Given below are two statements: Statement (I): According to the Law of Octaves, the elements were arranged in the increasing order of their atomic number. Statement (II): Meyer observed a periodically repeated pattern upon plotting physical properties of certain elements against their respective atomic numbers. In the light of the above statements, choose the correct answer from the options given below:

Solution & Explanation

### Related Formula Early historical periodic classification principles depended on atomic weight attributes: textPeriodic Property = f(textAtomic Weight) ### Core Logic Evaluating historical accuracy: - **Statement I**: Newlands' Law of Octaves arranged elements in the increasing order of their **atomic weights** (not atomic numbers). Hence, Statement I is false. - **Statement II**: Lothar Meyer plotted physical properties like atomic volume, melting point, and boiling point against **atomic weight** (not atomic number). Hence, Statement II is false. ### Step 1: Conclusion Since both statements incorrectly reference atomic number instead of atomic weight, both Statement I and Statement II are false. ### Pattern Recognition Almost all classical periodic classifiers (Newlands, Döbereiner, de Chancourtois, Lothar Meyer, Mendeleev) relied strictly on *atomic weight*. The pivot to *atomic number* occurred later with Moseley's X-ray studies and the Modern Periodic Law. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Reference Study Guides

Classification Of Elements And Periodicity In Properties

More Classification of Elements and Periodicity in Properties Previous-Year Questions — Page 6

Q84 jee_main_2024_30_jan_morning IUPAC Nomenclature for Elements > 100
If IUPAC name of an element is "Unununnium" then the element belongs to nth group of periodic table. The value of n is
Numerical Answer. Answer: 11 to 11

Solution

### Core Logic Decode the IUPAC name "Unununnium" into its atomic number. Un = 1 Un = 1 Un = 1 Suffix = ium So, the atomic number is 111. ### Step 1: Finding group number For d-block elements with Z > 100, the group number is the last two digits of the atomic number. Since Z = 111, the element belongs to group 11 (the Copper group). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q80 jee_main_2024_31_jan_evening Periodic Trends in Properties
Consider the following elements.
Group downarrowA^primeB^prime rightarrow Period
C^primeD^prime
Which of the following is/are true about A^prime, B^prime, C^prime and D^prime? A. Order of atomic radii: B^prime < A^prime < D^prime < C^prime B. Order of metallic character: B^prime < A^prime < D^prime < C^prime C. Size of the element: D^prime < C^prime < B^prime < A^prime D. Order of ionic radii: B^prime + < A^prime + < D^prime + < C^prime + Choose the correct answer from the options given below:
  • A. text1. A only
  • B. text2. A, B and D only
  • C. text3. A and B only
  • D. text4. B, C and D only

Solution

### Core Logic From the grid layout, A^prime and B^prime are in the same period (left to right), while C^prime and D^prime are below them in the next period. In general, moving along a period from left to right, size decreases and metallic character decreases. Moving down a group, size increases and metallic character increases. Atomic Radii / Size: B^prime < A^prime (across period) D^prime < C^prime (across period) A^prime < C^prime and B^prime < D^prime (down the group) Combining them: B^prime < A^prime < D^prime < C^prime. (Statement A is correct, C is incorrect). Metallic Character: B^prime < A^prime and D^prime < C^prime Overall trend follows size closely for these blocks: B^prime < A^prime < D^prime < C^prime. (Statement B is correct). Ionic Radii (+ ions): Follows the same basic trend as atomic radii: B^prime + < A^prime + < D^prime + < C^prime +. (Statement D is correct). ### Step 1: Final Selection Statements A, B, and D are correct. Statement C contradicts A, so it is incorrect. Option 2 is the right choice. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q70 jee_main_2024_31_jan_morning Electron Gain Enthalpy
The correct sequence of electron gain enthalpy of the elements listed below is A. Ar B. Br C. F D. S Choose the most appropriate from the options given below:
  • A. C > B > D > A
  • B. A > D > B > C
  • C. A > D > C > B
  • D. D > C > B > A

Solution

### Core Logic Electron gain enthalpy (Delta_egH) values are considered with their signs when ordering from largest to smallest algebraically.
ElementDelta_egH text (kJ/mol)
F-333
S-200
Br-325
Ar+96
Arranging these algebraically from highest to lowest: +96 > -200 > -325 > -333 Thus, Ar > S > Br > F, which corresponds to A > D > B > C. ### Pattern Recognition Noble gases always have positive electron gain enthalpies (highest algebraically). Halogens have highly negative values (lowest algebraically). Fluorine is highly negative, Bromine is slightly less negative. Sulfur sits in between. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

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