Home / Chemistry / Class XI / Classification of Elements and Periodicity in Properties
Classification of Elements and Periodicity in Properties — CUET Chemistry hero
Class XI ⚗️ Chemistry Ch 3 of 9

Classification of Elements and Periodicity in Properties

CUET unit: Classification of Elements and Periodicity in Properties

📌 Snapshot

  • Periodic trends in radius, ionisation enthalpy and electronegativity, explained with solved CUET examples.
  • How Mendeleev's gaps predicted germanium, why Moseley changed the law, and how the long form table is built.
  • Why each trend holds, with the exceptions CUET loves: Be and B, N and O, and Cl against F.
  • Block positions, IUPAC names above Z = 100, diagonal relationships and the acidic to basic oxide trend.

📖 Detailed Notes

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Chemistry Class XI, Chapter 3.

Could you predict an element nobody had seen?

In 1869 Dmitri Mendeleev arranged the known elements in a table, and where a pattern broke he left a gap and said an element would be found there. He even gave its properties. For the gap under silicon he predicted an atomic weight near 72, a density of 5.5 g cm⁻³ and an oxide of formula EO₂. In 1886 germanium was discovered with atomic weight 72.6, density 5.36 and the oxide GeO₂. A table that predicts the future is a powerful tool, and this chapter shows why it works. The answer is the electron configurations you met in Structure of Atom. Elements repeat their properties because they repeat their outer electron arrangement.

Why classify, and the first attempts

By 1800 only 31 elements were known, by 1865 there were 63, and today 118 have been named. Studying each element and its compounds one by one is hopeless, so chemists looked for patterns. Every attempt used atomic weight, because atoms of different elements were first compared by mass.

Scientist Idea Why it fell short
Döbereiner (early 1800s) Triads: in a group of three similar elements, the middle one has an atomic weight near the mean of the other two Only a few triads existed
De Chancourtois (1862) Elements listed by atomic weight on a spiral around a cylinder, and similar ones lined up Received little attention
Newlands (1865) Law of Octaves: every eighth element resembles the first, like musical notes Worked only up to calcium
Mendeleev and Lothar Meyer (1869) Properties are a periodic function of atomic weight Several elements out of order, and no explanation of why

Check a triad. Lithium (7), sodium (23) and potassium (39): the mean of the outer two is $\dfrac{7 + 39}{2} = 23$, exactly sodium's weight. Calcium, strontium, barium (40, 88, 137) give $\dfrac{40 + 137}{2} = 88.5$, and chlorine, bromine, iodine (35.5, 80, 127) give $81.3$, both close.

Mendeleev's table: gaps that predicted the future

Mendeleev put elements in order of increasing atomic weight but grouped them by chemical similarity, and when the two clashed, similarity won. Iodine (126.9) is lighter than tellurium (127.6), yet he put iodine after tellurium so that it sat in the same column as chlorine and bromine. He also left gaps, naming the missing elements eka-aluminium, eka-silicon and eka-boron (later found as gallium, germanium and scandium). The predictions were remarkably accurate.

Property Eka-aluminium (predicted) Gallium (found, 1875) Eka-silicon (predicted) Germanium (found, 1886)
Atomic weight 68 70 72 72.6
Density (g cm⁻³) 5.9 5.94 5.5 5.36
Oxide E₂O₃ Ga₂O₃ EO₂ GeO₂
Chloride ECl₃ GaCl₃ ECl₄ GeCl₄

The table had real defects, though. In order of atomic weight, potassium should come before argon, nickel before cobalt and iodine before tellurium, but their properties demanded the opposite order. Hydrogen had to be placed in two groups. There was no place for isotopes, which have different weights but the same chemistry, and no way to fit the lanthanoids. The cause was unknown until atomic structure was understood.

