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Practice each question, then open the answer or worked solution.
Explain the non-linear shape of H2 S and the non-planar shape of PCl3 using valence shell electron pair repulsion theory.
In H2S, the Sulphur atom is surrounded by four electron pairs (two bond pairs and two lone pairs).
These four electron pairs adopt tetrahedral geometry.
The repulsion between the lone pair electrons brings distortion in the shape of the H2S.
Thus, H2S is not linear in shape.
Using molecular orbital theory, compare the bond energy and magnetic character of O2+ and O2- species.
The Molecular Orbital configuration of O2+ and O-2 is given below:
O2+ (15): σ1s2 σ *1s2 σ 2s2 σ *2s2 σ 2pz2 π2px2 = π 2py2π *2px1
O2- (17): σ1s2 σ *1s2 σ 2s2 σ *2s2 σ 2pz2 π2px2 = π 2py2π *2px2= π*2py1
Bond order for O2+ = 10-5/2 = 2.5
Bond order for O-2 = 10-7/2 = 1.5
According to Molecular Orbital Theory, the greater the bond order greater is the bond energy.
Thus, O2+ is more stable than O2-
Explain the shape of BrF5.
In BrF5, the central atom is bromine which has the hybridization sp3d2.
Br atom has 7 valence electrons, out of which 5 are used to make pair with the F atoms, and two are used to make lone pair of electrons.
The lone pair and bond pair repel each other. Thus, the shape is square pyramidal.
Structures of molecules of two compounds are given below :
(a) Which of the two compounds will have intermolecular hydrogen bonding and which compound is expected to show intramolecular hydrogen bonding.
(b) The melting point of a compound depends on, among other things, the extent of hydrogen bonding. On this basis explain which of the above two compounds will show a higher melting point.
(c) The solubility of compounds in water depends on the power to form hydrogen bonds with water. Which of the above compounds will form a hydrogen bond with water easily and be more soluble in it?
(a) Compound 1 will have intramolecular hydrogen bonding in o-nitrophenol
Compound (II) will have intermolecular hydrogen bonding in p-nitrophenol.
(b) The compound (II) has a higher melting point because of the intermolecular bonding, a large number of molecules that will get attached.
(c) The compound (II) will be more soluble in water because it will form hydrogen bonding with the water molecules easily.
Why does the type of overlap given in the following figure not result in bond
formation?
In the figure (i) the area of the contact of ++ overlap is equal to the area of the +- overlap. The so-net overlap is zero.
In figure (ii) there is no overlap of the orbitals due to the different symmetry.
Explain why PCl5 is trigonal bipyramidal whereas IF5 is square pyramidal.
In PCl5, P has 5 valence electrons in orbitals and makes 5 bonds with 5 Cl atoms, it will share one of its electrons from 3s to 3d orbital, therefore the hybridization will be sp3d and the geometry will be trigonal bipyramidal.
IF5, the Iodine atom has 7 valence electrons in molecular orbitals it will form 5 bonds with 5 Cl atoms using 5 electrons from its molecular orbital, two electrons will form one lone pair on Iodine atom, which gives the square pyramidal geometry.
In both water and dimethyl ether (CH3 —Ο — CH3 ), the oxygen atom is the central atom and has the same hybridization, yet they have different bond angles. Which one has a greater bond angle? Give reason.
Dimethyl ether will have a greater bond angle. There will be more repulsion between bond pairs of CH3 groups attached in ether than between bond pairs of hydrogen atoms attached to oxygen in the water.
Write Lewis structure of the following compounds and show a formal charge
on each atom.
HNO3, NO2, H2SO4
The formal charge is calculated by
Formal charge = ½ [total no: of bonding or shared electrons]
The formal charge on the oxygen with single bond =6-6-2/2 = -1
The formal charge on the oxygen with double bond 6-4-4/2 = 0
The formal charge on nitrogen=5-2-6/2 = 0
The formal charge on oxygen 1 and 4 = 6-4-4/2= 0
The formal charge on oxygen 2 and 3 = 6-4-4/2=0
The formal charge on hydrogen 1 and 2 = 1-0-2/2=0
The formal charge on sulfur =6-0-12/2 = 0
The energy of σ2pz molecular orbital is greater than π2px and π2py molecular orbitals in nitrogen molecule. Write the complete sequence of energy levels in the increasing order of energy in the molecule. Compare the relative stability and the magnetic behaviour of the following species :
N2, N2+ , N2- , N22+
The general sequence of the energy level of the molecular orbital is
σ1s < σ*1s < σ2s< σ*2s < π2px = π2py < σ2pz
N2 σ1s2 σ*1s2 σ2s2 σ*2s2 π2p2x = π2p2y σ2p2z
N2+ σ1s2 σ*1s2 σ2s2 σ*2s2 π2p2x = π2p2y σ2p1z
N2- σ1s2 σ*1s2 σ2s2 σ*2s2 π2p2x = π2p2y σ2p2z σ2p2x
N22+ σ1s2 σ*1s2 σ2s2 σ*2s2 π2p2x = π2p2y
Bond order = ½[electrons in BMO – electrons in ABMO]
For N2 = 10-4/2 = 3
Bond order for N2+= 9-4/2 = 2.5
Bond order for N2-= 10-5/2 = 2.5
Bond order for N22+= 8-4/2 = 2
Thus, the order of stability is:
N2> N2- > N2+> N22+
What is the effect of the following processes on the bond order in N2 and O2?
(i) N2→ N2+ + e-
(ii) O2→ O2+ + e-
(i) N2 is having 14 electrons when it donates one electron, these electrons are removed from the Bonding molecular orbital. BO for N2 = 3
(ii) O2 has 16 electrons, 8 electrons in the molecular orbitals and 4 in the antibonding molecular orbitals.
BO for O2 = 2