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Discuss in brief sp3 hybridisation. Explain the formation of methane and ethane.


sp3 hybridisation: Electronic configuration of carbon (Z = 6) in the excited state is 1s2 2s1

This type of hybridization involves the mixing of all four filled orbitals i.e. 1s and 3p orbitals to form four new orbitals called sp3 hybrid orbitals of equivalent enthalpies and identical shapes. 


Fig. Representation of sp3 hybridization:
A single sp3 hybrid orbital.
The four sp3 hybrid orbitals are directed towards the four corners of a tetrahedral. The angle between two adjacent sp3 hybrid orbitals. Each sp3 hybrid orbital has 1/4 s-character and 3/4 p-character. sp3 hybridization is also known as tetrahedral hybridisation.

(i) The molecular orbital structure of methane: In methane molecule, carbon atom undergoes sp3 hybridisation. Each sp3 hybrid orbital overlaps with 1s orbital of hydrogen atom along the internuclear axis to form four 
σ bond.

Molecular orbital structure of methane

The four C – H bonds are directed towards the four corners of a regular tetrahedron. So methane has a tetrahedral structure. Each H – C – H bond angle is of 109°.28’. Each C – H bond length is 109 pm (1.09 Å).

(ii) Molecular orbital picture of ethane: In ethane molecule, both carbon atoms are in the sp3 hybrid state. In its formation, one hybrid orbital of one carbon atom overlaps with one sp3 hybrid orbital of a second carbon atom along the internuclear axis to form a sigma (σ) C – C bond. The remaining three sp3 hybrid orbitals of each carbon atom overlap with 1 s orbital of hydrogen atom axially to form six sigma C – H bonds.


     

The length of C-C bond in ethane is 154 pm (or 1- 54 Å) and that of each C - H bond is 109 pm (or 1-09 Å).


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With the help of VSEPR theory, explain the shape of: (i) NH3    (ii) H2O.


(i) NHMolecule: In ammonia molecule, the central nitrogen atom has five electrons in the valence shell. Three of these electrons are mutually shared with the electrons of three hydrogen atoms to form three N- H bonds as shown. 



Hence, the central N atom in NH
3 is surrounded by three bond pairs and one lone pair. The geometry expected for the molecule is tetrahedral since lone pair-bond pair repulsion is more than bond pair-bond pair repulsion. As a result, the lone pair of electrons will repel another pair strongly. Therefore three N–H bonds of NH3 are forced slightly closer.
This leads to decrease in H – N – H bond angles from a normal angle of a tetrahedron (109.5°) to 107°. The most favourable arrangement is distorted tetrahedral i.e. pyramidal. In this, nitrogen atom lies at the centre and three hydrogen atoms occupying the triangular base and the orbital with a lone pair of electrons from the apex of the pyramid.

(ii) H2
O: In a water molecule, the central oxygen atom  has six electrons in the valence shell. Two of these electrons are mutually shared with the electrons of two hydrogen atoms to form two O - H bonds. 


Hence, the central oxygen atom is 
H2O is surrounded by two bond pairs and two lone pairs. These four electron pairs adopt tetrahedral arrangement. The presence of two lone pairs brings



distortion in the geometry of the molecule. The lone pairs repel the bond pairs more effectively resulting in the decrease of H – O – H angle from 109.5° to 104.5°. The water molecule may be regarded as bent or angular or V-shaped.

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On the basis of VSEPR theory, discuss the geometry of the following covalent molecules: (i) BeF(ii) BF(iii) CH4.


(i) BeF2 : The electronic configuration of Be (Z = 4) is1s22s2. The central atom Be has 2 valence electrons. These two electrons are shared mutually with the electrons of two fluorine atoms to form two Be – F bonds.


To have a minimum repulsion in the electron pairs around the central atom, the geometry of the molecules is regular as well as linear. Bond angle in such cases is 180°.



(iii) BF3: The electronic configuration of B (Z = 5) is 1s2 2s2 2p1x. The central atom B has 3 valence electrons. These three valence electrons are shared mutually with the electrons of the fluorine atoms to form three B - F bonds.



Thus, B atom is surrounded by three bond pairs. These repel each other and go as far apart as possible so that there are no further repulsions. This is so if the three electron pairs are placed at 120° with respect to each other i.e. the most favourable arrangement is the triangular planar geometry.



(iii) CH4 The electronic configuration of C(Z =6) is 
The central atom has 4 valence electrons. These four valence electrons are shared mutually with the electrons of four hydrogen atoms to form four C - H bonds as shown.



Thus carbon atom is surrounded by four bond pairs. In order to minimise the inter-electron pair repulsions i.e. having a state of minimum enthalpy and hence maximum stability, these bond pairs should be as far apart from one another as possible. This is so if the four electron pairs are placed at an angle of 109.5° with respect to each other i.e. the most favourable arrangement is tetrahedral geometry. In other words, carbon is present in the centre of a regular tetrahedron and four bond pairs are directed to the four corners with bond angle-critical to 109.5°.



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On the basis of VSEPR theory, predict the shapes of given molecules: PCl5 and SF6.


(i) PCl5 molecule. The electronic configuration of central phosphorus atom is

It has five valence electrons. All the five electrons are mutually shared with the electrons of five chlorine atoms to form five P - Cl bonds as shown.



Hence, P atom is surrounded by five shared pairs of electrons. These repel each other and take up such positions and there is no further repulsion between them. The most favourable arrangement is trigonal bipyramidal.



(ii)  SF6 molecule: The electronic configuration of central sulphur atom is

It has six valence electrons. 
All the six valence electrons are mutually shared by the electrons of six fluorine atoms to form six S – F bonds.



Hence, S atom is surrounded by six shared pairs of electrons (six bond pairs). These repel each other and try to remain as far apart as possible so that there is no further repulsion between them. Under such conditions, the most favourable arrangement is octahedral.

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State and explain hybridisation.


The phenomenon of intermixing of atomic orbitals of slightly different enthalpies of an atom so as to redistribute their enthalpies to form the same number of new orbitals of equivalent enthalpies and identical shapes is called hybridization. The new orbitals, thus formed are called hybrid orbitals or hybridised orbitals.

Explanation : In order to understand hybridisation, let us take an example of carbon (Z= 6). Its ground state electronic configuration is,




Since it has two half filled orbitals, therefore, the valency of the carbon atom should be 2. But actually, carbon atom always exhibits a valency of four (tetravalent). To achieve this, an electron is promoted from 2s filled orbital to the vacant higher enthalpy 2p orbital. This is called excited state of a carbon atom.




In the excited state of carbon s and p, orbitals have different enthalpies. Consequently, four bonds of carbon must be of two types. Three of the bonds should be of one type (s - p bonds) while fourth bond should be a different type (s - s bond). However, experimental evidence indicates that all the four bonds in case of CH4 (methane) are equivalent. To explain the equivalence of all the four bonds in case of methane, the concept of hybridisation is used i.e. all the four orbitals in the valence shell of carbon may get mixed, redistribute enthalpies and give orbitals of new enthalpy and shape. These equivalent orbitals are called hybrid orbitals.

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