Moseley and the Modern Periodic Law

In 1913 Henry Moseley bombarded different metals with electrons and measured the frequency $\nu$ of the characteristic X-rays they gave off. A plot of $\sqrt{\nu}$ against atomic number $Z$ was a straight line, which means the relation is

$$\sqrt{\nu} = a\,(Z - b)$$

for constants $a$ and $b$ specific to the type of X-ray line. Plotted against atomic weight, the same data scatter. So atomic number, the number of protons, is the real ordering quantity. This gives the Modern Periodic Law: the physical and chemical properties of elements are periodic functions of their atomic numbers. Mendeleev's anomalies vanish at once, because Ar ($Z = 18$) does come before K ($Z = 19$).

How the long form of the table is built

The long form has 7 periods (rows) and 18 groups (columns, numbered 1 to 18 under the IUPAC system). The period number is the highest principal quantum number $n$ occupied. The number of elements in a period comes straight from the filling order: count the orbitals filled between one noble gas and the next, and double.

Period Orbitals filled Elements Starts with, ends with
1 1s 2 H, He
2 2s, 2p 8 Li, Ne
3 3s, 3p 8 Na, Ar
4 4s, 3d, 4p 18 K, Kr
5 5s, 4d, 5p 18 Rb, Xe
6 6s, 4f, 5d, 6p 32 Cs, Rn
7 7s, 5f, 6d, 7p 32 Fr, Og

Period 4, for example, has $1 + 5 + 3 = 9$ orbitals, and 9 orbitals hold 18 electrons, which is why the fourth period has 18 elements. The 14 lanthanoids (Ce to Lu) and 14 actinoids (Th to Lr) belong to periods 6 and 7 but are printed in two rows at the bottom to keep the table narrow.

To place an element, write its configuration, and then use these rules.

Block Group number Outer configuration
s Number of valence electrons (1 or 2) $ns^{1-2}$
p 10 + number of valence electrons ($ns$ and $np$) $ns^2np^{1-6}$
d Number of electrons in $(n-1)d$ and $ns$ $(n-1)d^{1-10}ns^{0-2}$
f Lanthanoids and actinoids (not given a group number) $(n-2)f^{1-14}(n-1)d^{0-1}ns^2$

Worked example: find the position. Locate the elements with $Z$ = 35, 26, 56, 82 and 29.

$Z$ Configuration Period Group Block
35 (Br) [Ar] 3d¹⁰ 4s² 4p⁵ 4 $10 + 2 + 5 = 17$ p
26 (Fe) [Ar] 3d⁶ 4s² 4 $6 + 2 = 8$ d
56 (Ba) [Xe] 6s² 6 2 s
82 (Pb) [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p² 6 $10 + 2 + 2 = 14$ p
29 (Cu) [Ar] 3d¹⁰ 4s¹ 4 $10 + 1 = 11$ d

The common wrong step: giving copper group 1 because it ends in 4s¹. The 3d electrons count towards the group number, so Cu is in group 11.

Beyond the textbook. For the exam: hydrogen ($1s^1$) is placed at the top of group 1, and helium ($1s^2$), although its configuration is s block, is placed in group 18 with the other noble gases because its shell is completely filled. In reality: hydrogen fits no group well. It gains an electron like the halogens and loses one like the alkali metals, yet it is a non metal. Some tables place it above the halogens and others show it on its own. Why it matters: it shows that a periodic table is a convenient arrangement of repeating properties, and a few elements will never fit perfectly.

Naming the elements above Z = 100

New heavy elements are made in laboratories and exist for fractions of a second. While discovery is disputed, IUPAC gives each a temporary systematic name built from its atomic number, using a root for each digit and ending in "ium".

Digit Root Symbol letter
0 nil n
1 un u
2 bi b
3 tri t
4 quad q
5 pent p
6 hex h
7 sept s
8 oct o
9 enn e

Worked example: systematic names. Name the elements with $Z$ = 120, 119 and 102, and give their three letter symbols.

