Chapter 3: Classification of Elements and Periodicity in Properties Case Study Questions | chemistry Class 11 CBSE

Case Study

Passage with linked questions

Case Set

Case Set 1

Case Analysis

Passage

A chemistry teacher shows her class a graph plotting atomic radius (in pm) against atomic number for elements of Period 2 and Period 3. The graph shows a clear downward trend within each period, with a sharp jump at the beginning of each new period. She points out that lithium has an atomic radius of 152 pm while fluorine has only 64 pm, and that sodium (186 pm) is much larger than lithium despite being in the same group but the period below. She then asks students to use the concepts of effective nuclear charge and shielding to explain these observations, and to predict the relative sizes of the ions formed when sodium and fluorine react to form sodium fluoride.

Question 1: Why does atomic radius decrease from lithium (152 pm) to fluorine (64 pm) across Period 2?

Across a period, electrons are added to the same principal energy level (n=2) while the nuclear charge increases with each element from Li (Z=3) to F (Z=9).
The increasing nuclear charge pulls all electrons in the same shell closer to the nucleus; the shielding by inner core electrons (1s2) remains nearly constant and does not compensate for the rising nuclear charge.
As a result, effective nuclear charge experienced by the outermost electrons increases across the period, contracting the electron cloud and reducing atomic radius progressively.

Question 2: Why is sodium (186 pm) larger than lithium (152 pm) even though sodium has a higher atomic number?

Sodium is in Period 3 (n=3) while lithium is in Period 2 (n=2); sodium's valence electron occupies the 3s orbital, which is a higher principal energy level farther from the nucleus than lithium's 2s orbital.
As we descend Group 1, each successive element adds a new principal energy level; inner electrons of the additional shell shield the outer electron from nuclear attraction, increasing atomic radius.
The combined effect of a higher principal quantum number and increased shielding by inner electrons outweighs the increase in nuclear charge, so sodium is larger than lithium.

Question 3: Predict the relative sizes of Na+, Na, F, and F- ions and atoms formed when sodium fluoride is produced, and explain the trend using nuclear charge and electron count.

Na has atomic radius 186 pm; when Na loses one electron to form Na+, the remaining 10 electrons are pulled more strongly by the same nuclear charge (Z=11), so Na+ (ionic radius ~95 pm) is much smaller than Na.
F has atomic radius 64 pm; when F gains one electron to form F-, the 10 electrons experience increased electron-electron repulsion with the same nuclear charge (Z=9), so F- (ionic radius ~136 pm) is larger than F.
Na+ and F- are isoelectronic (both have 10 electrons); Na+ has higher nuclear charge (Z=11) than F- (Z=9), so Na+ attracts the 10 electrons more strongly and is smaller than F-.
The overall size order is: Na > F- > Na+ > F; this demonstrates that cation formation shrinks an atom while anion formation expands it, and that for isoelectronic species a higher nuclear charge means smaller size.

ByteNotes

Case Study

Passage with linked questions

Case Set

Case Set 2

Case Analysis

Passage

In 1869, Dmitri Mendeleev published his Periodic Table arranging 63 known elements in increasing order of atomic weights. He noticed that gallium and germanium were missing and deliberately left gaps, naming them Eka-Aluminium and Eka-Silicon respectively. He predicted that Eka-Aluminium would have an atomic weight of 68, density 5.9 g/cm³, a low melting point, and its oxide formula would be E2O3. When gallium was discovered in 1875, its atomic weight was found to be 70, density 5.94 g/cm³, melting point 302.93 K, and its oxide formula was Ga2O3 — remarkably close to Mendeleev's predictions. Similarly, Eka-Silicon matched germanium's properties when it was discovered in 1886.

Question 1: What was the basis of Mendeleev's Periodic Law, and how did it allow him to predict properties of undiscovered elements?

Mendeleev's Periodic Law states that the properties of elements are a periodic function of their atomic weights; he arranged elements in order of increasing atomic weight so that elements with similar properties fell in the same vertical column (group).
By examining trends within a group and across periods, Mendeleev could interpolate the properties of missing elements occupying gaps in the table between known elements.
He named the gaps Eka-Aluminium and Eka-Silicon (eka meaning 'one beyond' in Sanskrit) and predicted specific values for atomic weight, density, melting point, and oxide/chloride formulas based on the average of surrounding elements.

Question 2: Mendeleev placed iodine after tellurium in his table even though iodine has a lower atomic weight than tellurium. Explain this anomaly and how the Modern Periodic Law resolved it.

