Periodic Classification of Elements for CTET Paper II (Class 6-8)

intermediate 18 min read

Concept

The periodic table is not just a chart on a classroom wall — it is one of the most powerful predictive tools in all of science. The core idea is deceptively simple: when elements are arranged in order of increasing atomic number, their properties repeat in a regular, predictable pattern. This repetition is called periodicity.

Think of it like the days of the week. Monday always follows Sunday, and the same character of a "Monday" — new beginnings, fresh energy, whatever you associate with it — repeats every seven days. In the periodic table, certain chemical personalities (reactive metals, inert gases, aggressive non-metals) keep reappearing as you move across the table and down the columns.

Before arriving at the modern table, scientists tried multiple classification schemes. Döbereiner noticed triads of elements with similar properties where the middle element's atomic mass was the average of the other two. Newlands tried arranging elements like musical notes (Law of Octaves). Mendeleev took the decisive step — he arranged 63 known elements by increasing atomic mass and left deliberate gaps for elements not yet discovered, predicting their properties in advance. The modern periodic table, built on atomic number rather than atomic mass (Henry Moseley's contribution), resolved the anomalies Mendeleev's version could not.

For CTET Paper II, you are preparing to teach this to Class 8 students. That means two things: you need conceptual accuracy (no fuzzy understanding), and you need to see the pedagogical structure — which idea comes first, what misconceptions students carry, and how the history of classification itself teaches scientific thinking.

The questions you will face test whether you can read an electronic configuration and derive the period and group, identify historical facts about early classification, and apply the periodic law to predict element properties. These are straightforward once the underlying logic is airtight.


Deep Dive

Historical Development: Why It Matters for CTET

CTET questions regularly ask about the history of classification because the NCERT Class 10 chapter "Periodic Classification of Elements" explicitly covers it. Do not skip this section.

Döbereiner's Triads (1817)

Johann Wolfgang Döbereiner identified groups of three elements with similar properties where:

Atomic mass of middle element ≈ (Atomic mass of first + Atomic mass of third) / 2

Classic triads to remember:

Limitation: Only a few triads worked. Most elements could not be fitted into triads.

Newlands' Law of Octaves (1865)

Newlands arranged elements by increasing atomic mass and observed that every eighth element had properties similar to the first — like the eight notes of a musical octave. This worked reasonably well up to calcium, but broke down for heavier elements. Noble gases had not been discovered, so his eighth-element pattern had structural holes.

Mendeleev's Periodic Table (1869)

Mendeleev's key achievements:

  1. Arranged 63 elements by increasing atomic mass, grouping similar properties in the same vertical column.
  2. Left gaps for undiscovered elements and predicted their properties (density, atomic mass, oxide formula).
  3. His three famous predictions:
    • Eka-boron → discovered as Scandium (Sc)
    • Eka-aluminium → discovered as Gallium (Ga)
    • Eka-silicon → discovered as Germanium (Ge) in 1886

Limitation: Could not explain why atomic mass ordering sometimes had to be violated (e.g., Cobalt/Nickel, Tellurium/Iodine). Noble gases, discovered later, needed to be accommodated awkwardly.

The Modern Periodic Table

The Modern Periodic Law (based on Moseley's work): Properties of elements are a periodic function of their atomic number.

This resolved all of Mendeleev's anomalies — atomic number, not mass, is the fundamental property.

Structure of the Modern Periodic Table

Periods (horizontal rows): There are 7 periods. The period number tells you how many electron shells (energy levels) an element has.

| Period | Elements | Max electrons in outer shell | |--------|----------|------------------------------| | 1 | H, He | 2 | | 2 | Li to Ne | 8 | | 3 | Na to Ar | 8 | | 4 | K to Kr | 18 |

Groups (vertical columns): There are 18 groups. The group number (for Groups 1, 2 and 13–18) tells you the number of valence electrons.

Key groups to know cold:

Reading Electronic Configuration: The Core Skill

Every CTET question about period and group reduces to this two-step read:

Step 1 — Count the shells → Period number

Electronic configuration 2, 8, 7 has 3 shells → Period 3

Electronic configuration 2, 8, 8, 1 has 4 shells → Period 4

Step 2 — Count valence electrons → Group number

For Groups 1–2 and 13–18: valence electrons = group number directly.

