/ Preparation Chemistry Ncert Solutions Class 11th
Classification of Elements and Periodicity in Properties NCERT Solutions is a comprehensive guide for the periodic table. It introduces the rationale behind grouping elements based on their properties. Class 11 Chemistry Ch 3 NCERT Solutions provide step-by-step answers to all NCERT textbook questions. It covers key concepts like trends in atomic size, modern periodic law, electron gain enthalpy, ionization enthalpy, and electronegativity.
NCERT Solutions for Class 11 Chemistry Chapter 3 help students to boost their understanding of the fundamentals and teach them how to apply concepts effectively. The solutions are ideal for CBSE Board exam preparation.
- Class 11 Classification of Elements and Periodicity in Properties: Key Topics, Weightage
- Important Formula of Class 11 Classification of Elements and Periodicity in Properties
- Class 11th Chemistry Classification of Elements and Periodicity NCERT Solution PDF: Download PDF for Free
- Classification of Elements and Periodicity in Properties Solutions
- NCERT Class XIth Chemistry Classification of Elements and Periodicity Covers
- NCERT Class 11 Chemistry Chapter 3 Exercise Solutions
- Class 11 Chemistry Assertion Questions
- Classification of Elements and Periodicity in Properties Questions and Answers
Class 11 Classification of Elements and Periodicity in Properties: Key Topics, Weightage
Class 11th Chapter 3 is a very important unit responsible for introducing crucial chemistry concepts to lay the foundational bricks for students. Students can check the topics covered in this chapter:
| Exercises | Topics Covered |
|---|---|
| 3.1 | Why do we need to classify elements |
| 3.2 | Genesis of Periodic Classification |
| 3.3 | Modern Periodic Law and the Present Form of the Periodic Table |
| 3.4 | Nomenclature of Elements with Atomic Numbers > 100 |
| 3.5 | Electronic Configurations of Elements and the Periodic Table |
| 3.6 | Electronic Configurations and Types of Elements: S-, P-, D-, F- Blocks |
| 3.7 | Periodic Trends in Properties of Elements |
Classification of Elements and Periodicity in Properties Weightage in NEET, JEE Main Exams
| Exam | Number of Questions | Weightage |
|---|---|---|
| NEET | 1-2 questions | 3% |
| JEE Main | 2-3 questions | 2% |
Important Formula of Class 11 Classification of Elements and Periodicity in Properties
Important Formulae/Theorems of Classification of Elements and Periodicity in Properties for CBSE and Competitive Exams
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Slater’s Rule Approximation: Effective Nuclear Charge ( )
where = Atomic number, = Shielding constant
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Atomic Radius Trend Explanation:
(Higher nuclear charge pulls electrons closer, reducing size)
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Ionization Energy Trend Explanation:
(Higher shielding decreases ionization energy)
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Electronegativity Trends (Pauling Scale):
(IE = Ionization Energy, EA = Electron Affinity)
Class 11th Chemistry Classification of Elements and Periodicity NCERT Solution PDF: Download PDF for Free
Download free PDF from the link below:
Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties NCERT Solution PDF: Download Free PDF
Classification of Elements and Periodicity in Properties Solutions
Students can check the Classification of elements class 11 NCERT solutions below. These are short-answer type questions.
| 3.1. What is the basic theme of organisation in the periodic table? |
| Answer: The basic theme of organisation of elements in the periodic table is to simplify, systematize and classify the elements based on their similarities in properties. This arrangement makes it easier for us to group the elements and makes it less confusing to present the periodic table. |
| 3.2. Which important property did Mendeleev use to classify the elements in this periodic table and did he stick to that? |
| Answer: Mendeleev arranged the elements horizontally, in periods and vertically, in groups in the increasing order of their atomic masses. This classification was done in such a way that the elements with similar properties fall in the same group. |
| 3.3. What is the basic difference in approach between Mendeleev’s Periodic Law and the Modern Periodic Law? |
| Answer: Mendeleev’s periodic law classifies the elements based on their atomic weight. Modern periodic law classifies the elements based on their atomic number. |
| 3.4. On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements. |
| Answer: The sixth period means that the highest principal quantum number = 6. The subsequent periods comprise of 8, 8, 18, 18 and 32 elements in the same order. The sixth period comprises of 32 elements where electrons enter the 6s, 4f, 5d and 6p orbitals. Number of orbitals present in 6s=1. 4f= 7, 5d= 5 and 6p= 3. The total number of orbitals is 16. A total of 32 electrons can be filled in these 32 orbitals; therefore, the sixth period should have a maximum of 32 elements. |
| 3.5. In terms of period and group where will you locate the element with z = 114? |
| Answer: Elements with atomic numbers from Z = 87 to Z = 114 are present in the 7th period of the periodic table. So, this elements lies in the Period – 7 and Group -14 of Block-p. |
| 3.6. Write the atomic number of the element present in the third period and seventeenth group of the periodic table. |
| Answer: The element is chlorine (Cl) with atomic number (Z) = 17. |
| 3.7. Which element do you think would have been named by (i) Lawrence Berkeley Laboratory (ii) Seaborg’s group? |
| Answer: (i) Lawrencium (Lr) with atomic number (z) = 103 (ii) Seaborgium (Sg) with atomic number (z) = 106. |
