Chapter 4 Chemical Bonding and Molecular Structure Solutions
Question - 21 : - Apart from tetrahedral geometry, anotherpossible geometry for CH4 is square planar with the four H atomsat the corners of the square and the C atom at its centre. Explain why CH4 is not squareplanar?
Answer - 21 : -
Electronic configurationof carbon atom:
6C: 1s2 2s2 2p2
In the excited state, theorbital picture of carbon can be represented as:
Hence, carbon atom undergoes sp3 hybridization in CH4 molecule and takes atetrahedral shape.
For a square planar shape, the hybridizationof the central atom has to be dsp2. However, an atom ofcarbon does not have d-orbitalsto undergo dsp2 hybridization.Hence, the structure of CH4 cannot be square planar.
Moreover,with a bond angle of 90° in square planar, the stability of CH4 will be very lessbecause of the repulsion existing between the bond pairs. Hence, VSEPR theoryalso supports a tetrahedral structure for CH4.
Question - 22 : - Explain why BeH2 molecule has a zerodipole moment although the Be–H bonds are polar.
Answer - 22 : -
The Lewis structure for BeH2 is as follows:
There is no lone pair at the central atom(Be) and there are two bond pairs. Hence, BeH2 is of the type AB2. It has a linearstructure.
Dipolemoments of each H–Be bond are equal and are in opposite directions. Therefore,they nullify each other. Hence, BeH2 molecule has zerodipole moment.
Question - 23 : - Which out of NH3 and NF3 has higher dipolemoment and why?
Answer - 23 : -
In both molecules i.e., NH3 and NF3, the central atom (N) hasa lone pair electron and there are three bond pairs. Hence, both molecules havea pyramidal shape. Since fluorine is more electronegative than hydrogen, it isexpected that the net dipole moment of NF3 is greater than NH3. However, the net dipolemoment of NH3 (1.46 D) is greater than that of NF3 (0.24 D).
This can be explained on the basis of thedirections of the dipole moments of each individual bond in NF3 and NH3. These directions can beshown as:
Thus, the resultant momentof the N–H bonds add up to the bond moment of the lone pair (the two being inthe same direction), whereas that of the three N – F bonds partly cancels themoment of the lone pair.
Hence,the net dipole moment of NF3 is less than that ofNH3.
Question - 24 : - What is meant by hybridisation of atomic orbitals?Describe the shapes of sp, sp2, sp3 hybrid orbitals.
Answer - 24 : -
Hybridization is definedas 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-orbitalhybridizes with two 2p-orbitals of carbon to form three new sp2 hybrid orbitals.
These hybrid orbitals haveminimum repulsion between their electron pairs and thus, are more stable.Hybridization helps indicate the geometry of the molecule.
Shape of sp hybridorbitals: sp hybridorbitals have a linear shape. They are formed by the intermixing of s and p orbitalsas:
Shape of sp2 hybrid orbitals:
sp2 hybrid orbitals are formedas a result of the intermixing of one s-orbital and two 2p-orbitals.The hybrid orbitals are oriented in a trigonal planar arrangement as:
Shape of sp3 hybrid orbitals:
Four sp3 hybrid orbitals areformed by intermixing one s-orbital with three p-orbitals. Thefour sp3 hybrid orbitals are arranged in theform of a tetrahedron as:
Question - 25 : - Describe the change inhybridisation (if any) of the Al atom in the following reaction.
Answer - 25 : -
The valence orbitalpicture of aluminium in the ground state can be represented as:
The orbital picture ofaluminium in the excited state can be represented as:
Hence, it undergoes sp2 hybridization togive a trigonal planar arrangement (in AlCl3).
Toform AlCl4–, the empty 3pz orbital also getsinvolved and the hybridization changes from sp2 to sp3. As a result, the shapegets changed to tetrahedral.
Question - 26 : - Is there any change in thehybridisation of B and N atoms as a result of the following reaction?
BF3 + NH3 → F3B.NH3
Answer - 26 : -
Boron atom in BF3 is sp2 hybridized. Theorbital picture of boron in the excited state can be shown as:
Nitrogen atom in NH3 is sp3 hybridized. Theorbital picture of nitrogen can be represented as:
Afterthe reaction has occurred, an adduct F3B⋅NH3 is formed ashybridization of ‘B’ changes to sp3. However, thehybridization of ‘N’ remains intact.
Question - 27 : - Draw diagrams showing the formation of adouble bond and a triple bond between carbon atoms in C2H4 and C2H2 molecules.
Answer - 27 : -
C2H4 :
The electronicconfiguration of C-atom in the excited state is:
In the formation of an ethane molecule (C2H4), one sp2 hybrid orbital ofcarbon overlaps a sp2 hybridized orbitalof another carbon atom, thereby forming a C-C sigma bond.
The remaining two sp2 orbitals of eachcarbon atom form a sp2–s sigma bondwith two hydrogen atoms. The unhybridized orbital of one carbon atom undergoessidewise overlap with the orbital of a similar kind present on another carbonatom to form a weak π-bond.
C2H2 :
In the formation of C2H2 molecule, eachC–atom is sp hybridized with two 2p-orbitals in anunhybridized state.
One sp orbital of eachcarbon atom overlaps with the other along the internuclear axis forming a C–Csigma bond. The second sp orbital of each C–atom overlaps ahalf-filled 1s-orbital to form a σ bond.
The two unhybridized 2p-orbitals ofthe first carbon undergo sidewise overlap with the 2p orbital ofanother carbon atom, thereby forming two pi (π) bonds between carbon atoms.Hence, the triple bond between two carbon atoms is made up of one sigma and twoπ-bonds.
Question - 28 : - What is the total numberof sigma and pi bonds in the following molecules?
(a) C2H2 (b) C2H4
Answer - 28 : -
A single bond is a resultof the axial overlap of bonding orbitals. Hence, it contributes a sigma bond. Amultiple bond (double or triple bond) is always formed as a result of thesidewise overlap of orbitals. A pi-bond is always present in it. A triple bondis a combination of two pi-bonds and one sigma bond.
Structure of C2H2 can be representedas:
Hence, there are three sigma and twopi-bonds in C2H2.
The structure of C2H4 can be representedas:
Hence,there are five sigma bonds and one pi-bond in C2H4.
Question - 29 : - Considering x-axis as the internuclear axiswhich out of the following will not form a sigma bond and why? (a) 1s and1s (b) 1s and 2px (c) 2py and 2py (d) 1s and2s.
Answer - 29 : -
2py and 2py orbitals will not aform a sigma bond. Taking x-axis as the internuclear axis, 2py and 2py orbitals willundergo lateral overlapping, thereby forming a pi (π) bond.
Question - 30 : - Which hybrid orbitals areused by carbon atoms in the following molecules?
CH3–CH3; (b) CH3–CH=CH2; (c) CH3-CH2-OH; (d) CH3-CHO (e) CH3COOH
Answer - 30 : -
(a)
Both C1 and C2 are sp3 hybridized.
(b)
C1 is sp3 hybridized, while C2 and C3 are sp2 hybridized.
(c)
Both C1 and C2 are sp3 hybridized.
(d)
C1 is sp3 hybridized and C2 is sp2 hybridized.
(e)
C1 is sp3 hybridized and C2 is sp2 hybridized.