1
In the triad Li, Na and K, the middle element is —
💡 Hint:
The masses are 6.9, 23.0 and 39.0.
Explanation:
Na is in the middle and its mass sits near the average.
Class 10 · Science · Chapter 5 · बिहार बोर्ड (BSEB)CBSE · NCERT 2026-27
Periodic Classification of Elements
Periodic Classification of Elements
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1. Read — 5.1 hydrogen · 5.2 isotopes · 5.3 cobalt-nickel · 5.4 group 1 · 5.5 period 2 · 5.6 valency · 5.7–5.8 size · 5.9 metals · 5.10 losing · 5.11 gaining, diagram, worked example, board tip
2. Check — each lesson has its own questions; the number follows the lesson
3. Mastery ★ — all of that lesson correct. Redo the wrong ones
4. The memory figure is the short strip of groups 1 and 17, with metals on the left and halogens on the right in periods 2 and 3.
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- 1 The first sorting and Dobereiner’s triads
- 2 Newlands’ law of octaves
- 3 Mendeleev’s law and Activity 5.1
- 4 The predictions, noble gases and Activity 5.2
- 5 The modern periodic law — Activity 5.3
- 6 Groups and periods — Activities 5.4 and 5.5
- 7 Valency — Activity 5.6
- 8 Atomic size — Activities 5.7 and 5.8
- 9 Metals and non-metals — Activities 5.9 to 5.11
- Chapter winner — every lesson at mastery ★
🎬 Shorts
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📄 PDF download — notes + answer key
The first sorting and Dobereiner’s triads
शुरुआत और डोबेराइनर के त्रिक · Section 5.1 · Tables 5.1 and 5.2
Metals and non-metals first, then groups of threeNotes
As more elements were found, the facts about them became hard to keep in order. The earliest sort was metals and non-metals. In 1817 the German chemist Dobereiner made sets of three elements with similar properties. He called each set a triad.
In rising atomic mass, the middle mass is roughly the average of the other two. Li 6.9, Na 23.0 and K 39.0 make a clean triad. Only three such triads could be picked from the elements known then, so the method could not cover every element.
The English print says 118 elements, of which 98 occur naturally. The Hindi print still says 114, and about 30 by the year 1800. In an exam, use the number printed in your own copy.
Worked exampleExample
Question: Check the three groups in Table 5.1. Which ones are triads?
Formula: middle mass ≈ (first + third) / 2
Substitution: Group B, Ca and Ba: (40.1 + 137.3) / 2 = 177.4 / 2 = 88.7. Sr is 87.6, close enough. It is a triad.
Group C, Cl and I: (35.5 + 126.9) / 2 = 162.4 / 2 = 81.2. Br is 79.9, close enough. It is a triad.
Group A, N and As: (14.0 + 74.9) / 2 = 44.45. P at 31.0 is far from that. Not a triad. The unit is u.
| Group | Elements and masses | Decision |
|---|---|---|
| A | N 14.0, P 31.0, As 74.9 | Not a triad |
| B | Ca 40.1, Sr 87.6, Ba 137.3 | Triad |
| C | Cl 35.5, Br 79.9, I 126.9 | Triad |
10-second revision
- A triad is three elements of similar properties
- The middle mass is about the average of the other two
- Only three triads were found, so the method stayed incomplete
Board tip · BSEBBoard tip
Write the average formula, then the numbers, then the unit u. Only “about equal” stays incomplete.
Board tip · CBSEBoard tip
Do not call group A a triad. 44.45 and 31.0 do not match.
Check your understandingall correct = mastery ★
2
N, P and As form a Dobereiner triad.
Explanation:
False — the average is 44.45 and the mass of P is 31.0.
3
Dobereiner made the triads in the year ______.
💡 Hint:
Early in the nineteenth century.
Explanation:
1817.
4
The earliest classification of the elements was —
💡 Hint:
The triads came after this attempt.
Explanation:
The known elements were first placed as metals and non-metals.
5
From Li 6.9 u and K 39.0 u, find the expected mass of Na. Show the formula and the unit.
Model answer (board format)
Middle mass = (6.9 + 39.0) / 2 = 45.9 / 2 = 22.95 u. The given mass of Na is 23.0 u, so this is a triad.
Check your answer:
Newlands’ law of octaves
न्यूलैंड्स का अष्टक नियम · Section 5.1.2 · Table 5.3
The eighth element resembles the firstNotes
In 1866 the English scientist Newlands arranged the known elements by rising atomic mass. He began at hydrogen and stopped at thorium, the 56th element. He saw the properties repeat the way musical notes do. The Indian notes are sa, re, ga, ma, pa, dha, ni. The eighth note becomes the first note of the next line.
Count like this: take Li as the first, then Be, B, C, N, O, F, and the eighth is Na. Na falls under the same note as Li. Be and Mg share a note too. This is the law of octaves.
The law held only up to calcium. After that, every eighth element did not resemble the first. Newlands assumed that nature held only 56 elements. Later elements did not fit the law.
Worked exampleExample
Question: Why is putting cobalt and nickel in one slot of Table 5.3 treated as a fault?
Answer: Co and Ni were put in one slot, and that slot sits on the same note as F, Cl and Br. The properties of Co and Ni are not like those halogens. Fe resembles Co and Ni, yet it was placed far away. So the octaves worked only for the lighter elements.