| 3.8. Why do elements in the same group have similar physical and chemical properties? |
| Answer: The elements in a group have same number of valence electrons and hence have similar physical and chemical properties. |
| 3.9. What does atomic radius and ionic radius really mean to you? |
| Answer: Atomic radius: It is the distance from the centre of nucleus to the outer most shell of electrons in the atom of any element. It incorporates both the covalent, metallic radius or van der Waal’s radius depending on whether the element is a non-metal or a metal. Ionic radius: It is the distance between the centre of the nucleus of an ion up to the point where it exerts its influence on the electron cloud of a canton or anion. |
| 3.10. How do atomic radii vary in a period and in a group? How do you explain the variation? |
| Answer: Across a period, the atomic radii decrease from left to right due to increase in effective nuclear charge from left to right across a period Within a group, atomic radius increases down the group due to continuous increases in the number of electronic shells or orbit numbers in the structure of atoms of the elements down a group. |
| 3.11. What do you understand by isoelectronic species? Name a species that will be iso electronic with each of the following atoms or ions. (i) F–(ii) Ar (iii) Mg2+(iv) Rb+ |
| Answer: Species (atoms/ions) which have same number of electrons are called isoelectronic species. The isoelectronic species out of the given atoms/ions are: (i) Na+ is isoelectronic to F- (iii) Na+ is isoelectronic to Mg2+ (ii) K+ is isoelectronic to Ar (iv) Sr2+ is isoelectronic to Rb+
|
| 3.12. Consider the following species: N3-, O2-, F–, Na+, Mg2+, Al3+ (a) What is common in them? (b) Arrange them in order of increasing ionic radii? |
| Answer: (a) All of them have 10 electrons each and are isoelectronic in nature. (b) In isoelectronic species, higher the nuclear charge, smaller will be the atomic or ionic radius. Al3+< Mg2+< Na+< F–< O2-< N3- |
| 3.13. Explain why cations are smaller and anions larger in radii than their parent atoms? |
| Answer: A cation is obtained by removing an electron from the outermost shell. The removal of electron(s) results in decrease of the size of the resulting ion than the parent atom because it has fewer electrons while its nuclearcharge remains the same. Anions are obtained by addition of electron(s) in the outermost shell. This results in increased repulsion among the electrons and a decrease in effective nuclear charge. |
| 3.14. What is the significance of the terms – isolated gaseous atom and ground state while defining the ionization enthalpy and electron gain enthalpy? [Hint:Requirements for comparison purposes] |
| Answer:
|
| 3.15. Energy of an electron in the ground state of the hydrogen atom is- 2.18 x 10-18 J.Calculate the ionization enthalpy of atomic hydrogen in terms of J mol-1. [Hint: Apply the idea of mole concept to derive the answer] |
| Answer: The ionisation enthalpy is for 1 mole atoms. Therefore, ground state energy of theatoms may be expressed as Eground state = (– 2.18 x 10-18 J) x(6.022 x 1023 mol-1) = –1.312 x 106 J mol-1 Ionisation enthalpy =E∞–Eground state = 0–(–1.312 x 106mol-1) |
| 3.16. Among the second period elements, the actual ionization enthalpies are in the order: Li
Explain why (i) Be has higher ∆ i H than B? (ii) O has lower ∆ i H than N and F? |
| Answer: (i) In the electronic configuration of Be (1s2 2s2) the outermost electron is present in 2s-orbital while in B (1s2 2s2 2p1) it is present in 2p-orbital. Since 2s – electrons are more strongly attracted by the nucleus than 2p-electrons, therefore, lesser amount of energy is required to knock out a 2p-electron than a 2s – electron. Therefore, Be has higher ∆i H than that of B. |
| 3.17. How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium? |
| Answer: Electronic configuration of Na and Mg are Na = 1s2 2s2 2p6 3s1 First electron in both cases has to be removed from 3s-orbital but the nuclear charge of Na (+ 11) is lower than that of Mg (+ 12) therefore first ionization energy of sodium is lower than that of magnesium. Na+ = 1s2 2s2 2p6 Therefore, second ionization enthalpy of sodium is higher than that of magnesium. |
| 3.18. What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down the group? |
| Answer: Atomic size: With the increase in atomic size, the number of electron shells increase. Therefore, the force that binds the electrons with the nucleus decreases. The ionization enthalpy thus decreases with the increase in atomic size. |
| 3.19. The first ionization enthalpy values (in kJ mol -1) of group 13 elements are:
|
| Answer: On moving down the group, the ionization enthalpy decreases. This is true for B and Al due to the bigger size of Al. The ionization enthalpy of Ga is unexpectedly higher than Al because Ga contains 10d electrons in inner shell whose shielding is less effective than that of s and p electrons. The outer electron is held fairly strongly by the nucleus. The ionization enthalpy increases slightly. A similar increase is observed from In to Tl due to presence of 14f electrons in the inner shell of Tl which have poor screening effect. |
| 3.20. Which of the following pairs of elements would have a more negative electron gain enthalpy? (i) O or F (ii) F or Cl. |
| Answer: (i) The electronic configuration of O and F are: O: 1s2 2s2 2p4 F: 1s2 2s2 2p5 F would readily accept an electron to attain the noble gas electronic configuration of Ne as compared to O. Hence, F has a more negative electron gain enthalpy. (ii) The electronic configuration of Cl and F are: Cl: [Ne] 3s2 3p5 F: 1s2 2s2 2p5 Due to a larger size, Cl can accommodate an additional electron. F, on the other hand, has a smaller atomic size and therefore the accommodation of another electron increases the inter-electronic repulsion thereby, decreasing the stability. Therefore, Cl has a higher negative electron gain enthalpy. |
