Application of Hess Law

Hess’s law finds its application in determining the heat of changes for reactions for which experimental determination is not possible. The various applications of Hess’s law can be understood from following headings;

1.   Calculation of Enthalpy of Formation

The enthalpy of formation of certain compounds can be determined by Hess’s Law, where direct determination is not possible experimentally.

Example: Calculate the enthalpy of formation of methane, given that the enthalpies of combustion of methane, graphite and hydrogen are 890.2 KJ, 393.4 KJ and 285.7 KJ mol^–1 respectively.

Solution:

Given

(i).          CH₄  +  2O₂   ––––––––>  CO₂  +  2H₂O  ;  DH₁  =  – 890.2 KJ/mol^–1

(ii).         C   +   O₂        ––––––––>  CO₂  ;  DH₂  =  393.4 KJ/mol^–1    

(iii).        H₂  +  ½ O_2­    ––––––––>  H₂O   ;  DH₃  =  285.7 KJ/mol^-1

To find

C  +   2H₂   ––––––––>  CH₄  ;  DH  =  ?

Multiplying (iii) by 2 and then add to (ii) we get,

C  +  O₂ + 2 (H₂  +  ½ O₂)    ––––––––>  CO₂ + 2H₂O  ;  DH  =  DH₂  +  2DH₃  =   – 393.4  +  2(– 285.7)  

C  +  O₂ + 2 H₂  + O₂             ––––––––>  CO₂ + 2H₂O    ;  DH  =  – 393.4 – 571.4

C  +  2 H₂  +  2O₂                    ––––––––>  CO₂ + 2H₂O    ;  DH  =  – 964.8 –––––––– (iv)

Subtract (i) from (iv), we get.    

C  +  2H₂ – CH₄  ––––––––> 0             ;  DH  =  – 964 + 890.2 KJ         

C  +  2H₂                     ––––––––> CH₄        ;  DH  =   – 74.6 KJ      

2.     Calculation of Enthalpy of Allotropic Transformation

      The transformation of an element from one allotropic form to another form involves a small change of heat. This heat can be calculated using Hess’s Law.

Example:  Calculate enthalpy change of transformation of sulphur (rhombic) to sulphur (monoclinic). Given than combustion of rhombic sulphur and monoclinic sulphur as – 71.1 K Cal. and – 71.7 K Cal respectively.

Solution:

Given

 (i).     S _(Rhombic)   +    O₂  (g) ––––––––>   SO₂ (g)  ;  DH  =  – 71.1 K Cal.

 (ii).    S _(monoclinic) +    O₂ (g)  ––––––––>   SO₂ (g)  ; DH   =  – 71.7 K Cal.

Subtract (ii) from (i) we get           

S _(Rhombic)   –   S _(monoclinic)  ––––––––>     0   ;   DH    =    – 71.1  +  71.7           

S _(Rhombic)   ––––––––> S _(monoclinic)    ;   DH  =  0.6 K Cal.

3.            Calculation of Calorific Value

Example:   Calculate the heat of glucose and its calorific value from following data:

(i).          C _(graphite)  +  O₂ (g)   ––––––––>  CO₂ (g)  ;  DH  =  – 395 KJ

(ii).         H₂ (g)  +     ––––––––>  H₂O (l)   ;  DH  =  – 269.4 KJ.

(iii).        6C _(graphite)  +  6H₂ (g)  +  3O₂ (g) ––––––––>  C₆H₁₂O₆ (s)  ;  DH  =  –1169.8 KJ.

Solution:

We have to find

C₆H₁₂O₆ (s) +  6O₂ (g)   ––––––––>   6CO₂ (g)   +   6H₂O (l)  ;  DH   =   ?

Multiplying (i) by 6 and (ii) by 6, then add.

 6C _(graphite) + 6O₂ (g) + 6H2 (g)  +  6 ´ ½ O₂ (g) ––––––––> 6CO₂ + 6H₂O (l) ; DH = –395 ´ 6 + (– 269.4 ´ 6)

(iv) 6C_graphite + 6H2 (g)  + 9O₂ (g)  ––––––––> 6CO₂ + 6H₂O (l)  ;  DH =–2370–1616.4 = – 3986.4 KJ

Subtract eq. (iii) from (iv), we get.

