Mendeleev chose hydrogen (H) and oxygen (O) as the basis for classification because these two elements form compounds (hydrides and oxides) with most other elements.

He selected them because they are the most widely combining elements — hydrogen is the lightest and oxygen is the most abundant reactive element.

The formulae of hydrides and oxides helped Mendeleev decide the group of an element. For example, if an element X formed an oxide XO and a hydride XH₄, it was placed in the group whose general formula matched. The valency of the element, shown by these formulae, determined its position in the periodic table.

Source: Chapter 5 (Periodic Classification of Elements), Mendeleev's Periodic Table section

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• 1 mark each for: naming the two elements (H and O), reason for choosing them, and how formulae determined position — three clear points needed.
• Examiners look for: naming both elements correctly, stating that oxides and hydrides were used, and linking the formula (valency) to group placement.
• Avoid vague statements — be specific that it was the formulae of oxides and hydrides that revealed valency and thus group number.
• Note: The source passages provided are from Chemistry Chapter 4 (Carbon), which does not cover Mendeleev. This answer is based on the correct CBSE Chapter 5 content. Always match your answer to the right chapter.

In a period (left to right): Nuclear charge increases but electrons are added to the same shell. The increased nuclear pull acts on electrons in the same energy level, pulling them closer. So atomic size decreases across a period.

In a group (top to bottom): Nuclear charge increases, but a new shell is added with each element. The outermost electrons are farther from the nucleus and are also shielded by inner shells. So atomic size increases down a group.

Thus, the opposing effect of addition of new shells in a group overrides the increased nuclear charge, causing different trends in the two cases.

The examiner expects two separate, reasoned explanations — one for period and one for group. The key contrast is: same shell vs. new shell added. Mentioning "shielding effect" for the group trend is a bonus but expected at this level. Do not just state the trend — always link it to the reason (nuclear charge vs. shell addition/shielding).

(a) Newland's Law of Octaves:
When elements are arranged in increasing order of their atomic masses, every eighth element has properties similar to the first element, just like the eighth note of a musical octave.

(b) Dobereiner's Triads:
Dobereiner grouped elements with similar properties into sets of three (triads), where the atomic mass of the middle element was approximately the arithmetic mean of the other two.
Example: In the triad Li (7), Na (23), K (39) — atomic mass of Na = (7+39)/2 = 23. ✓

(c) Limitations:

• Newland's Law of Octaves: It was applicable only up to calcium; elements discovered later did not fit the pattern.
• Dobereiner's Triads: Only a few elements could be grouped into triads; the law could not cover all known elements.

• The question is from Chapter 5 (Periodic Classification of Elements), not the provided passages — answer is based on standard NCERT Class 10 Science content.
• For (b), always give a specific triad with atomic masses and show the mean calculation — examiners expect the numerical verification.
• For (c), one limitation each is enough (½ mark each); keep them distinct — one for Octaves, one for Triads.
• Do not mix up the two scientists' contributions.

The table is associated with Newlands' Law of Octaves.

Two Important Features:

1. When elements are arranged in increasing order of atomic masses, every eighth element has properties similar to the first, like the eighth note of a musical scale (octave).
2. It was the first attempt to classify elements systematically, covering elements from hydrogen (H) to thorium (Th).

Two Anomalies:

1. It worked only up to calcium; after calcium, every eighth element did not possess properties similar to the first element.
2. Newlands adjusted two elements in one slot (e.g., Co and Ni, Ce and La) to fit the pattern, which was not justified.

The musical note pattern in the table is the direct clue — Newlands compared element repetition to musical octaves. Examiners expect you to name the law clearly, then give exactly two features and two anomalies (each as a distinct point). Common mistakes: confusing this with Mendeleev's law, or giving only one anomaly. Stick to the points above for full marks.

Criteria used by Mendeleev: Mendeleev arranged elements in increasing order of their atomic masses and grouped elements with similar chemical properties in the same vertical column (group).

Isotopes — comparison:

• In Mendeleev's Periodic Table, isotopes of an element would have to be placed in different positions (as they have different atomic masses), which was a drawback.
• In the Modern Periodic Table, isotopes are placed in the same position (same group and period) since elements are arranged by atomic number, which is the same for all isotopes of an element.

• The question has two parts — state the criterion AND compare isotope positions. Allocate roughly one line to Mendeleev's criterion and two lines to the comparison.
• Key terms examiners look for: atomic mass (Mendeleev), similar properties, atomic number (Modern), and the resolution of the isotope problem.
• A common mistake is forgetting to explicitly state that isotopes have the same atomic number but different atomic masses — this is the logical basis for the comparison.

(a) Two advantages of using atomic number as the basis of classification:

1. Atomic number is a unique, fixed property of each element, leaving no ambiguity in an element's position in the table.
2. It removes anomalies of Mendeleev's table (e.g., cobalt and nickel are placed correctly based on atomic number, not atomic mass).

(b) Electronic configurations:

• Element X (atomic number 13): 2, 8, 3
• Element Y (atomic number 20): 2, 8, 8, 2

• For part (a), examiners expect any two valid advantages — uniqueness of atomic number and removal of Mendeleev's anomalies are the most standard ones. Avoid vague answers.
• For part (b), simply fill shells in order (2, 8, 8, 2 for 20 electrons). These are straightforward but students often make errors with Y — remember the 3rd shell can hold up to 18, but it stabilises at 8 before the 4th shell begins filling.

(a) Element X has 1 electron in its outermost shell → Group 1. Element Y has 7 electrons in its outermost shell → Group 17.

(b) X donates 1 electron and Y accepts 1 electron. The molecular formula is XY (e.g., NaCl type). (Formula: XY)

(c) Element X is electropositive, as it tends to lose electrons and form a positive ion.

Source: Chapter 3, Section 3.3

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• Elements with 1 valence electron are in Group 1 (alkali metals); elements with 7 are in Group 17 (halogens).
• Since X needs to lose 1 electron and Y needs to gain 1, they combine in a 1:1 ratio → formula XY.
• Electropositive elements are those that lose electrons easily (metals/Group 1), so X is electropositive. Examiners expect you to justify briefly — "loses electrons" is the key phrase.