| 3.21. Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer. |
| Answer: When an electron is added to oxygen atom, energy is released to form a negative ion. This enthalpy change called the first electron gain enthalpy is thus negative. On adding another electron to the O− ion, it experiences repulsion from the anion due to which the instability of the ion increases. Thus, the addition of the second electron requires energy due to which the second electron gain enthalpy is positive. O + e− → O− ΔH1 = -141KJ O− + e− → O2− ΔH2 = 780 KJ |
| 3.22. What is basic difference between the terms electron gain enthalpy and electro negativity? |
| Answer: Electron gain enthalpy refers to tendency of an isolated gaseous atom to accept an additional electron to form a negative ion. Whereas electronegativity refers to tendency of the atom of an element to attract shared pair of electrons towards it in a covalent bond. |
| 3.23. How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds? |
| Ans. On Pauling scale, the electronegativity of nitrogen is 3.0 which indicates that it is quite electronegative. But it is not correct to say that the electronegativity of nitrogen in all the compounds is 3. It varies with the atoms that N is bonded with. For example, in NH3, N has a different electronegativity than N in NO2 |
| 3.24. Describe the theory associated with the radius of an atom as it: (a) gains an electron (b) loses an electron? |
| Answer: (a) Gain of an electron leads to the formation of an anion. The size of an anion will be larger than that of the parent atom because the addition of one or more electrons would result in increased repulsion among electrons and decrease in effective nuclear charge. (b) Loss of an electron from an atom results in the formation of a cation. A cation is smaller than its parent atom because it has former electrons while its nuclear charge remains the same. |
| 3.25. Would you expect the first ionization enthalpies of two isotopes of the same element to be the same or different? Justify your answer. |
| Answer: Isotopes have same number of electrons and protons, only the number of neutrons is different. The atomic number remains the same and only atomic mass differs. Hence the ionization energy of the isotopes of a chemical element remains the same. |
| 3.26. What are major differences between metals and non-metals? | ||||||||||
| Answer:
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| 3.27. Use periodic table to answer the following questions: (a) Identify the element with five electrons in the outer sub shell. (b) Identify the element that would tend to lose two electrons. (c) Identify the element that would tend to gain two electrons. |
| Answer: (a) Element belonging to nitrogen family (group 15) e.g., nitrogen. (b) Element belonging to alkaline earth family (group 2) e.g., magnesium. (c) Element belonging to oxygen family (group 16) e.g., oxygen |
| 3.28. The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that of group 17 is F > Cl > Br > I. Explain? |
| Answer: The elements of Group I have only one electron in their respective valence shells and thus have a strong tendency to lose this electron. The tendency to lose electrons in turn, depends upon the ionization enthalpy. Since the ionization enthalpy decreases down the group therefore, the reactivity of group 1 elements increases in the order Li < Na < K < Rb < Cs. In contrast, the elements of group 17 have seven electrons in their respective valence shells and thus have strong tendency to accept one more electron to make stable configuration. So for group 17, the electron gain enthalpy and electronegativity decreases down the group and thus the reactivity also decreases. |
| 3.29. Write the general electronic configuration of s-, p-, d-, and f-block elements? |
| Answer: (i) s-Block elements: ns1-2 where n = 2-7. (ii) p-Block elements: ns2 np1-6 where n = 2-6. (iii) d-Block elements: (n-1)d1-10 ns 0-2 where n = 4-7. (iv) f-Block elements: (n-2)f0-14 (n-1)d0-1 ns2where n = 6-7. |
| 3.30. Assign the position of the element having outer electronic configuration, (i) ns2 np4 for n = 3 (ii) (n – 1) d2 ns2 for n = 4 and (iii) (n – 2) f7 (n – 1) d1 ns2 for n = 6 in the periodic table? |
| Answer: (i) For n = 3, the element belong to 3rd period, p-block element. The electronic configuration is =1s2 2s2 2p6 3s2 3p4. The element name is sulphur. (ii) For n = 4, the element belongs to 4th period and since the valence shell has 4 electrons it belongs to group 4. Electronic configuration= 1s2 2s2 2p6 3s2 3p6 3d2 4s2, and the element name is Titanium (Ti). (iii) For n = 6, the element belongs to 6th period. Last electron goes to the f-orbital, element is from f-block. It belongs to group = 3 The element is gadolinium (z = 64) Complete electronic configuration =[Xe] 4 f7 5d1 6s2. |
| 3.31. The first (ΔiH1) and the second (ΔiH2) ionization enthalpies (in kJ mol–1) and the (ΔegH) electron gain enthalpy (in kJ mol–1) of a few elements are given below: Elements ΔH1 ΔH2 ΔegH I 520 7300 –60 II 419 3051 –48 III 1681 3374 –328 IV 1008 1846 –295 V 2372 5251 +48 VI 738 1451 –40 (b) the most reactive metal (e) the metal which can form a stable binary halide of the formula MX2 (X = halogen) (f) the metal which can form a predominantly stable covalent halide of the formula MX (X = halogen)? |
