IMU CET Merchant Navy Sponsorship Exam MCQs (Previous Year Questions) on Periodic Table and Periodicity are essential for understanding fundamental trends and concepts in chemistry. This topic plays an important role in the exam and includes key areas such as atomic size, ionization energy, electron affinity, and electronegativity. Practicing these PYQs helps students become familiar with the exam pattern, improve problem-solving speed, and strengthen conceptual clarity required for scoring well in the IMU CET.
IMU CET Merchant Navy Sponsorship Exam MCQs (Previous Year Questions) on Classification of Elements and Periodicity in Properties
1. Which of the following has firstly organised the elements according to same trend?
a. Newland
b. Mendeleef
c. Lothar Meyer
d. Dobereiner
Answer
1. (d) Dobereiner has firstly used triad system to organise the elements.
2. Johann Dobereiner gave the idea of trends among the physical and A…… of several groups of three elements. Here, A refers to
a. atomic number
b. atomic mass
c. chemical properties
d. electronic configuration
2. (c) Johann Dobereiner in early 1800’s was the first to consider the idea of trends among properties of elements. By 1829 he noted a similarity among the physical and chemical properties of several groups of three elements (triads).
3. Lothar Meyer plotted the physical properties such as atomic volume, melting point and A…… against atomic weight. Here, A refers to
a. mass
b. boiling point
c. surface tension
d. molecules
3. (b)
4. Lothar Meyer drew a graph showing the relation between
a. atomic number and atomic weight
b. atomic number and atomic size
c. atomic weight and atomic volume
d. atomic weight and atomic size
4. (c) Lothar Meyer drew a graph showing the relation between atomic weight and atomic volume.
5. Which of the following is incorrect explanation about Mendeleef’s periodic law?
a. Mendeleef arranged elements in horizontal rows only
b. Mendeleef arranged elements with increasing atomic weight
c. Mendeleef’s system of classifying elements was more elaborate than that of Lothar
d. Both (a) and (b)
5. (a) Mendeleef arranged elements in horizontal rows and vertical columns of a table in increasing order to their atomic weights.
6. Which basic concept was used by Mendeleef for organising the elements?
a. They organise the metals only
b. They organise the non-metals only
c. They organise both metals and non-metals according to increasing number of masses
d. None of the above
6. (c) Mendeleef observed that the periodic properties of an element is a periodic function of atomic masses.
7. Which element was named as eka-silicon in Mendeleef classification of elements?
a. Germanium
b. Gallium
c. Thallium
d. Selenium
7. (a) Germanium was called eka-silicon in Mendeleef’s time.
8. The statement that is not correct for the periodic classification of elements, is
a. the properties of the elements are the periodic function of their atomic number
b. non-metallic elements are lesser in number than metallic elements
c. the first ionisation energies of elements along a period do not vary in a regular manner with increase in atomic number
d. for transition elements, the d-subshells are filled with electrons monotonically with increase in atomic numbers
8. (d)
9. The period number in the long form of the periodic table is equal to
a. magnetic quantum number of any element of the period
b. atomic number of any element of the period
c. maximum principal quantum number of any element of the period
d. maximum azimuthal quantum number of any element of the period
9. (c)
10. Which of the following is incorrect?
a. Henry Moseley observed regularities in the characteristics X-ray spectra of elements
b. A plot of frequency of X-ray emitted against atomic number ($Z$) gives a straight line
c. The atomic number of an element is equal to the number of neutron(s) in a neutral atom
d. Electronic configuration of an atom determines the physical and chemical properties
10. (c) The atomic number of elements is equal to the number of electrons or number of protons in a neutral atom.