C₆H₁₂O₆ (s)  +  6O₂     ––––––––>  6CO₂  +  6H₂O (l)    ;     DH  =  2816.6 KJ.

Thus,

1 mole of C₆H₁₂O₆ liberates energy = – 2816.6 KJ.

i.e.   180 gm of C₆H₁₂O₆ liberates energy = – 2816.6 KJ.

\            1 gm of C₆H₁₂O₆ liberates energy =  -2816.6/180= 15.64 KJ.

Thus, calorific value of C₆H₁₂O₆ is 15.64 KJ.

Coordinate Bond or Dative Bond

Coordinate Bond may be defined as, "the bond formed between the two atoms or ions in which the contribution of shared pair of electron is made by a single atom while sharing is done by both". The atom which contributes the electron pair for bond formation is called Donor Atom while the one which accepts the electron pair is called as Acceptor Atom. The bond is shown by an arrow (→) pointing towards the acceptor atom. A few examples of coordinate bond are as follows;

(i). Ozone (O₃):

 
 

Resonating Structures of Few Molecules

Resonance

    When a single structure is unable to explain all the properties of a compound, a number of other structures are suggested to that substance but none of them is able to explain all the properties of the compound completely. The different structures have no real existence and are called Contributing, Canonical or Resonating Structures. This phenomenon is known as Resonance. The resonating structures are separated from each other by double headed arrows (↔).The actual structure known as Resonance Hybrid lies in between the resonating structure and explains all the properties of the compound.
Due to resonance, molecule gains stability or in other words, energy of the molecule is decreased. The amount of energy decreased due to resonance is known as Resonance Energy.

Prediction of Shape of Molecules by VSEPR Theory

Let us apply VSEPR theory to predict the shape of few molecules;

1.    Boron triflouride (BF₃)

    In BF₃ molecule, the Boron atom has atomic number 5 and has electronic configuration of 2, 3. Thus, Boron has 3 valence electrons. These electrons are shared mutually with three fluorine atoms to form three B–F bonds. So, in BF₃, Boron has no lone pair over it.

No. of bond pairs = 3

No. of lone pairs = 0

Total no. electron pairs = bp + lp

= 3 + 0 = 3

        So, according to VSEPR theory, BF₃ has triangular planar shape. This is because it has 3 electron pairs.

2.     Methane CH₄

In CH₄ molecule, the Carbon atom has atomic number 6 and has electronic configuration of 2, 4. Thus, Carbon has 4 valence electrons. These electrons are shared mutually with three hydrogen atoms to form four C–H bonds. So, in CH₄, Boron has no lone pair over it.

No. of bond pairs = 4

No. of lone pairs = 0

Total no. electron pairs = bp + lp

= 4 + 0 = 4

        So, according to VSEPR theory, CH₄ has Tetrahedral shape. This is because it has 4 electron pairs.

3.     Ammonia NH₃

        In NH₃ molecule, the Nitrogen atom has atomic number 7 and has electronic configuration of 2, 5. Thus, Nitrogen has 5 valence electrons. Out of these 3 electrons are shared mutually with three hydrogen atoms to form three N – H bonds and the remaining 2 electrons exist as a lone pair. So, in NH₃, Nitrogen has one lone pair over it.

No. of bond pairs = 3

No. of lone pairs = 1

Total no. electron pairs = bp + lp

= 3 + 1 = 4

            So, according to VSEPR theory, NH₃ should have tetrahedral shape due to presence of 4 electron pairs. The presence of lone pair causes greater repulsion to the bond pairs, as a result of which the three N–H bonds move slightly closer, thereby decreasing the normal tetrahedral angle of 109.5⁰ to 107⁰. So, Ammonia has irregular geometry.

    Since one of the tetrahedral position is occupied by lone pair, the shape of ammonia molecule is said to be Pyramidal.

4.     Water H₂O

    In H₂O molecule, the Oxygen atom has atomic number 8 and has electronic configuration of 2, 6. Thus, Oxygen has 6 valence electrons. Out of these 2 electrons are shared mutually with two hydrogen atoms to form two O–H bonds and the remaining 4 electrons exist as two lone pairs. So, in H₂O, Oxygen has two lone pairs over it.

