pair repulsion. But in ether, due to strong mutual repulsion between two alkyl groups bond angle becomes greater than 109028’.
Illustration : In following which central atom has different hybridization than other \(C{{l}_{2}}O, O{{F}_{2}}, {{H}_{2}}O, S{{O}_{2}}.\)
Sol. \(S{{O}_{2}}\); because it has \(\text{s}{{\text{p}}^{\text{2}}}\) hybridization other three have \(\text{s}{{\text{p}}^{\text{3}}}\) hybridization.
Forces of attraction (weaker bonds)
Types of H‑bonding:
Applications in:
[A] Abnormal behaviour of water. [B] Association of a molecule as in carboxylic acid.
[C] Dissociation of a polar species. [D] Abnormal melting point & boiling point.
[E] Enhanced solubility in water.
Types of Vander Waal’s Forces
(i) Ion dipole attraction: It is between an ion such as Na+ and a polar molecule such as HCl.
(ii) Dipole-Dipole attraction: It is in between two polar molecules such as HF and HCl.
(iii) Ion induced dipole attraction: In this case a neutral molecule is induced by an ion as a dipole.
(iv) Dipole Induced dipole attraction: In this case a neutral molecule is induced as a dipole by another dipole.
(v) Instantaneous dipole- induced dipole :
Instantaneous dipole Induced dipole attraction or London dispersion force between two non-polar molecules as in \(\text{C}{{\text{l}}_{\text{2}}},\) He etc.
Ionic bond > Covalent bond > Metallic bond > H-bond > Vander Waal bond.
Odd Electron Bond: These include one electron bond and three-electron bond e.g. \(He_{2}^{+}\). The dissociation energy of \(He_{2}^{+}\) is 60.0 cal., which means that it is one electron bond. The bond is only half as strong as a shared electron pair bond. The three-electron bond also exists in ferro-magnetic metals. The three-electron bond concept was given by Pauling. NO and NO2 are the example of odd molecule having three electron bonds e.g.
The three-electron bond is formed when the two atoms are identical or have nearly same electronegativity. The three-electron bond is also about half as strong as a normal bond.
Back Bonding: The interaction between an empty orbital and lone pair electron known as back bonding. p\(\pi\) – p\(\pi\) bonding is called back bonding. In \(B{{F}_{3}}\)molecule, the 2p orbitals of each F atom are fully filled and one of 2p orbital of the B atom is vacant. The two 2p-orbitals involved in the formation of B–F bond, therefore overlap laterally resulting in the transference of electron of the fluorine atom.
The bonding is effective since energy level of both orbital is almost same. An additional pp - pp bond is formed between Boron and Fluorine atom. As a result of the back donation of electrons from F to B, the deficiency of the Boron atom gets compensated and hence the Lewis acidic nature of \(B{{F}_{3}}\) decreases. The order of acidity is as follows:
\(B{{F}_{3}}\)
This type of bond has some double bond character and is known as dative or back bonding. All three-bond lengths are same in \(B{{F}_{3}}\) due to delocalisation.
For example the nitrogen in trimethyl amine and trisilyl amine has a lone pair electron at nitrogen but nitrogen in trimethyl amine has pyramidal shape while in trisilyl amine nitrogen has planer shape because in trimethyl amine there is repulsion between lone pair and bond pair that’s why the shape becomes pyramidal
In trisilyl amine, there is vacant d-orbital at silicon which overlaps with lone pair of nitrogen called back bonding hence geometry becomes planar.
Banana Bonding: This type of bonding is present in \({{B}_{2}}{{H}_{6}}.\) A bridge structure for diborane was proposed by Dilthey in 1921.
This structure shows that there are two types of hydrogen atoms – Terminals and bridging. 4-terminal hydrogen atoms can easily be replaced by methyl groups but when two bridging hydrogen atoms are attacked, the molecule is ruptured.
