NCERT Intext Solutions for Classification of Elements and Periodicity in Properties provide clear and detailed explanations of important chemistry concepts related to the modern periodic table and periodic trends. These solutions help students understand topics such as atomic radius, ionization enthalpy, electron gain enthalpy, electronegativity, valence electrons, metallic and non-metallic character, and periodic classification of elements. The chapter explains how the properties of elements change systematically across periods and down groups due to electronic configuration. These NCERT intext solutions are useful for CBSE board exams, competitive exams, and concept building in chemistry.
For complete preparation, also study NCERT Textbook Solutions: Classification of Elements and Periodicity in Properties Class 11 Chemistry
NCERT Intext Questions and Answers for Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties
NCERT Intext Questions and Solutions for Class 11 Chemistry Chapter Classification of Elements and Periodicity in Properties help students understand the modern periodic table and important periodic trends in a simple and conceptual manner. Students can use these intext solutions for board exam preparation, revision, and strengthening their chemistry concepts.
Related topics include Why Do Elements Show Periodicity?
Are NCERT intext questions important for Class 11 Chemistry exams?
Yes, NCERT intext questions are very important for Class 11 Chemistry exams because many conceptual and board examination questions are directly based on NCERT textbook exercises and examples.
Students should also study Why Exact Atomic Radius is Difficult to Determine ?
Can NCERT intext solutions help in competitive exam preparation?
Yes, NCERT intext solutions help students build strong basic concepts which are useful for competitive exams like JEE Main, NEET, and other entrance examinations.
This concept is linked with What is Metallic Radius ?
NCERT Intext 3.1 Question
What would be the IUPAC name and symbol for the element with atomic number 120?
Solution :
The systematic temporary name for the element with atomic number 120 (Z = 120) is Unbinilium, and its official three-letter symbol is Ubn.
IUPAC creates these placeholder names by translating each digit of the atomic number into a specific Latin or Greek root, putting them together, and adding the suffix “-ium” :
As per NCERT/IUPAC 1 means un, 2 means bi, 0 means nil
Assemble digits “1–2–0” i.e. un–bi–nil + suffix – ium. The complete name is unbinilium.
Temporary symbol : take first letters of the roots with proper capitalization i.e. Ubn
Since element 120 has not been officially synthesized and confirmed in a lab yet, it doesn’t have a permanent, traditional name (like Gold or Einsteinium). Until a research team successfully creates it and IUPAC approves a permanent name to honor a scientist or laboratory, Unbinilium (Ubn) is the universally accepted name used in chemistry textbooks and research papers.
Build strong concepts by studying Nomenclature System for the Elements with Atomic Numbers More than 100
NCERT Intext 3.2 Question
How would you justify presence of 18 elements in 5th period of Periodic Table ?
Solution :
For the 5th period, the highest principal quantum number is n = 5. According to the Aufbau principle, electrons fill subshells in the order of increasing energy. For this period, the relative energy order that electrons fill the 5s, then the 4d, and finally the 5p subshells before moving on to the 6th energy level.
The number of elements in a period corresponds exactly to the maximum number of electrons that these available subshells can accommodate :
5s subshell : Contains 1 orbital and holds a maximum of 2 electrons (2 elements : Rubidium to Strontium)
4d subshell : Contains 5 orbitals and holds a maximum of 10 electrons (10 elements : Yttrium to Cadmium)
5p subshell : Contains 3 orbitals and holds a maximum of 6 electrons (6 elements : Indium to Xenon)
Total Elements = 2 from 5s + 10 from 4d + 6 from 5p = 18 elements.
Because these three subshells can collectively hold a maximum of 18 electrons, the 5th period of the Periodic Table must contain exactly 18 elements (starting from Rubidium, Z = 37, and ending at Xenon, Z = 54).
Practice more questions from Modern Periodic Table Solved Examples, Conceptual Questions Answers, Practice Problems
NCERT Intext 3.3 Question
The elements Z = 117 and 120 have not yet been discovered. In which family/group would you place these elements and also give the electronic configuration in each case.