| Answer: (a) The element V has highest first ionization enthalpy (∆iH1) and positive electron gain enthalpy (∆egH) and hence it is the least reactive element. Since inert gases have positive ∆egH, therefore, the element-V must be an inert gas. The values of ∆iH1, ∆iH2 and ∆egH match that of He. (b) The element II which has the least first ionization enthalpy (∆iH1) and a low negative electron gain enthalpy (∆egH) is the most reactive metal. The values of ∆iH1, ∆iH2 and ∆egH match that of K (potassium). (c) The element III which has high first ionization enthalpy (∆iH1) and a very high negative electron gain enthalpy (∆egH) is the most reactive non-metal. The values of ∆iH1, ∆iH2 and ∆egH match that of F (fluorine). (d) The element IV has a high negative electron gain enthalpy (∆egH) but not so high first ionization enthalpy (∆egH). Therefore, it is the least reactive non-metal. The values of ∆iH1, ∆iH2 and ∆egH match that of I (Iodine). (e) The element VI has low first ionization enthalpy (∆iH1) but higher than that of alkali metals. Therefore, it appears that the element is an alkaline earth metal and hence will form binary halide of the formula MX2 (where X = halogen). The values of ∆i H1, ∆iH2 and ∆egH match that of Mg (magnesium). (f) The element I has low first ionization (∆iH1) but a very high second ionization enthalpy (∆iH2), therefore, it must be an alkali metal. Since the metal forms a predominantly stable covalent halide of the formula MX (X = halogen), therefore, the alkali metal must be least reactive. The values of ∆iH1, ∆iH2 and ∆egH match that of Li (lithium). |
| 3.32. Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements: (a) Lithium and oxygen (b) Magnesium and nitrogen (c) Aluminium and iodine (d) Silicon and oxygen (e) Phosphorous and fluorine (f) Element 71 and fluorine. |
| Answer: (a) LiO2 (Lithium oxide) (b) Mg3N2 (Magnesium nitride) (c) AlI3 (Aluminium iodide) (d) SiO2 (Silicon dioxide) (e) Phosphorous pentafluoride (f) Z = 71 The element is Lutenium (Lu). Electronic configuration [Xe] 4 f7 5d1 6s2. With fluorine it will form a binary compound = LuF3. |
| 3.33. In the modern periodic table, the period indicates the value of (a) atomic number (b) mass number (c) principal quantum number (d) azimuthal quantum number? |
| Answer: In the modern periodic table, each period begins with the filling of a new shell. Therefore, the period indicates the value of principal quantum number. Thus, option (c) is correct. |
| 3.34. Which of the following statements related to the modem periodic table is incorrect? (a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-subshell. (b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell. (c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell. (d) The block indicates value of azimuthal quantum number (l) for the last subshell that received electrons in building up the electronic configuration. |
| Answer: Statement (b) is incorrect. |
| 3.35. Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell? (a) Valence principal quantum number (n) (b) Nuclear charge (Z) (c) Nuclear mass (d) Number of core electrons |
| Answer: (c) Nuclear mass. |
| 3.36. The size of isoelectronic species-F–, Ne and Na+ is affected by (a) nuclear charge (Z) (b) valence principal quantum number (n) (c) electron-electron interaction in the outer orbitals (d) none of the factors because their size is the same |
| Answer: (a) Nuclear charge (Z). |
| 3.37. Which of the following statements is incorrect in relation to ionization enthalpy? (a) ionization enthalpy increases for each successive electron. (b) The greatest increase in ionization enthalpy is experienced on removal of electrons from core noble gas configuration. (c) End of valence electrons is marked by a big jump in ionization enthalpy. (d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value. |
| Answer: (d) is incorrect. |
| 3.38. Considering the elements B, Al, Mg and K, the correct order of their metallic character is: (a) B>Al>Mg>K (b) Al>Mg>B>K (c) Mg>Al>K>B (d) K>Mg>Al>B |
| Answer: In a period, metallic character decreases as we move from left to right. Therefore, metallic character of K, Mg and Al decreases in the order: K > Mg > Al. However, within a group, the metallic character, increases from top to bottom. Thus, Al is more metallic than B. Therefore, the correct sequence of decreasing metallic character is: K > Mg > Al > B, i.e., option (d) is correct |
| 3.39. Considering the elements B, C, N, F and Si, the correct order of their non-metallic character is: (a) B>C>Si>N>F (b) Si>C>B>N>F (c) F>N>C>B>Si (d) F>N>C>Si>B |
| Answer: In a period, the non-metallic character increases from left to right. Thus, among B, C, N and F, non-metallic character decreases in the order: F > N > C > B. However, within a group, non-metallic character decreases from top to bottom. Thus, C is more non-metallic than Si. Therefore, the correct sequence of decreasing non-metallic character is: F > N > C > B > Si, i.e., option (c) is correct. |
| 3.40. Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidising property is: (b) F > O > Cl> N (c) Cl> F > O > N (d) O > F > N > Cl |
| Answer: Within a period, the oxidising character increases from left to right. Therefore, among F, O and N, oxidising power decreases in the order: F > O > N. However, within a group, oxidising power decreases from top to bottom. Thus, F is a stronger oxidising agent than Cl. Further because O is more electronegative than Cl, therefore, O is a stronger oxidising agent than Cl. Thus, overall decreasing order of oxidising power is: F > O > Cl > N, i.e., option (b) is correct. |
NCERT Class XIth Chemistry Classification of Elements and Periodicity Covers
Below is the list of topics covered in Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties.