In Mendeleev's arrangement by increasing atomic weight, iodine (atomic weight ~127) should come before tellurium (~128), but iodine's chemical properties — resembling fluorine, chlorine, and bromine — placed it firmly in Group VII (halogens), not Group VI.
Mendeleev prioritised chemical similarity over strict atomic weight order, placing tellurium in Group VI and iodine in Group VII, believing the atomic weight measurements might be incorrect.
The Modern Periodic Law (Moseley, 1913) resolved this by arranging elements by atomic number: tellurium has Z=52 and iodine has Z=53, so iodine correctly follows tellurium without any anomaly.

Question 3: Compare the predicted properties of Eka-Silicon with the experimentally found properties of germanium. What does the accuracy of these predictions reveal about the validity of Mendeleev's periodic classification?

Mendeleev predicted Eka-Silicon would have atomic weight 72, density 5.5 g/cm³, high melting point, oxide formula EO2, and chloride formula ECl4; germanium was found to have atomic weight 72.6, density 5.36 g/cm³, melting point 1231 K, oxide GeO2, and chloride GeCl4 — an extremely close match.
The accuracy of these predictions demonstrated that Mendeleev's periodic classification was not arbitrary but captured a genuine law of nature: properties repeat periodically, allowing reliable interpolation of unknown values.
The predictive success gave Mendeleev's periodic table enormous scientific credibility and showed that the table could guide future research, not just organise existing knowledge.
This validated the concept of periodicity — that elements are not random but display systematic, predictable trends — and made the periodic table an essential tool in chemistry, ultimately stimulating continued research into new elements.

ByteNotes

Case Study

Passage with linked questions

Case Set

Case Set 3

Case Analysis

Passage

A laboratory technician is characterising four unknown elements labelled W, X, Y, and Z. The data collected is as follows: Element W has first ionization enthalpy 2372 kJ/mol and positive electron gain enthalpy; Element X has first ionization enthalpy 419 kJ/mol and is found in nature only in combined form; Element Y has highly negative electron gain enthalpy (-349 kJ/mol) and is a gas at room temperature; Element Z has first ionization enthalpy 738 kJ/mol and forms a stable MX2 type halide. The technician needs to classify these elements and predict their chemical behaviour using the Modern Periodic Table.

Question 1: Identify which block and likely group Element W belongs to, based on its ionization enthalpy and electron gain enthalpy values.

Element W has an extremely high first ionization enthalpy (2372 kJ/mol) and positive electron gain enthalpy; noble gases (Group 18) have the highest ionization enthalpies in each period because their valence shells are completely filled and very stable.
The positive electron gain enthalpy indicates that energy must be supplied to add an electron, consistent with noble gases where the added electron would have to enter the next higher principal quantum level, creating an unstable configuration.
Element W belongs to the p-block, Group 18 (noble gases); from the chapter's Fig. 3.5, helium has ΔiH ~2372 kJ/mol and neon ~2080 kJ/mol, so W is most likely helium.

Question 2: Element X has the first ionization enthalpy of 419 kJ/mol and is never found free in nature. Identify the element from the chapter data and explain why it is always found in combined form.

From Fig. 3.6(b) in the chapter, potassium (K) has a first ionization enthalpy of 419 kJ/mol; potassium is an alkali metal (Group 1, Period 4, s-block) with outer configuration 4s1.
The very low ionization enthalpy means the outermost 4s1 electron is very loosely held and easily lost; potassium reacts vigorously with water, oxygen, and other substances, readily forming ionic compounds.
Because of this high reactivity, elemental potassium instantly reacts with moisture and oxygen in the environment and is always found as compounds (e.g., KCl, KNO3) rather than as a free metal in nature.

Question 3: Element Z forms a stable MX2 halide and has a first ionization enthalpy of 738 kJ/mol. Identify the element from chapter data, determine its block position, and predict two additional properties it would exhibit.

From Fig. 3.6(a), beryllium (Be) has ΔiH = 899 kJ/mol and magnesium (Mg) is in Group 2; from Table 3.31 data in the chapter's exercises, element VI has ΔiH1=738 and ΔiH2=1451 kJ/mol forming MX2 halide — this corresponds to a Group 2 alkaline earth metal, likely calcium or magnesium (Mg: 737 kJ/mol from chapter context).
As a Group 2 (s-block) element, Z has configuration ns2; it loses both outermost electrons to form a 2+ ion, hence the stable MX2 formula where X is a halogen; second ionization enthalpy (1451) is much higher than first, consistent with losing both ns2 electrons before reaching a noble gas core.
Additional predicted properties: (1) Metallic character — Z is a reactive metal with high melting point and good electrical conductivity; (2) its compounds are predominantly ionic (with exceptions for beryllium, which shows covalent character due to anomalous small size).
The element's metallic character and reactivity increase as we go down Group 2; if Z is magnesium, it is less reactive than calcium but more reactive than beryllium; its oxide (MO) would be basic in nature.