2, 8, 7 → 7 valence electrons → Group 17

2, 8, 8, 2 → 2 valence electrons → Group 2

Trends in the Periodic Table

Across a period (left to right):

Down a group (top to bottom):


Memory Tricks & Shortcuts

patternShell Count = Period, Valence Count = Group

When you see an electronic configuration in a question, do exactly two counts and stop. First count: how many numbers are in the configuration? That is your period. Second count: what is the last number? That is your group (for Groups 1–2 and 13–18).

Example: Configuration 2, 8, 8, 1 → 4 numbers → Period 4; last number is 1 → Group 1. Done in 5 seconds.

Standard method (writing out shells and reasoning): ~45 seconds. This pattern: ~5 seconds.

patternDöbereiner in One Formula

For any Döbereiner Triad question, apply one formula instantly:

Middle element mass = (First + Third) / 2

You do not need to remember which triads are valid — the question gives you the numbers. Just add the two given masses, divide by 2, and match to the options.

Li (7) + K (39) = 46 → 46/2 = 23 = Na. Confirmed in 8 seconds versus re-reading the triad logic: ~30 seconds.

patternSame Group = Same Valence Electrons, Different Period = Different Shells

Questions about "same number of valence electrons but different periods" are asking: same group, different period. Just check which pair of elements is in the same vertical group but different horizontal rows.

Na (Group 1, Period 3) and K (Group 1, Period 4): same group (1 valence electron), different periods — this is the answer pattern every time. You can eliminate options by checking: if two elements are in the same period, they cannot share this property. Eliminate Na-Mg (same period 3), Cl-Ar (same period 3), Li-Be (same period 2) in under 10 seconds.

This elimination drops 4-option questions to a guaranteed answer in 3 steps instead of working through all options.

patternEka-Mendeleev: The Three Gaps

Mendeleev's three predicted elements appear in CTET options alongside decoys. The pattern is fixed — memorise exactly three pairs:

  • Eka-boron → Scandium (Sc)
  • Eka-aluminium → Gallium (Ga)
  • Eka-silicon → Germanium (Ge)

When you see "Eka-silicon" in a question, write Ge immediately — 3 seconds. The wrong options (Boron, Gallium, Scandium) are the other two correct pairs reshuffled. Knowing all three prevents the shuffle from catching you.

eliminationBasic Oxide + Reacts with Water = Group 1 Metal

A question describing an element as having a basic oxide and reacting with water to form an alkali is describing a Group 1 alkali metal by definition. Eliminate all non-metals immediately (C, N, O — their oxides are acidic or neutral). Eliminate Group 2 metals when the question specifies vigorous reaction with water (Be and Mg react very slowly with cold water). Narrows four options to one in 10 seconds versus working through oxide chemistry for each element: ~60 seconds.


Fast-Solving Framework

When you see a periodic table question in the exam, run this decision tree:

Is the question about historical classification? → Yes: Identify the scientist — Döbereiner (triads, arithmetic mean), Newlands (octaves, eighth element), Mendeleev (gaps, eka-elements, atomic mass ordering), Moseley/Modern (atomic number ordering). → Apply the one relevant formula or fact. Do not mix up scientists' contributions.

Is the question about finding period and group from atomic number or configuration? → Write the electronic configuration if given atomic number (use 2, 8, 8, 2 shell-filling rule for elements up to 20). → Count shells → Period. Count valence electrons → Group. Two steps, full stop.

Is the question about trends (reactivity, atomic size, metallic character)? → Across period: size decreases, metallic character decreases, non-metallic character increases. → Down group: size increases, metallic character increases (for metals), non-metallic reactivity decreases. → Match the described property to the correct direction. Eliminate options that go in the wrong direction.

Is the question pedagogical (how to teach this topic)? → Think: concrete to abstract, historical development as a narrative, hands-on classification activities before the formal table.


Solved PYQs

Why this question: Tests whether you can identify the "same group, different period" relationship — the single most common periodic table question type in CTET.