$Z$ Digits Roots Name Symbol
120 1, 2, 0 un + bi + nil + ium Unbinilium Ubn
119 1, 1, 9 un + un + enn + ium Ununennium Uue
102 1, 0, 2 un + nil + bi + ium Unnilbium Unb

The common wrong step: dropping the "nil" for zero, which turns 120 into "Unbium". When a root ending in "i" (bi, tri) meets "ium", only one "i" is kept, as in Unnilbium.

After the discoveries were confirmed, permanent names were approved.

$Z$ Name (symbol) $Z$ Name (symbol) $Z$ Name (symbol)
101 Mendelevium (Md) 107 Bohrium (Bh) 113 Nihonium (Nh)
102 Nobelium (No) 108 Hassium (Hs) 114 Flerovium (Fl)
103 Lawrencium (Lr) 109 Meitnerium (Mt) 115 Moscovium (Mc)
104 Rutherfordium (Rf) 110 Darmstadtium (Ds) 116 Livermorium (Lv)
105 Dubnium (Db) 111 Roentgenium (Rg) 117 Tennessine (Ts)
106 Seaborgium (Sg) 112 Copernicium (Cn) 118 Oganesson (Og)

The four blocks and the metal to non metal divide

Where an element sits depends on which subshell receives its last electron, and the block tells you how it behaves.

Block Groups Members Typical behaviour
s 1 and 2 Alkali metals (group 1), alkaline earth metals (group 2) Soft reactive metals with low ionisation enthalpy forming 1+ and 2+ ions, mostly ionic compounds, coloured flames
p 13 to 18 Boron, carbon and nitrogen families, chalcogens (16), halogens (17), noble gases (18) Mix of metals, metalloids and non metals. Groups 16 and 17 have very negative electron gain enthalpy
d 3 to 12 Transition elements Coloured ions, variable oxidation states, paramagnetism, catalysts
f Two bottom rows Lanthanoids and actinoids All metals. Actinoids are radioactive, and elements beyond uranium (transuranium) are man made

More than 78% of elements are metals, found on the left and centre. They are usually solid with high melting and boiling points, and are malleable, ductile and good conductors. Non metals sit at the top right, are often gases or low melting solids, and are poor conductors. A zig zag line separates them, and along it lie the metalloids (Si, Ge, As, Sb, Te) with intermediate properties. Metallic character falls from left to right across a period and rises down a group.

Beyond the textbook. For the exam: group 12 elements Zn, Cd and Hg are d block elements. In reality: the IUPAC definition of a transition element is one with an incomplete d subshell in the atom or in a common ion. Zn, Cd and Hg have full $d^{10}$ shells in the atom and in their usual +2 ions, so they do not behave as typical transition metals. Why it matters: you will meet this again in the d and f block chapter, where colour and magnetism need partly filled d orbitals.

The master idea: effective nuclear charge

Every periodic trend rests on one question: how strongly does the nucleus hold the outermost electron? The answer is the effective nuclear charge, the pull of the nucleus after the inner electrons have shielded part of it.

$$Z_{\text{eff}} = Z - S$$

where $S$ measures the shielding. Across a period, the electron added goes into the same shell, where electrons shield one another poorly, so $Z_{\text{eff}}$ rises and the atom shrinks. Down a group, a new shell is added, which is far from the nucleus and well shielded by all the inner shells, so the outer electron is held more loosely despite the higher $Z$. With these two facts you can predict almost every trend below.

Property Across a period (left to right) Down a group
Atomic radius Decreases Increases
Ionisation enthalpy Increases (with two exceptions) Decreases
Electron gain enthalpy Becomes more negative (with exceptions) Becomes less negative (with the F exception)
Electronegativity Increases Decreases
Metallic character Decreases Increases

Atomic radius

An atom has no sharp edge, so its size is defined from distances between nuclei, and the definition depends on how it is bonded.