$Z =$ number of $e^-$ / number of $p^+$
$A =$ number of protons $+$ number of neutrons
11. Match the Column I with Column II and select the correct answer by given codes.
| Column I (Number of periods) | Column II (Number of elements) |
|---|---|
| A. First period | 1. 14 |
| B. Third period | 2. 02 |
| C. Lanthanoids | 3. 14 |
| D. Actinoids | 4. 08 |
| 5. 04 |
Codes (A B C D)
a. 2 4 1 5
b. 2 4 1,3 3,1
c. 4 2 1 3
d. None of these
11. (b) A → 2, B → 4, C → 1,3, D → 3,1
A. First period contains 2 elements.
B. Third period contains 8 elements.
C, D. 14 elements of both sixth and seventh periods are known as lanthanoids and actinoids respectively.
12. Observe the following statements,
I. The physical and chemical properties of elements are the periodic functions of their electronic configuration.
II. Electronegativity of fluorine is less than the electronegativity of chlorine.
III. Electropositive nature decreases from top to bottom in a group.
The correct answer is
a. I, II and III
b. Only I
c. I and II
d. II and III
12. (b) The physical and chemical properties of elements are periodic functions of their electronic configuration.
Electronegativity of fluorine is more than that of chlorine.
Electropositive nature increases from top to bottom in a group.
13. In an element $_ZM^{19}$, there are 10 neutrons in the nucleus. It belongs to
a. s-block
b. d-block
c. f-block
d. None of the above
13. (d) $Z = 19 – 10 = 9$
EC = $1s^2 2s^2 2p^6 3s^2 3p^5$, hence, it belongs to p-block.
14. The third alkaline earth metal ion contains number of electrons and protons as
a. $e^-$ 20, $p^+$ 20
b. $e^-$ 18, $p^+$ 18
c. $e^-$ 18, $p^+$ 20
d. $e^-$ 19, $p^+$ 20
14. (c) Ca$^{2+}$ ; $e^- = 18$, $p^+ = 20$
15. A metal having electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2$ is in
a. s-block
b. d-block
c. p-block
d. None of these
15. (b) Zn ($3d^{10} 4s^2$) is d-block element.
16. Which of the following orbitals are in the process of filling in the 6th period?
a. 6s, 6p, 6d, 6f
b. 6s, 5f, 6d, 6p
c. 6s, 4f, 5d, 6p
d. 5s, 5p, 5d
16. (c) The order in which orbitals are filled, is $6s < 4f < 5d < 6p$
17. The configuration to second excited state of the element isoelectronic with O$_2$ or P$^-$ or Cl$^+$ is
a. $[Ne] 3s^2 3p_x^2 3p_y^2 3p_z^2$
b. $[Ne] 3s^2$
c. $[Ne] 3s^1 3p_x^1 3p_y^1 3p_z^1 3d_{xy}^1 3d_{yz}^1$
d. $[Ne] 3s^2 3p_x^1 3p_y^1 3p_z^1$
17. (c) In O$_2$ or P$^-$ or Cl$^-$ the electrons present are 18. The element is sulphur ($Z = 16$). The electronic configuration of sulphur in second excited state is given by (c).
18. The element having atomic number 33 lies in the group
a. 16
b. 14
c. 15
d. 13
18. (c) Arsenic is in group 15.
19. General configuration of ultimate and penultimate shell is $ns^2 np^6 (n-1)d^x$. If $n=4$ and $x=5$, the number of protons in the nucleus would be
a. < 24
b. 25
c. 24
d. > 25
19. (b) The element having outer electronic configuration $3s^2 3p^6 3d^5 4s^2$, has atomic number 25. Therefore, the number of protons is 25.
20. Generally, the valency of noble gases is
a. two
b. three
c. one
d. zero
20. (d) They are chemically inert (inactive).
21. Identify the least stable ion among the following.
a. Li$^-$
b. Be$^-$
c. B$^-$
d. C$^-$
21. (b) Be has fully filled 2s-subshell ($1s^2 2s^2$) and therefore, show least tendency to accept an electron. Thus, Be$^-$ is least stable.
