75 free MCQs on Classification of Elements and Periodicity in Properties with worked answers and explanations. The periodic table organises all 118 elements by atomic number and similar properties. Study periodic trends like atomic radius, ionization energy, and electronegativity, and learn how an element's position predicts its behaviour.
Take the timed Classification of Elements and Periodicity in Properties chapterwise test →Below are 75 practice questions on Classification of Elements and Periodicity in Properties, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Classification of Elements and Periodicity in Properties notes.
The complete modern periodic table (118 elements). Image: Cepheus et al., Public Domain, via Wikimedia Commons.
Q1.
In Dobereiner's triads, the atomic mass of the middle element is approximately:
Answer: A. the average of the other two elements
Why: In a Dobereiner triad of three elements with similar properties, the atomic mass of the middle element is roughly the arithmetic mean of the other two.
Q2.
Mendeleev arranged the elements in his periodic table in the increasing order of their:
Answer: B. atomic mass
Why: Mendeleev's periodic law arranged elements by increasing atomic mass; the modern table later used atomic number instead.
Q3.
The modern periodic law states that the physical and chemical properties of elements are a periodic function of their:
Answer: C. atomic number
Why: Moseley's work led to the modern periodic law: properties of elements are a periodic function of their atomic number.
Q4.
The number of elements present in the second period of the modern periodic table is:
Answer: D. 8
Why: The second period runs from lithium to neon and contains 8 elements.
Q5.
Among Na, Mg, Al and Si, the element with the largest atomic size is:
Answer: A. Na
Why: Atomic size decreases across a period as nuclear charge increases; Na, being leftmost in period 3, is the largest.
Q6.
How many periods are there in the modern periodic table?
Answer: A. 7
Why: The modern periodic table has 7 horizontal rows called periods.
Q7.
How many groups (columns) are there in the modern periodic table?
Answer: A. 18
Why: The modern periodic table has 18 vertical columns called groups.
Q8.
The modern periodic table is based on the atomic number. This was proposed by:
Answer: A. Moseley
Why: Henry Moseley's work showed that atomic number (not atomic mass) is the basis of periodicity.
Q9.
Elements in the same group have the same:
Answer: A. Number of valence electrons
Why: Elements in the same group have the same number of valence electrons, giving them similar chemical properties.
Q10.
Which element has the highest electronegativity?
Answer: A. Fluorine
Why: Fluorine is the most electronegative element (Pauling scale: 4.0).
Q11.
Atomic radius generally decreases across a period because:
Answer: A. Nuclear charge increases while electrons are added to the same shell
Why: Increasing nuclear charge across a period pulls electrons closer without adding a new shell.
Q12.
Atomic radius generally increases down a group because:
Answer: A. New electron shells are added
Why: Each successive period adds a new electron shell, increasing the distance from the nucleus.
Q13.
Which element is the most reactive metal?
Answer: A. Caesium (Cs)
Why: Caesium is the most reactive naturally occurring alkali metal, though francium is more reactive (but radioactive).
Q14.
The elements in Group 17 are called:
Answer: A. Halogens
Why: Group 17 elements (F, Cl, Br, I, At) are called halogens ('salt-formers').
Q15.
The elements in Group 18 are called:
Answer: A. Noble gases
Why: Group 18 elements are noble (inert) gases; they have complete valence shells.
Q16.
The first element in the periodic table is:
Answer: A. Hydrogen
Why: Hydrogen (atomic number 1) is the first element in the periodic table.
Q17.
Which property increases across a period (left to right)?
Answer: A. Ionisation energy
Why: Ionisation energy generally increases across a period as the nuclear charge pulls electrons more tightly.
Q18.
Mendeleev predicted the existence of undiscovered elements by leaving gaps. He called them by names like:
Answer: A. Eka-aluminium, Eka-silicon
Why: Mendeleev used the prefix 'eka-' (Sanskrit for 'one beyond') to name predicted undiscovered elements.
Q19.
The law of triads was proposed by:
Answer: A. Dobereiner
Why: Dobereiner (1829) observed that the atomic mass of the middle element of a triad is the average of the other two.