    No. of bond pairs = 2

    No. of lone pairs = 2

    Total no. electron pairs = bp + lp

     = 2 + 2 = 4

        So, according to VSEPR theory, H₂O
should have tetrahedral shape due to presence of 4 electron pairs. The presence of two lone pairs causes greater repulsion to the bond pairs, as a result of which the two O–H bonds move closer, thereby decreasing the normal tetrahedral angle of 109.5⁰ to 104⁰. So, Water molecule has irregular geometry.

    Since two of the tetrahedral positions are occupied by lone pairs, the shape of water molecule is said to be Bent or V-shaped.

Valence Shell Electron Pair Repulsion (VSEPR) Theory

    Sidgwick & Powell in 1940 formulated a theory based on the repulsions between electron pairs, known as Valence shell electron pair repulsion theory. The various postulates of this theory are as follows;

i).    The unpaired electrons in the valence shell of central atom form bond pairs with surrounding atoms while paired electrons remain as lone pairs.

ii).    The electron pairs surrounding the central atom repel each other. Consequently, they stay as far apart as possible in space to attain stability.

iii).    The geometry and shape of the molecule depends upon the number of electron pairs (bond pair as well as lone pair) around the central atom. The geometrical arrangement of electron pairs with different number of electron pairs around central atom are as follows;

iv).    The force of repulsion is minimum when the electron pairs are far away from one another.

v).    The force of repulsion among the bond pairs and the lone pairs is not same. The order of force of repulsion is as follows;

    lp – lp >lp – bp > bp – bp

    A molecule has Regular geometry if it has no lone pair around it, but if it has lone pairs over it, the molecule has Irregular geometry.

Types of Hybridisation

There are three common types of hybridization in compounds;

1.    Tetrahedral or sp³ Hybridization

        The mixing of one 's' and three 'p' orbitals to form four equivalent hybrid orbitals is called sp³ hybridization. Due to mutual repulsion of electrons in these four orbitals, sp³ hybrid orbitals try to keep themselves as far away as possible from each other. The arrangement in space that keeps them farthest apart is that of a tetrahedron. Thus, the four hybrid orbitals are directed towards the four corners of a regular tetrahedron, making an angle of 109.5⁰ with each other. Because of their tetrahedral orientation, this hybridization is also called tetrahedral hybridization.

For Example:–

    In methane, the central atom is carbon, with the electronic configuration 1s², 2s² 2p_x¹2p_y¹.

    Carbon atom (At. No. 6)

    Carbon atom in the ground state has only two unpaired electrons but in the excited state one 2s electron is promoted to vacant 2p_z orbital. Thus there are four unpaired electrons, one each in 2s, 2p_x, 2p_y and 2p_z orbitals. These four orbitals hybridize and form four equivalent sp³ hybrid orbitals directed towards the four corners of a regular tetrahedron with bond angle 109.5^o.

    Each sp³ hybrid orbital, overlaps 1s orbital of hydrogen atom and forms four C – H bonds in CH₄ molecule. Thus each C–H bond involves sp³ – s overlap and methane molecule has tetrahedral structure with H – C – H bond angle 109.5^o.

    Formation of methane (CH₄) molecule:

2.     Triagonal or sp² hybridization

        The mixing of one s & two p orbitals to form three new equivalent hybrid orbitals to form three new equivalent hybrid orbitals is called sp² hybridization. The three hybrid orbitals lie in one plane, making an angle of 120⁰ with each other to avoid mutual repulsion.

    For example

    Formation of ethylene (C₂H₂) molecule:

        During the formation of ethylene molecule, each carbon atom undergoes sp² hybridization.

        In the excited state of carbon atom, out of the four orbitals, each containing an unpaired electron, only three orbitals i.e., one 2s and two 2p orbitals (2p_x and 2p_y) undergo sp² hybridization. The resulting three hybrid orbitals are planar inclined at an angle of 120° and lie in xy – plane while the 2p_z atomic orbital which has not taken part in hybridization, lies perpendicular to the hybrid orbitals.

        One of the sp² hybrid orbital of a carbon atom overlaps with one sp² hybrid orbital of the other carbon atom by head on collision and forms σ bond between the two carbon atoms. The remaining two hybrid orbitals of each carbon atom overlap with 1s
orbitals of hydrogen atoms to form σ bonds. Thus, each carbon atom is left with one unhybridized 2p_z orbital with lobes above and below the plane of hybrid orbitals. These two 2p_z orbitals of the two carbon atoms overlap laterally and form a pi (π) bond between the two carbon atoms.

