CBSE Class 11 Chemistry Structure Of Atom Worksheet Set 02

Structure Of Atom MCQ Questions with Answers Class 11 Chemistry

Question- The shortest wavelength in hydrogen spectrum of Lyman series when RH = 109678 cm^–1 is

(a) 1002.7Å

(b) 1215.67Å

(c) 1127.30Å

(d) 911.7Å

Answer-(d)

Question-The radius of hydrogen atom in the ground state is 0.53 Å.The radius of Li²⁺ ion (atomic number = 3) in a similar state is

(a) 0.17 Å

(b) 0.265 Å

(c) 0.53 Å

(d) 1.06 Å

Answer-(a)

Question-The energy of an electron in second Bohr orbit of hydrogen atom is :

(a) –5.44 × 10^–19eV

(b) –5.44 × 10^–19cal

(c) –5.44 × 10^–19kJ

(d) –5.44 × 10^–19J

Answer-(d)

Question-What does negative sign in the electronic energy for hydrogen atom convey.

(a) Energy of electron when n = ∞

(b) The energy of electron in the atom is lower than the energy of a free electron in motion

(c) The energy of electron in the atom is lower than the energy of a free electron of rest

(d) The energy of electron decreases as it moves away from nucleus

Answer-(c)

Question- Which of the following statements do not form a part of Bohr’s model of hydrogen atom ?

(a) Energy of the electrons in the orbits are quantized

(b) The electron in the orbit nearest the nucleus has the lowest energy

(c) Electrons revolve in different orbits around the nucleus

(d) The position and velocity of the electrons in the orbit cannot be determined simultaneously.

Answer-(d)

Question-The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097×10⁷ m^–1)

(a) 406 nm

(b) 192 nm

(c) 91 nm

(d) 9.1×10^–8 nm

Answer-(c)

Question-If the energy difference between the ground state of an atom and its excited state is 4.4 × 10^–4 J, the wavelength of photon required to produce the transition :

(a) 2.26 × 10^–12 m

(b) 1.13 × 10^–12 m

(c) 4.52 × 10^–16 m

(d) 4.52 × 10^–12 m

Answer-(d)

Question-According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon ?

(a) n = 6 to n = 1

(b) n = 5 to n = 4

(c) n = 6 to n = 5

(d) n = 5 to n = 3

Answer-(c)

Question-The mass of a photon with a wavelength equal to 1.54 × 10^–8 cm is

(a) 0.8268 × 10^–34 kg

(b) 1.2876 × 10^–33 kg

(c) 1.4285 × 10^–32 kg

(d) 1.8884 × 10^–32 kg

Answer-(c)

Question-Bohr model can explain :

(a) the solar spectrum

(b) the spectrum of hydrogen molecule

(c) spectrum of any atom or ion containing one electron only

(d) the spectrum of hydrogen atom only

Answer-(c)

Question-In a hydrogen atom, if energy of an electron in ground state is 13.6. eV, then that in the 2nd excited state is

(a) 1.51 eV

(b) 3.4 eV

(c) 6.04 eV

(d) 13.6 eV.

Answer-(a)

Question-The line spectrum of He⁺ ion will resemble that of

(a) hydrogen atom

(b) Li⁺ ion

(c) helium atom

(d) lithium atom

Answer-(a)

Question-Bohr’s theory can be applied to which of the following ions.

(a) Na⁺

(b) Be²⁺

(c) Li⁺

(d) Li²⁺

Answer-(d)

Question-If r is the radius of the first orbit, the radius of nth orbit of H-atom is given by

(a) rn²

(b) rn

(c) r/n

(d) r² n²

Answer-(a)

Question-Bohr’s model is not able to account for which of the following.

(a) Stability of atom.

(b) Spectrum of neutral helium atom.

(c) Energy of free electron at rest.

(d) Calculation of radii of the stationary states.

Answer-(b)

Question-If electron, hydrogen, helium and neon nuclei are all moving with the velocity of light, then the wavelength associated with these particles are in the order

(a) Electron > hydrogen > helium > neon

(b) Electron > helium > hydrogen > neon

(c) Electron < hydrogen < helium < neon

(d) Neon < hydrogen < helium < electron

Answer-(a)

Question- Heisenberg's uncertainity principle is applicable to

(a) atoms only

(b) electron only

(c) nucleus only

(d) any moving object

Answer-(d)

Question-In which of the following Bohr’s stationary state, the electron will be at maximum distance from the nucleus ?

(a) IInd

(b) Ist

(c) Vth

(d) IIIrd

Answer-(c)

Question-The orientation of an atomic orbital is governed by

(a) Spin quantum number

(b) Magnetic quantum number

(c) Principal quantum number

(d) Azimuthal quantum number

Answer-(b)

Q1 - Find the wavelength of radiation emitted when an electron from infinity falls to stationary state 1 in a hydrogen atom. (\(R_H = 1.09 \times 10^7 \text{ m}^{-1}\))
Answer: According to Rydberg's formula for hydrogen atom spectrum:
\(\bar{\nu} = \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\)
Here, initial state \(n_2 = \infty\), final state \(n_1 = 1\), and \(R_H = 1.09 \times 10^7 \text{ m}^{-1}\).
\(\frac{1}{\lambda} = 1.09 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right)\)
\(\frac{1}{\lambda} = 1.09 \times 10^7 \times (1 - 0) = 1.09 \times 10^7 \text{ m}^{-1}\)
\(\lambda = \frac{1}{1.09 \times 10^7 \text{ m}^{-1}} = 9.17 \times 10^{-8} \text{ m} = 91.7 \text{ nm}\)
In simple words: The wavelength is calculated using the Rydberg formula by substituting the initial and final energy levels.