ByteNotes

Case Study

Passage with linked questions

Case Set

Case Set 4

Case Analysis

Passage

Henry Moseley, an English physicist, conducted experiments in 1913 in which he bombarded different elements with electrons and analysed the X-rays emitted. He found that when he plotted the square root of the frequency of emitted X-rays (√ν) against the atomic number (Z) of the element, he obtained a straight line. However, when he plotted √ν against atomic mass, the plot was not a straight line. This led him to propose that atomic number, not atomic mass, is the more fundamental property of an element. This discovery fundamentally changed the basis of periodic classification and resolved several anomalies present in Mendeleev's table, such as the placement of tellurium before iodine.

Question 1: State the Modern Periodic Law that emerged from Moseley's work and explain how it differs from Mendeleev's Periodic Law.

The Modern Periodic Law states: the physical and chemical properties of elements are periodic functions of their atomic numbers (not atomic weights as in Mendeleev's law).
Mendeleev's law used atomic weight as the basis of classification, which caused anomalies where elements had to be placed out of strict weight order to maintain chemical similarity in groups (e.g., Te before I).
Moseley's law uses atomic number (number of protons = nuclear charge) as the basis; since atomic number is a whole number that increases by exactly 1 from element to element, it eliminates all anomalies and provides a more fundamental and theoretically sound basis for classification.

Question 2: Explain the connection between atomic number, electronic configuration, and the periodic law as described in the chapter.

Atomic number equals the nuclear charge (number of protons) and also the number of electrons in a neutral atom; it directly determines how electrons are arranged in orbitals (electronic configuration).
The Periodic Law is essentially a consequence of the periodic variation in electronic configurations: elements in the same period have incrementally increasing electron counts, while elements in the same group have the same type and number of outermost electrons.
Since electronic configuration determines all physical and chemical properties of an element (bonding, reactivity, ionization enthalpy, etc.), the periodicity of properties is a natural consequence of the periodicity of electronic configurations with increasing atomic number.

Question 3: Moseley's work resolved the Te-I anomaly in Mendeleev's table. Explain this anomaly, how atomic number resolved it, and discuss one other anomaly that the Modern Periodic Law resolved compared to Mendeleev's arrangement.

Anomaly 1 (Te-I): Tellurium (atomic mass ~128) has a higher atomic mass than iodine (~127), so in Mendeleev's mass-based arrangement, Te should come after I; but Te belongs in Group VI and I in Group VII chemically. Mendeleev placed them out of mass order. With atomic numbers Z(Te)=52 and Z(I)=53, iodine correctly follows tellurium without any anomaly.
Anomaly 2 (Ar-K): Argon (atomic mass 39.9) is heavier than potassium (atomic mass 39.1), again placing them out of order in Mendeleev's table; by atomic number, Z(Ar)=18 and Z(K)=19, so argon correctly precedes potassium in Period 3 and Period 4 respectively.
The Modern Periodic Law also resolved the Co-Ni anomaly: cobalt (atomic mass 58.9) appears before nickel (58.7) by mass, but Z(Co)=27 and Z(Ni)=28 places them correctly.
Overall, Moseley's discovery that atomic number is the fundamental property eliminated all such inversions and gave the periodic table a rigorous theoretical foundation tied to nuclear structure.

ByteNotes

Case Study

Passage with linked questions

Case Set

Case Set 5

Case Analysis

Passage

During a school project, a student studies the first ionization enthalpies (in kJ/mol) of the second period elements: Li=520, Be=899, B=801, C=1086, N=1402, O=1314, F=1681, Ne=2080. The student notices that while the general trend shows an increase from Li to Ne, there are two unexpected dips: one at boron (B) compared to beryllium (Be), and another at oxygen (O) compared to nitrogen (N). The teacher explains that these dips are not errors but are caused by specific features of electronic structure — differences in orbital types and electron pairing — and are important anomalies that frequently appear in board examinations.

Question 1: State the general trend in first ionization enthalpy across Period 2 and explain the overall reason for this trend.

The general trend shows that first ionization enthalpy increases from lithium (520 kJ/mol) to neon (2080 kJ/mol) across Period 2, with two exceptions at boron and oxygen.
As we move across the period, electrons are added to the same principal quantum level (n=2) while nuclear charge increases from Z=3 to Z=10; the shielding provided by the inner 1s2 core remains nearly constant.
The increasing nuclear charge outweighs the constant shielding, so effective nuclear charge increases across the period, pulling valence electrons closer and requiring progressively more energy for their removal.

