Multiple Choice QuestionsDuring the change of O2 to O2- ion, the electron adds on which one of the following orbitals?
π* orbital
π orbital
σ* orbital
σ orbital
A.
π* orbital
During the change of O2 to O2-the electron adds on π* orbital. Molecular orbital configuration of O2 (total e- =16)
Given the reaction between two gases represented by A2 and B2 to give the compound AB (g).
A2(g) +B2 (g) ⇌ 2AB (g)
At equilibrium the concentration
of A2 = 3.0 x 10-3 M
of B2 = 4.2 x 10-3 M
of AB = 2.8 x 10-3 M
If the reaction takes place in a sealed vessel at 527oC, then the value of Kc will be
2.0
1.9
0.62
4.5
C.
0.62
A2 (g) +B2 (g) ⇌ 2AB (g)
The equilibrium constant is given by
Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them.
NO< O2-<C22- < He2+
O2- <NO< C22- < He2+
C22- < He2+ <O2- <NO
< He2+ < O2- <NO< C22-
D.
< He2+ < O2- <NO< C22-
Equal volumes of two monoatomic gases, A and B, at same temperature and pressure are mixed. The ratio of specific heats (Cp/Cv) of the mixture will be
0.83
1.50
3.3
1.67
D.
1.67
A certain gas takes three times as long to effuse out as helium. Its molecular mass will be
27 u
36 u
64 u
9 u
B.
36 u
From Graham's diffusion law,
In the replacement reaction
The reaction will be most favourable if M happens to be
Na
K
Rb
Li
C.
Rb
For the reaction,
The reaction will be faster with Rb because lattice energy of RbF is less than LiF, NaF, KF.
For real gases van der Waal's equation is written as,
where, a and b are van der Waal's constants.
Two sets of gases are
I. O2, CO2, H2 and He
II. CH4, O2, and H2
The gases given in set- I increasing order of 'b' and gases given in set -II in decreasing order of 'a' are arranged below. Select the correct order from the following.
(I) He < H2 < CO2 < O2
(II) CH4> H2> O2
(I) O2 < He < H2 < CO2
(II) H2> O2 > CH4
(I) O2 < He < O2 < CO2
(II) CH4 > O2 > H2
(I) H2 > O2 <He < CO2
(II) O2 > CH4 > H2
C.
(I) O2 < He < O2 < CO2
(II) CH4 > O2 > H2
From experimental values of a and b, sequence of Van der Waals constant (b) is
|
Gas |
H2 |
He |
O2 |
CO2 |
|
Value of b |
0.0273 |
0.0318 |
0.0273 |
0.0427 |
Hence correct order is
H2 < He < O2 < CO2
And sequence of van der Waals constant (a) is
|
Gas |
CH4 |
O2 |
H2 |
|
Value of a |
2.25 |
1.36 |
0.244 |
Hence, correct order is
(II) CH4 >O2> H2
Give that the equilibrium constant for the reaction,
2SO2 (g) + O2 (g) ⇌ 2SO3 (g)
has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction t the same temperature?
SO3 (g) ⇌ SO2 (g) +1/2 O2 (g)
1.8 x 10-3
3.6 x 10-3
6.0 x 10-2
1.3 x 10-5
C.
6.0 x 10-2
2SO2 (g) +O2 (g) ⇌ 2SO3 (g)
Equilibrium constant for this reaction,
A structure of a mixed oxide is cubic close packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide is
ABO2
A2BO2
A2B3O4
AB2O2
D.
AB2O2
According to ccp,
Number of O2- ions = 4
So, tetrahedral void = 8
and octahedral void = 4
Since A ions occupied 1/4th of the tetrahedral void.
Therefore,
Number of A ions = 1/4 x 8 = 2
Again, B ions occupied all octahedral void.
Therefore, Number of B ions = 4
A: B:O = 2:4:4
= 1:2:2
Structure of oxide= AB2O2
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