Unit 9: Classification of Elements and Periodicity in Properties — Online MCQ Test
CHEMISTRY · NEET (All) · NEET
Practice Unit 9: Classification of Elements and Periodicity in Properties with a free chapter-wise online MCQ test.
This chapter covers: Modem periodic law and present form of the periodic table s p. d and f block elements periodic trends in properties of elements atomic and ionic radii ionization enthalpy electron....
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Unit 9: Classification of Elements and Periodicity in Properties — Important Questions & Answers
Which among the following electronic configurations belong to main group elements? A. [Ne]3s¹ B. [Ar]3d³4s² C. [Kr]4d¹⁰5s²5p⁵ D. [Ar]3d¹⁰4s¹ E. [Rn]5f⁰6d²7s²
- A. B and E only
- B. A and C only
- C. D and E only
- D. A, C and D only
Answer: B. A and C only
Main group elements have s or p orbitals as the valence shell (s-block and p-block). A ([Ne]3s¹) and C ([Kr]4d¹⁰5s²5p⁵) are main group elements. B, D, and E are transition or inner-transition metals.
Main group elements have s or p orbitals as the valence shell (s-block and p-block). A ([Ne]3s¹) and C ([Kr]4d¹⁰5s²5p⁵) are main group elements. B, D, and E are transition or inner-transition metals.
In which of the following options the order of arrangement does not agree with the variation of property indicated against it?
- A. B < C < N < O (increasing first ionisation enthalpy)
- B. I < Br < Cl < F (increasing electron gain enthalpy)
- C. Li < Na < K < Rb (increasing metallic radius)
- D. Al3+ < Mg2+ < Na+ < F- (increasing ionic size)
Answer: A. B < C < N < O (increasing first ionisation enthalpy)
First IE of Nitrogen is higher than Oxygen due to stable half-filled 2p subshell (2p3). Electron gain enthalpy of Fluorine is less negative than Chlorine due to small size and inter-electronic repulsion. Both A and B are technically incorrect statements.
First IE of Nitrogen is higher than Oxygen due to stable half-filled 2p subshell (2p3). Electron gain enthalpy of Fluorine is less negative than Chlorine due to small size and inter-electronic repulsion. Both A and B are technically incorrect statements.
Which one of the following orders in correct for the bond dissociation enthalpy of halogen molecules?
- A. Cl2 > Br2 > F2 > I2
- B. Br2 > I2 > F2 > Cl2
- C. F2 > Cl2 > Br2 > I2
- D. I2 > Br2 > Cl2 > F2
Answer: A. Cl2 > Br2 > F2 > I2
Due to the small size of the fluorine atom, the lone pairs have significant inter-electronic repulsion, weakening the F-F bond. Thus, Cl2 has the highest bond dissociation enthalpy.
Due to the small size of the fluorine atom, the lone pairs have significant inter-electronic repulsion, weakening the F-F bond. Thus, Cl2 has the highest bond dissociation enthalpy.
Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules ?
- A. Cl2 > Br2 > F2 > I2
- B. Br2 > I2 > F2 > Cl2
- C. F2 > Cl2 > Br2 > I2
- D. I2 > Br2 > Cl2 > F2
Answer: A. Cl2 > Br2 > F2 > I2
The bond dissociation enthalpy of F2 is lower than that of Cl2 and Br2 due to the small size of the fluorine atom, which leads to strong inter-electronic repulsion between the lone pairs.
The bond dissociation enthalpy of F2 is lower than that of Cl2 and Br2 due to the small size of the fluorine atom, which leads to strong inter-electronic repulsion between the lone pairs.
Arrange the following elements in increasing order of electronegativity: N, O, F, C, Si Choose the correct answer from the options given below:
- A. O < F < N < C < Si
- B. F < O < N < C < Si
- C. Si < C < N < O < F
- D. Si < C < O < N < F
Answer: C. Si < C < N < O < F
Electronegativity increases across a period and decreases down a group. Thus Si < C < N < O < F.
Electronegativity increases across a period and decreases down a group. Thus Si < C < N < O < F.
Arrange the following elements in increasing order of first ionization enthalpy: Li, Be, B, C, N. Choose the correct answer from options given below:
- A. Li < Be < C < B < N
- B. Li < Be < N < B < C
- C. Li < Be < B < C < N
- D. Li < B < Be < C < N
Answer: C. Li < Be < B < C < N
Across Period 2, ionization enthalpy generally increases, with the known exception B < Be due to the stable filled 2s subshell in Be. Hence Li < Be < B < C < N.
Across Period 2, ionization enthalpy generally increases, with the known exception B < Be due to the stable filled 2s subshell in Be. Hence Li < Be < B < C < N.
The element Z = 114 has been discovered recently. It will belong to which of the following family group and electronic configuration?
- A. Carbon family, [Rn] 5f14 6d10 7s2 7p2
- B. Oxygen family, [Rn] 5f14 6d10 7s2 7p4
- C. Nitrogen family, [Rn] 5f14 6d10 7s2 7p6
- D. Halogen family, [Rn] 5f14 6d10 7s2 7p5
Answer: A. Carbon family, [Rn] 5f14 6d10 7s2 7p2
Element with Z = 114 (Flerovium) belongs to Group 14 (Carbon family) with the outer electronic configuration ns2 np2.
Element with Z = 114 (Flerovium) belongs to Group 14 (Carbon family) with the outer electronic configuration ns2 np2.
Ionic mobility of which of the following alkali metal ions is lowest when aqueous solution of their salts are put under an electric field?
- A. K
- B. Rb
- C. Li
- D. Na
Answer: C. Li
Li+ has the smallest size and the highest charge density among alkali metals, leading to the greatest degree of hydration. This makes the hydrated ion the largest and slowest in an aqueous electric field.
Li+ has the smallest size and the highest charge density among alkali metals, leading to the greatest degree of hydration. This makes the hydrated ion the largest and slowest in an aqueous electric field.
Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is 1s² 2s² 2p³, the simplest formula for this compound is
- A. Mg₂X₃
- B. Mg₃X₂
- C. Mg₂X
- D. MgX₂
Answer: B. Mg₃X₂
Element X with configuration 1s² 2s² 2p³ is nitrogen (N), which has 5 valence electrons and forms X³⁻ ion (gaining 3 electrons to complete octet). Magnesium forms Mg²⁺ ion. To balance charges: 3 Mg²⁺ provides +6 charge; 2 N³⁻ provides −6 charge. So the formula is Mg₃X₂ (analogous to Mg₃N₂).
Element X with configuration 1s² 2s² 2p³ is nitrogen (N), which has 5 valence electrons and forms X³⁻ ion (gaining 3 electrons to complete octet). Magnesium forms Mg²⁺ ion. To balance charges: 3 Mg²⁺ provides +6 charge; 2 N³⁻ provides −6 charge. So the formula is Mg₃X₂ (analogous to Mg₃N₂).
Which of the following is an amphoteric hydroxide?
- A. Be(OH)2
- B. Sr(OH)2
- C. Ca(OH)2
- D. Mg(OH)2
Answer: A. Be(OH)2
Beryllium hydroxide, Be(OH)2, is amphoteric as it reacts with both acids and bases.
Beryllium hydroxide, Be(OH)2, is amphoteric as it reacts with both acids and bases.