Solution :
Element with Atomic Number Z = 117 will lie in Group 17 (halogen family). Group 17 elements have the general electronic configuration ns²np⁵, i.e., they end with 7 valence electrons. Its Electronic Configuration is [Rn] 5f14 6d10 7s2 7p5. Thus, element 117 will behave like a halogen (similar to Iodine and Astatine).
Element with Atomic Number Z = 120 will lie in Group 2 (alkaline earth metals). Group 2 elements have the general electronic configuration ns², i.e., they end with 2 valence electrons. Its Electronic Configuration is [Uuo] 8s2 where [Uuo] represents the noble gas with Z = 118 (Oganesson). Hence, element 120 would resemble Calcium, Strontium, Barium, and Radium in its chemical behavior.
Enhance your preparation with Division of Periodic Table into s,p,d,f blocks : Prediction of Period, Group and Block of Elements
NCERT Intext 3.4 Question
Considering the atomic number and position in the periodic table, arrange the following elements in the increasing order of metallic character : Si, Be, Mg, Na, P.
Solution :
Metallic character refers to the tendency of an element to lose electrons and form positive ions (electropositive nature). Metallic character increases down a group (due to larger atomic size and lower ionization energy) and decreases across a period from left to right (due to smaller atomic size and higher ionization energy).
P (Phosphorus) belongs to Group 15, Period 3 is Non-metal.
Si (Silicon) belongs to Group 14, Period 3 is Metalloid (less metallic than Be or Mg).
Be (Beryllium) belongs to Group 2, Period 2 is Alkaline earth metal (least metallic in Group 2).
Mg (Magnesium) belongs to Group 2, Period 3 is Alkaline earth metal (more metallic than Be).
Na (Sodium) belongs to Group 1, Period 3 is Alkali metal (highly metallic).
Arranging by Metallic Character : Least metallic P (non-metal) < Si (metalloid) < Be (metal, but small size, low metallic nature among Group 2) < Mg (more metallic than Be) < Most metallic Na (alkali metal)
Strengthen your fundamentals with How is electronegativity related to metallic and non-metallic character?
NCERT Intext 3.5 Question
Which of the following species will have the largest and the smallest size? Mg, Mg²⁺, Al, Al³⁺.
Solution :
Magnesium (Mg, Z = 12) and Aluminum (Al, Z = 13) belong to the same period (Period 3). Atomic radius decreases from left to right across a period because the nuclear charge increases while the electrons are added to the same principal shell. Therefore, Magnesium is larger than Aluminum : Mg > Al.
Both Mg2+ and Al3+ have lost electrons to achieve a stable noble gas configuration, leaving both with exactly 10 electrons. Species with the same number of electrons are called isoelectronic. For isoelectronic species, the higher the nuclear charge (number of protons), the stronger the nucleus pulls the remaining electrons inward, making the ionic radius smaller. Al3+ has 13 protons pulling 10 electrons, while Mg2+ has only 12 protons pulling 10 electrons. Therefore, Al3+ experiences a stronger pull and is smaller than Mg2+ : Mg2+ > Al3+.
When an atom loses electrons to form a cation, its size always decreases drastically because it loses an entire outer shell and experiences a higher effective nuclear charge. Combining both rules, the overall order of increasing size is :
Al3+ < Mg2+ < Al < Mg
Mg is the largest because it retains its valence shell (n = 3) and has a lower nuclear charge than Aluminum.
Al3+ is the smallest because it has lost its outer shell (n = 3 to n=2) and possesses the highest nuclear charge (+13) to pull its remaining electrons closest to the nucleus.
Practice more questions from Conceptual Questions Based On Atomic Radii, Ionic Radii and Isoelectronic Ions Trends of Periodic Table
NCERT Intext 3.6 Question
The first ionization enthalpy $(\Delta_i H)$ values of the third period elements, Na, Mg and Si are respectively 496, 737 and 786 kJ/mol. Predict whether the first $(\Delta_i H)$ value for Al will be more close to 575 or 760 kJ/mol ? Justify your answer.