        Thus, in ethylene molecule, there is a double bond between the two carbon atoms, out of which one is σ bond involving sp²-sp²
overlap and the other is a π bond involving p-p overlap. In all, in the ethylene molecule, there are five σ bonds
and one π bond.

3.    Diagonal or sp Hybridization

        The mixing of one s and one p – orbital to form two equivalent hybrid orbitals is called sp hybridization. The hybrid orbitals lie as far apart as possible from each other to minimize the forces of repulsion. Hence, they point in opposite directions, making an angle of 180⁰.

    For example

    Formation of acetylene (C₂H₂) molecule:

        In acetylene molecule, each carbon atom undergoes sp hybridization.

        In the excited state of carbon atom, there are four half filled orbitals. Out of these, two orbitals, i.e., 2s and 2p_x undergo sp hybridization and form two equivalent linear hybrid orbitals. The other two orbitals 2p_y and 2p_z that have not taken part in hybridization remain at right angles to the hybrid orbitals. One of the hybrid orbitals of each carbon atom overlaps 1s orbital of H atom and form a σ bond. The remaining hybrid orbital of the two C atoms overlap to forms a σ bond between the two carbon atoms. Thus, each C atom is left with two unhybridized p orbitals (2p_y and 2p_z) which are mutually perpendicular to H–C–C–H axis. These p orbitals overlap sideways and form two pi bonds between the two carbon atoms.

        Thus, in acetylene molecule, there is a triple bond between the two carbon atoms, out of which one is σ bond involving sp-sp overlap and the other two are pi-bonds involving p-p overlap. In all, in the acetylene molecule, there are three σ bonds and two π bonds.

Hybridization

NEED FOR CONCEPT OF HYBRIDIZATION

Structures of simple molecules like H₂, F₂ and HF etc can be readily explained by the electron pair theory. But the formation of the compound of Beryllium, Boron & carbon cannot be that easily explained on the basis of same theory.

The electronic configuration of Be, b & c are

        ₄Be = 1s², 2s²

        ₅B = 1s², 2s² 2p_x¹

        ₆C = 1s², 2s² 2p_x¹2p_y¹

    On the basis of electron pair theory, Beryllium should be zero valent and behave like noble gas since it has no unpaired electron, boron with one unpaired electron (2P_x^l) should be monovalent and carbon with two unpaired electrons (2p_x¹, 2p_y¹) should be divalent. But Be forms compounds like BeH₂, BeF₂
in which it is bivalent, B forms compounds
like
BF₃, BCl₃ in which it is trivalent and C forms compounds like CH₄, CCl₄, in which it is tetravalent. Thus, it is not possible to explain the valencies of the elements like Be, B and C, on the basis of the number of unpaired electrons in the ground state of their atoms. Hence, in order to explain the bi, tri and tetravalency of Be, B and C respectively and to explain the observed geometry of their molecules, a new concept, called Hybridization is introduced by Pauling. This concept, though hypothetical is extremely useful.

CONCEPT

    "Hybridization may be defined as the phenomenon of mixing up of orbitals of an atom of nearly equal energy, giving rise to entirely new orbitals equal in number to the mixing orbitals & having same energies & identical shapes". The new set of orbitals so formed is known as Hybrid Orbitals. In order to minimize the repulsion among hybrid orbitals these rearrange themselves in space in definite orientation.

Distinction between Sigma and PI Bond

S. No.	SIGMA (s) BOND	PI (p) BOND
1.	The bond is formed by the axial overlap of the atomic orbitals.	The bond is formed by the sidewise overlap of the atomic orbitals.
2.	The overlap is quite larger and the bond is quite strong.	The overlap is less and the bond is comparatively weak.
3.	s-orbitals can take part in the sigma bond formation.	s-orbitals don't take part in the bond formation.
4.	Only one lobe of the p-orbital is involved in the overlap.	Both lobes of the p-orbitals are involved in the overlap.
5.	The electron cloud of the molecular orbital is symmetrical around the internuclear axis.	The electron cloud is unsymmetrical.
6.	The bond can be formed even in the absence of p bond.	The bond can't be formed in the absence of s bond.
7.	Free rotation of the atoms around s bond is possible.	Free rotation of the atoms around pi bond is not possible.

