C-1
Vidyasagar Classes
CHEMICAL BONDING
1
Sodium sulphate is soluble in water while barium sulphate
is not. explain
A.
In Na 2 SO 4 , q 1 × q 2 =1 × 2. While in BaSO 4 , q 1 × q 2
=2 × 2 = 4. Since lattice energy ∝ q 1 q 2 , lattice
energy of BaSO 4 is about twice that of Na 2 SO 4 . In
Na 2 SO 4 , lattice energy is less than hydration energy,
hence soluble. In BaSO 4 lattice energy > hydration
energy, hence insoluble.
2.
Anhydrous AlCl 3 is covalent from the data given
below predict whether it would remain covalent or become
ionic in aqueous solution. Ionization energy for AlCl 3 =
5137 KJ/mole. ∆H hydration for Al+3 is – 4665 kJ/mole.
∆H hudration for Cl– is – 381 kJ/mole.
Solution:
When hydration energy is greater than ionization energy,
the bond is ionic. When ionization energy is greater than
hydration energy the bond is covalent.
Hydration energy : – 4665 – 3 × 381 kJ
Ionization energy : + 5137 kJ
∴ Hydration energy + ionization energy = – 671 kJ
Hydration energy is greater than ionization energy
∴ AlCl 3 becomes ionic in aqueous solution.
3.
is?
The correct order of O – O bond length O 2 , H 2 O 2 and O 3
(A) BF 3 is symmetrical molecule. It has zero dipole moment.
Oxygen being more electronegative than S, bond moment of O –
H is more than S – H. So, the dipole moments are in the order of
BF 3 < H 2 S < H 2 O
(B) The anion size in increasing order is
Cl– < Br– < I–
Hence, LiCl is least covalent and Lil most. The order is
LiCl < LiBr < LiI
(C) Cation charge increases in the order
Na+ < Mg2+ < Al3+
Thus, Al3+ ion has maximum polarization effect and Na+
ion has least. Thus, the decreasing order of melting point is
NaCl < MgCl 2 < AlCl 3
6.
NH +4 has bond angle identical to CH 4 but NH 3 has
different bond angle; explain with proper reasoning.
Solution:
In NH +4 there are four bond pairs and no lone pair like CH 4 .
However, NH 3 has only three bond pairs and one lone pair,
hence, its bond angle is less due to lone pair-bond pair repulsion.
7.
Is there any change in the hybrid state of B and N atoms as
a result of the following reaction?
BF 3 + NH 3 → H 3 N+ – B–F 3
Solution:
During the combination of NH 3 and BF 3 . N atom is donor and B
atom of BF 3 is acceptor. The hybrid state of n in NH 3 is sp3
wheras that of B in BF 3 is sp2. In the compound H 3 N+ – B – F 3 ,
both B and N atoms is surrounded by 4 bond pairs. The hybrid
state of N remains as sp3 but that of B atom changes from sp2 to
sp3.
What is the shape and hybridization of anion of Ba(Br 3 ) 2 ?
Solution:
30.
O2 < O3 < H2O2
In O 2 double bond is present between oxygen atom.
O 3 exhibit resonance, the bond length is in between O = O and O
–O
In H 2 O 2 only O – O is present
Solution:
Ba(Br 3 ) 2
Br 3− is sp3d-hybridized with 2 bond pairs 3 lone pairs and linear.
4.
Explain why BeH 2 molecule has a zero dipole moment
although the Be – H bonds are polar.
Solution:
Dipole moment is vector quantity, due to 180° bond angle the net
dipole moment becomes zero.
5.
Arrange the following in order of increasing (A) dipole
moment: H 2 O, H 2 S, BF 3 ; (B) covalent character: LiCl, LiBr, Lil;
(C) melting point: NaCl, MgCl 2 , AlCl 3 .
Solution:
Hints to Classwork chemistry 1:Some basic concepts
We teach success
STD. XI
C-2
Vidyasagar Classes
8.
Write the electronic configuration and calculate the bond
H +2
order of
, H 2 and He 2 . Bond length in H
H 2 . Explain why?
+
2
is longer than in
Superoxide ion, O −2 :
O −2 : KK (σ2s)2(σ*2s)2(σ2p z )2(π2p x )2(π2p y )2(π*2p x )2(π*2p y )1
1
8−5
=1
2
2
Solution:
Bond order =
The number of electrons, their configuration, etc. for the given
species are
Species
No. of
Configuration
N2
Na
Bond
electrons
order =
Peroxide ion, O 22− :
1
[N b
2
– Na]
+
2
1
(σ1s)1
1
0
1
2
H2
He 2
2
4
(σ1s)2
2
2
0
2
1
0
H
* 2
(σ1s)2(σ1s)
O 22− : KK (σ2s)2(σ*2s)2(σ2p z )2(π2p x )2(π2p y )2(π*2p x )2(π*2p y )2
Bond order =
8−6
=1
2
Now, superoxide ion (O −2 ) has larger bond order than peroxide
ion (O 22− ), therefore, bond length of peroxide ion will be larger.
11.
Why is the nergy of π2p x and π2p y MOs lower than σ2p z
MO in N 2 molecule?
Solution:
The bond length in H +2 is longer than in H 2 because in H +2 only
one electron is present to shield the two nuclei from mutual
repulsion. In H 2 there are two electrons to hold the two nuclei
This is because of intermixing of 2s and 2p z orbitals because of
their close proximity. Due to intermixing σ2p z MO becomes
higher in energy than π2p x and π2p y MOs.
thus nuclear repulsion is less than that in H +2 . Hence, nuclear
separation in H +2 is more than in H 2 .
12. H 2 S has more vapour pressure than H 2 O under similar
conditions. Why?
9.
Solution:
Why He 2 does not exist?
Due to H-bonding in H 2 O.
Solution:
b
Energy
The energy level diagram for He 2 is similar to that for H 2
except that it has two more electrons. These occupy the
antibonding σ1*s , orbital, giving He 2 a bond order of zero.
That is, the two electrons in the bonding orbital of He 2
would be more stable than in the separate atoms. But the
two electrons in the antibonding orbital would be less stable
than in the separate atoms. These effects cancel, so the
molecule would be no more stable than the separate atoms.
The bond order is zero, and the molecule would not exist.
In fact, He 2 is unknown.
σ 1*s
1s
He
AO
10.
σ1s
He2
MO
1s
He
AO
Which of the two peroxide ion or superoxide ion has larger
bond length?
Solution:
The bond length of a molecule is related to its bond order. The
larger the bond order, the smaller will be the bond length.
Hints to Classwork chemistry 1:Some basic concepts
We teach success
STD. XI