Topics List:
- Why do we need to classify Elements?
- Genesis of Periodic Classification
- Modern Periodic Law and the Present Form of the Periodic Table
- Nomenclature of Elements with Atomic Numbers > 100
- Electronic Configurations of Elements and the Periodic Table
- Electronic Configurations and Types of Elements: s-, p-, d-, f- Blocks
- The s-Block Elements
- The p-Block Elements
- The f-Block Elements (Inner-Transition Elements)
- The d-Block Elements (Transition Elements)
- Metals, Non-metals and Metalloids
- Periodic Trends in Properties of Elements
- Periodic Trends in Properties of Elements
- Periodic Trends in Chemical Properties
- Periodic Trends and Chemical Reactivity
NCERT Class 11 Chemistry Chapter 3 Exercise Solutions
Students can check the Classification of elements class 11 NCERT solutions below. These are short-answer type questions.
| 3.1. What is the basic theme of organisation in the periodic table? |
| Answer: The basic theme of organisation of elements in the periodic table is to simplify, systematize and classify the elements based on their similarities in properties. This arrangement makes it easier for us to group the elements and makes it less confusing to present the periodic table. |
| 3.2. Which important property did Mendeleev use to classify the elements in this periodic table and did he stick to that? |
| Answer: Mendeleev arranged the elements horizontally, in periods and vertically, in groups in the increasing order of their atomic masses. This classification was done in such a way that the elements with similar properties fall in the same group. |
| 3.3. What is the basic difference in approach between Mendeleev’s Periodic Law and the Modern Periodic Law? |
| Answer: Mendeleev’s periodic law classifies the elements based on their atomic weight. Modern periodic law classifies the elements based on their atomic number. |
| 3.4. On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements. |
| Answer: The sixth period means that the highest principal quantum number = 6. The subsequent periods comprise of 8, 8, 18, 18 and 32 elements in the same order. The sixth period comprises of 32 elements where electrons enter the 6s, 4f, 5d and 6p orbitals. Number of orbitals present in 6s=1. 4f= 7, 5d= 5 and 6p= 3. The total number of orbitals is 16. A total of 32 electrons can be filled in these 32 orbitals; therefore, the sixth period should have a maximum of 32 elements. |
| 3.5. In terms of period and group where will you locate the element with z = 114? |
| Answer: Elements with atomic numbers from Z = 87 to Z = 114 are present in the 7th period of the periodic table. So, this elements lies in the Period – 7 and Group -14 of Block-p. |
| 3.6. Write the atomic number of the element present in the third period and seventeenth group of the periodic table. |
| Answer: The element is chlorine (Cl) with atomic number (Z) = 17. |
| 3.7. Which element do you think would have been named by (i) Lawrence Berkeley Laboratory (ii) Seaborg’s group? |
| Answer: (i) Lawrencium (Lr) with atomic number (z) = 103 (ii) Seaborgium (Sg) with atomic number (z) = 106. |
| 3.8. Why do elements in the same group have similar physical and chemical properties? |
| Answer: The elements in a group have same number of valence electrons and hence have similar physical and chemical properties. |
| 3.9. What does atomic radius and ionic radius really mean to you? |
| Answer: Atomic radius: It is the distance from the centre of nucleus to the outer most shell of electrons in the atom of any element. It incorporates both the covalent, metallic radius or van der Waal’s radius depending on whether the element is a non-metal or a metal. Ionic radius: It is the distance between the centre of the nucleus of an ion up to the point where it exerts its influence on the electron cloud of a canton or anion. |
| 3.10. How do atomic radii vary in a period and in a group? How do you explain the variation? |
| Answer: Across a period, the atomic radii decrease from left to right due to increase in effective nuclear charge from left to right across a period Within a group, atomic radius increases down the group due to continuous increases in the number of electronic shells or orbit numbers in the structure of atoms of the elements down a group. |
| 3.11. What do you understand by isoelectronic species? Name a species that will be iso electronic with each of the following atoms or ions. (i) F–(ii) Ar (iii) Mg2+(iv) Rb+ |
| Answer: Species (atoms/ions) which have same number of electrons are called isoelectronic species. The isoelectronic species out of the given atoms/ions are: (i) Na+ is isoelectronic to F- (iii) Na+ is isoelectronic to Mg2+ (ii) K+ is isoelectronic to Ar (iv) Sr2+ is isoelectronic to Rb+
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| 3.12. Consider the following species: N3-, O2-, F–, Na+, Mg2+, Al3+ (a) What is common in them? (b) Arrange them in order of increasing ionic radii? |
| Answer: (a) All of them have 10 electrons each and are isoelectronic in nature. (b) In isoelectronic species, higher the nuclear charge, smaller will be the atomic or ionic radius. Al3+< Mg2+< Na+< F–< O2-< N3- |