Types of Covalent Bond

    It has already been discussed that the formation of a covalent bond involves the overlapping of half filled atomic orbitals. The covalent bond can be classified into two different categories depending upon the type of overlapping. These are:

    (a).    Sigma covalent bond.

    (b).    Pi covalent bond.

(a).    Sigma (σ) Bond

        This type of covalent bond is formed by the axial overlapping of half filled atomic orbitals. The atomic orbitals overlap along the inter-nuclear axis and involve end to end or head on overlap. The electron cloud formed as a result of axial overlap is cylindrically symmetrical about inter-nuclear axis. The electrons constituting sigma bond are called Sigma Electrons. There can be three types of axial overlap as discussed below:

• s-s overlap
  
      It involves mutual overlap of half filled s-orbitals of the atoms approaching to form a bond. The bond formed so called s-s σ bond.
• s-p overlap
      It involves mutual overlap of half filled s-orbitals of the one atom with half filled p-orbital of the other. The bond so formed is called s-p σ bond.
  p-p overlap
  
  

      It involves mutual overlap of half filled p-orbitals of the two atoms. The bond so formed is called p-p σ bond.

  

  
          The s-s, s-p and p-p overlaps have been shown diagrammatically as under:
  (b).    Pi (π) Bond
  
           This type of covalent bond is formed by the lateral or sidewise overlap of the atomic orbitals. The orbital overlap takes place in such a way that their axis is parallel to each other but perpendicular to the inter-nuclear axis. The pi bond consists of two charged clouds above and below the plane of the atoms involved in the bond formation. The electrons involved in the π bond formation are called π-electrons.
   
  
  

  

  
  

Valence Bond Theory (Modern Approach of Covalent Bond)

        This theory was developed by Heitler and London in 1927. Later, Pauling and Slater modified this theory. The various postulates of this theory are as follows;

(i).    Covalent bonds are formed by the overlapping of half filled atomic orbitals present in the valence shell of the atoms participating in bonding.

(ii).    The orbitals undergoing overlapping must have electrons with opposite spins.

(iii).    Overlapping of atomic orbitals results in the decrease of energy and formation of covalent bond.

(iv).    The strength of a covalent bond depends on the extent of overlapping. The greater the overlapping more is the energy released and consequently, stronger will be the covalent bond.

Let us now apply the valence bond theory to the formation of few covalent molecules;

    (a).    Formation of fluorine (F₂) molecule

        Fluorine molecule is formed by the overlap of two half filled p-orbitals of two fluorine atoms as shown below:

    (b).    Formation of hydrogen fluoride (HF) molecule

    Hydrogen fluoride molecule is formed by the overlapping of 1s-orbital of hydrogen atom with one of 2p-orbital of fluorine atom which is half filled as shown below:

    (c).    Formation of oxygen (O₂) molecule

    Oxygen molecule is formed when two half filled 2p-orbitals of each oxygen atom overlap with the 2p-orbitals, of the other oxygen atom to form a double bond.

    (d).    Formation of water (H₂O) molecule

        Water molecule is formed when two half-filled 2p-orbitals of oxygen atom overlap with two half filled 1s-orbitals of two hydrogen atoms.

    (e).    Formation of nitrogen (N₂) molecule

        Nitrogen molecule is formed when the three half filled 2p-orbitals of one nitrogen atom overlap with three similar orbitals of the other nitrogen atom to constitute a triple bond.

    (f).    Formation of ammonia (NH₃) molecule

        Ammonia molecule is formed when three half filled p-orbitals of nitrogen atom overlap with three half filled 1s-orbitals of three hydrogen atoms.  

Applications of Dipole Moment

Some of the applications of dipole moment are given below:

1.     DISTINCTION BETWEEN POLAR AND NON-POLAR MOLECULES

        The molecules having dipole moment are called polar molecules whereas molecules having zero dipole moment are said to be non-polar molecules.

    For example:

(i).    Molecules such as H₂, N₂, O₂, Cl₂ etc. have non-polar bonds and zero value of dipole moment.