Previous Year Questionपिछले वर्ष का प्रश्न
Which of the following pairs of elements has the SAME number of valence electrons but belongs to DIFFERENT periods?
निम्नलिखित में से किस जोड़े के तत्वों में संयोजकता इलेक्ट्रॉनों की संख्या समान है लेकिन वे अलग-अलग आवर्तों से संबंधित हैं?
  1. Na and Mg
  2. Cl and Ar
  3. Na and K
  4. Li and Be
  1. Na और Mg
  2. Cl और Ar
  3. Na और K
  4. Li और Be
Solutionसमाधान
Na (Period 3, Group 1) and K (Period 4, Group 1) both have 1 valence electron but are in different periods. Na and Mg are in the same period. Cl and Ar are in the same period. Li and Be are in the same period.
Na (आवर्त 3, समूह 1) और K (आवर्त 4, समूह 1) दोनों में 1 संयोजकता इलेक्ट्रॉन है लेकिन वे अलग-अलग आवर्तों में हैं। Na और Mg एक ही आवर्त में हैं। Cl और Ar एक ही आवर्त में हैं। Li और Be एक ही आवर्त में हैं।

Solving path: Check each pair for same period versus same group. Na (Period 3) and Mg (Period 3) — same period, eliminate. Cl (Period 3) and Ar (Period 3) — same period, eliminate. Li (Period 2) and Be (Period 2) — same period, eliminate. Na (Period 3, Group 1) and K (Period 4, Group 1) — different periods, same group, same valence electrons (1 each). Answer: Na and K.


Why this question: Döbereiner's arithmetic mean formula applied directly — the cleanest possible test of whether you know the law.

Previous Year Questionपिछले वर्ष का प्रश्न
Döbereiner's Law of Triads states that the atomic mass of the middle element is approximately the arithmetic mean of the other two. For the triad Li (7), Na, K (39), what is the approximate atomic mass of Na?
डोबेराइनर के त्रिक नियम के अनुसार मध्य तत्व का परमाणु द्रव्यमान अन्य दो का अंकगणितीय माध्य होता है। त्रिक Li (7), Na, K (39) के लिए Na का अनुमानित परमाणु द्रव्यमान क्या है?
  1. 19
  2. 31
  3. 23
  4. 27
  1. 19
  2. 31
  3. 23
  4. 27
Solutionसमाधान
According to Döbereiner's Law of Triads: atomic mass of middle element = (atomic mass of first + atomic mass of third) / 2 = (7 + 39) / 2 = 46 / 2 = 23. The atomic mass of Na is indeed 23.
डोबेराइनर के त्रिक नियम के अनुसार: मध्य तत्व का परमाणु द्रव्यमान = (पहले तत्व का द्रव्यमान + तीसरे तत्व का द्रव्यमान) / 2 = (7 + 39) / 2 = 46 / 2 = 23। Na का परमाणु द्रव्यमान वास्तव में 23 है।

Solving path: (7 + 39) / 2 = 46 / 2 = 23. Match to options — 23 is option C. Done. Do not overthink; the question is a single arithmetic step.


Why this question: Tests property-to-element reasoning, which requires knowing Group 1 characteristics and distinguishing them from Group 2 and non-metals.

Previous Year Questionपिछले वर्ष का प्रश्न
An element has the following properties: it is in Period 2, its oxide is basic, and it reacts vigorously with water to form an alkali. Which element is it most likely to be?
एक तत्व में निम्नलिखित गुण हैं: यह आवर्त 2 में है, इसका ऑक्साइड क्षारीय है, और यह पानी के साथ तीव्र प्रतिक्रिया करके एक क्षार बनाता है। यह तत्व सबसे संभावित रूप से कौन सा है?
  1. Nitrogen (N)
  2. Lithium (Li)
  3. Carbon (C)
  4. Beryllium (Be)
  1. नाइट्रोजन (N)
  2. लिथियम (Li)
  3. कार्बन (C)
  4. बेरीलियम (Be)
Solutionसमाधान
Lithium (Period 2, Group 1) has a basic oxide (Li₂O) and reacts with water to form LiOH (an alkali) and hydrogen gas. Beryllium forms an amphoteric oxide. Carbon and Nitrogen are non-metals whose oxides are acidic or neutral, not basic.
लिथियम (आवर्त 2, समूह 1) का ऑक्साइड क्षारीय (Li₂O) है और पानी के साथ प्रतिक्रिया करके LiOH (क्षार) और हाइड्रोजन गैस बनाता है। बेरीलियम उभयधर्मी ऑक्साइड बनाता है। कार्बन और नाइट्रोजन अधातु हैं जिनके ऑक्साइड अम्लीय या उदासीन होते हैं।

Solving path: "Basic oxide" eliminates C and N (acidic/neutral oxides). "Reacts vigorously with water to form alkali" eliminates Be (amphoteric oxide, slow water reaction). Period 2 leaves Li and Be — Be is already eliminated. Answer: Lithium (Li). The reaction is 2Li + 2H₂O → 2LiOH + H₂.


