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NEET CHEMISTRYClassification of Elements and Periodicity in PropertiesMedium

Question

Identify the wrong statement in the following:

A

Amongst isoelectronic species, smaller the positive charge on the cation, smaller is the ionic radius

B

Amongst isoelectronic species, greater the negative charge on the anion, larger is the ionic radius

C

Atomic radius of the elements increases as one moves down the first group of the Periodic Table

D

Atomic radius of the elements decreases as one moves across from left to right in the 2nd period of the Periodic Table

Step-by-Step Solution

Let's analyze the statements based on periodic trends and isoelectronic species:

  1. Isoelectronic Species: These are species with the same number of electrons. Their size is governed by the nuclear charge (ZZ). A higher nuclear charge exerts a stronger attraction on the electrons, shrinking the radius .
  • Cations: Comparing isoelectronic cations (e.g., Na+Na^+ and Mg2+Mg^{2+}), the one with the smaller positive charge (Na+Na^+) has fewer protons (Z=11Z=11) than the one with the larger charge (Mg2+Mg^{2+}, Z=12Z=12). Fewer protons mean less attraction, resulting in a larger radius. Therefore, the statement "smaller the positive charge... smaller is the ionic radius" is incorrect.
  • Anions: A greater negative charge means fewer protons relative to electrons, leading to reduced effective nuclear charge and increased electron-electron repulsion, which expands the radius. This statement is correct .
  1. Group Trend: Atomic radius increases down a group (e.g., Group 1) due to the addition of electron shells shielding the outer electrons . This statement is correct.
  2. Period Trend: Atomic radius decreases across a period (e.g., 2nd period) due to increasing effective nuclear charge pulling valence electrons closer . This statement is correct.

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 Propertiesidentifystatementfollowing

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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⁺
EasySolve

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⁻
EasySolve

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
EasySolve

Be²⁺ is isoelectronic with which of the following ions?

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.$Ca^{2+} < Ar < K^+ < Cl^- < S^{2-}$
MediumSolve

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
MediumSolve

Which of the following arrangements is correct for increasing order of electronegativity?

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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The correct order of decreasing second ionization enthalpy of Ti(22), V(23), Cr(24) and Mn(25) is:

A.Cr > Mn > V > Ti
B.V > Mn > Cr > Ti
C.Mn > Cr > Ti > V
D.Ti > V > Cr > Mn
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