log1/2-log5/6+log3/4-log7/8( convert it into single logarithm)
Answers
Answer:
(b) 1
(c) 2
(d) 3
(e) 4
8.
Which one of the following statements is false?
(a) Valence bond theory and molecular orbital theory can be described as two different views of the same thing.
(b) When one considers the molecular orbitals resulting from the overlap of any two specific atomic orbitals, the bonding orbitals are always lower in energy than the antibonding orbitals.
(c) Molecular orbitals are generally described as being more delocalized than hybridized atomic orbitals.
(d) One of the shortcomings of molecular orbital theory is its inability to account for a triple bond in the nitrogen molecule, N2.
(e) One of the shortcomings of valence bond theory is its inability to account for the paramagnetism of the oxygen molecule, O2.
9.
Antibonding molecular orbitals are produced by
(a) constructive interaction of atomic orbitals.
(b) destructive interaction of atomic orbitals.
(c) the overlap of the atomic orbitals of two negative ions
(d) all of these
(e) none of these
10.
Which statement regarding stable heteronuclear diatomic molecules is false?
(a) All have bond orders greater than zero.
(b) The antibonding molecular orbitals have more of the character of the more electropositive element than of the more electronegative element.
(c) Their molecular orbital diagrams are more symmetrical than those of homonuclear diatomic molecules.
(d) The bonding molecular orbitals have more of the character of the more electronegative element than of the less electronegative element.
(e) The greater is the difference in energy between two overlapping atomic orbitals, the more polar the resulting bond is, due to electrons occupying the resulting bonding molecular orbital.
Answers:
1. (a) 2. (a) 3. (c) 4. (e) 5. (b) 6. (d) 7. (c) 8. (d) 9. (b) 10. (c)The frequency
ν
ν of a characteristic X-ray of an element is related to its atomic number
Z
Z by
√
ν
=
a
(
Z
−
b
)
,
ν=a(Z−b),
where
a
a and
b
b are constants called proportionality and screening (or shielding) constants. For
K
K series, the value of
a
a is
√
3
R
c
/
4
3Rc/4 and that of
b
b is 1. Here
R
R is Rydberg's constant and
c
c is speed of light (as in Bohr's model). For
L
L series, the value of
a
a is
√
5
R
c
/
36
5Rc/36 and
b
b is 7.4. The relation and values of
a
a and
b
b are experimentally determined by Henry Moseley.
Moseley's Law Formula.
Relation between the frequency of characteristic x-ray and the atomic number Z. The line intersect the Z axis at
Z
=
b
Z=b (b is 1 for K series and it is 7.4 for L series).
Solved Problems on Moseley's Law
Problem from IIT JEE 2003
Characteristic X-rays of frequency
4.2
×
10
18
4.2×1018 Hz are produced when transitions from
L
L-shell to
K
K-shell take place in a certain target material. Use Moseley's law to determine the atomic number of the target material. (Rydberg's constant
=
1.1
×
10
7
m
−
1
=1.1×107m−1.)
Solution:
The characteristic X-ray is emitted when an electron in
L
L shell makes a transition to the vacant state in
K
K shell. In Moseley's equation,
√
ν
=
a
(
Z
−
b
)
,
ν=a(Z−b),
the parameter
b
≈
1
b≈1 for this transition because electron from
L
L shell finds nuclear charge
Z
e
Ze shielded by remaining one electron in
K
K shell i.e., effective nuclear charge is
(
Z
−
1
)
e
(Z−1)e. Thus, by substituting values,
1
λ
=
ν
c
=
4.2
×
10
18
3
×
10
8
=
R
(
Z
−
1
)
2
[
1
n
2
1
−
1
n
2
2
]
=
1.1
×
10
7
(
Z
−
1
)
2
[
1
1
2
−
1
2
2
]
,
1λ=νc=4.2×10183×108=R(Z−1)2[1n12−1n22]=1.1×107(Z−1)2[112−122],
which gives,
Z
=
42
Z=42. Moseley's law played key role in arrangement of elements in the periodic table and to find many new (missing) elements.
Problem from IIT JEE 2014
If
λ
C
u
λCu is the wavelength of
K
α
Kα X-ray line of copper (atomic number 29) and
λ
M
o
λMo is the wavelength of the
K
α
Kα X-ray line of molybdenum (atomic number 42), the the ratio
λ
C
u
/
λ
M
o
λCu/λMo is close to
1.99
2.14
0.50
0.48
Solution:
The wavelength of
K
α
Kα X-ray line is related to atomic number
Z
Z by Moseley's Formula
1
λ
=
R
(
Z
−
1
)
2
[
1
1
2
−
1
2
2
]
=
3
4
R
(
Z
−
1
)
2
.
1λ=R(Z−1)2[112−122]=34R(Z−1)2.
Substitute the value of
Z
Z to get
λ
C
u
λ
M
o
=
(
Z
M
o
−
1
)
2
(
Z
C
u
−
1
)
2
=
(
41
)
2
(
28
)
2
=
2.14.
λCuλMo=(ZMo−1)2(ZCu−1)2=(41)2(28)2=2.14.
The elements with higher atomic number (molybdenum in this example) gives high energy X-rays (short wavelengths).
Problem from IIT JEE 2008
Which of the following statements is wrong in the context of X-rays generated from a X-ray tube?
Wavelength of characteristic X-rays decreases when the atomic number of the target increases.
Cut-off wavelength of the continuous X-rays depends on the atomic number of the target.
Intensity of the characteristic X-rays depends on the electrical power given to the X-ray tube.
Cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the X-ray tube.
Solution:
The frequency
ν
ν of characteristic X-rays is related to atomic number
Z
Z by Moseley's law,
√
ν
=di
Answer:
Step-by-step explanation:
By change of base.
--> Log3/log4 × Log4/log5 ×log5/log6 × log6/log7 × log7/log8 × log8/log9 × log9/log9
We get:
Log3/log9
Log 3 / log 3^2
Log3 /2 log 3 = 1/2
=1/2