Why this question: Read electronic configuration from atomic number and place correctly — a core CTET skill for teaching Class 8.

Previous Year Questionपिछले वर्ष का प्रश्न
A teacher asks students to identify the element with atomic number 19. A student correctly states its period and group. Which answer is CORRECT?
एक शिक्षक छात्रों से परमाणु क्रमांक 19 वाले तत्व की पहचान करने के लिए कहते हैं। एक छात्र इसके आवर्त और समूह को सही बताता है। कौन सा उत्तर सही है?
  1. Period 4, Group 1
  2. Period 3, Group 1
  3. Period 4, Group 2
  4. Period 3, Group 9
  1. आवर्त 4, समूह 1
  2. आवर्त 3, समूह 1
  3. आवर्त 4, समूह 2
  4. आवर्त 3, समूह 9
Solutionसमाधान
Atomic number 19 is Potassium (K) with configuration 2, 8, 8, 1. It has 4 electron shells → Period 4. It has 1 valence electron → Group 1 (alkali metals).
परमाणु क्रमांक 19 पोटेशियम (K) है जिसका विन्यास 2, 8, 8, 1 है। इसमें 4 इलेक्ट्रॉन कोश हैं → आवर्त 4। इसमें 1 संयोजकता इलेक्ट्रॉन है → समूह 1 (क्षार धातु)।

Solving path: Atomic number 19 → configuration: 2 (fills shell 1), 8 (fills shell 2), 8 (fills shell 3), 1 (shell 4). Four shells → Period 4. One valence electron → Group 1. Answer: Period 4, Group 1.


Why this question: Tests Mendeleev's eka-element predictions, a fact-based question that appears regularly in CTET across years.

Previous Year Questionपिछले वर्ष का प्रश्न
Mendeleev left gaps in his Periodic Table for undiscovered elements and predicted their properties. The element 'Eka-silicon' predicted by him was later discovered as:
मेंडेलीव ने अनदेखे तत्वों के लिए अपनी आवर्त सारणी में खाली स्थान छोड़े और उनके गुणों की भविष्यवाणी की। उनके द्वारा भविष्यवाणी किए गए 'एका-सिलिकॉन' की खोज बाद में किस तत्व के रूप में हुई?
  1. Germanium
  2. Boron
  3. Gallium
  4. Scandium
  1. जर्मेनियम
  2. बोरॉन
  3. गैलियम
  4. स्कैंडियम
Solutionसमाधान
Mendeleev predicted three missing elements: Eka-boron (discovered as Scandium), Eka-aluminium (discovered as Gallium), and Eka-silicon (discovered as Germanium in 1886 by Clemens Winkler). The predicted properties of Eka-silicon closely matched Germanium's actual properties, validating Mendeleev's Periodic Law.
मेंडेलीव ने तीन तत्वों की भविष्यवाणी की थी: एका-बोरॉन (स्कैंडियम के रूप में खोजा गया), एका-एलुमिनियम (गैलियम के रूप में खोजा गया), और एका-सिलिकॉन (1886 में क्लेमेंस विंकलर द्वारा जर्मेनियम के रूप में खोजा गया)। एका-सिलिकॉन के भविष्यवाणित गुण जर्मेनियम के वास्तविक गुणों से काफी मिलते थे, जिसने मेंडेलीव के नियम को सही साबित किया।

Solving path: The three eka-elements and their modern names: Eka-boron = Scandium, Eka-aluminium = Gallium, Eka-silicon = Germanium. The question asks specifically for Eka-silicon → Germanium. The other options (Boron, Gallium, Scandium) are decoys built from the other two correct eka-pairs.


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