22. The electronic configuration of an element is $1s^2 2s^2 2p^6$, $3s^2 3p^3$. The atomic number and the group number of the element X which is just below the above element in the periodic table respectively, are
a. 23 and 5
b. 23 and 15
c. 33 and 15
d. 33 and 5
22. (c) Atomic number of the given element = 15
Group number = 10 + 5 (valence electrons) = 15
Period = 3
Atomic number of the element just below = 33
23. What will be the IUPAC name of element having $Z=106$?
a. Unnilquadium
b. Unnilhexium
c. Unnilheptium
d. Ununhexium
23. (b) For $Z = 106$, Root of 1 — un, Root of 0 — nil, Root of 6 — hex + ium, IUPAC name = unnilhexium
24. The one with the largest ionic size is
a. O$^{2-}$
b. Mg$^{2+}$
c. F$^-$
d. Na$^+$
24. (a) O$^{2-}$, F$^-$, Mg$^{2+}$ and Na$^+$ are isoelectronic species (each having 10$e^-$). The size of isoelectronic species decreases with increase in nuclear charge (number of protons). Hence, correct order of size is O$^{2-}$ > F$^-$ > Na$^+$ > Mg$^{2+}$
25. Fluorine and neon have atomic radii in angstrom given by
a. 1.60, 1.60
b. 0.72, 0.72
c. 0.72, 1.60
d. None of these
25. (c) Noble gases are largest amongst their periods due to vander walls radii.
26. Which of the following alkali metals has smallest size?
a. Cs
b. Rb
c. Na
d. K
26. (c) Atomic radii increases down the group
27. Which of the following ionic species has largest size?
a. Rb$^+$(aq)
b. Li$^+$(g)
c. Na$^+$(aq)
d. Li$^+$(aq)
27. (d)
The correct answer is (d) $\text{Li}^+(\text{aq})$.
While this might seem surprising at first glance, the question distinguishes between gaseous ions and aqueous (hydrated) ions. To understand why, we have to look at the process of hydration.
1. Gaseous vs. Aqueous Size
In the gas phase, size follows the standard periodic trend: as you go down a group, the number of shells increases, so the ionic radius increases.
Gas phase order: $\text{Li}^+ < \text{Na}^+ < \text{Rb}^+$
However, when these ions are placed in water ($\text{aq}$), they attract water molecules. This creates a hydration shell around the ion. The “size” of the species is now the size of the ion plus its surrounding water layer.
2. The Hydration Rule
The degree of hydration depends on Charge Density ($\text{Charge} / \text{Size}$).
Small Ions ($\text{Li}^+$): Because the lithium ion is naturally very small, its positive charge is highly concentrated. This high charge density creates a very strong electrostatic attraction for the oxygen atoms (the negative end) of water molecules.
Large Ions ($\text{Rb}^+$): The rubidium ion is much larger, so its charge is “spread out” (low charge density). It attracts fewer water molecules and holds them less tightly.
Because $\text{Li}^+$ has the highest charge density, it undergoes extensive hydration. It ends up surrounded by a massive, thick shell of water molecules, making its effective radius (the hydrated radius) the largest in the group.
28. Chloride ion and potassium ion are isoelectronic, then
a. their sizes are same
b. Cl$^-$ ion is bigger than K$^+$ ion
c. K$^+$ ion is relatively bigger
d. their sizes depend on other cation and anion
28. (b)