| 3.13. Explain why cations are smaller and anions larger in radii than their parent atoms? |
| Answer: A cation is obtained by removing an electron from the outermost shell. The removal of electron(s) results in decrease of the size of the resulting ion than the parent atom because it has fewer electrons while its nuclearcharge remains the same. Anions are obtained by addition of electron(s) in the outermost shell. This results in increased repulsion among the electrons and a decrease in effective nuclear charge. |
| 3.14. What is the significance of the terms – isolated gaseous atom and ground state while defining the ionization enthalpy and electron gain enthalpy? [Hint:Requirements for comparison purposes] |
| Answer: Significance of the term ‘isolated gaseous atom’. The atoms in the gaseous state are far separated in the sense that they do not have any mutual attractive and repulsive interactions. These are therefore regarded as isolated atoms. In this state, the value of ionization enthalpy and electron gain enthalpy are not influenced by the presence of the other atoms. It is not possible to express these when the atoms are in the liquid or solid state due to the presence of inter atomic forces. Significance of the term ‘ground state’. Ground state of the atom represents the normal – energy state of an atom. It means electrons in a particular atom are in the lowest energy state and they neither lose nor gain electron. Both ionisation enthalpy and electron gain enthalpy are generally expressed with respect to the ground state of an atom only. |
| 3.15. Energy of an electron in the ground state of the hydrogen atom is- 2.18 x 10-18 J.Calculate the ionization enthalpy of atomic hydrogen in terms of J mol-1. [Hint: Apply the idea of mole concept to derive the answer] |
| Answer: The ionisation enthalpy is for 1 mole atoms. Therefore, ground state energy of theatoms may be expressed as Eground state = (– 2.18 x 10-18 J) x(6.022 x 1023 mol-1) = –1.312 x 106 J mol-1 Ionisation enthalpy =E∞–Eground state = 0–(–1.312 x 106mol-1) = 1.312 x 106 J mol-1. |
| 3.16. Among the second period elements, the actual ionization enthalpies are in the order: Li
Explain why (i) (ii) O has lower ∆i H than N and F? |
| Answer: (i) In the electronic configuration of Be (1s2 2s2) the outermost electron is present in 2s-orbital while in B (1s2 2s2 2p1) it is present in 2p-orbital. Since 2s – electrons are more strongly attracted by the nucleus than 2p-electrons, therefore, lesser amount of energy is required to knock out a 2p-electron than a 2s – electron. Therefore, Be has higher ∆i H than that of B. |
| 3.17. How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium? |
| Answer: Electronic configuration of Na and Mg are Na = 1s2 2s2 2p6 3s1 First electron in both cases has to be removed from 3s-orbital but the nuclear charge of Na (+ 11) is lower than that of Mg (+ 12) therefore first ionization energy of sodium is lower than that of magnesium. Na+ = 1s2 2s2 2p6 Therefore, second ionization enthalpy of sodium is higher than that of magnesium. |
| 3.18. What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down the group? |
| Answer: Atomic size: With the increase in atomic size, the number of electron shells increase. Therefore, the force that binds the electrons with the nucleus decreases. The ionization enthalpy thus decreases with the increase in atomic size. |
| 3.19. The first ionization enthalpy values (in kJ mol -1) of group 13 elements are: B Al Ga In Tl |
| Answer: On moving down the group, the ionization enthalpy decreases. This is true for B and Al due to the bigger size of Al. The ionization enthalpy of Ga is unexpectedly higher than Al because Ga contains 10d electrons in inner shell whose shielding is less effective than that of s and p electrons. The outer electron is held fairly strongly by the nucleus. The ionization enthalpy increases slightly. A similar increase is observed from In to Tl due to presence of 14f electrons in the inner shell of Tl which have poor screening effect. |
| 3.20. Which of the following pairs of elements would have a more negative electron gain enthalpy? (i) O or F (ii) F or Cl. |
| Answer: (i) The electronic configuration of O and F are: O: 1s2 2s2 2p4 F: 1s2 2s2 2p5 F would readily accept an electron to attain the noble gas electronic configuration of Ne as compared to O. Hence, F has a more negative electron gain enthalpy. (ii) The electronic configuration of Cl and F are: Cl: [Ne] 3s2 3p5 F: 1s2 2s2 2p5 Due to a larger size, Cl can accommodate an additional electron. F, on the other hand, has a smaller atomic size and therefore the accommodation of another electron increases the inter-electronic repulsion thereby, decreasing the stability. Therefore, Cl has a higher negative electron gain enthalpy. |
| 3.21. Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer. |