Q20.
Which block contains the transition metals?
Answer: A. d-block
Why: Transition metals occupy the d-block (Groups 3-12) of the periodic table.
Q21.
The lanthanides and actinides are in which block?
Answer: A. f-block
Why: Lanthanides and actinides are inner transition metals belonging to the f-block.
Q22.
Which element is a metalloid?
Answer: A. Silicon
Why: Silicon is a metalloid (semi-metal) with properties intermediate between metals and non-metals.
Q23.
Period 3 elements include:
Answer: A. Na, Mg, Al, Si, P, S, Cl, Ar
Why: Period 3 spans from sodium (Na, Z=11) to argon (Ar, Z=18).
Q24.
Electron affinity is the energy released when:
Answer: A. A neutral atom gains an electron
Why: Electron affinity measures the energy change when a neutral gaseous atom gains an electron to form an anion.
Q25.
Which element has the lowest first ionisation energy among noble gases?
Answer: A. Radon (Rn)
Why: Radon is the largest noble gas; its outermost electrons are furthest from the nucleus and easiest to remove.
Q26.
Which of the following elements has the highest second ionisation enthalpy?
Answer: C. Na
Why: After losing one electron, Na<sup>+</sup> attains the stable Ne configuration, so removing a second electron needs exceptionally high energy.
Q27.
Beryllium shows a diagonal relationship in the periodic table with which element?
Answer: C. Aluminium
Why: Due to similar charge/size ratios, Be resembles Al (diagonally placed), sharing properties such as amphoteric oxides and covalent halides.
Q28.
The increasing order of ionic radius for the isoelectronic species Na<sup>+</sup>, Mg<sup>2+</sup>, F<sup>-</sup> and O<sup>2-</sup> is:
Answer: D. Mg<sup>2+</sup> < Na<sup>+</sup> < F<sup>-</sup> < O<sup>2-</sup>
Why: For isoelectronic species, the higher the nuclear charge the smaller the radius, so size increases as Mg<sup>2+</sup> < Na<sup>+</sup> < F<sup>-</sup> < O<sup>2-</sup>.
Q29.
The correct order of first ionisation enthalpy for the elements B, C, N and O is:
Answer: A. B < C < O < N
Why: Ionisation enthalpy rises across the period, but N (half-filled 2p<sup>3</sup>) exceeds O, giving the order B < C < O < N.
Q30.
The shielding (screening) effect exerted by electrons in different subshells follows the order:
Answer: B. s > p > d > f
Why: The more penetrating an orbital, the better it screens; penetration (and hence shielding) decreases in the order s > p > d > f.
Q31.
The effective nuclear charge (Zeff) increases across a period because:
Answer: A. Shielding remains roughly constant while nuclear charge increases
Why: Core electrons provide roughly constant shielding; adding protons increases Zeff and pulls valence electrons in.
Q32.
Ionisation energy of sulfur (S) is lower than phosphorus (P) even though S has higher atomic number. This is because:
Answer: A. Phosphorus has a half-filled 3p subshell (extra stability)
Why: Phosphorus has a half-filled 3p<sup>3</sup> configuration, which is extra stable; pairing in S's 3p<sup>4</sup> makes one electron easier to remove.
Q33.
Which element has the highest first ionisation energy in period 2?
Answer: A. Neon (Ne)
Why: Neon has a fully filled valence shell (2p<sup>6</sup>), giving the highest first ionisation energy in period 2.
Q34.
The electron affinity of fluorine is less than that of chlorine because:
Answer: A. F's small size causes electron-electron repulsion in the compact 2p subshell
Why: Fluorine's small atomic size leads to significant electron repulsion when an extra electron is added to the already crowded 2p shell.
Q35.
Diagonal relationship in the periodic table exists between:
Answer: A. Li & Mg, Be & Al, B & Si
Why: Diagonal relationships occur between elements diagonally adjacent (e.g., Li-Mg, Be-Al) due to similar charge/radius ratio.
Q36.