29. Which one of the following is correct increasing order of size?
a. Mg < Na$^+$ < F$^-$ < Al
b. Na$^+$ < Al < Mg < F$^-$
c. Na$^+$ < F$^-$ < Al$^{3+}$ < Mg$^{2+}$
d. Na$^+$ < F$^-$ < Mg$^{2+}$ < Al$^{3+}$
29. (b) Na$^+$ and F$^-$ ions are isoelectronic, therefore F$^-$ has the largest and Na$^+$ has the lowest size. Further, Al with higher nuclear charge has lower size than Mg. Thus, the overall order is Na$^+$ < Al < Mg < F$^-$
30. Ionic radii vary in
a. inverse proportion to the effective nuclear charge
b. inverse proportion to the square of effective nuclear charge
c. inverse proportion to the screening effect
d. direct proportion to the square of the screening effect
30. (a)
31. The first ionisation energy of beryllium is more than that of boron because
a. boron has higher nuclear charge
b. boron has only one electron in p-subshell
c. atomic size of boron is less than that of beryllium
d. atomic size of boron is more than that of beryllium
31. (b) Boron has only one electron in p-subshell, hence it can be lost easily, while beryllium has comparatively stable configuration ($1s^2 2s^2$), hence, cannot give an electron easily, consequently its IE is greater.
32. Among the following, the element with highest ionisation potential is
a. boron
b. carbon
c. oxygen
d. nitrogen
32. (d) Nitrogen has stable half-filled orbital.
33. The isoelectronic ion having lowest ionisation energy is
a. S$^{2-}$
b. Ca$^{2+}$
c. K$^+$
d. Cl$^-$
33. (b)
34. The values in electron-volt per atom which represent the first ionisation energy of oxygen and nitrogen atom respectively are
a. 13.6 and 14.6
b. 14.6 and 13.6
c. 14.6 and 14.6
d. 13.6 and 13.6
34. (a) Ionisation energy of N > ionisation energy of oxygen, since, nitrogen has stable configuration due to half-filled 2p-orbitals.
35. The pair in which the ionisation energy of first species is less than that of second is
a. N, P
b. Be, Be$^+$
c. S, P
d. N, N$^-$
35. (c) P has higher ionisation energy due to stable half-filled configuration.
36. How many Cs-atoms can be converted to Cs$^+$ ion by 1 J energy if IE$_1$ for Cs is 376 kJ mol$^{-1}$?
a. $1.60 \times 10^{23}$
b. $1.60 \times 10^{15}$
c. $1.60 \times 10^{18}$
d. $16.0 \times 10^{26}$
36. (c) 376 × 10$^3$ J energy produces 6.023 × 10$^{23}$ ions.
∴ 1 J energy will produce $\frac{6.023 \times 10^{23}}{3.76 \times 10^5} = 1.60 \times 10^{18}$ Cs$^+$ ions
37. How many joules of energy must be absorbed to convert Li to Li$^+$, all the atoms present in 1.00 mg of gaseous Li? IE$_1$ of Li is 520.3 kJ mol$^{-1}$ (Li = 7).
a. 0.00743 kJ
b. 0.520 kJ
c. 520 kJ
d. 0.0743 kJ
37. (d) 1.00 mg = $\frac{1 \times 10^{-3}}{7}$ mol Li
1 mole of Li is converted by 520.3 kJ.
$\frac{1 \times 10^{-3}}{7}$ mole is converted to Li$^+$ by $\frac{520.3 \times 1 \times 10^{-3}}{7}$ kJ = 0.0743 kJ
38. The incorrect statements among the following is
a. the first ionisation potential of Al is less than the first ionisation potential of Mg
b. the second ionisation potential of Mg is greater than the second ionisation potential of Na
c. the first ionisation potential of Na is less than the first ionisation potential of Mg
d. the third ionisation potential of Mg is greater than the third ionisation potential of Al
38. (b) IE$_2$ of Mg is lower than that of Na because in case of Mg$^{+}$, 3s electron has to be removed while in case of Na$^+$, an electron from the stable inert gas configuration (neon) has to be removed.
