CBSE Gujarat Board Haryana Board
Class 10 Class 12
Discuss in brief sp3 hybridisation. Explain the formation of methane and ethane.
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).
Draw the Lewis structures for the following molecules and ions:
Lewis structure of the given molecule and ions are,
Discuss in brief sp2 hybridization (hybridization in C = C bond). Discuss the molecular orbital structure of ethylene (first member of alkene).
Or
Draw diagrams showing the formation of a double bond between carbon atoms in C2H4.
.
s-character and 
) bond. The 
bond consists of two 
electron clouds which lie above and below the plane of carbon and hydrogen atoms.
bond between carbon-carbon atom. The bond length of carbon-carbon double bond of ethylene is 134pm (1–34 A).Consider two hydrogen atoms A and B with electron eA and eB respectively. HArepresents the nucleus of hydrogen atom A and HB represents the nucleus of hydrogen atom B.
When the two hydrogen atoms approach each other, the following two forces come into existence:
(a) Attractive interactions in between:
(i) the nucleus HA an electron eB and
(ii) the nucleus HB and electron eA
(b) Repulsive interactions in between:
(i) electron eA and electron eB and
(ii) nucleus HA and nucleus HB.
Since attractive forces overpower the repulsive forces, as a result, the enthalpy of the system decreases and a molecule of hydrogen is formed.
Enthalpy diagram: When two hydrogen atoms are at an infinite distance from each other, there is no interaction between them and therefore, the enthalpy of the system is assumed to be zero in this state (stage-A). As the two atoms start coming closer to each other, the potential enthalpy continues to decrease (stage B). Ultimately a stage is reached when the enthalpy of the system becomes minimum and hydrogen atoms are said to be bonded together to form a stable H2 molecule (state C).
The internuclear distance r0 between two hydrogen atoms at this stage is referred to as bond length. In the case of the hydrogen molecule, the bond length is 74 pm. It should be noted that two hydrogen atoms can not be brought at a distance lesser than rQ (i.e. 74 pm) because the potential enthalpy of the system increases and curve shows an upward trend (dotted lines) and molecule becomes unstable.
Which hybrid orbitals are used by carbon atoms in the following molecules ?
(a) CH3 – CH3
(b) CH3 – CH = CH2
(c) CH3 – CH2 - OH
(d) CH3 - CHO
(e) CH3COOH
Discuss the shape of the following molecules using VSEPR model:
BeCl2, BCl3, SiCl4, AsF5, H2S, PH3
BeCl2:
Lewis dot structure Cl: Be : Cl. The central atom (Be) has only two bond pairs and no lone pair. Hence shape is linear.
BCl3:
The central atom (B) has only three bond pairs and no lone pair. Hence shape is triangular planar.
SiCl4:
The central atom (Si) has four bond pairs and no lone pair. Hence the shape is tetrahedral.
AsF5:
The central atoms (As) has five bond pairs and no lon∈ pair. Hence, the shape is trigonal bipyramidal.
H2S:
The central atom (S) has two bond pairs and two lone pairs. Hence, the shape is Bent or V-shaped.
PH3:
The central atom (P) has three bond pairs and two lone pairs. Hence, the shape is. Bent or V-shaped.
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