Exam Tip: Always check units carefully; remember that \(1 \text{ m} = 10^9 \text{ nm}\) when converting wavelengths.

Q2 - Explain Hund's rule with the help of example of nitrogen.
Answer: Hund's rule of maximum multiplicity states that pairing of electrons in the orbitals belonging to the same subshell (p, d, or f) does not take place until each orbital belonging to that subshell has got one electron each, i.e., it is singly occupied.
Example: Nitrogen atom has an atomic number of 7. Its electronic configuration is \(1s^2 2s^2 2p^3\).
In the \(2p\) subshell, there are three degenerate orbitals (\(2p_x, 2p_y, 2p_z\)). According to Hund's rule, the three electrons in the \(2p\) subshell will occupy separate orbitals with parallel spins rather than pairing up in a single orbital.
In simple words: Fill all empty orbitals in a subshell with one electron each before pairing them up.

Exam Tip: Mention the term 'degenerate orbitals' to secure full marks in this definition.

Q3 - Derive de Broglie's relationship. What is its significance?
Answer: Derivation:
According to Planck's quantum theory, the energy of a photon is given by:
\(E = h\nu = \frac{hc}{\lambda}\) ---(1) (where \(c\) is velocity of light)
According to Einstein's mass-energy relationship:
\(E = mc^2\) ---(2)
Equating equation (1) and (2) for a photon of light:
\(\frac{hc}{\lambda} = mc^2\)
\(\frac{h}{\lambda} = mc\)
\(\lambda = \frac{h}{mc}\)
For any material particle of mass \(m\) moving with velocity \(v\), de Broglie replaced \(c\) with \(v\):
\(\lambda = \frac{h}{mv} = \frac{h}{p}\) (where \(p\) is momentum)
Significance: de Broglie's relationship establishes the dual wave-particle nature of matter. It is significant for microscopic particles like electrons, protons, etc., for which wavelengths are measurable, whereas for macroscopic objects, wavelengths are too small to be observed.
In simple words: Moving microscopic particles possess both particle and wave characteristics.

Exam Tip: Clearly state how Einstein's and Planck's equations are combined during the derivation.

Q4 - The frequency of a radiation is \(6 \times 10^{14}\) cycles / sec. Find out the wavelength of radiation in nanometer.
Answer: Given frequency \(\nu = 6 \times 10^{14} \text{ s}^{-1}\)
Speed of light \(c = 3 \times 10^8 \text{ m/s}\)
The relationship between wavelength (\(\lambda\)), frequency (\(\nu\)), and speed of light (\(c\)) is:
\(\lambda = \frac{c}{\nu}\)
\(\lambda = \frac{3 \times 10^8 \text{ m/s}}{6 \times 10^{14} \text{ s}^{-1}}\)
\(\lambda = 0.5 \times 10^{-6} \text{ m} = 5 \times 10^{-7} \text{ m}\)
Converting meters to nanometers (\(1 \text{ m} = 10^9 \text{ nm}\)):
\(\lambda = 5 \times 10^{-7} \text{ m} \times 10^9 \text{ nm/m} = 500 \text{ nm}\)
In simple words: Divide the speed of light by the frequency to find the wavelength in meters, then convert it to nanometers.

Exam Tip: Always remember the unit conversion factor between meters and nanometers (\(10^9\)).

Q5 - An isotope of atomic mass 27 has 14 neutrons in the nucleus. What is the atomic number, name and symbol of element.
Answer: Atomic mass (\(A\)) = 27
Number of neutrons (\(n\)) = 14
Atomic number (\(Z\)) = Atomic mass (\(A\)) - Number of neutrons (\(n\))
\(Z = 27 - 14 = 13\)
Atomic number = 13
Name of the element = Aluminium
Symbol of the element = Al
In simple words: Subtract the number of neutrons from the atomic mass to get the atomic number, which identifies the element.

Exam Tip: Memorize the first 20 elements of the periodic table along with their atomic numbers and symbols.

Q6 - Write the designation for orbitals with following quantum numbers-
(a) \(n = 3, l = 1\)
(b) \(n = 5, l = 2\)
Answer:
(a) For \(n = 3\) and \(l = 1\): The principal quantum number is 3, and azimuthal quantum number \(l = 1\) corresponds to the \(p\) orbital. Therefore, the orbital designation is \(3p\).
(b) For \(n = 5\) and \(l = 2\): The principal quantum number is 5, and azimuthal quantum number \(l = 2\) corresponds to the \(d\) orbital. Therefore, the orbital designation is \(5d\).
In simple words: Azimuthal quantum number values \(l = 0, 1, 2, 3\) correspond to \(s, p, d, f\) orbitals respectively.

Exam Tip: Remember the code: \(l=0 \to s\), \(l=1 \to p\), \(l=2 \to d\), \(l=3 \to f\).

Q8 - Which element does not have any neutron in it?
Answer: Ordinary hydrogen (Protium, \(_1^1\text{H}\)) does not have any neutron in its nucleus.
In simple words: Protium is the only stable nuclide with zero neutrons.

Exam Tip: Specify 'Protium' or ordinary hydrogen to be scientifically precise.

Q9 - Give two examples from everyday life where cathode ray tubes are used?
Answer: Two everyday examples where cathode ray tubes (CRTs) are used are:
1. Traditional television sets (CRT televisions).
2. Older computer monitors and oscilloscope displays.
In simple words: Old-time TVs and computer monitors use cathode ray tubes to display images.

Exam Tip: Keep examples practical and well-recognized, such as old televisions or computer monitors.