39. Ionisation enthalpy of Na would be same as
a. value of electron affinity of Na$^+$
b. value of electronegativity of Na
c. value of ionisation potential of Mg
d. value of electron affinity of Na
39. (a)
Na → Na$^+$ + e$^-$ ; IE
Na$^+$ + e$^-$ → Na ; –EA
IE = –EA
40. Which one of the following statements is incorrect in relation to ionisation enthalpy?
a. Ionisation enthalpy increases for each successive electrons
b. The greatest increase in ionisation enthalpy is experienced on removal of electron from core noble gas configuration
c. End of valence electrons is marked by a big jump in ionisation enthalpy
d. Removal of electron from orbitals bearing lower n value is easier than from orbitals having higher n value
40. (d) Higher n value has lower value of IE due to large size
41. The formation of the oxide ion, O$^{2-}$(g) from oxygen atom requires first an exothermic and then an endothermic step as shown below
$$ \ce{O(g) + e^- -> O^-(g)}; \quad \Delta H = -141 \, \text{kJ mol}^{-1} $$
$$ \ce{O^-(g) + e^- -> O^{2-}(g)}; \quad \Delta H = +780 \, \text{kJ mol}^{-1} $$
Thus, process of formation of O$^{2-}$ in gas phase is unfavourable even though O$^{2-}$ is isoelectronic with neon. It is due to the fact that,
a. oxygen is more electronegative
b. addition of electron in oxygen results in larger size of the ion
c. electron repulsion outweighs the stability gained by achieving noble gas configuration
d. O$^-$ ion has comparatively smaller size than oxygen atom
41. (c) The process of formation of O$^{2-}$ in gas phase is unfavourable even though O$^{2-}$ is isoelectronic with neon because electron repulsion outweighs the stability gained by achieving noble gas configuration.
42. Which one of the following arrangements represent the correct order of electron gain enthalpy (with negative sign) of the given atomic species?
a. Cl < F < S < O
b. O < S < F < Cl
c. S < O < Cl < F
d. F < Cl < O < S
42. (b)
43. Electronic configurations of four elements A, B, C and D are given below
A. $1s^2 2s^2 2p^6$
B. $1s^2 2s^2 2p^4$
C. $1s^2 2s^2 2p^6 3s^1$
D. $1s^2 2s^2 2p^5$
Which of the following is the correct increasing order of tendency to gain electron?
a. A < C < B < D
b. A < B < C < D
c. D < B < C < A
d. D < A < B < C
43. (b) np$^5$ and ns$^1$ elements possess a large tendency to accept the electron than fully and half-filled elements
44. In which of following order of arrangement does not agree with variation of property indicated against it?
a. Al$^{3+}$ < Mg$^{2+}$ < Na$^+$ < F$^-$ (Increasing ionic size)
b. B < C < N > O (Increasing first ionisation enthalpy)
c. I < Br < Cl < F (Increasing electron gain enthalpy)
d. Li < Na < K < Rb (Increasing metallic radius)
44. (c)
45. The electron affinity values (in kJ mol$^{-1}$) of three halogens X, Y and Z are –349, –333 and –325 respectively. Then, X, Y and Z respectively are
a. F$_2$, Cl$_2$ and Br$_2$
b. Cl$_2$, F$_2$ and Br$_2$
c. Cl$_2$, Br$_2$ and F$_2$
d. Br$_2$, Cl$_2$ and F$_2$
45. (b) The electron affinity for halogens,
Cl$_2$ = 348.5 → X
F$_2$ = 332.6 → 333 → Y
Br$_2$ = 324.7 → 325 → Z
X has the highest value of electron affinity. Therefore, the correct order of electron affinity is Cl$_2$ > F$_2$ > Br$_2$
46. Which one of the following statements is false?
a. The electron affinity of chlorine is less than that of fluorine
b. The electronegativity of fluorine is more than that of chlorine
c. The electron affinity of bromine is less than that of chlorine
d. The electronegativity of chlorine is more than that of bromine
46. (a) The electron affinity of fluorine is less than that of chlorine, due to very small size of fluorine in which negative charge is highly concentrated and repels the incoming electrons.
47. Which of the following element has highest electronegativity?
a. Oxygen
b. Chlorine
c. Fluorine
d. Nitrogen
47. (c)
48. In C, N, O and F, the electronegativity
a. increases from carbon to fluorine
b. decreases from carbon to oxygen and then increases
c. decreases from carbon to fluorine
d. increases from carbon to oxygen and then decreases
48. (a) In a period, it increases from left to right.
