Theories of Covalent Bond and Shapes of Molecules Chemistry Class 11 notes 2021 for kpk, fbise, Punajb and sindh board.
Chemistry Notes Cass 11 Chapter 4 for kpk 2021
Q.3) CO2 molecule has linear but H2O molecule has angular structure.
Molecular geometry means that lone pairs repel ligands out, changing the geometry of the molecule. As there are no lone pairs on the carbon in CO2, so the CO2 molecule stays unbent.
Water, on the other hand, has two solo pairs on the central oxygen atom. Since the only pairs repel each other, the O-H bonds are turned away and as a effect , an angular structure is formed.
Q.4) Explain hybridization with reference to sp3, sp2 and sp modes of hybridization.
“The process in which the orbitals having different energies and different shapes intermix to form new hybrid orbitals of the same energy and same shapes are called hybridization”. Hybridization leads to the entirely new shape and orientation of the valence orbitals of an atom. It holds significant importance in determining the shape and geometry of molecules.
Depending upon the number and nature of the orbitals participating in hybridization, a different type of hybridization take place. sp3, sp2 and sp modes of hybridization are discussed below.
“The mixing of one s and three p orbitals to form four equivalent sp3 hybrid orbitals of same energy and shape is called sp3 hybridization”. For example, the carbon atom is sp3 hybridized in CH4 molecule.
In methane CH4, carbon atom forms four identical bonds by using four equivalent hybrid orbitals. These orbitals are the result of sp3 hybridization i.e. one s and three p orbitals are mixed to form four sp3 hybrid orbitals. Each hybrid orbital is composed of 25% s and 75% p character. The angle between any two such hybrid orbitals is 109.5o.
Each sp3 hybrid orbital overlaps with 1s orbital of hydrogen atom to form four C-H bonds.
“The mixing of one s and two p orbitals to form three equivalent sp2 hybrid orbitals is known as sp2 hybridization”.
There are four half-filled orbitals in each carbon atom in ethene. Three of them are hybridized while the fourth one is unhybridized which forms Pi bond between two carbon atoms. In sp2 hybridization one s and two p orbitals of carbon mixed to form three sp2 hybrid orbitals. Each sp2 orbital has 33% s character and 66% p character. These three sp2 hybrid orbitals are in one plane at 120o angle. One of three sp2 orbitals of each carbon overlap each other to form C-C bond. The other two sp2 orbitals of each carbon overlap with 1s orbital of two hydrogen atoms to form two C-H bonds and the unhybridized p orbitals overlap side wise to form π bond as shown in figure below.
“When one s and one p orbitals combine to give two hybrid orbitals of the same energy and shape, it is known as sp hybridization”.
In acetylene carbon atom forms two identical hybrid orbitals by mixing one s and one p at an angle of 180o. Two p orbitals remain unhybridized. Each sp orbital has 50% s and 50% p character. One sp orbital of each carbon overlaps each other to form C-C bond and other sp hybrid orbitals of each carbon overlap with overlap with 1s orbital of hydrogen to form C-H bond while the two unhybridized “p” orbitals overlap sidewise to form two π bonds as shown in the figure below.
Q.5) Define the dipole moment. How does it help to find out the geometry of a molecule?
The product of the magnitude of charges and the distance between them is called a dipole moment. The polarity of a molecule is expressed in terms of its dipole moment. Mathematically it is expressed as:
μ= q x r
Where q is the charge and r is the distance between the charges.
Geometry of Molecules
Dipole moment can provide important information about the geometry of molecules. If there are two or more possible geometries for a molecule, the correct one can be identified from the study of its dipole moment e.g. water.
Water has two possible structures, linear and angular.
In structure (a) the bond moments of two O-H bonds are equal and in opposite direction. They will cancel each other. The net dipole moment will be zero. In structure (b) the bond moments are added vectorially and gave net dipole moment. Experimental data shows that water has 1.84D dipole moment. This shows that water has an angular structure (b) and not a linear one (a).
CO has a dipole moment while CO2 does not have any dipole moment. It suggests that CO2 has a linear structure where dipoles being equal and opposite, cancel out the effect of each other.
Q.6) Explain the bonding in the following molecules with the help of molecular orbital theory.
i) Bonding in He2
Helium atoms have the electronic configuration 1s2. For the formation of He2 molecule, the molecular orbitals required are σ(1s) and σ*(1s) which have to accommodate four electrons. Two electrons go to the lower energy bonding molecular orbital σ(1s) and two electrons go to the higher energy bonding molecular orbital σ*(1s). The molecule of helium is represented by the equation.
2He (1s2) → He2 [σ(1s2), σ*(1s2)]
There are two electrons in each of the bonding and antibonding orbitals giving zero bond order i.e. this molecule does not exist.
Bond order=1/2(No. of electrons in bonding molecular orbitals – No. of electros in antibonding molecular orbitals)=1/2(1-1)=0
ii) Bonding in H2
Hydrogen molecule is formed by the overlap of two 1s atomic orbitals of two hydrogen atoms. They give two molecular orbitals σ(1s) and σ*(1s). The molecule has electrons to be filled in these two molecular orbitals according to Aufbau principle. Both electrons go to the lower energy bonding molecular orbital and antibonding molecular orbital remains vacant. The electronic configuration of the molecule is represented by the equation.
2H (1s1) → H2[σ(1s2), σ*(1so)]
Bond order=1/2(No. of electrons in bonding molecular orbitals-No. of electrons in antibonding molecular orbitals)=1/2(2-0)=1
Bond order suggests that there will be a single covalent bond between two hydrogen atoms.
iii) Bonding in O2
Each oxygen atom contributes six electrons to O2 molecule from its valence shell. The two participating oxygen atoms (1s2, 2s2, 2px2, 2py1, 2pz1) contribute a total of 12 valence electrons. We have to fill 12 electrons in eight molecular orbitals. Eight electrons are placed in bonding molecular orbitals while four electrons are placed in anti-bonding molecular orbitals.
Π*2py and Π*2pz are half filled according to Hund’s rule and have unpaired electrons. The presence of these two unpaired electrons imparts paramagnetic property to oxygen molecule. Both VSEPR and VBT fail to show unpaired electrons in oxygen molecule. This is the greatest success of the MOT. The molecule of O2 is represented by the equation.
2O (1s2, 2s2, 2p4) → O2 [σ(2s2), σ*(2s2), σ2px2, Π2py2, Π2pz2, Π*2py1, Π*2pz1, Π*2px0]
Bond order=1/2(No. of electrons in BMOs–No of electrons in ABMOs)=1/2(8-4)=2
Thus two bonds are formed between oxygen atoms in the molecule of oxygen.