According to molecule orbital theory each of the two boron atoms is in sp3.hybrid state of th four hybrid orbitals, three have one electron each while the fourth is empty. Two of the four orbitals of each boron atom overlap with two terminal hydrogen atoms forming two normal B-H\(\sigma\) bonds one of the remaining hybrid orbitals (either filled or empty) of one of the other boron atom overlap to form a delocalised orbital covering the three nuclei with a pair of electrons. Such a bond is known as three centres two electron bond as shown in fig.
Similar overlapping occurs in one hydrogen atom (bridging) and fourth hybrid orbital of each boron atom. The formation of diborane molecule is depicted below.
On account of repulsion between the two hydrogen nuclei, the delocalised orbitals of bridges are drifted away from each other giving the shape of a banana. The three centre two electron bonds are also known as banana bond.
Non-Equivalent Hybrid Orbital (Schrodinger Concept)
(i)
(ii) \(\cos \theta =-\frac{1}{i}=\frac{S}{S-1}=\frac{S}{S-1}=\frac{P-1}{P}\),i = hybridisation index.
(iii) Sfs = 1 \({{f}_{p}}\) = fractional s-character
\(\Sigma {{f}_{p}}=\left[ \begin{matrix} 1 \\ 2 \\ 3 \\ \end{matrix} \right]\)
\({{f}_{s}}=\frac{1}{i+1}-\frac{\cos \theta }{\cos \theta -1}\) \(\Rightarrow\) \(i=\frac{{{f}_{p}}}{{{f}_{s}}}\)
\({{f}_{p}}\) = fraction of p-character.
(iv)
Drago's Rule
According to the Drago's rule, if the central atom of a molecule belong to \(II{{I}^{rd}}\) period or below and it is bonded to substituent having electronegativity less than or equal to 2.5 then the lone pair of S-orbital lies in stereochemically inactive s-orbital and bond pairs are formed by overlapping of almost pure p-orbitals as p-orbital are at 900 to another hence the angle between bond pair approches towards 900 Drago's Rule is applicable for elements of Vth and VIth group.
Mathematically \(\cos \theta =\frac{S}{S-1}=\frac{P-1}{P}\) where s = fraction of s-character
If q\(\rightarrow\) 900 p = fraction of p-character
S \(\rightarrow\) 0
P \(\rightarrow\) 1 (pure p-orbital)
Period
|
V group
|
VI group
|
II
|
\(\ddot{N}{{H}_{3}}\) 1070 (sp3-1s)
|
\({{H}_{2}}\ddot{O}:\) 1040(sp3-1s)
|
III
|
\(\ddot{P}{{H}_{3}}\) 93.80 (3p-1s)
|
\({{H}_{2}}\ddot{S}:\) 920(3p-1s)
|
IV
|
\(\ddot{A}s{{H}_{3}}\) 91.80 (4p-1s)
|
\({{H}_{2}}\ddot{S}e:\) 920(4p-1s)
|
V
|
\(\ddot{S}b{{H}_{3}}\) 91.30 (5p-1s)
|
\({{H}_{2}}Te\) 89.50(5p-1s)
|
BENT'S RULE: Only applicable when central atom having same Hybridisation.
1.According to this Bent's rule the more electronegative substituents bond's itself to a hybrid orbital that contains more p character and less S-character. While less electronegative substituent bond itself to a hybrid orbital that contains more s-character & less p-character.
2.Lone pair is always placed in the hybrid orbital where there is more S-character.
3.Although Nitrogen & Oxygen are more electronegative. If they are bonded by multiple bonds then they are placed in the hybrid orbital that contains more s-character.
4.Bent's Rule is consistant with the VSEPR theory.
Application of Bent's Rule
[A] \(PC{{l}_{3}}{{F}_{2}}\) \(PC{{l}_{2}}{{F}_{3}}\)
In the above formulae,
V = Number of monovalent atoms or groups attached to the central atom
G = Number of outer shell electrons in ground state of the central atom
a = Magnitude of charge on anion
c = Magnitude of charge on cation
X = number of hybrid orbitals
Calculate the value of X and decide the hybridization state of central atom as follows :