10-second revision
- 1866, from hydrogen to thorium, 56 elements
- Counting from Li, the eighth is Na
- The law breaks after calcium
Board tip · BSEBBoard tip
The count of “eighth” includes the starting element. Na is not the seventh after Li; Na is the eighth in the count that starts at Li.
Board tip · CBSEBoard tip
Write three limits: only up to Ca, only 56 elements assumed, and unlike elements placed on one note.
Check your understandingall correct = mastery ★
1
In Newlands’ octaves, the next element on the note of Li is —
💡 Hint:
The eighth in the count.
Explanation:
Li, Be, B, C, N, O, F, Na. The eighth is Na.
2
The law of octaves fitted every element even after calcium.
Explanation:
False — after calcium every eighth element did not resemble the first.
3
Newlands assumed that the number of elements was ______.
💡 Hint:
Thorium was the last in that count.
Explanation:
56.
4
Write two limitations of Newlands’ octaves.
Model answer (board format)
The law held only up to calcium. Co and Ni were placed in one slot with F, Cl and Br, although the properties differ.
Check your answer:
Mendeleev’s law and Activity 5.1
मेंडलीफ का नियम और क्रियाकलाप 5.1 · Section 5.2 · Table 5.4 · Activity 5.1
Both the mass and the formula of the compoundNotes
When Mendeleev began, 63 elements were known. He wrote the properties on 63 cards and pinned similar cards together on a wall. Two things decided the place: rising atomic mass, and the formula of the compound with oxygen and with hydrogen. Both of those are very reactive.
The law: the properties of elements are a periodic function of their atomic masses. Vertical columns are groups and horizontal rows are periods. The table was published in a German journal in 1872. At the head of a group, R stands for any element of that group. CH4 of carbon is written RH4, and CO2 is written RO2.
Sometimes a heavier element had to come before a lighter one so that the properties would match. Co has mass 58.9 and Ni has 58.7, yet Co comes first. Another such place is Te (127.60), which comes before I (126.90).
| Group | Oxide | Hydride |
|---|---|---|
| I | R2O | RH |
| II | RO | RH2 |
| III | R2O3 | RH3 |
| IV | RO2 | RH4 |
| V | R2O5 | RH3 |
| VI | RO3 | RH2 |
| VII | R2O7 | RH |
| VIII | RO4 | Not in the table |
Activity 5.1 — where to put hydrogenActivity
Hydrogen resembles the alkali metals and also the halogen family. Like an alkali metal it joins with halogens, oxygen and sulphur: HCl with NaCl, H2O with Na2O, H2S with Na2S.
Like a halogen it also exists as a diatomic molecule, H2, and it forms covalent compounds with both metals and non-metals.
Hydrogen has no single fixed place in Mendeleev’s table. That is the first limitation. Asking for one group and one period does not give one answer.
Worked exampleExample
Question: Write the formulae of the oxides of K, C, Al, Si and Ba.
Answer: K is in group I, oxide R2O, so K2O. C is in group IV, RO2, so CO2. Al is in group III, R2O3, so Al2O3. Si is in group IV, RO2, so SiO2. Ba is in group II, RO, so BaO.
10-second revision
- The law says a periodic function of atomic mass
- Oxides from R2O to RO4, hydrides from RH to RH4
- Hydrogen has no single fixed place
Board tip · BSEBBoard tip
Take the oxide formula from the group. Do not write K2O and KO as the same. The group I oxide is R2O.
Board tip · CBSEBoard tip
Both Co–Ni and Te–I put the heavier element first. An answer with only one of them looks thin.
Check your understandingall correct = mastery ★
1
Mendeleev’s periodic law links the properties to —
💡 Hint:
The modern law is the one that uses number.
Explanation:
Properties are a periodic function of atomic mass.
2
The oxide of carbon, CO2, is written in Mendeleev’s notation as —
💡 Hint:
The hydride is RH4.
Explanation:
CO2 is written RO2. CH4 is written RH4.
3
In Mendeleev’s table cobalt comes before nickel even though Co is heavier.
Explanation:
True — the masses are 58.9 and 58.7, and the order is reversed so the properties match.
4
The number of elements known when Mendeleev worked was ______.
💡 Hint:
That many cards were made.
Explanation:
63.
5
The conclusion of Activity 5.1 is —
💡 Hint:
It resembles both alkali metals and halogens.
Explanation:
No single group and period could be fixed. That is the first limitation.
6
Using Mendeleev’s group formulae, write the oxides of Al and Si.
Model answer (board format)
Al is in group III, oxide R2O3, so Al2O3. Si is in group IV, oxide RO2, so SiO2.
Check your answer:
The predictions, noble gases and Activity 5.2
अनुमान, उत्कृष्ट गैसें और क्रियाकलाप 5.2 · Table 5.5 · Activity 5.2 · limitations
A gap became a prediction, not a faultNotes
Mendeleev left some houses empty. The Sanskrit word eka means one. The empty house was named by adding eka to the element above it. Later, scandium matched eka-boron, gallium matched eka-aluminium, and germanium matched eka-silicon.
The book prints numbers only for eka-aluminium. The predicted mass is 68, and gallium is 69.7. The oxides are E2O3 and Ga2O3. The chlorides are ECl3 and GaCl3. No second number-table is printed for eka-silicon. That gap was in group IV, and germanium filled it. The group IV oxide is written RO2.
Helium, neon and argon were found very late because they are inert and extremely scarce in air. They were given a new group and the old order was not disturbed.
