The Lewis structure for is as follows:
There is no lone pair at the central atom (Be) and there are two bond pairs. Hence, is of the type . It has a linear structure.
Dipole moments of each bond are equal and are in opposite directions. Therefore, they nullify each other. Hence, molecule has zero dipole moment.
Q: 23. Which out of and has higher dipole moment and why?
In both molecules i.e., and the central atom (N) has a lone pair electron and there are three bond pairs. Hence, both molecules have a pyramidal shape. Since fluorine is more electronegative than hydrogen, it is expected that the net dipole moment of is greater than However, the net dipole moment of (1.46 D) is greater than that of (0.24 D).
This can be explained on the basis of the directions of the dipole moments of each individual bond in and These directions can be shown as:
Thus, the resultant moment of the N – H bonds add up to the bond moment of the lone pair (the two being in the same direction), whereas that of the three N – F bonds partly cancels the moment of the lone pair.
Hence, the net dipole moment of is less than that of
Q: 24. What is meant by hybridization of atomic orbitals? Describe the shapes of hybrid orbitals.
Hybridization is defined as an intermixing of a set of atomic orbitals of slightly different energies, thereby forming a new set of orbitals having equivalent energies and shapes. For example, one 2s-orbital hybridizes with two 2p-orbitals of carbon to form three new hybrid orbitals.
These hybrid orbitals have minimum repulsion between their electron pairs and thus, are more stable. Hybridization helps indicate the geometry of the molecule.
Shape of sp hybrid orbitals: sp hybrid orbitals have a linear shape. They are formed by the intermixing of s and p orbitals as:
Shape of hybrid orbitals:
hybrid orbitals are formed as a result of the intermixing of one s-orbital and two 2p- orbitals. The hybrid orbitals are oriented in a trigonal planar arrangement as:
Four hybrid orbitals are formed by intermixing one s-orbital with three p-orbitals. The four hybrid orbitals are arranged in the form of a tetrahedron as:
Q: 25. Describe the change in hybridization (if any) of the Al atom in the following reaction.
The valence orbital picture of aluminium in the ground state can be represented as:
The orbital picture of aluminium in the excited state can be represented as:
Hence, it undergoes hybridization to give a trigonal planar arrangement To form the empty orbital also gets involved and the hybridization changes from As a result, the shape gets changed to tetrahedral.
Q: 26 Is there any change in the hybridisation of B and N atoms as a result of the following reaction?
Boron in hybridized. The orbital picture of boron in the excited state can be shown as:
Nitrogen atom in is hybridized. The orbital picture of nitrogen can be represented as:
After the reaction has occurred, an adduct is formed as hybridization of ‘B’ changes to However, the hybridization of ‘N’ remains intact.