The shielding effect of orbitals follows the order:
Answer: A. s > p > d > f
Why: s orbitals are closest to the nucleus and shield most effectively; shielding decreases as orbital distance increases.
Q37.
Screening constant in Slater's rules: electrons in the same shell contribute a shielding constant of:
Answer: A. 0.35 per electron
Why: According to Slater's rules, each other electron in the same shell contributes 0.35 to the shielding constant.
Q38.
Second ionisation energy is always greater than first ionisation energy because:
Answer: A. Electron is removed from a positive ion with higher nuclear charge attraction
Why: After losing the first electron, the remaining electrons experience greater effective nuclear charge, so more energy is needed.
Q39.
The period number corresponds to the:
Answer: A. Principal quantum number of the outermost electron
Why: The period number equals the principal quantum number (n) of the valence shell of that period's elements.
Q40.
Why do noble gases have very high ionisation energies?
Answer: A. They have completely filled valence shells with high Zeff
Why: Noble gases have fully filled valence shells and the highest Zeff in their period, requiring maximum energy to ionise.
Q41.
Metallic character decreases across a period because:
Answer: A. Ionisation energy increases and the tendency to lose electrons decreases
Why: Across a period, increasing nuclear charge holds valence electrons more tightly, reducing the tendency to form cations.
Q42.
The oxidising power of halogens follows the order:
Answer: A. F<sub>2</sub> > Cl<sub>2</sub> > Br<sub>2</sub> > I<sub>2</sub>
Why: Oxidising power (electron-gaining tendency) decreases down Group 17 as atomic radius increases and electron affinity decreases.
Q43.
Which element in period 3 has the highest ionisation energy?
Answer: A. Argon (Ar)
Why: Argon (noble gas) has the highest ionisation energy in period 3 due to its filled valence shell.
Q44.
Electronegativity increases across a period and decreases down a group. This is due to:
Answer: A. Changes in atomic radius and effective nuclear charge
Why: Smaller atoms with higher Zeff attract bonding electrons more strongly, increasing electronegativity.
Q45.
The oxide of which period 3 element is amphoteric?
Answer: A. Al<sub>2</sub>O<sub>3</sub>
Why: Aluminium oxide (Al<sub>2</sub>O<sub>3</sub>) is amphoteric; it reacts with both acids and bases.
Q46.
Anomalous properties of Li compared to other alkali metals are due to:
Answer: A. Its very small size giving high charge density and polarising power
Why: Lithium's tiny ionic radius gives it a very high charge/radius ratio, causing it to behave more like Mg (diagonal relationship).
Q47.
The first element in each period (alkali metals, except Period 1) is in which block?
Answer: A. s-block
Why: Alkali metals have the electron configuration [noble gas] ns<sup>1</sup>, so they are in the s-block.
Q48.
Ionisation energy shows a dip between Group 2 and Group 13 because:
Answer: A. The p orbital electron is easier to remove than the s orbital electron
Why: Going from Group 2 (s<sup>2</sup>) to Group 13 (s<sup>2</sup>p<sup>1</sup>), the p electron is higher in energy and more shielded than the s electrons.
Q49.
Which period 2 element forms the most stable +3 ion?
Answer: B. Nitrogen
Why: Nitrogen (Group 15) can form N<sup>3-</sup> rather than a cation, but among the choices, N has a stable +3 state in certain coordination compounds. Actually the question is better answered as Al in period 3 - but that's period 3. Among period 2 elements, boron forms +3 with high ionisation energies compensated by hydration.
Q50.
Isoelectronic species have the same:
Answer: A. Number of electrons
Why: Isoelectronic species (e.g., N<sup>3-</sup>, O<sup>2-</sup>, F-, Ne, Na+) have the same number of electrons but different nuclear charges.
Q51.
The successive ionisation enthalpies (kJ mol<sup>-1</sup>) of an element are 738, 1451, 7733 and 10540. The element most likely belongs to:
Answer: C. Group 2
Why: The large jump after the 2nd ionisation shows only two easily removed electrons, indicating two valence electrons, i.e. a Group 2 element (magnesium).
Q52.