Activity 5.2 — Cl-35 and Cl-37Activity
Take the isotopes of chlorine, Cl-35 and Cl-37. The masses differ. Would they get two different houses? The chemical properties are the same. Would both stay in one place?
Mendeleev’s law rested on mass, so isotopes became a challenge. A further limit is that the mass does not rise by a steady step from one element to the next. From the masses alone you cannot say how many elements might still be found between two heavy ones.
Worked exampleExample
Question: Why could Mendeleev not settle Cl-35 and Cl-37, and what does the modern table do?
Answer: The properties are the same, so two houses would split a chemical family. The masses differ, so a mass law also cannot see them as one house. The modern table looks at atomic number 17. Both have the same Z, so they share one place.
10-second revision
- Eka-boron scandium, eka-aluminium gallium, eka-silicon germanium
- E2O3 became Ga2O3 for gallium
- Isotopes were more than the mass law could hold
Board tip · BSEBBoard tip
Remember the three gallium rows: mass 68 and 69.7, oxides E2O3 and Ga2O3, chlorides ECl3 and GaCl3.
Board tip · CBSEBoard tip
Do not write that the noble gases “broke the old order”. A new group was added and the old groups stayed.
Check your understandingall correct = mastery ★
1
Who later sat in the place of eka-aluminium?
💡 Hint:
The oxide is Ga2O3.
Explanation:
Gallium. Scandium is eka-boron and germanium is eka-silicon.
2
Cl-35 and Cl-37 have different chemical properties.
Explanation:
False — isotopes have the same chemical properties and different masses.
3
Eka-silicon was later found as ______.
💡 Hint:
The empty house in group IV.
Explanation:
Germanium.
4
How did the noble gases fit into Mendeleev’s table?
💡 Hint:
They are inert and were found late.
Explanation:
A new group was added and the elements already placed did not move.
5
Write two limitations of Mendeleev’s classification, one of them about isotopes.
Model answer (board format)
Hydrogen was not given one fixed place. Isotopes have the same properties and different masses, so a mass law could not hold them. The mass also does not rise by a steady step.
Check your answer:
The modern periodic law — Activity 5.3
आधुनिक आवर्त नियम — क्रियाकलाप 5.3 · Section 5.3 · Moseley 1913 · Activity 5.3
Atomic number in place of atomic massNotes
In 1913 Henry Moseley showed that atomic number is a more basic property than atomic mass. The atomic number is the count of protons in the nucleus, and it rises by one at the next element.
The modern periodic law: the properties of elements are a periodic function of their atomic number. The table made by rising atomic number Z is the modern periodic table. Predictions of properties became sharper.
Activity 5.3 — the old snags in the new orderActivity
The places of cobalt and nickel are now fixed by atomic number. Co has Z = 27 and Ni has Z = 28. There is no need to force the heavier Co in front. The order is already right with Z.
The place of isotopes is also fixed by Z. Both Cl-35 and Cl-37 have Z = 17, so they share one house.
An element with Z = 1.5 cannot sit between hydrogen and helium, because protons increase as whole numbers. Hydrogen is still awkward in the modern table. In period 1 it can be thought of in group 1 or in group 17. The reason opens in the next activities.
Worked exampleExample
Question: Why is Z = 1.5 impossible?
Answer: Atomic number is a count of protons. It rises by 1 from one element to the next. There is no whole number between 1 and 2, so no element sits between hydrogen and helium.
10-second revision
- The modern law is a periodic function of atomic number
- Co has Z 27, Ni has Z 28
- Isotopes of the same Z share one house
Board tip · BSEBBoard tip
The Mendeleev sentence says mass. The modern sentence says number. Keep the two words apart in the answer.
Board tip · CBSEBoard tip
In the Z = 1.5 question write “protons are whole numbers”. Only “no” is not enough.
Check your understandingall correct = mastery ★
1
The basis of the modern periodic law is —
💡 Hint:
Moseley, 1913.
Explanation:
Properties are a periodic function of atomic number.
2
In the modern table Cl-35 and Cl-37 are given two different groups.
Explanation:
False — both have atomic number 17, so they share one place.
3
The atomic number of cobalt is ______.
💡 Hint:
Nickel is 28.
Explanation:
27. Nickel is 28, so Ni comes later.
4
How did the modern table remove both the cobalt-nickel snag and the isotope snag?
Model answer (board format)
The order is now by atomic number. Co has Z 27 and Ni has Z 28, so the masses need not be reversed. Cl-35 and Cl-37 both have Z 17, so both stay in one place.
Check your answer:
Groups and periods — Activities 5.4 and 5.5
समूह और आवर्त — क्रियाकलाप 5.4 और 5.5 · Section 5.3.1 · Activities 5.4 and 5.5
18 groups, 7 periodsNotes
The modern table has 18 vertical groups and 7 horizontal periods. Elements of one group have the same number of electrons in the outermost shell. The number of shells rises as you go down. Elements of one period have the same number of occupied shells. From left to right, if the atomic number rises by one, the valence electrons also rise by one.
Mendeleev’s use of compound formulae was a good basis, because elements with the same valence electrons form the same kind of bond and fall into one group.
Activity 5.4 — the first three of group 1Activity
The elements of group 1 are H, Li, Na, K, Rb, Cs and Fr. The first three, in order of atomic number, are H, Li and Na.
Configurations: H is 1. Li is 2, 1. Na is 2, 8, 1. All three have 1 valence electron. That is the similarity. The number of shells rises downward.