Solution :
Generally, ionization enthalpy increases as you move from left to right across a period due to an increase in nuclear charge. Following this general trend, the ionization enthalpy of Aluminum (Z = 13) is in between Magnesium (Z = 12) and Silicon (Z = 14). However, an exception occurs between Magnesium and Aluminum because of their electronic configurations. The electronic configurations of both elements is :
Mg (Z = 12): $1s^2 2s^2 2p^6 3s^2$
Al (Z = 13): $1s^2 2s^2 2p^6 3s^2 3p^1$
An electron in a 3s orbital more closer to the nucleus than an electron in a 3p orbital. Therefore, Magnesium’s 3s valence electrons are pulled more tightly by the nucleus. The 3p electron of Aluminum is effectively shielded from the nuclear charge by the inner core electrons plus its own filled 3s2 subshell.
Because the 3p1 electron in Aluminum experiences a weaker effective nuclear charge and is less stable in its position, it requires less energy to remove than a 3s electron from Magnesium. The first ionization enthalpy of Al will be more close to 575 kJ/mol because the 3p electron is more shielded and easier to remove compared to the 3s electron of Mg.
Since $\Delta_i H$ for Mg is 737 kJ/mol, the value for Al must be lower than Magnesium’s value, not higher. Therefore, the actual value for Aluminum must be 575 kJ/mol.
Similar topics for practice include Why does Ionization Enthalpy Increase from left to right in a Period ?
NCERT Intext 3.7 Question
Which of the following will have the most negative electron gain enthalpy and which the least negative?
P, S, Cl, F. Explain your answer.
Solution :
Chlorine has the most negative electron gain enthalpy among the given elements. Generally, electron gain enthalpy becomes more negative as you move from left to right across a period because the increasing nuclear charge and decreasing atomic size allow the nucleus to pull an incoming electron more strongly.
Consequently, the halogen group (Group 17) releases the most energy when gaining an electron. While you might expect Fluorine to have a more negative value than Chlorine because it lie above it in the group and is smaller, its smallest atomic radius actually creates an anomaly.
The incoming electron in Fluorine must enter a highly compact 2p subshell where it experiences intense electron-electron repulsion from the existing electrons. In Chlorine, the incoming electron enters the larger 3p subshell, where the electrons are spread over a greater volume, resulting in significantly less repulsion and a more exothermic, negative electron gain enthalpy.
Phosphorus has the least negative electron gain enthalpy in this group. Since it sits in Group 15, it is located furthest to the left among the third-period elements given, meaning it has the lowest nuclear charge and the largest atomic size of the three. Furthermore, Phosphorus possesses a stable, half-filled 3p3 electronic configuration. Because half-filled subshells are inherently stable due to symmetrical electron distribution and high exchange energy, Phosphorus has least tendency to accept an extra electron. Forcing an electron into this stable configuration requires overcoming this stability, which drastically reduces the amount of energy released during the process.
Sulfur falls right in the middle of the trend. It has a more negative electron gain enthalpy than Phosphorus because it has a higher nuclear charge and lacks the extra stability of a half-filled shell.
However, it is less negative than Fluorine and Chlorine because it is in Group 16 and experiences a weaker nuclear pull than the halogens. Hence Chlorine is the most negative electron gain enthalpy because of its high nuclear charge and spacious valence shell, while Phosphorus is the least negative electron gain enthalpy due to its larger size and stable, half-filled electronic configuration.
Similar topics for practice include Why do halogens (group 17) have highly negative electron gain enthalpy values ?
NCERT Intext 3.8 Question
Using the Periodic Table, predict the formulas of compounds which might be formed by the following pairs of elements:
(a) silicon and bromine
(b) aluminium and sulphur
Solution :
For the first pair, silicon and bromine, silicon is in Group 14 of the periodic table (configuration : 3s2 3p2), meaning it has four valence electrons. Bromine is a halogen located in Group 17 (configuration : 4s2 4p5), which means it has seven valence electrons and requires just one more electron to complete its stable octet.