| Answer: When an electron is added to oxygen atom, energy is released to form a negative ion. This enthalpy change called the first electron gain enthalpy is thus negative. On adding another electron to the O− ion, it experiences repulsion from the anion due to which the instability of the ion increases. Thus, the addition of the second electron requires energy due to which the second electron gain enthalpy is positive. O + e− → O− ΔH1 = -141KJ O− + e− → O2− ΔH2 = 780 KJ |
| 3.22. What is basic difference between the terms electron gain enthalpy and electro negativity? |
| Answer: Electron gain enthalpy refers to tendency of an isolated gaseous atom to accept an additional electron to form a negative ion. Whereas electronegativity refers to tendency of the atom of an element to attract shared pair of electrons towards it in a covalent bond. |
| 3.23. How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds? |
| Ans. On Pauling scale, the electronegativity of nitrogen is 3.0 which indicates that it is quite electronegative. But it is not correct to say that the electronegativity of nitrogen in all the compounds is 3. It varies with the atoms that N is bonded with. For example, in NH3, N has a different electronegativity than N in NO2 |
| 3.24. Describe the theory associated with the radius of an atom as it: (a) Gains an electron (b) Loses an electron? |
| Answer: (a) Gain of an electron leads to the formation of an anion. The size of an anion will be larger than that of the parent atom because the addition of one or more electrons would result in increased repulsion among electrons and decrease in effective nuclear charge. (b) Loss of an electron from an atom results in the formation of a cation. A cation is smaller than its parent atom because it has former electrons while its nuclear charge remains the same. |
| 3.25. Would you expect the first ionization enthalpies of two isotopes of the same element to be the same or different? Justify your answer. |
| Answer: Isotopes have same number of electrons and protons, only the number of neutrons is different. The atomic number remains the same and only atomic mass differs. Hence the ionization energy of the isotopes of a chemical element remains the same. |
| 3.26. What are major differences between metals and non-metals? | ||||||||||
| Answer:
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| 3.27. Use periodic table to answer the following questions: (a) Identify the element with five electrons in the outer sub shell. (b) Identify the element that would tend to lose two electrons. (c) Identify the element that would tend to gain two electrons. |
| Answer: (a) Element belonging to nitrogen family (group 15) e.g., nitrogen. (b) Element belonging to alkaline earth family (group 2) e.g., magnesium. (c) Element belonging to oxygen family (group 16) e.g., oxygen |
| 3.28. The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that of group 17 is F > Cl > Br > I. Explain? |
| Answer: The elements of Group I have only one electron in their respective valence shells and thus have a strong tendency to lose this electron. The tendency to lose electrons in turn, depends upon the ionization enthalpy. Since the ionization enthalpy decreases down the group therefore, the reactivity of group 1 elements increases in the order Li < Na < K < Rb < Cs. In contrast, the elements of group 17 have seven electrons in their respective valence shells and thus have strong tendency to accept one more electron to make stable configuration. So for group 17, the electron gain enthalpy and electronegativity decreases down the group and thus the reactivity also decreases. |
| 3.29. Write the general electronic configuration of s-, p-, d-, and f-block elements? |
| Answer: (i) s-Block elements: ns1-2 where n = 2-7. (ii) p-Block elements: ns2 np1-6 where n = 2-6. (iii) d-Block elements: (n-1)d1-10 ns 0-2 where n = 4-7. (iv) f-Block elements: (n-2)f0-14 (n-1)d0-1 ns2where n = 6-7. |
| 3.30. Assign the position of the element having outer electronic configuration, (i) ns2 np4 for n = 3 (ii) (n – 1) d2 ns2 for n = 4 and (iii) (n – 2) f7 (n – 1) d1 ns2 for n = 6 in the periodic table? |
| Answer: (i) For n = 3, the element belong to 3rd period, p-block element.
(ii) For n = 4, the element belongs to 4th period and since the valence shell has 4 electrons it belongs to group 4.
(iii) For n = 6, the element belongs to 6th period. Last electron goes to the f-orbital, element is from f-block. It belongs to group = 3 |
| 3.31. The first (ΔiH1) and the second (ΔiH2) ionization enthalpies (in kJ mol–1) and the (ΔegH) electron gain enthalpy (in kJ mol–1) of a few elements are given below: Elements ΔH1 ΔH2 ΔegH I 520 7300 –60 II 419 3051 –48 III 1681 3374 –328 IV 1008 1846 –295 V 2372 5251 +48 VI 738 1451 –40 (b) The most reactive metal (e) The metal which can form a stable binary halide of the formula MX2 (X = halogen) (f) The metal which can form a predominantly stable covalent halide of the formula MX (X = halogen)? |
| Answer: (a) The element V has highest first ionization enthalpy (∆iH1) and positive electron gain enthalpy (∆egH) and hence it is the least reactive element. Since inert gases have positive ∆egH, therefore, the element-V must be an inert gas. The values of ∆iH1, ∆iH2 and ∆egH match that of He.