Hydrogen is still awkward. One valence electron takes it toward group 1. Being one electron short of a full outer shell reminds us of group 17. The period is still the first one.
Activity 5.5 — the eight elements of period 2Activity
Period 2: Li, Be, B, C, N, O, F, Ne. The configurations run from 2,1 to 2,8. The valence electrons are not the same. The shells are two and two. F is in group 17, configuration 2, 7. Cl is also in group 17, configuration 2, 8, 7. Both have 7 electrons in the outer shell.
Period 3: Na, Mg, Al, Si, P, S, Cl, Ar. Their electrons sit in the K, L and M shells. Each period shows a new shell being filled.
Worked exampleExample
Question: Find the number of elements in the first three periods from 2n².
Formula: maximum electrons in a shell = 2n². n is the number of the shell from the nucleus.
K, n = 1: 2 × (1)² = 2. The first period has 2 elements.
L, n = 2: 2 × (2)² = 8. The second period has 8 elements.
M, n = 3: 2 × (3)² = 18. But the outermost shell does not hold more than 8 electrons, so the third period also has 8 elements. The fourth period has 18 elements.
10-second revision
- Group = the same valence electrons
- Period = the same occupied shells
- Period 1 has 2 elements, periods 2 and 3 have 8
Board tip · BSEBBoard tip
Write 7 outer electrons for both F and Cl. Only the name “group 17” stays incomplete.
Board tip · CBSEBoard tip
2n² at n = 3 gives 18, yet period 3 has 8. Skipping that step loses marks.
Check your understandingall correct = mastery ★
1
The number of groups in the modern table is —
💡 Hint:
The periods are 7.
Explanation:
18 groups and 7 periods.
2
Elements of one group have the same number of valence electrons.
Explanation:
True — only the number of shells rises as you go down.
3
The number of elements in the second period is ______.
💡 Hint:
The L shell, 2 × 4.
Explanation:
8.
4
The electronic configuration of Na is —
💡 Hint:
Group 1, period 3.
Explanation:
2, 8, 1. There is one valence electron.
5
F and Cl are in one group because —
💡 Hint:
Configurations 2, 7 and 2, 8, 7.
Explanation:
Both have 7 electrons in the outermost shell. Cl is in period 3.
6
Write 2n² for the K and L shells and say how many elements the first two periods have.
Model answer (board format)
Maximum electrons = 2n². For K, n = 1, 2 × 1 = 2, so period 1 has 2 elements. For L, n = 2, 2 × 4 = 8, so period 2 has 8 elements.
Check your answer:
Valency — Activity 5.6
संयोजकता — क्रियाकलाप 5.6 · Section 5.3.2 · Activity 5.6
The outer electrons decide the valencyNotes
Valency comes from the electrons of the outer shell. If the valence electrons are 4 or fewer, the valency is that number. If they are more than 4, the valency is found by subtracting from 8. A full outer shell gives valency 0.
Across a period the valency rises from 1 to 4, then falls to 0. Down a group the valency does not change, because the valence electrons stay the same.
Activity 5.6 — the valency of Mg and SActivity
Magnesium has atomic number 12. The configuration is 2, 8, 2. The valence electrons are 2, which is not more than 4. The valency is 2.
Sulphur has atomic number 16. The configuration is 2, 8, 6. The valence electrons are 6. The valency is 8 − 6 = 2.
The same rule runs over the first twenty elements. In period 2, from Li to Ne, the valencies are 1, 2, 3, 4, 3, 2, 1, 0. In period 3, from Na to Ar, the order is the same. Down one group the number stays the same.
Worked exampleExample
Question: Give the valency and the group of the element with configuration 2, 8, 7.
Calculation: valence electrons = 7. Valency = 8 − 7 = 1. Seven outer electrons are the mark of group 17. Atomic number 2 + 8 + 7 = 17, which is chlorine.
10-second revision
- If 4 or fewer, valency = valence electrons
- If more than 4, valency = 8 − valence electrons
- Valency stays the same down a group
Board tip · BSEBBoard tip
Do not write the valency of S as 6. There are 6 valence electrons, and the valency is 2.
Board tip · CBSEBoard tip
Valency across a period is not one steady climb. It falls after 4. It is 0 at Ne and Ar.
Check your understandingall correct = mastery ★
1
The valency of Mg, atomic number 12, is —
💡 Hint:
Configuration 2, 8, 2.
Explanation:
There are 2 valence electrons, so the valency is 2.
2
Valency changes as you go down a group.
Explanation:
False — the valence electrons stay the same, so the valency stays the same.
3
The valency of sulphur (Z = 16) is ______.
💡 Hint:
8 − 6.
Explanation:
2. The configuration is 2, 8, 6.
4
In period 2, the valency of nitrogen after carbon —
💡 Hint:
N has configuration 2, 5.
Explanation:
The valency of C is 4. N has 5 valence electrons, so the valency is 8 − 5 = 3.
5
Find the valency of Z = 12 and Z = 16 from the configuration. Show the steps.
Model answer (board format)
Z = 12 has configuration 2, 8, 2. Valence electrons are 2, so the valency is 2. Z = 16 has configuration 2, 8, 6. Valency = 8 − 6 = 2.
