Model Test Questions Chemistry Test # 6 for Chapter # 2 (Atomic Structure)
Short Questions
Answers
Q1. What are Limitations
of Bohr's Atomic Model?
Q2. Differentiate
between shell and sub shell with examples?
Q3.What is maximum number of electrons that can be accommodate in's'
subshell?
Q4. How many electrons will be in L shell of an atom having atomic number 11?
Q5. In the distribution of electrons of an atom, which shell filled first and
why?
Q6. If both K and L shells of an atom are
completely filled, what is the total number of electrons are present in them?
Q7. An atom has 5 electrons in M shell than:
(a) Find out its atomic number?
(b) Write Electronic configuration of atom?
(c) Name the element of atom?
Q8. Describe wave
particle duality of electron of De Broglie Hypothesis?
Long Questions Answers
Q9. Prove that modern theory of De Broglie is related with Einstein and
Plank's equations.
Q10. Describe the schrodinger
atomic model.
Q11.Explain how Bohr's
atomic model is different from Rutherford atomic model.
OR
State
postulates of Bohr’s atomic model.
Q12. Write down electronic configuration of B, F, N, Na, P, Cl, Ca, K+,
O2-, S2-, Mg2+, Cl-.
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Solutions of Questions
Q1. What are Limitations of Bohr's Atomic Model?
Answer
1. It
failed to explain the Zeeman Effect (effect of magnetic field on the spectra of
atoms).
2. It
also failed to explain the Stark effect (effect of electric field on the
spectra of atoms).
3. It
is against the Heisenberg Uncertainty Principle.
4. It
could not explain the spectra obtained from larger atoms.
5. It
explained the mono-electronic species like H+, Li2+, B3+.
Q2. Differentiate between shell and sub shell with examples?
Q8. Describe wave particle duality of electron of De Broglie Hypothesis?
Answer
In 1923 Lois
De Broglie extend the wave particle duality to electron, and propose a
hypothesis that all matter has particle as well as wave nature at the
submicroscopic level.
De Broglie
combined the Einstein and Planck equations to derive its de-Broglei’ equation.
The wave nature of a particle is quantified by De Broglie wavelength defined as
l = h/p where p is the momentum of the
particle.
All matter (material) particles
in motion have a dual character exhibiting both particle and wave nature i.e.
electrons, protons, neutrons, atoms, and molecules possess the characteristics
of both the material particle and a wave. This is called wave-particle Duality
in matter.
de-Broglie derived a
mathematical equation known as de-Broglie’s equation which relates the
wavelength (l) of
the material particle (electron) of mass m moving with velocity v to its
momentum (mv = p):
Q9. Prove that modern theory of De Broglie is related with Einstein and Plank's equations.
Answer
Q10. Describe the schrodinger atomic model.
Answer
In 1926 Erwin
Schrödinger, an Austrian physicist, took the Bohr’s atomic model one step
forward. Schrödinger used mathematical equations to describe the likelihood of
finding an electron in a certain position. This atomic model is known as the
quantum mechanical model of the atom.
The quantum mechanical model determines that electron can be find in various location around the nucleus. He found electrons are in orbit as an electron cloud.
Basic Postulates
1. The quantum mechanical model
determines that electron can be found in various location around the nucleus as an electron cloud (Electrons
in orbits are found as an electron cloud).
2. Each energy subshell in an orbit has
different shapes which determine the presence of electron.
3. Different subshells or orbitals are
named as s, p, d and f with different shapes. e.g. s-orbital is spherical and p-orbital is
dumbbell shaped.
4. The numbers and kind of atomic
orbitals depends on the energy subshell.
According to
quantum mechanical model probability of finding an electron within certain
volume of space surrounding the nucleus can be represented as a fuzzy cloud.
The cloud is denser the probability of finding
electron is high which are called atomic orbitals.
Q11. Explain how Bohr's atomic model is different from Rutherford atomic model.
OR
State postulates of Bohr’s atomic model.
Answer
Introduction
After Planck and Einstein’s discoveries,
a Danish Physicist Neil Bohr in 1913 offered a theoretical explanation of line
spectra of hydrogen atom and proposed a new model for structure of atom based
on Planck’s Quantum theory of Max Planck.
Basic Postulates
Postulate # 1
The atom has a number of fixed energy orbits or energy levels called stationary orbit’s or permissible orbits in which electron revolves around the nucleus. Bohr adopted Planck’s idea that energies are quantized. These orbits or shells have certain fixed amount of energy and are named as K, L, M, N. Electrons in atoms move only in certain allowed energy levels in which they are completely stable and will not emit or absorb energy continuously and therefore will not spiral (fall) into the nucleus.
Postulate # 2
When an electron jumps from lower energy level (E1) to higher energy level (E2), it absorbs a definite discrete quantity of energy called quantum of energy. When electron jumps from higher energy level (E2) to lower energy level (E1), it emits energy (as a light in a definite discrete quantity called quantum of energy). A quantum of energy is directly proportional to the frequency of the radiation. Since the process of emission of energy is not continuous, so a line spectrum is produced by atoms.
Postulate # 3
an electron cannot revolve in an arbitrary orbits but only those orbits are permissible (which are called allowed orbits or stationary states) in which angular momentum of electrons is an integral (whole number) multiple of h/2p.
mvr = nh/2p.
[Where; n = Principal quantum number =
number of orbits = 1, 2, 3 …., h = Planck’s constant]
Answer
Boron (5B) = 5 eˉ =
1s2, 2s2 2p1
Carbon (6C) = 6 eˉ =
1s2, 2s2 2p2
Nitrogen (7N) = 7 eˉ =
1s2, 2s2 2p3
Fluorine (9F) = 9 eˉ = 1s2,
2s2 2p5
Sodium (11Na) = 11 eˉ =
1s2, 2s2 2p6, 3s1
Phosphorus (15P)
= 15 eˉ= 1s2, 2s2 2p6, 3s2 3p3
Chlorine (17Cl) = 17 eˉ = 1s2, 2s2 2p6,
3s2 3p5
Calcium (20Ca) = 20 eˉ = 1s2, 2s2 2p6,
3s2 3p6, 4s2
K+ = 19 –1 = 18 eˉ = 1s2, 2s2 2p6, 3s2 3p6
O2− = 8 + 2 =10 eˉ =
1s2, 2s2 2p6
S2− = 16 + 2 = 18
eˉ = 1s2, 2s2 2p6,
3s2 3p6
Mg2+= 12 – 2 = 10 eˉ = 1s2, 2s2 2p6
Cl− = 17 + 1= 18
eˉ = 1s2, 2s2 2p6,
3s2 3p6
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