Type Definition Example
Covalent radius Half the distance between the nuclei of two like atoms joined by a single covalent bond Cl–Cl is 198 pm, so $r = 99$ pm
Metallic radius Half the distance between the cores in a metallic crystal Cu–Cu is 256 pm, so $r = 128$ pm
van der Waals radius Half the distance between nuclei of two non bonded atoms of neighbouring molecules, for noble gases Always larger than the covalent radius
Element (period 2) Radius (pm) IE₁ (kJ mol⁻¹) Electronegativity
Li 152 520 1.0
Be 111 899 1.5
B 88 801 2.0
C 77 1086 2.5
N 70 1402 3.0
O 66 1314 3.5
F 64 1681 4.0
Ne not defined 2081 not defined
Element (period 3) Radius (pm) IE₁ (kJ mol⁻¹) Electronegativity
Na 186 496 0.9
Mg 160 738 1.2
Al 143 578 1.5
Si 117 787 1.8
P 110 1012 2.1
S 104 1000 2.5
Cl 99 1251 3.0
Element (group 1) Radius (pm) IE₁ (kJ mol⁻¹) Electronegativity
Li 152 520 1.0
Na 186 496 0.9
K 231 419 0.8
Rb 244 403 0.8
Cs 262 376 0.7
Element (group 17) Radius (pm) Electron gain enthalpy (kJ mol⁻¹) Electronegativity
F 64 −328 4.0
Cl 99 −349 3.0
Br 114 −325 2.8
I 133 −295 2.5
At not listed −270 2.2

These four tables hold most of the data you will need for the rest of the chapter. Radii fall steadily across a period and rise down a group. The covalent and metallic radii are listed together here, and the noble gas values (van der Waals) are left out because they are measured differently.

Worked example: radius from bond length. The Br–Br bond length is 228 pm. Estimate the covalent radius of bromine, and then the C–Br bond length (covalent radius of C is 77 pm).

Step What we do Result
1 Radius is half the bond length $r_{\text{Br}} = \dfrac{228}{2} = 114$ pm
2 In a heteronuclear bond the bond length is the sum of the radii $77 + 114 = 191$ pm

The common wrong step: comparing a noble gas radius with its neighbours without noticing it is a van der Waals radius. Neon looks bigger than fluorine only because it is measured differently.

Ionic radius

An ion's size depends on its charge, and the rule is simple. A cation is smaller than its atom because it has lost electrons (often a whole shell), while the same nuclear charge now pulls fewer electrons. An anion is larger than its atom because the extra electrons repel each other and $Z_{\text{eff}}$ per electron falls. Na is 186 pm but Na⁺ is about 100 pm. Cl is 99 pm but Cl⁻ is 181 pm. The more positive the charge, the smaller the cation, so Fe³⁺ is smaller than Fe²⁺.

Species with the same number of electrons are isoelectronic. Among these, size is controlled by nuclear charge alone, and more protons pull the same electrons closer.

Worked example: rank isoelectronic ions. Arrange O²⁻, F⁻, Na⁺, Mg²⁺ and Al³⁺ in order of decreasing radius.

Step What we do Result
1 Check the electron count All have 10 electrons
2 List nuclear charge $Z$ = 8, 9, 11, 12, 13
3 More protons means a smaller ion O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺

Pause and predict: Which is larger, Fe²⁺ or Fe³⁺?

Check your answer: Fe²⁺. Both have the same nuclear charge of 26 protons, but Fe³⁺ has one electron fewer, so the same pull acts on fewer electrons that repel each other less, and the ion shrinks. The more positive the charge on a cation of one element, the smaller it is.

Ionisation enthalpy

The first ionisation enthalpy is the energy needed to remove the most loosely held electron from an isolated gaseous atom. It is always positive, and it is the measure of how well an element can form a cation.

X(g) → X⁺(g) + e⁻ X⁺(g) → X²⁺(g) + e⁻ (second)

Removing each further electron is harder than the one before, because the electron is leaving an ion of higher positive charge: IE₃ > IE₂ > IE₁. A sudden huge jump shows that the next electron comes from an inner, tightly held shell.