Check your answer:
Atomic size — Activities 5.7 and 5.8
परमाणु आकार — क्रियाकलाप 5.7 और 5.8 · Section 5.3.2 · Activities 5.7 and 5.8
The distance from the nucleus to the outer shellNotes
Atomic size is the radius of an isolated atom: from the centre of the nucleus to the outermost shell. The radius of hydrogen is 37 pm. 1 pm = 10⁻¹² m.
From left to right in a period the nuclear charge rises. It pulls the electrons toward the nucleus and the size falls. Down a group a new shell is added. The outer electrons move farther from the nucleus, so the size grows even though the charge rises.
Activity 5.7 — period 2 radii in falling orderActivity
The given radii are not in the order of the period. B 88 pm, Be 111 pm, O 66 pm, N 74 pm, Li 152 pm, C 77 pm.
Falling order: Li 152, Be 111, B 88, C 77, N 74, O 66. That is the left-to-right order of period 2. In this list the largest atom is Li and the smallest is O. The radius falls from left to right.
Activity 5.8 — group 1 radii in rising orderActivity
Na 186 pm, Li 152 pm, Rb 244 pm, Cs 262 pm, K 231 pm.
Rising order: Li 152, Na 186, K 231, Rb 244, Cs 262. The smallest is Li and the largest is Cs. Size grows down the group because a new shell is added.
Worked exampleExample
Question: Write the 37 pm radius of hydrogen in metres.
Formula: size in metres = (radius in pm) × 10⁻¹²
Substitution: 37 × 10⁻¹² m = 3.7 × 10⁻¹¹ m.
In the same unit, the falling order of period 2 runs from Li to O, and the rising order of group 1 runs from Li to Cs.
10-second revision
- Size falls from left to right in a period
- Size grows down a group
- The radius of H is 37 pm, and 1 pm = 10⁻¹² m
Board tip · BSEBBoard tip
Do not copy the 5.7 list as it is printed. The falling order is Li, Be, B, C, N, O.
Board tip · CBSEBoard tip
The reasons differ. Nuclear pull in a period. A new shell in a group. Do not merge them into one sentence.
Check your understandingall correct = mastery ★
1
In the given period-2 list, the largest atom is —
💡 Hint:
152 pm.
Explanation:
Li, 152 pm. The smallest is O, 66 pm.
2
The largest atom in group 1, from the given list, is —
💡 Hint:
262 pm.
Explanation:
Cs, 262 pm. The smallest is Li, 152 pm.
3
Atomic radius increases from left to right in a period.
Explanation:
False — the radius falls as the nuclear charge rises.
4
The radius of a hydrogen atom is ______ pm.
💡 Hint:
1 pm = 10⁻¹² m.
Explanation:
37 pm.
5
The reason size grows down a group is —
💡 Hint:
The outer electrons move farther from the nucleus.
Explanation:
A new shell is added, so the distance from the nucleus to the outer electron grows.
6
Write the given period-2 radii in falling order and say why the radius falls across a period.
Model answer (board format)
Li 152, Be 111, B 88, C 77, N 74, O 66 pm. From left to right the nuclear charge rises and pulls the electrons closer, so the size falls.
Check your answer:
Metals and non-metals — Activities 5.9 to 5.11
धातु और अधातु — क्रियाकलाप 5.9 से 5.11 · Activities 5.9 · 5.10 · 5.11
Metals on the left, non-metals on the right, a zig-zag betweenNotes
In period 3, Na and Mg on the left are metals. Sulphur and chlorine on the right are non-metals. Silicon in the middle shows some properties of both. It is a metalloid.
Along the zig-zag line, boron, silicon, germanium, arsenic, antimony, tellurium and polonium are metalloids. Metals lose electrons, so they are electropositive. Non-metals gain electrons. Oxides of metals are basic and oxides of non-metals are generally acidic.
Activity 5.9 — sort period 3Activity
Na, Mg and Al are metals and sit toward the left of the table. Si is a metalloid. P, S and Cl are non-metals and sit toward the right. Ar is a noble gas. Metals on the left, non-metals on the right.
Activity 5.10 — the tendency to lose electronsActivity
Across a period the effective nuclear charge rises, so losing an electron becomes harder. Down a group the outer electron is farther from the nucleus, the pull weakens, and the electron is lost more easily.
So metallic character falls across a period and rises down a group.
Activity 5.11 — the tendency to gain electronsActivity
From left to right across a period, the tendency to gain electrons rises. Down a group that tendency falls. Non-metals are found on the right and toward the top. These trends also tell the nature of the oxide: the oxide of a metal is basic, and the oxide of a non-metal is generally acidic.
Worked exampleExample
Question: Among Ga, Ge, As, Se and Be, which should have the most metallic character?
Answer: Metallic character is greater toward the left and downward. Be sits high in period 2. Ge, As and Se are to the right of Ga. Ga is the element of this list that sits farthest down and to the left, so it has the most metallic character.
10-second revision
- Metals on the left, non-metals on the right, Si a metalloid
- Metallic character falls across a period and rises down a group
- Metal oxides are basic, non-metal oxides are usually acidic
Board tip · BSEBBoard tip
The metalloid list has seven names. Silicon alone is the period-3 example, not the whole line.
Board tip · CBSEBoard tip
Losing and gaining electrons run in opposite directions. Metallic character follows the losing direction.
Check your understandingall correct = mastery ★
1
The metalloid in period 3 is —
💡 Hint:
Between the metals and the non-metals.
Explanation:
Silicon is the metalloid. Na is a metal, S and Cl are non-metals.
