CH4 , NH3 , and H2O have same hybridisation but different geometries and bond angles . Explain the geometries and bond angles of all the molecules and give reason for the same .
Carbon, nitrogen and oxygen have 4, 5 and 6 valence electrons with the electronic configurations in ground and excited state as shown in the figure. They all undergo sp3 hybridization but have different numbers of lone pair of electrons, 0 in C, 1 in N and 2 in O. The lone pairs are not available for bonding and so C forms four bonds with H in CH4, N forms three with H in NH3 and O forms two bonds with H in H2O.
The arrangement of electron pairs is tetrahedral for all three molecules. The molecular geometry or shape is determined by position of bond pairs and not lone pair of electrons. In methane, there are four identical bonds which are oriented symmetrically in space in tetrahedral shape with a bond angle of 109.5o.
In case of NH3, there are only 3 bonds and they give a trigonal planar shape to the molecule.
Presence of a lone pair makes the molecule unsymmetrical. The bonding electrons are present at the center of the two atoms involved in the bonding whereas lone pairs remain closer to the central atom to which they belong. Interelectronic repulsion between a lone pair and bond pair is more than two bond pairs and this pushes the bond pairs closer to each other reducing the bond angle in ammonia to 107o.
Bond length is the sum of the atomic radii of the elements forming the bond. In methane it is carbon and hydrogen while in ammonia it is nitrogen and hydrogen. Nitrogen contains an extra proton in its nucleus than carbon and is able to attract the electrons more strongly towards itself thereby reducing the atomic radii and hence the bond length.
In water, O contains two lone pairs and geometry is determined by two bonds and hence the bent shape of the molecule. The two lone pairs of electrons repel each other and the two bond pairs and this pushes the bonds in water more closer and reduces bond angle to 104.5o.
Oxygen has an extra proton than nitrogen and this reduces the bond length due to a stronger attractive pull of the nucleus on the electrons.
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The arrangement of electron pairs is tetrahedral for all three molecules. The molecular geometry or shape is determined by position of bond pairs and not lone pair of electrons. In methane, there are four identical bonds which are oriented symmetrically in space in tetrahedral shape with a bond angle of 109.5o.
In case of NH3, there are only 3 bonds and they give a trigonal planar shape to the molecule.
Presence of a lone pair makes the molecule unsymmetrical. The bonding electrons are present at the center of the two atoms involved in the bonding whereas lone pairs remain closer to the central atom to which they belong. Interelectronic repulsion between a lone pair and bond pair is more than two bond pairs and this pushes the bond pairs closer to each other reducing the bond angle in ammonia to 107o.
Bond length is the sum of the atomic radii of the elements forming the bond. In methane it is carbon and hydrogen while in ammonia it is nitrogen and hydrogen. Nitrogen contains an extra proton in its nucleus than carbon and is able to attract the electrons more strongly towards itself thereby reducing the atomic radii and hence the bond length.
In water, O contains two lone pairs and geometry is determined by two bonds and hence the bent shape of the molecule. The two lone pairs of electrons repel each other and the two bond pairs and this pushes the bonds in water more closer and reduces bond angle to 104.5o.
Oxygen has an extra proton than nitrogen and this reduces the bond length due to a stronger attractive pull of the nucleus on the electrons.