(ii).     Molecules such as CO₂, BF₃, CH₄, etc., have polar bonds but zero value of dipole moment.   

(iii).    Molecules such as HF, HCl, NH₃, H₂O, have polar bonds and their dipole moment is greater than zero. In other words, they are polar molecules.

2.     IONIC CHARACTER IN A MOLECULE

        Dipole moment gives an idea about the ionic character in a bond or a molecule. For example, let us calculate the percentage of ionic character in HCl molecule. Experiments have shown that the dipole moment of HCl is 1.03 D and its bond length is 1.275Å. Now for 100% ionic character the charge developed on H and Cl atoms would be 4.8 ´ 10^–10 e. s. u. Therefore, dipole moment in case of 100% ionic character is given as:

        µ = 4.8 × 10^–10 × 1.275 × 10^–8 e. s. u.-cm

              = 6.12 × 10^–18 e. s. u.-cm = 6.12 D

            The observed dipole moment

                µ = 1.03 D

        Therefore, Percentage Ionic character = 1.03/6.12 × 100 = 16.83%

    In general, larger the value of dipole moment is, more will be the ionic character.

3.     SHAPES OF THE MOLECULES

        The value of dipole moment helps to predict the shape of the molecule. For example, experiments show that the dipole moment of BeF₂ is zero. This is possible only if bond dipoles of two Be – F bonds cancel each other. In other words, the two bond dipoles must be oriented in opposite directions. This is possible if the molecule is linear. Similarly, water has a dipole moment of 1.83 D. Thus, its molecule cannot be linear because the bond dipoles do not cancel each other. Thus, the molecule must have an angular shape.

Dipole Moment

    It has been pointed out earlier that a covalent bond between two atoms acquires a partial polar character if the values of electro negativity of the two bonded atoms differ. The two charged ends of the bond behave as electrical dipole and the degree of polarity is measured in terms of dipole moment. Dipole moment is defined as the product of the magnitude of charge on anyone of the atoms and the distance between them.

Dipole moment is represented by a Greek letter 'µ'. It can be expressed mathematically, as:

µ = e × d

        Where, e = charge on anyone of the atoms

         d = distance between the atoms.

    Since the charge 'e'
is of the order of 10^–10 e. s. u. & 'd'
is of the order of 10^–8 cm, then 'µ' which is the product of 'd'
and 'e'
is of the order of 10^–18, e. s. u.-cm.
This unit is called Debye
and is represented by D. Thus,

                1D = 1 ´ 10^–18 e. s. u.-cm.

        For example, dipole moment of HCl is 1.03 ´ 10^–18 e. s. u.-cm and is expressed as 1.03 D.

    Dipole moment can be determined experimentally and its value can give an idea of the polar character of a molecule. It is a vector quantity as it has a direction as well as magnitude. The direction of dipole moment is usually represented by an arrow → pointing from positive end towards the negative end.

Polar and Non Polar Covalent Bond

In the H₂ or Cl₂molecule, the two electrons constituting the covalent bond are equally shared by the two identical nuclei. Due to even distribution of electrons between the two nuclei, the molecule remains neutral. Such a bond is called Non-Polar Covalent Bond. However, when two different atoms are joined by a covalent bond as in HCl, the electron pair is not shared equally.

Due to a greater attraction of one nucleus (Cl) for the electrons, the shared pair is displaced towards it. This makes one end of the bond partially positive (δ⁺) and the other partially negative (δ).

    A covalent bond in which electrons are shared unequally and the bonded atoms acquire a partial positive and negative charge, is called a Polar Covalent Bond.

    A molecule having partial positive and negative charge separated by a distance is commonly referred to as a dipole
(two poles). The dipole of a bond is indicated by an arrow from positive to negative end with a crossed tail as shown above in HCL molecule.

    Since two atoms of different atoms do not have exactly the same attraction for electrons in a bond, all bonds between unlike atoms are polar to some extent. The amount of polarity of a bond is determined by the difference of electro negativity (or tendency to attract electrons) of the two bonded atoms. The greater the difference of electro negativity between two atoms, greater the polarity. As a matter of fact, if this difference is around 1.9 & 2.9, the bond is generally ionic, meaning there by, that one atom has gained complete control of electron pair in the bond.