2
Metallic character increases from left to right across a period.
Explanation:
False — losing an electron becomes harder, so metallic character falls.
3
Oxides of non-metals are generally ______.
💡 Hint:
Oxides of metals are basic.
Explanation:
Acidic.
4
How does the tendency to lose electrons change down a group and across a period? Add one sentence on metallic character.
Model answer (board format)
Down a group the outer electron is farther away, so the tendency to lose it rises. Across a period the effective nuclear charge rises, so losing an electron becomes harder. Metallic character rises down a group and falls across a period.
Check your answer:
❓ Full question bank — with answers and explanations — 67 questions
Pick a type. The 45 lesson checks are separate — each lesson has as many as its topic needs. All correct earns mastery ★.
BSEB modelpractice · not annual60 tried · 0 correctMultiple choiceMCQ140 tried · 0 correctTrue / falseTrue / false80 tried · 0 correctFill in the blanksFill80 tried · 0 correctMatchMatch20 tried · 0 correctAssertion–reasonAssertion–reason50 tried · 0 correctBalancingBalancing40 tried · 0 correctIdentifyIdentify20 tried · 0 correctShort and longVSA, SA, LA120 tried · 0 correctCBSE-stylenot a PYQ60 tried · 0 correct
No question is marked as a verified past paper. The BSEB set is a model for practice. CBSE items are CBSE-style, not a copy of any year’s paper.
Multiple choice
0/14Pick one option. A wrong try brings a hint.
1
The middle mass of a Dobereiner triad is —
💡 Hint:
(First + third) / 2.
Explanation:
The middle mass is roughly the average of the other two.
2
The group in Table 5.1 that is not a triad is —
💡 Hint:
The average is 44.45, and P has mass 31.
Explanation:
N, P and As do not form a triad.
3
Which element did Newlands take as the last?
💡 Hint:
The 56th element.
Explanation:
Thorium, the 56th element.
4
The law of octaves held well —
💡 Hint:
The lighter elements.
Explanation:
Up to calcium. After that every eighth element did not resemble the first.
5
Mendeleev’s table was published —
💡 Hint:
A German journal.
Explanation:
In 1872.
6
The oxide of group I is written —
💡 Hint:
K2O has this form.
Explanation:
R2O. The hydride is RH.
7
The oxide of Ba is —
💡 Hint:
Group II, RO.
Explanation:
BaO.
8
Eka-boron was later recognised as —
💡 Hint:
Gallium is eka-aluminium.
Explanation:
Scandium.
9
The number of periods in the modern table is —
💡 Hint:
The groups are 18.
Explanation:
7 periods.
10
The third period has —
💡 Hint:
The M shell can hold 18, the outer shell holds 8.
Explanation:
8 elements.
11
The valency of the element with configuration 2, 8, 2 is —
💡 Hint:
This is Mg.
Explanation:
2.
12
From left to right in a period, atomic size —
💡 Hint:
The nuclear pull rises.
Explanation:
It decreases.
13
The statement that is not correct on going from left to right is —
💡 Hint:
Losing an electron becomes harder.
Explanation:
Losing electrons more easily is the false statement. The tendency falls.
14
X forms a chloride XCl2, a solid with a high melting point. X is most likely in the group of —
💡 Hint:
Valency 2.
Explanation:
Mg. The chloride XCl2 matches group II.
True or false
0/81
Dobereiner could make a triad for every element known at that time.
Explanation:
False — only three triads were identified.
2
In Newlands’ table Na falls on the same note as Li.
Explanation:
True — Na is the eighth in the count.
3
Te has a smaller atomic mass than I, which is why Te was placed first.
Explanation:
False — Te is 127.60 and I is 126.90. The heavier Te was placed first because of the properties.
4
The noble gases were found early because they are very reactive.
Explanation:
False — they are inert and extremely scarce in air, so they were found late.
5
The modern table has 18 groups.
Explanation:
True — and there are 7 periods.
6
From Li to Ne in period 2, the number of valence electrons is the same.
Explanation:
False — the valence electrons rise from 1 to 8. The shells are the same.
7
Atomic size increases down a group.
Explanation:
True — a new shell is added.
8
Oxides of metals are generally basic.
Explanation:
True — oxides of non-metals are generally acidic.
Fill in the blanks
0/81
The average of the masses of Li and K is about ______ u, close to Na.
💡 Hint:
22.95 u.
Explanation:
(6.9 + 39.0) / 2.
2
Newlands’ law is of the year ______.
💡 Hint:
1866.
Explanation:
Dobereiner is 1817.
3
In Mendeleev’s law the properties are a periodic function of atomic ______.
💡 Hint:
Atomic mass.
Explanation:
The modern law says number.
4
The measured mass of gallium is ______.
💡 Hint:
69.7.
Explanation:
The prediction was 68.
5
The modern law is tied to Moseley’s work in ______.
💡 Hint:
1913.
Explanation:
Atomic number.
6
The formula for the maximum number of electrons in a shell is ______.
💡 Hint:
2n².
Explanation:
n is the shell number.
7
F and Cl of group 17 have ______ valence electrons.
💡 Hint:
7.
Explanation:
Configurations 2,7 and 2,8,7.
8
In the given group-1 list, the element with the smallest atom is ______.
💡 Hint:
Li.
Explanation:
152 pm.
Match
0/21
Match the scientist with the basis he used.