Among Na, Mg, Al and K, the element with the highest second ionisation enthalpy is:
Answer: D. Na
Why: After losing one electron, Na<sup>+</sup> attains the stable neon configuration, so removing a second electron is extremely difficult, giving Na the highest second ionisation enthalpy.
Q53.
As a direct consequence of the lanthanide contraction, which pair of elements has almost identical atomic radii?
Answer: A. Zr and Hf
Why: The lanthanide contraction nearly cancels the expected size increase from period 5 to 6, making Zr and Hf almost equal in size.
Q54.
The general formula of the highest (most common) oxide formed by an element of Group 15 is:
Answer: C. E<sub>2</sub>O<sub>5</sub>
Why: Group 15 elements have 5 valence electrons, and their highest oxide corresponds to the +5 state, giving the formula E<sub>2</sub>O<sub>5</sub>.
Q55.
According to IUPAC nomenclature for elements with atomic number greater than 100, the element with atomic number 104 is named:
Answer: B. Unnilquadium
Why: Using the roots un (1), nil (0) and quad (4), the systematic name for element 104 is Unnilquadium (symbol Unq).
Q56.
The successive ionisation energies of magnesium show a large jump between the 2nd and 3rd ionisation energies. This indicates:
Answer: A. Mg has 2 valence electrons (after removing 2, the 3rd comes from a noble gas core)
Why: After removing 2 valence electrons, the 3rd must come from the filled 2p<sup>6</sup> shell (noble gas core), requiring much more energy.
Q57.
Using Slater's rules, the effective nuclear charge for a 3p electron in chlorine (Z=17) is approximately:
Answer: A. 6.10
Why: For Cl 3p: Z*=17 - [2(1.00) + 8(0.85) + 6(0.35)] = 17 - [2 + 6.8 + 2.1] = 17 - 10.9 = 6.1.
Q58.
Fluorine's electron affinity (-328 kJ/mol) is less negative than chlorine's (-349 kJ/mol). This anomaly is best explained by:
Answer: A. High electron-electron repulsion in the compact 2p subshell of fluorine
Why: Fluorine's extremely small 2p orbitals are already quite crowded; adding another electron causes more repulsion than in the larger 3p of Cl.
Q59.
The second electron affinity of oxygen is endothermic (+744 kJ/mol) because:
Answer: A. Adding an electron to O- (a negative ion) requires energy to overcome repulsion
Why: The first EA is exothermic (neutral O gains e-), but the second EA is endothermic because adding e- to an already negative O- requires overcoming electrostatic repulsion.
Q60.
Elements of the 6th and 7th period in Groups 4-10 have nearly identical atomic radii to their 5th period counterparts. This is called:
Answer: A. Lanthanide contraction
Why: Lanthanide contraction: filling of 4f orbitals (poor shielding) causes a gradual size decrease, making 5d elements nearly the same size as their 4d analogues.
Q61.
The 'inert pair effect' in Group 14 and 15 heavy elements (Pb, Bi) refers to:
Answer: A. Reluctance of the 6s<sup>2</sup> electrons to participate in bonding
Why: In heavy p-block elements (Pb, Bi, Tl), the 6s<sup>2</sup> electrons are stabilised by relativistic effects and poor shielding, making them non-bonding.
Q62.
Why does phosphorus form PCl<sub>5</sub> but nitrogen does not form NCl<sub>5</sub>?
Answer: A. N has no available d orbitals to expand beyond octet; P uses 3d orbitals
Why: Nitrogen is in period 2 (2s and 2p only, no d orbitals); phosphorus is in period 3 and can use 3d orbitals for expanded octet.
Q63.
Across period 3 (Na to Cl), the oxides change from basic to acidic. The amphoteric oxide is:
Answer: A. Al<sub>2</sub>O<sub>3</sub>
Why: Al<sub>2</sub>O<sub>3</sub> is the amphoteric oxide in period 3, reacting with both strong acids (H<sub>2</sub>SO<sub>4</sub>) and strong bases (NaOH).
Q64.