49. Electronegativity of F on Pauling scale is 4.0. What will be its value on Mulliken scale?
a. 10.0
b. 11.2
c. 8.54
d. 16
49. (b) Electronegativity on Mulliken scale and Pauling scale are given by
$X_M = 2.8 \times X_P = 2.8 \times 4.0 = 11.2$
Usually, Mulliken’s value of electronegativity are 2.8 times of Pauling value.
50. Two elements whose electronegativities are 1.2 and 3.0 respectively, the bond formed between them would be
a. ionic
b. covalent
c. coordinate
d. metallic
50. (a) When the electronegativity difference is more than 1.7, the bond formed will be ionic in nature.
51. What will be the electronegativity of carbon at Pauling scale? Given that, $E_{\ce{H-H}} = 104.2$ kcal mol$^{-1}$, $E_{\ce{C-C}} = 83.1$ kcal mol$^{-1}$, $E_{\ce{C-H}} = 98.8$ kcal mol$^{-1}$. [Electronegativity of hydrogen = 2.1]
a. 0.498
b. 0.598
c. 2.134
d. 2.597
51. (d) $X_C – X_H = 0.208 \sqrt{\Delta}$
where, $\Delta = E_{C-H} – \frac{E_{C-C} + E_{H-H}}{2}$
$\Delta = 98.8 – \frac{83.1 + 104.2}{2} = 5.75$
$X_C – 2.1 = 0.208 \times 5.75 = 1.197$
$X_C = 2.1 + 0.497 = 2.597$
52. Which one of the following is incorrect?
a. Non-metals have strong tendency to gain electron
b. Electronegativity is directly related to non-metallic properties of elements
c. Electronegativity is inversely proportional to the metallic properties of elements
d. Increase in electronegativity down the group is accompanied by a decrease in non-metallic properties
52. (d) Decrease in electronegativity down the group is accompanied by a decrease in non-metallic properties.
53. Pauling’s equation for determining the electronegativity of an element, is
$$(X_A, X_B \text{ are electronegativity values of elements A and B respectively, } $$ $$\Delta \text{ represents the polarity of A—B bond.})$$
a. $X_A – X_B = 0.208 \sqrt{\Delta}$
b. $X_A – X_B = 0.208 \Delta$
c. $X_A – X_B = 0.208 \Delta^2$
d. $X_A – X_B = \Delta$
53. (a) Pauling’s equation
$X_A – X_B = 0.208 \sqrt{\Delta}$
where, $X_A$ and $X_B$ are electronegativity values of elements A and B and $\Delta = E_{A-B} – \frac{E_{A-A} + E_{B-B}}{2}$
Here, $E_{A-B}$, $E_{A-A}$ and $E_{B-B}$ are the bond energies of the molecules AB, A$_2$ and B$_2$ respectively.
54. Match the Column I with Column II and select the correct answer by given codes.
| Column I (Elements) | Column II (Properties) |
|---|---|
| A. Li$^+$ < Al$^{3+}$ < Mg$^{2+}$ < K$^+$ | 1. EA (Electron affinity) |
| B. Li$^+$ > Al$^{3+}$ > Mg$^{2+}$ > K$^+$ | 2. Ionic radii |
| C. Cl > F > Br > I | 3. EN (Electronegativity) |
| D. F > Cl > Br > I | 4. ENC |
Codes
a. 2 4 3 1
b. 2 4 1 3
c. 4 2 3 1
d. 4 2 1 3
54. (b) A → 2, B → 4, C → 1, D → 3
A. Li$^+$ < Al$^{3+}$ < Mg$^{2+}$ < K$^+$
The cation with the greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus. Anion with the greater negative charge will have the larger radius.