Column B: A. Atomic number · B. Octaves · C. Triads · D. Atomic mass
1. Dobereiner
2. Newlands
3. Mendeleev
4. Moseley
Explanation:
Dobereiner triads, Newlands octaves, Mendeleev mass, Moseley number.
2
Match Mendeleev’s group formula.
Column B: A. RO · B. RH4 · C. R2O · D. R2O3
1. Group I oxide
2. Group IV hydride
3. Group III oxide
4. Group II oxide
Explanation:
The group I oxide is R2O, the group IV hydride is RH4, the group III oxide is R2O3, and the group II oxide is RO.
Assertion–reason
0/5Check both statements, then see whether the reason explains the assertion.
1
Assertion (A): Mendeleev placed some heavier elements before lighter ones.
Reason (R): The order had to be reversed so that elements of similar properties stayed in one group.
Explanation:
Both are true and R explains A. Co–Ni and Te–I are such pairs.
2
Assertion (A): The discovery of gallium matches the prediction for eka-aluminium.
Reason (R): The noble gases were already filled into the table before Mendeleev’s work.
Explanation:
A is true. R is false — the noble gases were found much later.
3
Assertion (A): Cl-35 and Cl-37 are placed in two groups in the modern table.
Reason (R): Both have atomic number 17.
Explanation:
A is false. R is true — the same Z means one place.
4
Assertion (A): Atomic size decreases from left to right in a period.
Reason (R): The nuclear charge rises and pulls the electrons closer to the nucleus.
Explanation:
Both are true and R is the correct reason.
5
Assertion (A): Metallic character increases down a group.
Reason (R): Mendeleev’s table was published in 1872.
Explanation:
Both are true, but R does not explain the metallic character.
Balance the equation
0/4These form the oxides from Mendeleev’s formulae. The element symbol stands where R stood. A blank coefficient is 1.
1
The group I oxide is R2O, so the oxide of potassium is K2O. Balance the formation. A blank coefficient means 1.
K + O2 → K2O
Explanation: 4K + O2 → 2K2O
2
The group IV oxide is RO2. Balance the formation of CO2 from carbon. A blank coefficient means 1.
C + O2 → CO2
Explanation: C + O2 → CO2
3
The group III oxide is R2O3. Balance the formation of Al2O3. A blank coefficient means 1.
Al + O2 → Al2O3
Explanation: 4Al + 3O2 → 2Al2O3
4
The group II oxide is RO. Balance the formation of BaO. A blank coefficient means 1.
Ba + O2 → BaO
Explanation: 2Ba + O2 → 2BaO
Classify
0/21
Place each trio as a triad or not.
Li, Na, K
N, P, As
Ca, Sr, Ba
Cl, Br, I
Explanation:
Li–Na–K, Ca–Sr–Ba and Cl–Br–I are triads. The average for N–P–As does not meet P.
2
Place each period-3 element in its type.
Na
Si
S
Cl
Explanation:
Na is a metal, Si a metalloid, and S and Cl non-metals.
Very short answer
0/51
Define a Dobereiner triad in one line.
Model answer (board format)
Three elements of similar properties, whose middle atomic mass lies near the average of the other two, are called a triad.
Check your answer:
2
State the modern periodic law.
Model answer (board format)
The properties of elements are a periodic function of their atomic number.
Check your answer:
3
Write the oxide and chloride formulae of eka-aluminium and gallium.
Model answer (board format)
The oxide of eka-aluminium is E2O3 and the chloride is ECl3. The oxide of gallium is Ga2O3 and the chloride is GaCl3.
Check your answer:
4
What is a group?
Model answer (board format)
A vertical column of the modern table is a group. Elements of one group have the same number of valence electrons.
Check your answer:
5
What is a metalloid? Give one example.
Model answer (board format)
An element that shows some properties of both metals and non-metals is a metalloid. Silicon is an example.
Check your answer:
Short answer
0/41
From Ca 40.1 and Ba 137.3, find the expected mass of Sr. Is group B a triad?
Model answer (board format)
Expected mass = (40.1 + 137.3) / 2 = 177.4 / 2 = 88.7 u. The mass of Sr is 87.6 u, which is close. So Ca, Sr and Ba form a triad.
Check your answer:
2
Write three limitations of Newlands’ octaves.
Model answer (board format)
The law held only up to calcium. He assumed that only 56 elements exist. Co and Ni were placed in one slot with elements of different properties, and Fe was placed far from Co and Ni.
Check your answer:
3
What criteria did Mendeleev use for classification? Also write the meaning of R.
Model answer (board format)
Rising atomic mass and chemical properties, especially the formulae of the oxide and the hydride. R stands for any element of the group. For example, CH4 is written RH4 and CO2 is written RO2.
Check your answer:
4
Calcium, atomic number 20, has neighbours 12, 19, 21 and 38. Which resemble it in properties, and why?
Model answer (board format)
Ca has configuration 2, 8, 8, 2. Z = 12 is magnesium, configuration 2, 8, 2. Z = 38 is strontium, with 2 outer electrons. Both are in the same group. Z = 19 has one valence electron and Z = 21 has a different configuration, so they are not like Ca.
Check your answer:
Long answer
0/31
Write three achievements and two limitations of Mendeleev. Include the evidence of eka-aluminium.
Model answer (board format)
Achievements: he arranged elements by mass and properties; from the gaps he predicted eka-boron, eka-aluminium and eka-silicon; a new group was added for the noble gases. Evidence: eka-aluminium mass 68 and gallium 69.7, oxides E2O3 and Ga2O3, chlorides ECl3 and GaCl3. Limitations: hydrogen had no fixed place, and isotopes plus irregular masses challenged the law.