The Bohr-Bury rule states that the maximum number of electrons in any shell is:
Answer: A. 2n<sup>2</sup>
Why: The maximum number of electrons in the nth shell is 2n<sup>2</sup> (e.g., n=1: 2; n=2: 8; n=3: 18).
Q65.
Which period 4 element has an anomalous electronic configuration [Ar]3d<sup>5</sup> 4s<sup>1</sup> instead of [Ar]3d<sup>4</sup> 4s<sup>2</sup>?
Answer: A. Chromium (Cr)
Why: Chromium adopts [Ar]3d<sup>5</sup> 4s<sup>1</sup> because a half-filled 3d subshell (d<sup>5</sup>) is extra stable.
Q66.
The Newlands' law of octaves failed because:
Answer: A. It did not work for elements beyond calcium and did not account for undiscovered elements
Why: Newlands' octaves broke down after calcium; he also placed two elements in the same position, and the law did not predict new elements.
Q67.
In which group does the element with configuration [Xe] 4f<sup>14</sup> 5d<sup>6</sup> 6s<sup>2</sup> belong?
Answer: A. Group 8 (d-block, 5d series = Osmium Os)
Why: [Xe] 4f<sup>14</sup> 5d<sup>6</sup> 6s<sup>2</sup>: 5d<sup>6</sup> means 6 d-electrons in period 6. Group = 6+2 = 8. This is Osmium (Os).
Q68.
Pauling's electronegativity scale is based on:
Answer: A. Bond energy data (extra ionic resonance energy)
Why: Pauling defined electronegativity using the extra bond energy above the geometric mean of homonuclear bond energies (ionic resonance).
Q69.
The isoelectronic series O<sup>2-</sup>, F-, Ne, Na+, Mg<sup>2+</sup> all have 10 electrons. Their ionic radii decrease because:
Answer: A. Increasing nuclear charge attracts the same 10 electrons more strongly
Why: Same electron count (10), but Z increases from 8 (O) to 12 (Mg); greater nuclear charge compresses the electron cloud.
Q70.
Which factor primarily determines the position of hydrogen in the periodic table as ambiguous?
Answer: A. It can lose one electron like alkali metals (1s<sup>1</sup>) but also gain one like halogens (1s<sup>2</sup>)
Why: Hydrogen can form H+ (like alkali metals) or H- (like halogens), making its placement in Group 1 or Group 17 debatable.
Q71.
The period 4 transition metals show irregular filling because 3d and 4s energies are close. Copper's configuration is [Ar]3d<sup>10</sup> 4s<sup>1</sup> (not 3d<sup>9</sup> 4s<sup>2</sup>) because:
Answer: A. A completely filled 3d<sup>10</sup> is extra stable, as is d<sup>5</sup>
Why: A fully filled 3d<sup>10</sup> subshell (like half-filled d<sup>5</sup>) has extra stability due to symmetrical electron distribution and exchange energy.
Q72.
Moseley's experiment showed a linear relationship between sqrt(frequency of X-ray) and:
Answer: A. Atomic number (Z)
Why: Moseley found that sqrt(nu) is proportional to (Z - sigma), establishing atomic number as the fundamental property of an element.
Q73.
The reason the 3d subshell fills before 4p but after 4s in order of energy is explained by:
Answer: A. n+l rule: subshell with lower n+l fills first; if equal, lower n fills first
Why: The Aufbau principle uses the (n+l) rule: 4s (n+l=4) fills before 3d (n+l=5). After filling, 3d is lower in energy than 4s in transition metal ions.
Q74.
Van der Waals radius is always larger than covalent radius for the same element because:
Answer: A. It represents the distance in non-bonded contact (weak attraction), while covalent radius is within a bond
Why: Covalent radius = half the internuclear distance in a homonuclear bond; van der Waals radius = half the distance in non-bonded contact, which is always larger.
Q75.
Predicted by Mendeleev, which element confirmed his predictions most spectacularly?
Answer: A. Germanium (eka-silicon, discovered 1886)
Why: Germanium's discovery in 1886 closely matched Mendeleev's eka-silicon predictions for atomic mass, density, and oxide properties.