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NEET CHEMISTRYClassification of Elements and Periodicity in PropertiesMedium

Question

Identify the incorrect statement:

A

The oxidation state and coordination number (or covalency) of Al in [AlCl(H₂O)₅]²⁺ are +3 and 6, respectively.

B

Na₂O is a basic oxide and Cl₂O₇ is an acidic oxide.

C

The following four species are called isoelectronic species: O²⁻, F⁻, Na⁺ and Mg²⁺.

D

Among the four species Mg, Al, Mg²⁺ and Al³⁺, the smallest one is Al.

Step-by-Step Solution

We analyse each statement based on periodic trends described in NCERT:

  1. Correct: In the complex [AlCl(H2O)5]2+[AlCl(H_2O)_5]^{2+}, Aluminum is bonded to 1 chloride ion and 5 water molecules, making the coordination number 6. The charge is calculated as x+(1)+5(0)=+2x + (-1) + 5(0) = +2, so x=+3x = +3. This statement is factually correct (See NCERT Class 11 Problem 3.9).
  2. Correct: Na2ONa_2O is an oxide of an alkali metal and is strongly basic. Cl2O7Cl_2O_7 is an oxide of a halogen in a high oxidation state and is strongly acidic. (See NCERT Class 11 Section 3.7.3).
  3. Correct: Isoelectronic species have the same number of electrons. O2O^{2-} (8+2=10), FF^- (9+1=10), Na+Na^+ (11-1=10), and Mg2+Mg^{2+} (12-2=10) all have 10 electrons. (See NCERT Class 11 Section 3.7.1).
  4. Incorrect: Cations are always smaller than their parent atoms due to increased effective nuclear charge holding the remaining electrons. Thus, Al3+<AlAl^{3+} < Al and Mg2+<MgMg^{2+} < Mg. Between the isoelectronic ions Mg2+Mg^{2+} and Al3+Al^{3+}, Al3+Al^{3+} has a higher nuclear charge (Z=13) than Mg2+Mg^{2+} (Z=12), pulling the electron cloud tighter. Therefore, Al3+Al^{3+} is smaller than Mg2+Mg^{2+}. Generally, atomic size decreases across a period (Al<MgAl < Mg). The overall order of size is Mg>Al>Mg2+>Al3+Mg > Al > Mg^{2+} > Al^{3+}. The smallest species is Al3+Al^{3+}, not Al. (See NCERT Class 11 Problem 3.5).

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 Propertiesidentifyincorrectstatement

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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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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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