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NEET CHEMISTRYClassification of Elements and Periodicity in PropertiesEasy

Question

Identify the correct order of the size of the following species: Ca²⁺, K⁺, Ar, S²⁻, Cl⁻

A

Ca²⁺ < K⁺ < Ar < S²⁻ < Cl⁻

B

Ca²⁺ < K⁺ < Ar < Cl⁻ < S²⁻

C

Ar < Ca²⁺ < K⁺ < Cl⁻ < S²⁻

D

Ca²⁺ < Ar < K⁺ < Cl⁻ < S²⁻

Step-by-Step Solution

The species listed (Ca²⁺, K⁺, Ar, S²⁻, Cl⁻) are isoelectronic, meaning they all possess the same number of electrons (18 electrons).

  1. Concept: For isoelectronic species, the radius is inversely proportional to the effective nuclear charge (atomic number, Z). As the number of protons increases, the attraction on the same number of electrons increases, pulling the electron cloud closer and reducing the size.
  2. Analysis of Atomic Numbers (Z): Sulphur (S): Z = 16 Chlorine (Cl): Z = 17 Argon (Ar): Z = 18 Potassium (K): Z = 19
  • Calcium (Ca): Z = 20
  1. Trend:
  • Largest Radius: S²⁻ (Lowest Z, weakest attraction).
  • Smallest Radius: Ca²⁺ (Highest Z, strongest attraction).
  1. Order: Increasing size follows decreasing atomic number: Ca²⁺ < K⁺ < Ar < Cl⁻ < S²⁻.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Classification of Elements and Periodicity in Properties. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYClassification of Elements and Periodicity in Propertiesidentifycorrectfollowingspecies

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The correct sequence of increasing radii is:

A.Ar < K⁺ < Ca²⁺
B.Ca²⁺ < Ar < K⁺
C.Ca²⁺ < K⁺ < Ar
D.K⁺ < Ar < Ca²⁺
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Which one of the following represents all isoelectronic species?

A.Na⁺, Cl⁻, O⁻, NO⁺
B.N₂O, N₂O₄, NO⁺, NO
C.Na⁺, Mg²⁺, O⁻, F⁻
D.Ca²⁺, Ar, K⁺, Cl⁻
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The value of electron gain enthalpy of $Na^+$, if $IE_1$ of Na = 5.1 eV, is:

A.+10.2 eV
B.–5.1 eV
C.–10.2 eV
D.+2.55 eV
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A.H⁺
B.Li⁺
C.Na⁺
D.Mg²⁺
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Identify the correct order of the size of the following species: $Ca^{2+}, K^+, Ar, S^{2-}, Cl^-$

A.$Ca^{2+} < K^+ < Ar < S^{2-} < Cl^-$
B.$Ca^{2+} < K^+ < Ar < Cl^- < S^{2-}$
C.$Ar < Ca^{2+} < K^+ < Cl^- < S^{2-}$
D.$Ca^{2+} < Ar < K^+ < Cl^- < S^{2-}$
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The formation of the oxide ion O²⁻(g) from oxygen atom requires first an exothermic and then an endothermic step as shown below: O(g) + e⁻ → O⁻(g); ΔH° = -141 kJ mol⁻¹ O⁻(g) + e⁻ → O²⁻(g); ΔH° = +780 kJ mol⁻¹ Thus, the process of formation of O²⁻ in gas phase is unfavourable even though O²⁻ is isoelectronic with neon. It is due to the fact that:

A.Electron repulsion outweighs the stability gained by achieving noble gas configuration
B.O⁻ ion has comparatively smaller size than oxygen atom
C.Oxygen is more electronegative
D.Addition of electron in oxygen result in large size of the ion
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A.Si < C < O < N < F
B.O < F < N < C < Si
C.F < O < N < C < Si
D.Si < C < N < O < F
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A.Cr > Mn > V > Ti
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D.Ti > V > Cr > Mn
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