Home/Past Papers/Sindh / DUHS/Sindh Mdcat Exclusive Course Chemistry Atomic Structure

Sindh Mdcat Exclusive Course Chemistry Atomic Structure — Solved Past Paper with Answers

All 20 MCQs from Sindh Mdcat Exclusive Course Chemistry Atomic Structure, solved with the correct answer highlighted and a full explanation for every question. This is a free MDCAT Sindh / DUHS past paper — no signup, no ads. Practise it interactively in timed mode, drill more with free MDCAT MCQs, or browse all Sindh / DUHS papers.

Attempt this paper interactively →

Q1. The presence of several fine lines in a line spectrum shows the presence of:

  • A. Shells
  • B. Energy levels
  • C. Sub shells
  • D. All of these options are correct

Explanation: A. Shells (Incorrect): Fine lines in a line spectrum are indicative of transitions between energy levels within an atom, not the presence of shells, which represent broad electron distribution regions. B. Energy levels (Incorrect): Fine lines in a line spectrum indeed indicate the presence of energy levels within an atom. However, it does not directly show the presence of sub-shells, which are further divisions within energy levels. C. (correct) It is because of sub energy levels.Previously the energy of sub energy level was considered to be equal but it actually differs by about 10−4 ev. Hence fine spectral lines are produced.D. This option is not correct.

Why the other options are wrong
  • A. Shells (Incorrect): Fine lines in a line spectrum are indicative of transitions between energy levels within an atom, not the presence of shells, which represent broad electron distribution regions.
  • B. Energy levels (Incorrect): Fine lines in a line spectrum indeed indicate the presence of energy levels within an atom. However, it does not directly show the presence of sub-shells, which are further divisions within energy levels.
  • D. This option is not correct.

Q2. Which of the following does not show variable valency?

  • A. Mn
  • B. Fe
  • C. Zn
  • D. Cr

Explanation: The electronic configuration of zinc is [Ar]3d10 4s2 It can only lose two electrons from the 4s orbital and show +2 oxidation state. It does not show variable valency. However, all other transition metals like Mn, Fe, Co have partially filled d orbitals and hence, show variable valency.

Why the other options are wrong
  • A. Mn (Manganese) and Fe (Iron) are transition metals that can exhibit variable valency. For example, manganese can have a valency of +2, +3, +4, +5, +6 or +7, while iron can have a valency of +2 or +3.
  • B. Mn (Manganese) and Fe (Iron) are transition metals that can exhibit variable valency. For example, manganese can have a valency of +2, +3, +4, +5, +6 or +7, while iron can have a valency of +2 or +3.

Q3. Name the orbital n = 5; l = 2; m = 0.

  • A. 5p
  • B. 5d
  • C. 5s
  • D. 5f

Explanation: Quantum numbers are the basis of the sequence, arrangement, shape, and orientation of the elements in periodic table. N = 5 indicates the 5th energy level (5th period of the periodic table or the 5th principal quantum number) L = 2 indicates the shape of the orbital (2 corresponds d-orbital-shaped quantum number) M = 0 indicates the orientation of the orbital (z - axis magnetic quantum number)

Q4. Which of the following shows the Planck's quantum theory?

  • A. E =hcv
  • B. E = c/v
  • C. E = hv
  • D. None of these options are correct

Explanation: In chemistry, Frequency is denoted by vSo, the Planck's quantum theory isE = hv where h is the Planck's constantIt can be rewritten asE = hc/λ since frequency(v) = c/λ where c is the speed and λ is the wavelength.

Why the other options are wrong
  • A. In chemistry, Frequency is denoted by vSo, the Planck's quantum theory isE = hv where h is the Planck's constantIt can be rewritten asE = hc/λ since frequency(v) = c/λ where c is the speed and λ is the wavelength.
  • B. In chemistry, Frequency is denoted by vSo, the Planck's quantum theory isE = hv where h is the Planck's constantIt can be rewritten asE = hc/λ since frequency(v) = c/λ where c is the speed and λ is the wavelength.
  • D. In chemistry, Frequency is denoted by vSo, the Planck's quantum theory isE = hv where h is the Planck's constantIt can be rewritten asE = hc/λ since frequency(v) = c/λ where c is the speed and λ is the wavelength.

