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 23 of 31
- A. Li, Na, K
- B. Be, Mg, Ca
- C. F, Cl, Br
- D. O, S, Se
Explanation: The correct answer is Li, Na, K. These elements are alkali metals, located in Group 1 of the periodic table, and are characterized by…
Correct answer: Li, Na, K- A. ZnO
- B. Al2O3
- C. PbO
- D. SO2
Explanation: Amphoteric oxides can react with both acids and bases. ZnO, Al2O3, and PbO exhibit this property, making them amphoteric.
Correct answer: SO2- A. NH3
- B. PH3
- C. AsH3
- D. SbH3
Explanation: The basicity of hydrides decreases down the group in the periodic table.
Correct answer: NH3- A. C
- B. N
- C. Be
- D. O
Explanation: The most electropositive element among the given options is beryllium (Be).
Correct answer: Be- A. Carbon
- B. Silicon
- C. Hydrogen
- D. Fluorine
Explanation: The correct answer is Hydrogen. Hydrogen is a part of water, one of the simplest and most abundant compounds, and forms a vast number of…
Correct answer: Hydrogen- A. Na
- B. Mg
- C. Al
- D. Si
Explanation: The metallic character of an element is determined by its ability to lose electrons easily.
Correct answer: Na- A. MgCO3
- B. CaCO3
- C. SrCO3
- D. BaCO3
Explanation: The decomposition temperature of carbonates increases as you move down Group 2 in the periodic table.
Correct answer: BaCO3- A. Cr6+<Cr3+<Cr2+
- B. Mn2+<Mn2+<Mn7+
- C. Pb4-<Pb2+
- D. Both (a) and (c)
Explanation: As nuclear charge decreases, ionic radii increases.
Correct answer: Both (a) and (c)- A. Ba>Sr>Mg : Atomic Radius
- B. F>O>N: First ionization energy
- C. Cl>F>I: Electron affinity
- D. O>Se>Te : Electro negativity
Explanation: F>N>O is correct order of first ionization energy.
Correct answer: F>O>N: First ionization energy- A. An endothermic process
- B. An exothermic process
- C. Neutral process
- D. Energy wasted
Explanation: When a second electron is added to a uni-negative ion, the incoming electron is repelled by the already present negative charge, and…
Correct answer: An endothermic process- A. [Ne] 3s2 3p1
- B. [Ne] 3s2 3p3
- C. [Ne] 3s2 3p2
- D. [Ar] 3d10 4s2 4p1
Explanation: Ionization energy increases as a shell fills. Half filled or completely filled orbitals are very stable and so it is difficult to ionize…
Correct answer: [Ne] 3s2 3p3- A. (A)<(B)<(C)
- B. (C)<(B)<(A)
- C. (A)<(C)<(B)
- D. (B)<(A)<(C)
Explanation: A, B and C are magnesium, aluminium and silicon. Magnesium form ionic oxide, MgO; Aluminium forms amphoteric oxide, Al2O3 and silicon…
Correct answer: (A)<(B)<(C)- A. Reducing power in aqueous solution is maximum for Lithium metal.
- B. electron affinity order O+<O<O-22<O-2
- C. pH of aqueous solution LiCl>BeCl2>MgCl2>AlCl3
Explanation: pH of aqueous solution LiCl<BeCl2<MgCl2<AlCL3 Therefore, option C is correct.
Correct answer: pH of aqueous solution LiCl>BeCl2>MgCl2>AlCl3- A. d-block.
- B. f-block.
- C. s-block.
- D. p-block.
Explanation: This electronic configuration is of Cr. Cr is a d-block element. Therefore, option A is correct.
Correct answer: d-block.- A. 1.36, 1.40, 1.71
- B. 1.36, 1.71, 1.40
- C. 1.71, 1.40, 1.36
- D. 1.71, 1.36, 1.40
Explanation: N-3 > O-2 > F-1.71 1.40 1.37Å Hence, option C is correct.
Correct answer: 1.71, 1.40, 1.36- A. 3, 11, 19, 37
- B. 5, 13, 21, 39
- C. 7, 15, 31, 49
- D. 5, 13, 31, 49
Explanation: The elements of group III A are: ⇒ 5B, 13Al, 31Ga, 49ln, 81Tl Therefore option D is correct.
Correct answer: 5, 13, 31, 49- A. Ni<Pd<Pt
- B. Ti<Zr<Hf
- C. Ti <Zr <Pb
- D. All are correct.
Explanation: Due to lanthanide contraction size of Pd ≈ Pt and size of Zr ≈ Hf (size of 5d series elements is ≈ size of 4d series elements).
Correct answer: Ti<Zr<Hf- A. I+<I <I- ⇒ order of size.
- B. Se <S <Cl ⇒ order of ionisation energy.
- C. F<O<N ⇒ order of oxidizing property
- D. Na <Mg <Al ⇒ order of electronegativity.
Explanation: N<O<F is correct order of oxidizing property, as oxidizing property increases in a period. Therefore option C is correct.
Correct answer: F<O<N ⇒ order of oxidizing property- A. Greater metallic character
- B. Larger atomic size
- C. Strong reducing agent
- D. Less electropositive
Explanation: The factors upon which the ionization energy of an atom mainly depends are magnitude of nuclear charge, size of the atom, and the…
Correct answer: Less electropositive- A. Al
- B. Si
- C. P
- D. S
Explanation: Sulfur is located below phosphorus in the periodic table, in the same group (Group 16).
Correct answer: S