Free Periodic Properties and Trends MCQs with Answers
606 Periodic Properties and Trends MCQs from Chemistry, each with the correct answer and a written explanation of why it is correct. Free and unlimited, with no account needed.
Periodic properties arise from electron configuration and effective nuclear charge, producing trends in atomic and ionic radius, ionization energy, electron affinity, electronegativity, metallic character and reactivity. Comparisons run across periods and down groups, with attention to common exceptions. The topic also relates these trends to the behavior of s-block and p-block elements.
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606 questions · page 26 of 31
- A. BaSO4
- B. SrSO4
- C. CaSO4
- D. MgSO4
Explanation: The solubility of Group 2 sulfates decreases as you move down the group from magnesium to barium.
Correct answer: MgSO4- A. Fe>Sc>Rb>Br>Te>F>Ca
- B. Ca>Rb>Sc>Fe>Te>F>Br
- C. Rb>Ca>Sc>Fe>Br>Te>F
- D. Rb>Ca>Sc>Fe>Te>Br>F
Explanation: Alkali and alkaline earth metals are most electropositive. Alkali metals are more electropositive than alkaline earth metals.
Correct answer: Rb>Ca>Sc>Fe>Te>Br>F- A. S2->Cl->K+>Ca2+
- B. Ca2+>K+>Cl->S2-
- C. Cl->S-2>Ca2+>K+
- D. K+>Cl->Ca2+>S2-
Explanation: Size of isoelectronic decreases with increase in atomic number. Therefore, option A is the correct answer.
Correct answer: S2->Cl->K+>Ca2+- A. Alkali metals are at the maxima and noble gases at the minima.
- B. Noble gases are at the maxima and alkali metals at the minima.
- C. Transition element are at the maxima.
- D. Minima and maxima do not show any regular behaviour.
Explanation: In a period, alkali metals have the lowest and noble gases have the maximum ionisation energy.Hence, (b) is correct.
Correct answer: Noble gases are at the maxima and alkali metals at the minima.505. According to modern periodic law, variations in the properties of elements is related to their:
- A. atomic weight.
- B. nuclear weight.
- C. atomic numbers.
- D. neutron-proton ratios.
Explanation: The Modern periodic law states "The chemical and physical properties of elements are a periodic function of their atomic numbers".
Correct answer: atomic numbers.- A. Mg(g)→ Mg+ (g) + e-
- B. Mg+(g)→ Mg2+(g) + e-
- C. Na(g) → Na+ (g)+ e-
- D. Na+(g) → Na+2(g) + e-
Explanation: Na+ has stable electronic configuration so removal of electron form Na+ to form Na+2 ion requires maximum energy.
Correct answer: Na+(g) → Na+2(g) + e-- A. Generally reducing character of elements increases in period.
- B. Generally oxidising character of elements increases in period.
- C. Generally, basic character of oxides decreases in a group.
- D. All are correct.
Explanation: Oxidising character of elements increases in a period. Therefore option B is correct.
Correct answer: Generally oxidising character of elements increases in period.- A. It forms an basic oxide.
- B. It belongs to II A group.
- C. It belongs to IV period.
- D. It forms an acidic oxide.
Explanation: The element is calcium and hence its oxide (CaO) is basic in nature. So option D is correct.
Correct answer: It forms an acidic oxide.- A. d5,d3,d1,d4 increasing magnetic moment.
- B. MO, M2O , MO2, M2O5 - decreasing basic strength.
- C. Sc, V, Cr, Mn - increasing number of oxidation states.
- D. Co2+,Fe3+,Cr3+,Sc3+ - increasing stability.
Explanation: d1 ,d3 ,d4 ,d5 - increasing magnetic moment.
Correct answer: d5,d3,d1,d4 increasing magnetic moment.- A. 1s2s22sp63s23p1
- B. 1s22s22p63s2
- C. 1s22s2sp63s1
- D. 1s22s2sp63s23p2
Explanation: Sodium has electronic configuration as 1s2 2s2 2p6 3s1. The last electron or the valence electron is in s orbital.
Correct answer: 1s22s2sp63s1- A. Ionization enthalpy increases for each successive electron.
- B. The greatest increase in ionization enthalpy is experienced on removal of electron from core of noble gas configuration.
- C. End of valence electrons is marked by a big jump in ionization enthalpy.
- D. Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
Explanation: Removal of Electron from orbital bearing lower n value is difficult than from orbitals having higher n values. Option D is correct.
Correct answer: Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.- A. 3, 11, 19, 37.
- B. 5, 13, 21, 39.
- C. 7, 15, 31, 49.
- D. 5, 13, 37, 49.
Explanation: Option D shows elements of group IIIA (5-Boron,13-Aluminum, 31-Gallium, 49-Indium. Hence option D is correct.
Correct answer: 5, 13, 37, 49.- A. IA.
- B. III A.
- C. III B.
- D. Zero Group.
Explanation: Group III B includes lanthanides and actinides and the total number of elements in this group = 32. Hence correct option is C.
Correct answer: III B.- A. Small size
- B. High electronegativity
- C. Low ionization energy
- D. Large atomic radius
Explanation: Lithium and beryllium's small size and high charge density cause strong polarizing effects, leading to covalent bond formation and their…
Correct answer: Small size- A. The properties of elements are periodic functions of their atomic numbers.
- B. Non-metallic elements are fewer in number than metallic elements.
- C. For transition elements, the 3d-orbitals are filled with electrons after 3p-orbitals and before 4s-orbitals.
- D. The first ionisation enthalpies of elements generally increase with an increase in atomic number as we move along a period.
Explanation: Option C is the correct answer because it incorrectly states the order of filling for transition elements.
Correct answer: For transition elements, the 3d-orbitals are filled with electrons after 3p-orbitals and before 4s-orbitals.- A. Increases
- B. Decreases
- C. Constant
- D. None
Explanation: Across a period, added electrons enter the same principal shell, so the inner-shell electrons causing most shielding remain almost…
Correct answer: Constant- A. F > N > O > C
- B. F > N < O > C
- C. F < N < O < C
- D. F > N > O < C
Explanation: Fluorine is obviously the most electronegative element (4.0) followed by Oxygen (3.44), then Nitrogen(3.04) and finally Carbon(2.55).
Correct answer: F > N < O > C- A. I < Br < F < Cl
- B. I < Cl < F < Br
- C. I < F < Br < Cl
- D. F < Cl < Br < I
Explanation: Option A is correct as Cl has the highest E.A in its group. Although Fluorine is expected to have the highest E.A but due to its small…
Correct answer: I < Br < F < Cl- A. I1 < I2
- B. I1 > I2
- C. I1 = I2
- D. None of these options
Explanation: Option A is correct as Successive ionization energy increases due to increasing Z.
Correct answer: I1 < I2- A. X3Y2
- B. X3Y6
- C. X2Y3
- D. None of these
Explanation: Element X will have a valency of +3 while Y will have a valency of -2. Thus the most likely formula of the compound is X2Y3.
Correct answer: X2Y3