Worked example: group from successive ionisation enthalpies. An element has successive ionisation enthalpies (kJ mol⁻¹) of 738, 1451, 7733 and 10543. Which group is it in?

Step What we do Result
1 Look at the ratios between successive values 1451/738 is 2.0, 7733/1451 is 5.3
2 The big jump comes after the second electron Two valence electrons, then a noble gas core
3 Two valence electrons Group 2 (the data are for magnesium)

Trend. Across a period, ionisation enthalpy rises because $Z_{\text{eff}}$ rises and the radius shrinks. The peaks are at noble gases (full shells) and the lows at alkali metals (one loosely held electron). Down a group it falls, because the outer electron is farther and better shielded. Group 1 shows this clearly, with IE₁ falling from 520 kJ mol⁻¹ for Li to 376 kJ mol⁻¹ for Cs.

Look again at the period 2 and period 3 columns of the data tables. Each period follows the rule except at two places. The full period 2 order is Li < B < Be < C < O < N < F < Ne. Two dips explain the exceptions. Boron's electron comes from a 2p orbital, which is farther out and shielded by the 2s electrons, so it leaves more easily than the 2s electron of beryllium. In oxygen, the fourth 2p electron must pair with another in the same orbital, and the repulsion helps it escape. Nitrogen's $2p^3$ is half filled, with three unpaired electrons, which is a stable arrangement.

Pause and predict: Why is the second ionisation enthalpy of sodium (4562 kJ mol⁻¹) so much higher than that of magnesium (1451 kJ mol⁻¹)?

Check your answer: Removing the second electron from Na⁺ means breaking into the stable $2p^6$ core. In Mg⁺ the second electron is still the outer 3s electron. A noble gas core is far more tightly held.

Electron gain enthalpy

Electron gain enthalpy is the enthalpy change when a gaseous atom takes up an electron.

X(g) + e⁻ → X⁻(g)

It is negative when energy is released. The more negative it is, the more readily the atom accepts an electron. Halogens are the most negative, because one electron completes their octet.

In the group 17 table above, chlorine, not fluorine, is the most negative. Fluorine is so small that its 2p subshell is crowded, and the incoming electron is repelled strongly by the electrons already there. In the larger 3p subshell of chlorine the repulsion is less. The same effect makes O (−141 kJ mol⁻¹) less negative than S (−200 kJ mol⁻¹). Noble gases have positive values (He +48, Ne +116, Ar +96 kJ mol⁻¹), because their shells are full and the new electron must go into the next shell. Nitrogen, with its stable half filled $2p^3$, has a value close to zero, much less negative than carbon (−122 kJ mol⁻¹).

The common wrong step: treating electron gain enthalpy and ionisation enthalpy as opposites that must follow the same order. They share trends but the exceptions are different.

Electronegativity

Electronegativity is the tendency of an atom bonded to another to attract the shared electrons towards itself. Unlike ionisation enthalpy, it is a relative scale and cannot be measured directly. The most widely used is the Pauling scale, in which fluorine, the most electronegative element, is fixed at 4.0.

The data tables show it rising steadily across each period (Li 1.0 to F 4.0, Na 0.9 to Cl 3.0) and falling down a group (F 4.0 to At 2.2, Li 1.0 to Cs 0.7). Electronegativity is directly related to non metallic character and inversely to metallic character. The three similar looking terms are different quantities.

Term Isolated or bonded? Measures
Ionisation enthalpy Isolated gaseous atom Energy to remove an electron
Electron gain enthalpy Isolated gaseous atom Energy change on gaining an electron
Electronegativity Atom in a molecule Pull on a shared pair

Valence and oxidation state

Valence is the combining capacity of an element. For groups 1, 2, 13 and 14 it equals the number of valence electrons, and for groups 15, 16 and 17 it is $8 - \text{valence electrons}$, so it is 3, 2 and 1. Transition elements show variable valence. The oxidation state is the charge an atom would have if the shared electrons went to the more electronegative atom. In OF₂, fluorine is more electronegative, so F is −1 and O is +2. In Na₂O, Na is +1 and O is −2.