Positive charge ∝ 1/ionic radius
Negative charge ∝ ionic radius
B. Greater positive charge, increases ENC in case of isoelectronic species while for same group elements, ENC decreases down the group.
C. Cl > F > Br > I
Electron affinity of Cl is highest in halogen family.
D. F > Cl > Br > I
Electronegativity of fluorine (F) is higher than that of Cl, Br and I.
55. The set that contains pairs of elements that do not belong to same group but show chemical resemblance is
a. B, Al
b. Be, Al
c. Hf, Zr
d. K, Rb
55. (b)
56. Which pair of elements has same chemical properties?
a. 13, 22
b. 3, 11
c. 4, 24
d. 2, 4
56. (b) The pair which belongs to same group, i.e. in which both the elements have same outer electronic configuration has same chemical properties.
$_3$Li : $1s^2 2s^1$
$_{11}$Na : $1s^2 2s^2 2p^6 3s^1$
57. The element of second period which forms most acidic oxide is
a. carbon
b. boron
c. fluorine
d. nitrogen
57. (c)
58. In periodic table, melting point/boiling point increases down the group in which of the following group?
a. Group 13
b. Group 2
c. Group 17
d. Group 1
58. (c)
59. Considering the elements B, C, N, F and Si, the correct order of their non-metallic character is
a. B > C > Si > N > F
b. Si > C > B > N > F
c. F > N > C > B > Si
d. F > N > C > Si > B
59. (d)
60. Considering the elements B, Al, Mg, and K, the correct order of their metallic character is
a. B > Al > Mg > K
b. Al > Mg > B > K
c. Mg > Al > K > B
d. K > Mg > Al > B
60. (d)
IMUCET ARCHIVES
1. Elements/ions having same number of electrons are known as isoelectronic species. Arrange the following elements in correct order of atomic/ionic radii and choose the correct option from the four choices given below
O$^{2-}$, Na$^+$, Mg$^{2+}$, F$^-$, Al$^{3+}$
a. Al$^{3+}$ < Mg$^{2+}$ < Na$^+$ < F$^-$ < O$^{2-}$
b. Al$^{3+}$ < Na$^+$ < Mg$^{2+}$ < F$^-$ < O$^{2-}$
c. Al$^{3+}$ < Mg$^{2+}$ < Na$^+$ < F$^-$ < O$^{2-}$
d. None of the above
1. (a) Elements/ions having equal number of electrons are known as isoelectronic species. Among isoelectronic species, cations having highest charge are smallest while anion having highest charge are largest.
Cation < Neutral atom < Anion
Hence, correct choice is Al$^{3+}$ < Mg$^{2+}$ < Na$^+$ < F$^-$ < O$^{2-}$
2. For the properties mentioned, the correct trend for the different species is in
a. Strength as Lewis acid — BCl$_3$ > AlCl$_3$ > GaCl$_3$
b. Inert pair effect — Al > Ga > In
c. Oxidising property — Al$^{3+}$ > In$^{3+}$ > Tl$^{3+}$
d. First ionisation enthalpy — B > Al > Tl
2. (d) As we know on moving down the group, first ionisation enthalpy decreases top to bottom, therefore order of first ionisation enthalpy for group 13 elements is B > Al > Ga > In > Tl
3. Which of the following choices represent the correct order of first ionisation enthalpy?
a. B < C < N < O < F
b. B > C > N > O > F
c. B < C < N > O < F
d. B < C < N > O > F
3. (d) Ionisation energy is the minimum amount of energy required to remove the outermost electron from an isolated gaseous atom. Quantitatively, it depends on the attraction between electron present on outermost shell and nucleus. Greater the interaction between outermost electron and nucleus, higher will be its ionisation enthalpy. So, correct order of first IE must be B < C < N < O < F But due to extra stable half-filled electronic configuration of p-orbital, N has more value of first ionisation enthalpy than oxygen hence, correct order is B < C < N > O > F
4. Electron gain enthalpy of fluorine is lower than that of chlorine is due to
a. smaller size
b. smaller nuclear charge
c. difference in their electronic configurations
d. its highest reactivity
4. (a) Because of the small size of F, electron-electron repulsions present in its relatively compact 2p-subshell, do not easily allow the addition of an extra electron. On the other hand, Cl because of its comparatively bigger size than F, allows the addition of an extra electron more easily. Thus, the EA of Cl is higher than that of F.