Q5. The mass of proton is _ times heavier than _.

  • A. 1889, electrons
  • B. 1886, electrons
  • C. 1836, protons
  • D. 1836, electrons

Explanation: Remember that the mass of a proton is 1836 times heavier than that of an electron.

Why the other options are wrong
  • A. This option is incorrect.
  • B. This option is incorrect.
  • C. This option is incorrect.

Q6. The shielding effect of inner electrons is responsible for:

  • A. Decreasing ionization energy
  • B. Having no effect on ionization energy
  • C. Increasing ionization energy
  • D. Increasing electronegativity

Explanation: The shielding effect of inner electrons causes a decrease in attraction between the nucleus and the outer electrons of the atom. Hence, it is easier to remove the outer electrons, which decreases the ionization potential. Therefore, options B and C are incorrect Option D is incorrect because 'Electronegativity' is a measure of an atom's ability to attract shared electrons to itself, and due to the decrease in attraction between the nucleus and the outer electrons, electronegativity will also decrease.

Q7. Maximum number of electrons in a subshell is given by:

  • A. 2 (2l - 1)
  • B. 2 (2l + 1)
  • C. 2(l + 1)
  • D. 2l + 1

Explanation: l = azimuthal quantum numberl values: s=0 , p=1, d=2, f=3Number of orbitals in a subshell: 2l+1You can check this for the 1st shell: 2(0) +1= 1Number of electrons in a shell = 2 x number of orbitals (as each orbital has a maximum of 2 electrons)Therefore,2(2l+1)

Q8. Orbitals having the same energy are called:

  • A. Hybrid orbitals
  • B. Degenerate orbitals
  • C. Valence orbitals
  • D. Sub-orbitals

Explanation: Degenerate orbitals are the orbitals of the same subshell having equal energies. When the electrons of the orbitals are not influenced by any external factors like an electric field or magnetic field, they have the same energies.

Why the other options are wrong
  • A. In chemistry, orbital hybridisation is the concept of mixing atomic orbitals to form new hybrid orbitals suitable for the pairing of electrons to form chemical bonds in valence bond theory.
  • C. The valence shell is the set of orbitals which are energetically accessible for accepting electrons to form chemical bonds. For main-group elements, the valence shell consists of the ns and np orbitals in the outermost electron shell.
  • D. Sub orbitals are the orbitals with the S P D and F shells that have their own unique quantum numbers and spin patterns when electrons are treated as waves.

Q9. Quantum number values for 2p orbitals are:

  • A. n=2,l = 1
  • B. n=2, l= 2
  • C. n=2, l=0
  • D. n=1, l = 0

Explanation: There are two types of quantum numbers: azimuthal quantum number(l) and principal quantum number(n). The principal quantum number,n, is actually the number of shells. Azimuthal quantum number, l, is actually the number of subshells. According to n-1 rule for the azimuthal quantum number, l will be equal to 1. “2p”, where 2 represents the “n=2” and p represents the second subshell hence “l=1”.

Why the other options are wrong
  • B. It represents 2d orbital.
  • C. It represents 2s orbital.
  • D. It represents 1s orbital.

Q10. Which of the following is the electronic configuration of Potassium(K) Atomic number= 19K?

  • A. 1s2, 2s2, 2p6, 3s2, 3p6, 4s2
  • B. 1s2, 2s2, 2p6, 3s2, 3p6, 3s2, 3d1
  • C. 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d5
  • D. 1s2, 2s2, 2p6, 3s2, 3p6, 4s1
  • E. 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10

Explanation: Option D is the correct option because this is the only option that has 19 electrons by adding manually.

Why the other options are wrong
  • A. It is incorrect as the electronic configuration is wrong.
  • B. It is incorrect as the electronic configuration is wrong.
  • C. It is incorrect as the electronic configuration is wrong
  • E. It is incorrect as the electronic configuration is wrong.

Q11. Which of the following statements is correct?