Check your answer:
2
Which limitations of Mendeleev did the modern table remove? What fixes a group and a period? Find the number of elements in periods 1 to 3 from 2n².
Model answer (board format)
By atomic number, Co (27) comes before Ni (28), and isotopes of the same Z share one house. Z = 1.5 is impossible. A group is fixed by the same valence electrons and a period by the same shells. 2n² gives 2 at n = 1, 8 at n = 2, and 18 at n = 3, but the outer shell stops at 8, so periods 1, 2 and 3 have 2, 8 and 8 elements.
Check your answer:
3
Write the orders from Activities 5.7 and 5.8. Then say why size changes in a period and in a group. Add one sentence on the direction of metallic character.
Model answer (board format)
Period 2, falling order: Li 152, Be 111, B 88, C 77, N 74, O 66 pm. Group 1, rising order: Li 152, Na 186, K 231, Rb 244, Cs 262 pm. In a period the nuclear charge pulls the electrons in, so the size falls. In a group a new shell is added, so the size grows. Metallic character falls across a period and rises down a group.
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BSEB model paper · practice
0/6This model set is for practice. It is not a question from any year’s annual examination. Annual questions will be added only when a source page is available.
1
The elements known in Mendeleev’s time were —
💡 Hint:
That many cards.
Explanation:
63.
2
In the modern table hydrogen can be thought of in —
💡 Hint:
One electron, or one short.
Explanation:
Group 1 or group 17, period 1.
3
Eka means ______ in Sanskrit.
💡 Hint:
One.
Explanation:
It was added to the name of the empty house.
4
In Activity 5.1, why is hydrogen linked both to alkali metals and to halogens?
Model answer (board format)
Like alkali metals it forms HCl, H2O and H2S, close to the formulae NaCl, Na2O and Na2S. Like halogens it forms the diatomic molecule H2 and also covalent compounds. So one fixed place does not appear.
Check your answer:
5
The group IV oxide is RO2. Balance the formation of SiO2. A blank coefficient means 1.
Si + O2 → SiO2
Explanation: Si + O2 → SiO2
6
A student says that the octaves and the modern table are the same law. Correct this with three differences. Include one point from Activity 5.3.
Model answer (board format)
The octaves rest on atomic mass and on every eighth element, and they held only up to calcium. The modern table rests on atomic number, with 18 groups and 7 periods. Activity 5.3: Co has Z 27 and Ni has Z 28, so the order is settled, and an element with Z = 1.5 is impossible.
Check your answer:
CBSE-style questions
0/6These are competency-based practice questions. They are not copies of a CBSE paper.
1
A student writes the average 44.45 without a unit and calls group A a triad. The correct decision is —
💡 Hint:
The formula is (14.0 + 74.9) / 2.
Explanation:
44.45 u does not match 31.0 u of P. Group A is not a triad.
2
Li is 152 pm and Cs is 262 pm. The correct size order in group 1 is —
💡 Hint:
A new shell downward.
Explanation:
Li 152, Na 186, K 231, Rb 244, Cs 262 pm.
3
Assertion (A): The element with configuration 2, 8, 7 behaves like fluorine.
Reason (R): Both have 7 electrons in the outer shell.
Explanation:
Both are true and R is the correct reason. The element is chlorine, group 17.
4
Assertion (A): The tendency to gain electrons rises toward the right of a period.
Reason (R): Metallic character also rises in that same direction.
Explanation:
A is true. R is false — metallic character falls toward the right.
5
An element has configuration 2, 8, 8, 2. Give its period, the valence clue to its group, and two elements that should resemble it chemically.
Model answer (board format)
There are four shells, so it is period 4. There are 2 valence electrons, so the valency is 2 and it sits in the group of magnesium. Be and Mg, or Mg and Sr, are of the same kind. The element is calcium, Z = 20.
Check your answer:
6
Nitrogen (Z = 7) and phosphorus (Z = 15) are in the same group. Write the configurations and say which is more electronegative.
Model answer (board format)
N has configuration 2, 5 and P has 2, 8, 5. Both have 5 valence electrons. N is more electronegative because it is smaller and the valence electrons are closer to the nucleus. The tendency falls down the group.
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🧠 What you learned + equation sheet
What you learned
- In a Dobereiner triad the middle mass is roughly the average of the other two.
- Newlands’ octaves worked for the lighter elements, and not after calcium.
- Mendeleev’s law uses atomic mass, with the oxide and hydride formulae as a basis.
- Eka-aluminium became gallium and eka-silicon became germanium. Hydrogen had no fixed seat.
- The modern law uses atomic number. There are 18 groups and 7 periods.
- Across a period, size and metallic character fall; down a group both rise.
| What | Keep this |
|---|---|
| Li–Na–K | (6.9 + 39.0) / 2 = 22.95 u |
| Octave limit | works up to Ca; Co and Ni share a slot |
| Eka-aluminium | mass 68; oxide E2O3; chloride ECl3 |
| Modern law | periodic function of atomic number |
| Shell limit | 2n²; period 1 has 2, periods 2 and 3 have 8 |
| Atomic size | falls left to right; rises down a group |
The notes are original writing. The textbook was used only for activity order and numbers. “Verified” will be used only on a question that has a source page.