To complete its octet, each bromine atom needs 1 electron. Silicon can share its 4 valence electrons with 4 bromine atoms, forming 4 covalent bonds. Therefore, the compound formed will be SiBr4 , chemically known as silicon tetrabromide.
For the second pair, aluminum and sulfur, aluminum belongs to Group 13 (configuration: 3s2 3p1) with three valence electrons, while sulfur is in Group 16 (configuration : 3s2 3p4) with six valence electrons. Aluminum is a metal that prefers to lose its three valence electrons to achieve a stable noble gas configuration, forming an Al3+ cation. Conversely, sulfur is a non-metal that needs to gain two electrons to complete its outer octet, forming an S2- anion. Aluminium tends to lose 3 electrons to achieve noble gas configuration, while sulphur needs 2 electrons to complete its octet. Thus, two Al atoms together lose 6 electrons which can be accepted by three S atoms (each gaining 2 electrons). Hence, the compound formed will be Al2S3 known as aluminum sulfide.
Enhance your preparation with NCERT Exemplar Solutions of Chapter 3 Classification of Elements and Periodicity in Properties for Class 11 Chemistry
NCERT Intext 3.9 Question
Are the oxidation state and covalency of Al in [AlCl(H2O)5]2+ the same?
Solution :
No, the oxidation state and the covalency of aluminum Al in this complex are not the same. To find the oxidation state of Al :
Let the oxidation state of Al be $x$.
Chloride ion contributes charge -1, and each water molecule is neutral. The overall charge on the complex is +2.
So,
$$ x + (-1) + 0 = +2 $$
$$ x – 1 = +2$$
$$ x = +3$$
Therefore, the oxidation state of Al is +3.
Covalency, on the other hand, is the total number of chemical bonds the aluminum atom forms with its neighbors. It does not care about charges or oxidation numbers. In this coordination complex, the central aluminum atom is directly bonded to one chloride ion and five separate water molecules using coordinate covalent bonds. By adding these up, aluminum forms a total of six bonds (1 + 5 = 6). Therefore, the covalency of aluminum in this complex is 6.
Hence, the oxidation state of aluminum is +3 because of the electronic charge balance, while its covalency is 6 because it is physically bonded to six surrounding groups.
Students should also study electronic configuration and bonding concepts to strengthen their understanding for CBSE, NEET, and JEE examinations.
NCERT Intext 3.10 Question
Show by a chemical reaction with water that Na2O is a basic oxide and Cl2O7 is an acidic oxide.
To check whether an oxide is acidic or basic, we can dissolve it in water and examine the chemical properties of the resulting solution. When an oxide reacts with water to yield a base, it is classified as a basic oxide. Conversely, if it reacts with water to yield an acid, it is classified as an acidic oxide.
When sodium oxide (Na2O), which is a metallic oxide, reacts with water, it undergoes a chemical combination to form sodium hydroxide. The balanced chemical equation for this reaction is :
$$\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH}$$
Because sodium hydroxide (NaOH) is a strong base that completely dissociates in water to release hydroxide ions, this reaction proves that sodium oxide is a basic oxide. If you test this resulting solution with a strip of red litmus paper, the paper will turn blue, confirming its basic nature.
On the other hand, when dichlorine heptoxide (Cl2O7), which is a non-metallic oxide, is dissolved in water, it reacts to form perchloric acid. The balanced chemical equation for this reaction is :
$$\text{Cl}_2\text{O}_7 + \text{H}_2\text{O} \rightarrow 2\text{HClO}_4$$
Since perchloric acid (HClO4) is a highly corrosive, strong acid that releases hydrogen ions in solution, this reaction shows that dichlorine heptoxide is an acidic oxide. Testing this particular solution with a strip of blue litmus paper will cause it to turn red, which serves as a classic chemical confirmation of its acidic nature.
Strengthen your fundamentals with Periodic Trends and Chemical Reactivity: Metallic and Non-Metallic, Basic, Acidic Character Across a Period and Down a Group