Second period anomalies and diagonal relationships

The first member of each group (Li, Be, B, C, N, O, F) differs from the rest of its group. The reasons are small size, high charge to radius ratio, high electronegativity and only four valence orbitals (2s and three 2p), with no d orbitals. This puts a ceiling on covalency.

Period Maximum covalency Why Examples
2 4 Only 2s and 2p are available CH₄, NH₄⁺, BF₄⁻
3 and below 6 or more d orbitals can be used, so the octet can expand PCl₅, SF₆, AlF₆³⁻

That is why phosphorus makes PCl₃ and PCl₅ while nitrogen makes only NCl₃. The first members also form strong multiple bonds ($p\pi$–$p\pi$) such as C=C, N≡N and C=O.

Moving one step down and one step right, elements in a diagonal pair behave alike, because the increase in charge is offset by the increase in size. The two key pairs are lithium and magnesium, and beryllium and aluminium.

Pair Shared behaviour Example
Li and Mg Form a normal oxide with oxygen. React directly with nitrogen 6Li + N₂ → 2Li₃N and 3Mg + N₂ → Mg₃N₂
Li and Mg Nitrides give ammonia with water Mg₃N₂ + 6H₂O → 3Mg(OH)₂ + 2NH₃
Be and Al Amphoteric oxides and hydroxides BeO + 2NaOH → Na₂BeO₂ + H₂O
Be and Al Form nitrides 3Be + N₂ → Be₃N₂

Reactivity and the nature of oxides

Reactivity is highest at the two ends of a period: at the left, metals lose electrons easily (Na reacts with water, 2Na + 2H₂O → 2NaOH + H₂), and at the right, non metals gain them easily. Reactivity is lowest at the centre. The nature of the oxide changes steadily across period 3, from strongly basic to strongly acidic.

Oxide Nature
Na₂O Basic
MgO Basic
Al₂O₃ Amphoteric
SiO₂ Acidic
P₄O₁₀ Acidic
SO₃ Acidic
Cl₂O₇ Strongly acidic

Basic oxide with water: Na₂O + H₂O → 2NaOH. Acidic oxide with water: Cl₂O₇ + H₂O → 2HClO₄. An amphoteric oxide reacts with both acid and base:

Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O and Al₂O₃ + 2NaOH + 3H₂O → 2Na[Al(OH)₄]

Neutral oxides, such as CO, NO and N₂O, react with neither. Metallic character grows down a group, so oxides become more basic as you descend.

Exam traps

The trap Why it's tempting The truth
"The Modern Periodic Law is based on atomic mass." Mendeleev's law was Moseley showed the basis is atomic number
"Fluorine has the most negative electron gain enthalpy." It is the most electronegative Chlorine (−349) beats fluorine (−328) because F is so small
"IE₁ of boron is higher than that of beryllium." Boron has more protons Be > B, since the 2s electron is more tightly held than the 2p electron
"IE₁ of oxygen is higher than that of nitrogen." Oxygen has more protons N > O, because the half filled 2p³ of nitrogen is stable
"Noble gases have negative electron gain enthalpy." They are next to the halogens Their values are positive, because the electron must enter a new shell
"A cation is larger than its parent atom." The nucleus is the same A cation is smaller. An anion is larger
"Lanthanoids belong to the d block." They sit under the d block The 4f subshell is filling, so they are f block
"Helium is a p block element." It is placed with the p block noble gases Its configuration is $1s^2$, so electronically it is s block, placed in group 18

