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Chemical Bonding and Molecular Structure

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Chemistry Part I

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Chemistry

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Class 10 Class 12

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
2 straight p subscript straight x superscript 1 space 2 straight p subscript straight y superscript 1 space 2 straight p subscript straight z superscript 1.
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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Draw the Lewis structures for the following molecules and ions:
straight H subscript 2 straight S comma space space SiCl subscript 4 comma space space BeF subscript 2 comma space space CO subscript 3 superscript 2 minus end superscript comma space space HCOOH


Lewis structure of the given molecule and ions are,

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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.

 


sp2 hybridisation. The electronic configuration of carbon (Z = 6) in the excited state is
1 straight s squared space 2 straight s to the power of 1 space 2 straight p subscript straight x superscript 1 space space 2 straight p subscript straight y superscript 1 space 2 straight p subscript straight z superscript 1.
In this type of hybridization one- s and two P-orbitals of the valence shell of carbon atom take part in hybridization go give three new sp2 hybrid orbitals. These sp2 hybrid orbitals lie in a plane and are directed towards the corners of an equilateral triangle with a carbon atom in the centre. The unhybridized 2p1 orbital lies perpendicular to the three hybridised orbitals.



Representation of sp2 hybridization sp2 hybridization is also known as trigonal hybridisation. Each sp1 hybrid orbital has 1 third s-character and 2 over 3 space straight p minus character.
The molecular orbital structure of ethylene: In ethene molecule, each carbon atom undergoes sp2 hybridisation. One sp2 hybrid orbital of one carbon atom overlaps axially with one sp2 hybrid orbital of the other carbon atom to form sigma (σ) C - C bond. The other two sp2 hybrid orbitals of each carbon atom overlap axially with its orbital of the hydrogen atom to form sigma (σ) C - H bonds. The unhybridized p-orbitals of the two carbon atoms overlap sidewise with each other to form weak pi (straight pi) bond. The straight pi bond consists of two straight pi electron clouds which lie above and below the plane of carbon and hydrogen atoms.


(a) Formation of ethylene     (b) Molecular orbital structure molecule of ethylene

Thus, ethylene molecule consists of four sigma C – H bonds, one sigma C - C bond and one straight pi bond between carbon-carbon atom. The bond length of carbon-carbon double bond of ethylene is 134pm (1–34 A).
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Explain the formation of H2 molecule on basis of valence bond theory.
Or
In the light of attractive and repulsive forces, show that a molecule of hydrogen is formed.

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 r
0 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.

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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



Both carbon atoms use sp3 hybrid orbitals. 




Both carbon atoms (C1, C2) use a sp3 hybrid orbital.

(d) CH3CHO


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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.



H
2S:

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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