  • A. Faraday's experiment proved the existence of electrons
  • B. Crookes tube experiment shows the presence of electrons and protons in the atoms
  • C. Radioactivity confirms the presence of electrons and protons
  • D. Chadwick experiment shows the presence of neutrons

Explanation: Option D is correct. Chadwick bombarded a sample of beryllium with alpha particles and observed a new form of radiation that was not affected by electric fields. He concluded that this radiation consisted of neutral particles, which he named neutrons. Neutrons have no electric charge but have a mass similar to protons.

Why the other options are wrong
  • A. This statement is incorrect. Faraday's experiments, particularly his work on electromagnetism and electrochemistry, did not directly prove the existence of electrons. His experiments focused on the relationship between electricity and magnetism, as well as the phenomenon of electrolysis. The discovery of electrons came later through the work of J.J. Thomson and his cathode ray experiments.
  • B. This statement is incorrect. The Crookes tube experiment, conducted by Sir William Crookes, aimed to study the properties of cathode rays. Although the experiment provided evidence for the existence of negatively charged particles (later identified as electrons), it did not demonstrate the presence of protons in atoms. The discovery of protons as constituents of the atomic nucleus came later through the experiments conducted by Ernest Rutherford.
  • C. This statement is incorrect. Radioactivity, the spontaneous emission of particles or radiation from certain elements, does not directly confirm the presence of electrons and protons. Radioactivity is primarily associated with the decay or transmutation of atomic nuclei, involving processes such as alpha decay, beta decay, or gamma radiation. These phenomena are related to the instability of atomic nuclei and the release of specific particles or energy, but they do not directly provide evidence for the presence of electrons and protons.

Q12. "In an atom no two electrons can have the same set of four Quantum numbers" is stated by:

  • A. Heisenberg’s uncertainty principle
  • B. Aufbau Principle
  • C. Pauli's Exclusion principle
  • D. Hund's Rule
  • E. (n+1) Rule

Explanation: The set of numbers used to describe the position and energy of the electron in an atom are called quantum numbers. There are four quantum numbers namely, principal, azimuthal, magnetic and spin quantum numbers. Pauli's exclusion principle states that in an atom no two electrons can have the same set of four quantum numbers. Let us take example of electrons in Helium, both have all quantum numbers same except spin quantum number, one is clockwise and other is anti-clockwise.

Why the other options are wrong
  • A. Heisenberg gave the uncertainty principle states that we cannot know both the position and speed of a particle simutaneously, such as a photon or electron, with perfect accuracy; the more we nail down the particle's position, the less we know about its speed and vice versa.
  • B. The Aufbau principle states that electrons fill lower-energy atomic orbitals before filling higher-energy ones.
  • D. Hund's rule: every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin.
  • E. The order in which the energy of orbitals increases can be determined with the help of the (n+I) rule, where the sum of the principal and azimuthal quantum numbers determines the energy level of the orbital. Lower (n+l) values correspond to lower orbital energies.

Q13. The radii of the second orbit of the hydrogen atom calculated from Bohr's model is:

  • A. 0.529 Å
  • B. 4.8 Å
  • C. 2.116 Å
  • D. 3.4 Å
  • E. 1 Å

Explanation: R = 0.529 0.529 x n2/z 0.529 x 4Å 2.116 Å

Why the other options are wrong
  • A. This value, 0.529, is actually a constant used in Bohr's model to relate the radius of an electron orbit to the principal quantum number (n). It is not the specific value for the radius of the second orbit. The correct calculation for the second orbit using Bohr's model gives a value of approximately 2.116 angstroms, not 0.529.
  • B. 4.8 is not the correct radius for the second orbit of the hydrogen atom according to Bohr's model. The radius calculated using Bohr's model for the second orbit is approximately 2.116 angstroms, not 4.8.
  • D. 3.4 is not the correct radius for the second orbit according to Bohr's model. The accurate value is approximately 2.116 angstroms, not 3.4.
  • E. The value 1 is too small to be the radius of the second orbit in Bohr's model. The radius calculated for the second orbit is approximately 2.116 angstroms, not 1.

Q14. The neutron was discovered by:

  • A. Goldstein
  • B. Rutherford
  • C. J.J Thomson
  • D. Chadwick

Explanation: Thomson discovered electrons in 1897, Chadwick discovered neutrons in 1932 and Goldstein discovered canal ray in 1886.