5. The atomic numbers of elements A, B, C and D are $Z-1$, $Z$, $Z+1$ and $Z+2$ respectively. If B is a noble gas, choose the correct statement among the following statements:
I. A has higher electron affinity.
II. C exists in +2 oxidation state.
III. D is an alkaline earth metal.
a. I and II
b. II and III
c. I and III
d. I, II and III
5. (c) A : $Z-1$
B : $Z$ → Noble gas (outer subshells are s or p)
C : $Z+1$
D : $Z+2$
Noble gas with outer s-subshell → He ($1s^2$)
Noble gas with outer p-subshell → $ns^2 np^6$ like [Ne]
In periodic table, electron affinity is highest in chlorine of group 17 with electronic configuration of $3s^2 3p^5$.
Noble gas = $3s^2 3p^6$ [Ar]
B = [Ar] → $Z = 3s^2 3p^6$ (Noble gas)
A = $Z-1$ = $3s^2 3p^5$ (Halogen family)
C = $Z+1$ = $3s^2 3p^6 4s^1$ (Alkali metal)
D = $Z+2$ = $3s^2 3p^6 4s^2$ (Alkaline earth metal)
FAQs – IMU CET Merchant Navy Sponsorship Exam MCQs (PYQs): Classification of Elements and Periodicity in Properties
What does this MCQ section cover?
This section covers IMU CET Previous Year Questions (MCQs) from the chapter Classification of Elements and Periodicity in Properties, including important topics like atomic radius, ionization energy, electron affinity, and electronegativity.
Who has prepared this study material?
This content is prepared by Er Neeraj Anand, Chemistry Anand Classes, under Anand Classes, and is published by Anand Technical Publishers.
Why are PYQs important for IMU CET preparation?
PYQs help students understand the exam pattern, frequently asked concepts, and difficulty level. They also improve speed, accuracy, and confidence in solving MCQs.
Are solutions provided with the MCQs?
Yes, all questions come with clear and step-by-step solutions to help students understand the correct approach and concepts.
Which topics are most important in this chapter for IMU CET?
Important topics include:
Atomic and ionic radii
Ionization energy
Electron affinity
Electronegativity
Periodic trends
Is this material suitable for beginners?
Yes, the content is designed in a simple and structured way, making it suitable for both beginners and advanced learners.
How should students prepare using these MCQs?
Students should first revise the theory, then solve MCQs topic-wise, analyze mistakes, and practice regularly to improve performance.
Does this content follow the IMU CET syllabus?
Yes, all questions are selected according to the latest IMU CET syllabus and exam pattern.
Can practicing these PYQs improve exam scores?
Yes, regular practice of PYQs helps in better understanding of concepts and increases the chances of scoring high marks.
Is this useful for other exams as well?
Yes, it is also helpful for exams like JEE Main, NEET, and other entrance tests where similar concepts are asked.
What makes this resource unique?
This material focuses on concept clarity, exam-oriented questions, and structured learning, making it highly effective for competitive exams.
Where can students find more study material?
Students can explore other Class 11 Chemistry chapters, notes, and topic-wise MCQs and PYQs available on this website for complete preparation.
Summary
To build a strong foundation in this chapter, students should also study Classification of Elements and Periodicity in Properties along with Modern Periodic Law and Modern Periodic Table (Groups, Periods, Blocks). It is equally important to understand Structure of Atom, as periodic trends are directly related to electronic configuration. For complete preparation, explore Class 11 Chemistry Chapter Notes and practice additional topic-wise MCQs and PYQs available on this website.