Free s-Block and p-Block Elements MCQs with Answers
1,359 s-Block and p-Block Elements MCQs from Chemistry, each with the correct answer and a written explanation of why it is correct. Free and unlimited, with no account needed.
The s and p blocks are studied through their valence-shell configurations, periodic trends and characteristic chemical properties. Work covers the reactions of Group I and Group II elements, the behaviour of Group IV elements, and how atomic size, ionization energy, electronegativity and metallic character change across periods and down groups.
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- A. Decreases
- B. Increases
- C. I st increase and then decrease
- D. I st decrease and then increase
Explanation: Ionic radius decreases moving from left to right across a row or period.
Correct answer: Decreases- A. decreases down the groups
- B. increases down the groups
- C. remains same across the periods
- D. increases across the periods
Explanation: As we go down the group, the size of atoms increases. Due to this the nuclear force of attraction on the electrons in the valence shell…
Correct answer: increases down the groups- A. 2H2O + NOCI+Cl2
- B. H2 + NOCI+ 2HOCI
- C. H2O+ NOCI+ 2HCl
- D. 2H2O+NOCI+ 2Cl
Explanation: HNO3 and HCL mix in the ratio of 1:3 to give aqua regia that is 2H2O + NOCl + 2Cl.
Correct answer: 2H2O+NOCI+ 2Cl- A. Na2B2O7.10H2O
- B. Na2B4O7 . H2O
- C. Na2B4O7 . 10 H2O
- D. Na2B2O5 . 10 H2O
Explanation: The chemical formula of Tincal is Na2B4O7 10H2O This requires just a factual recall.
Correct answer: Na2B4O7 . 10 H2O- A. I-A, V-A and VIll-A
- B. I-A, IV A and Vl-A
- C. I-A, II-A and VIl-A
- D. I-A, IV-A and VIl-A
Explanation: Hydrogen resembles the elements of group of I A as it has one valence electron and can lose and electron to form a cation, it also…
Correct answer: I-A, IV-A and VIl-A- A. Down the Group
- B. Along period
- C. Along d- block
- D. All of these
Explanation: The covalent character 'decreases in groups' as we go from top to bottom.
Correct answer: Down the Group- A. All blocks
- B. The p-block elements
- C. The d and f block elements
- D. None of the blocks
Explanation: In the case of d-block elements due to the presence of electrons at d orbitals, is closer to the outermost shell of the metal.
Correct answer: The d and f block elements- A. The number of electrons used in bonding
- B. The number of orbits holding electrons
- C. The (proton) atomic number
- D. The relative atomic number
Explanation: A. Lithium uses 1 electron in bonding, Beryllium 2, Boron 3, Carbon 4, Nitrogen 3, Oxygen 2, Fluorine 1 and Ne being a noble gas doesn't…
Correct answer: The number of electrons used in bonding- A. An element which has high electronegativity always has high electron gain enthalpy.
- B. Electron gain enthalpy is the property of an isolated atom.
- C. Electronegativity is the property of bonded atoms.
- D. Both electronegativity and electron gain enthalpy are usually directly related to nuclear charge and inversely related to atomic size.
Explanation: Option A is the correct answer because it incorrectly states that high electronegativity always corresponds to high electron gain…
Correct answer: An element which has high electronegativity always has high electron gain enthalpy.- A. Generally, the reducing character of elements increases across a period.
- B. Generally, the oxidizing character of elements increases across a period.
- C. Generally, the basic character of oxides decreases down a group.
- D. All are correct.
Explanation: The correct answer is Option B: 'Generally, the oxidizing character of elements increases across a period.' This is because as you move…
Correct answer: Generally, the oxidizing character of elements increases across a period.- A. IE1 of Li > IE1 of Be
- B. IE1 of Be > IE1 of B
- C. IE1 of Li < IE1 of Ca
- D. IE1 of He > IE1 of Ne
Explanation: The incorrect statement is Option A: IE1 of Li > IE1 of Be. This is incorrect because ionization energy increases across a period from…
Correct answer: IE1 of Li > IE1 of Be- A. Number of electrons
- B. Atomic number
- C. Number of valence electrons
- D. Electronic configurations
Explanation: The correct answer is that elements in the same vertical group of the periodic table have the same number of valence electrons.
Correct answer: Number of valence electrons- A. The element with the highest ionization energy (IE) belongs to group 18.
- B. In each period, the element with the lowest ionization energy belongs to group 1.
- C. In each period, the element with the highest ionization energy is a noble gas.
- D. In the second period, as we move from left to right, ionization energy increases regularly.
Explanation: The correct answer is Option D. While moving across a period from left to right, the general trend is an increase in ionization energy due…
Correct answer: In the second period, as we move from left to right, ionization energy increases regularly.- A. A: F > Cl > O > S
- B. B: S > Cl > O > F
- C. C: F > O > Cl > S
- D. D: Cl > F > O > S
Explanation: The correct order of increasing electronegativity is F > O > Cl > S. Electronegativity increases across a period from left to right and…
Correct answer: C: F > O > Cl > S- A. Among halogens, oxidizing behavior increases down the group
- B. Among alkali metals, reducing character increases down the group
- C. Fluorine is the most electronegative element
- D. Lithium is the hardest metal among alkali metals.
Explanation: The correct answer is Option A. In the halogen group, oxidizing behavior decreases down the group because the elements become less…
Correct answer: Among halogens, oxidizing behavior increases down the group- 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.1120. 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.