Question 2: Explain why boron (801 kJ/mol) has a lower first ionization enthalpy than beryllium (899 kJ/mol) despite having a higher atomic number.

Beryllium (Z=4) has configuration 1s²2s²; the electron removed during ionization is a 2s electron, which has greater penetration toward the nucleus (higher electron density near nucleus) and is more tightly held.
Boron (Z=5) has configuration 1s²2s²2p¹; the electron removed during ionization is a 2p electron; the 2p orbital is higher in energy and more effectively shielded from the nucleus by the 2s² electrons than the 2s orbital of beryllium.
Therefore, despite boron having a higher nuclear charge, its outermost 2p electron is more loosely held than beryllium's 2s electron, giving boron a lower first ionization enthalpy — an anomaly in the general increasing trend.

Question 3: Explain why oxygen (1314 kJ/mol) has a lower first ionization enthalpy than nitrogen (1402 kJ/mol), and predict whether a similar anomaly would occur between phosphorus and sulphur in Period 3. Justify your prediction.

Nitrogen (Z=7) has configuration 1s²2s²2p³; the three 2p electrons are distributed one each in the three 2p orbitals (Hund's rule), giving a stable half-filled configuration with no electron-electron repulsion within any single 2p orbital.
Oxygen (Z=8) has configuration 1s²2s²2p⁴; the fourth 2p electron must pair with an existing electron in one of the 2p orbitals, causing increased electron-electron repulsion within that orbital; this repulsion makes it easier to remove the paired electron.
Consequently, oxygen's first ionization enthalpy (1314 kJ/mol) is lower than nitrogen's (1402 kJ/mol) despite oxygen having higher nuclear charge — the repulsion effect outweighs the nuclear charge increase.
A similar anomaly would occur between phosphorus (P: 3p³, half-filled) and sulphur (S: 3p⁴, one paired electron) in Period 3; indeed, the first ionization enthalpy of P (1012 kJ/mol) is higher than that of S (1000 kJ/mol) for exactly the same reason — the half-filled 3p³ configuration of P is extra stable.

ByteNotes

Premium Access

Unlock all Case Study sets

You are viewing 5 free case study sets from this chapter. Upgrade to Premium to unlock the remaining 10 sets.

Upgrade to Premium10 questions locked

Chapter sections

Chapter 3: Classification of Elements and Periodicity in Properties | Case Study Questions

Passage

Question 1: Why does atomic radius decrease from lithium (152 pm) to fluorine (64 pm) across Period 2?

Question 2: Why is sodium (186 pm) larger than lithium (152 pm) even though sodium has a higher atomic number?

Question 3: Predict the relative sizes of Na+, Na, F, and F- ions and atoms formed when sodium fluoride is produced, and explain the trend using nuclear charge and electron count.

Passage

Question 1: What was the basis of Mendeleev's Periodic Law, and how did it allow him to predict properties of undiscovered elements?

Question 2: Mendeleev placed iodine after tellurium in his table even though iodine has a lower atomic weight than tellurium. Explain this anomaly and how the Modern Periodic Law resolved it.

Question 3: Compare the predicted properties of Eka-Silicon with the experimentally found properties of germanium. What does the accuracy of these predictions reveal about the validity of Mendeleev's periodic classification?

Passage

Question 1: Identify which block and likely group Element W belongs to, based on its ionization enthalpy and electron gain enthalpy values.

Question 2: Element X has the first ionization enthalpy of 419 kJ/mol and is never found free in nature. Identify the element from the chapter data and explain why it is always found in combined form.

Question 3: Element Z forms a stable MX2 halide and has a first ionization enthalpy of 738 kJ/mol. Identify the element from chapter data, determine its block position, and predict two additional properties it would exhibit.

Passage

Question 1: State the Modern Periodic Law that emerged from Moseley's work and explain how it differs from Mendeleev's Periodic Law.

Question 2: Explain the connection between atomic number, electronic configuration, and the periodic law as described in the chapter.

Question 3: Moseley's work resolved the Te-I anomaly in Mendeleev's table. Explain this anomaly, how atomic number resolved it, and discuss one other anomaly that the Modern Periodic Law resolved compared to Mendeleev's arrangement.

Passage

Question 1: State the general trend in first ionization enthalpy across Period 2 and explain the overall reason for this trend.

Question 2: Explain why boron (801 kJ/mol) has a lower first ionization enthalpy than beryllium (899 kJ/mol) despite having a higher atomic number.

Question 3: Explain why oxygen (1314 kJ/mol) has a lower first ionization enthalpy than nitrogen (1402 kJ/mol), and predict whether a similar anomaly would occur between phosphorus and sulphur in Period 3. Justify your prediction.

Unlock all Case Study sets