Trend and reaction sheet

Quantity Rule or value When to use
Group of a p block element $10 + (\text{electrons in } ns \text{ and } np)$ Locating an element from its configuration
Group of a d block element Electrons in $(n-1)d$ plus $ns$ Same
Period Highest $n$ occupied Same
Covalent radius $r = \tfrac{1}{2}d(\text{X–X})$ Radius from a bond length
Isoelectronic radius More protons means smaller O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺
Period 2 ionisation order Li < B < Be < C < O < N < F < Ne Ranking IE₁
Halogen electron gain Cl > F > Br > I Ranking electron gain enthalpy
Valence Groups 15 to 17: $8 - \text{valence electrons}$ Predicting formulas
Oxide trend Na₂O (basic) to Cl₂O₇ (acidic) Nature of an oxide

60-second revision

Topic Lock this in
Early classifications Döbereiner triads → Newlands octaves → Mendeleev (atomic weight) → Moseley (atomic number)
Mendeleev's gaps Eka-aluminium = Ga, eka-silicon = Ge, eka-boron = Sc. Oxides E₂O₃ and EO₂
Table layout 7 periods, 18 groups. Periods hold 2, 8, 8, 18, 18, 32, 32
Position Period = highest $n$. Group (p block) = 10 + valence electrons
IUPAC roots nil, un, bi, tri, quad, pent, hex, sept, oct, enn. Z = 120 is Unbinilium
Radius Falls across a period, rises down a group. Cation < atom < anion
Ionisation enthalpy Rises across, falls down. Exceptions: Be > B, N > O
Electron gain Cl > F. Positive for noble gases
Electronegativity Pauling, F = 4.0. Rises across, falls down
Period 2 anomaly Maximum covalency 4. Diagonal pairs Li–Mg and Be–Al
Oxides Basic → amphoteric → acidic across the period

Where this chapter connects

  • Before this: Structure of Atom gave the electron configurations and filling rules that decide every position in the table.
  • Next: Chemical Bonding and Molecular Structure uses ionisation enthalpy, electron gain enthalpy and electronegativity to decide whether atoms form ionic or covalent bonds.
  • Class XII link: The d and f Block Elements explains variable oxidation states, colour and the lanthanoid trends that begin here.

🎯 Practice MCQs

First 3 questions free · create a free account to unlock the rest — answers & explanations included, no payment needed

Q1. Assertion (A): Mendeleev placed iodine after tellurium even though iodine has the lower atomic weight. Reason (R): He grouped elements by similarity of chemical properties, even where this broke the order of atomic weight.

▸ Show answer & explanation

Answer: A

Tellurium (127.6) is heavier than iodine (126.9), yet iodine resembles chlorine and bromine, so Mendeleev put it in the halogen column. The reason states exactly why he did this. Options B, C and D each deny a true statement.

Q2. Which of the following correctly matches Mendeleev's prediction for eka-silicon with the properties of germanium?

▸ Show answer & explanation

Answer: A

Eka-silicon was predicted with atomic weight 72, oxide EO₂ and chloride ECl₄, and germanium has 72.6, GeO₂ and GeCl₄. Options B and C use the formulas predicted for eka-aluminium (E₂O₃, ECl₃), and D uses its atomic weight, 68.

Q3. The systematic IUPAC name and symbol of the element with atomic number 119 are

▸ Show answer & explanation

Answer: A

The digits 1, 1, 9 give the roots un, un and enn, so the name is Ununennium and the symbol is Uue. Option B corresponds to 109 (digits 1, 0, 9), C to 120 and D to 118.

🔒 9 more practice MCQs

Create a free account to unlock every MCQ in this chapter — answers and explanations included. No payment needed.

Already registered? Just log in and they'll all appear here.

📊 Previous-Year Questions

Practise with real CUET Chemistry previous-year papers — every question solved, with the correct answer and a step-by-step explanation.

View solved CUET PYQ papers →

Ready to drill Chemistry?

Unlock all MCQs, chapter tests, mocks & PYQs for ₹199/year.

Get UniDrill Pro