Why the other options are wrong
  • A. German physicist, Goldstein is known for his work on electrical phenomena in gases and on cathode rays, thus this is incorrect option.
  • B. Early in the century, Ernest Rutherford developed a crude model of the atom, based on the gold foil experiment and discovered the existence of nucleus, hence this option is incorrect
  • C. J.J Thomson is credited with the discovery of the electrons,hence this option is incorrect.

Q15. The ground state of an atom corresponds to a state of :

  • A. Maximum energy
  • B. Minimum energy
  • C. Negative energy
  • D. Positive energy

Explanation: The nucleus of an atom is surrounded by electrons that occupy shells or orbitals of varying energy levels. The ground state of an electron, the energy level it normally occupies, is the state of lowest energy for that electron.

Why the other options are wrong
  • A. Ground state is the state of lowest energy and most stability hence, option A is incorrect.
  • C. Minimum energy is the most appropriate option.
  • D. Minimum energy is the most appropriate option.

Q16. Shape of the orbital occupied by an electron is defined by:

  • A. Principal quantum number
  • B. Azimuthal quantum number
  • C. Spin quantum number
  • D. Magnetic quantum number

Explanation: The shape of the orbital can be s,p,d, or f. This shape is determined by the azimuthal quantum number. The spin quantum number tells us about the type of spin electron possessed in an orbital. The principal quantum number tells us about the no.of the shell, not the shape of the orbital.

Why the other options are wrong
  • A. The principal quantum number determines the main energy level or shell in which an electron resides.
  • C. Spin quantum number describes the intrinsic angular momentum of a particle.
  • D. Magnetic quantum number indicates the orientation of an orbital within a subshell.

Q17. The Balmer series lies in which region of the electromagnetic spectrum?

  • A. Ultraviolet
  • B. Visible
  • C. Infrared
  • D. Microwave

Explanation: The Balmer series lies in the visible region of the electromagnetic spectrum.

Why the other options are wrong
  • A. Lyman series of hydrogen atoms lie in the Ultraviolet region.
  • C. Paschen series lies in the near Infrared region.
  • D. Bracket, Pfund as well as Humphrey series lie in the far Infrared region of the electromagnetic spectrum.

Q18. If uncertainty in the position of an electron is zero, the uncertainty in it's momentum is?

  • A. 1
  • B. 0
  • C. 2 x π x r
  • D. >h/4π
  • E. Infinite

Explanation: According to Heisenberg's uncertainty principle, it is impossible to measure position x and momentum p simultaneously. Use ∆x∆p ≥ h/4π as ∆x → 0, ∆p → ∞ Here, neither uncertainty can be zero and if the position is zero the uncertainty in momentum becomes infinite.

Why the other options are wrong
  • A. As per the explanation,this option is incorrect.
  • B. As per the explanation,this option is incorrect.
  • C. As per the explanation,this option is incorrect.
  • D. As per the explanation,this option is incorrect.

Q19. Choose the value of the Rydberg constant among the following values:

  • A. 1.09678 x 107 m-1
  • B. 1.602 x 10-19 C
  • C. 1.7588 x 1011 C
  • D. 1.007 x 107 m-1

Explanation: You’ll have to rote learn this. Memorise the power and units.

Why the other options are wrong
  • B. B is the value of charge of an electron, not Rydberg constant.
  • C. These two are also not the true value of Rydberg Constant.
  • D. These two are also not the true value of Rydberg Constant.

Q20. Diamond is a bad conductor because it:

  • A. Has a tight structure
  • B. Has high density
  • C. Has no free electron in the crystal
  • D. Is transparent to light

Explanation: A good conductor has a large presence of free electrons, as they are not bonded strongly to the nucleus, and allow conduction of charges to take place.

Why the other options are wrong
  • A. Diamond has a tight structure as the carbons are held together by strong covalent bonds which makes it very hard making it useful for cutting, but it does not affect the conduction.
  • B. Diamond is very dense due to closely packed carbon atoms, however it doesn't effect the conduction of diamonds.
  • D. The bonds in diamond are held in such a tight structure that allows the light to pass around making it look transparent.

More Sindh / DUHS Solved Papers