(b) The element II which has the least first ionization enthalpy (∆iH1) and a low negative electron gain enthalpy (∆egH) is the most reactive metal. The values of ∆iH1, ∆iH2 and ∆egH match that of K (potassium). (c) The element III which has high first ionization enthalpy (∆iH1) and a very high negative electron gain enthalpy (∆egH) is the most reactive non-metal. The values of ∆iH1, ∆iH2 and ∆egH match that of F (fluorine). (d) The element IV has a high negative electron gain enthalpy (∆egH) but not so high first ionization enthalpy (∆egH). Therefore, it is the least reactive non-metal. The values of ∆iH1, ∆iH2 and ∆egH match that of I (Iodine). (e) The element VI has low first ionization enthalpy (∆iH1) but higher than that of alkali metals. Therefore, it appears that the element is an alkaline earth metal and hence will form binary halide of the formula MX2 (where X = halogen). The values of ∆i H1, ∆iH2 and ∆egH match that of Mg (magnesium). (f) The element I has low first ionization (∆iH1) but a very high second ionization enthalpy (∆iH2), therefore, it must be an alkali metal. Since the metal forms a predominantly stable covalent halide of the formula MX (X = halogen), therefore, the alkali metal must be least reactive. The values of ∆iH1, ∆iH2 and ∆egH match that of Li (lithium). |
| 3.32. Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements: (a) Lithium and oxygen (b) Magnesium and nitrogen (c) Aluminium and iodine (d) Silicon and oxygen (e) Phosphorous and fluorine (f) Element 71 and fluorine. |
| Answer: (a) LiO2 (Lithium oxide) (b) Mg3N2 (Magnesium nitride) (c) AlI3 (Aluminium iodide) (d) SiO2 (Silicon dioxide) (e) Phosphorous pentafluoride (f) Z = 71 The element is Lutenium (Lu). Electronic configuration [Xe] 4 f7 5d1 6s2. With fluorine it will form a binary compound = LuF3. |
| 3.33. In the modern periodic table, the period indicates the value of (a) Atomic number (b) Mass number (c) Principal quantum number (d) Azimuthal quantum number? |
| Answer: In the modern periodic table, each period begins with the filling of a new shell. Therefore, the period indicates the value of principal quantum number. Thus, option (c) is correct. |
| 3.34. Which of the following statements related to the modem periodic table is incorrect? (a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-subshell. (b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell. (c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell. (d) The block indicates value of azimuthal quantum number (l) for the last subshell that received electrons in building up the electronic configuration. |
| Answer: Statement (b) is incorrect. |
| 3.35. Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell? (a) Valence principal quantum number (n) (b) Nuclear charge (Z) (c) Nuclear mass (d) Number of core electrons |
| Answer: (c) Nuclear mass. |
| 3.36. The size of isoelectronic species-F–, Ne and Na+ is affected by (a) Nuclear charge (Z) (b) Valence principal quantum number (n) (c) Electron-electron interaction in the outer orbitals (d) None of the factors because their size is the same |
| Answer: (a) Nuclear charge (Z). |
| 3.37. Which of the following statements is incorrect in relation to ionization enthalpy? (a) Ionization enthalpy increases for each successive electron. (b) The greatest increase in ionization enthalpy is experienced on removal of electrons from core noble gas configuration. (c) End of valence electrons is marked by a big jump in ionization enthalpy. (d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value. |
| Answer: (d) is incorrect. |
| 3.38. Considering the elements B, Al, Mg and K, the correct order of their metallic character is: (a) B>Al>Mg>K (b) Al>Mg>B>K (c) Mg>Al>K>B (d) K>Mg>Al>B |
| Answer: In a period, metallic character decreases as we move from left to right. Therefore, metallic character of K, Mg and Al decreases in the order: K > Mg > Al. However, within a group, the metallic character, increases from top to bottom. Thus, Al is more metallic than B. Therefore, the correct sequence of decreasing metallic character is: K > Mg > Al > B, i.e., option (d) is correct |
| 3.39. Considering the elements B, C, N, F and Si, the correct order of their non-metallic character is: (a) B>C>Si>N>F (b) Si>C>B>N>F (c) F>N>C>B>Si (d) F>N>C>Si>B |
| Answer: In a period, the non-metallic character increases from left to right. Thus, among B, C, N and F, non-metallic character decreases in the order: F > N > C > B. However, within a group, non-metallic character decreases from top to bottom. Thus, C is more non-metallic than Si. Therefore, the correct sequence of decreasing non-metallic character is: F > N > C > B > Si, i.e., option (c) is correct. |
| 3.40. Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidising property is: (b) F > O > Cl> N (c) Cl> F > O > N (d) O > F > N > Cl |
| Answer: Within a period, the oxidising character increases from left to right. Therefore, among F, O and N, oxidising power decreases in the order: F > O > N. However, within a group, oxidising power decreases from top to bottom. Thus, F is a stronger oxidising agent than Cl. Further because O is more electronegative than Cl, therefore, O is a stronger oxidising agent than Cl. Thus, overall decreasing order of oxidising power is: F > O > Cl > N, i.e., option (b) is correct. |
Class 11 Chemistry Assertion Questions
Below are the NCERT solutions for periodic classification of elements class 11. These types of questions are asked in competitive exams. Students who wish to appear for a competitive test must attempt these questions. Solving these questions will boost confidence in students.
| 3.1. Assertion: The size of an anion will be larger than that of the parent atom. Reason: The addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge |
| Ans. (a) |
| 3.2. Assertion: Elements of the same group exhibit similar chemical properties. Reason: Elements of the same group have similar valence shell electronic configuration. |
| Ans. (a) Elements of the same group have similar valence shell electronic configuration and, therefore, exhibit similar chemical properties. However, the elements of the same period have an incrementally increasing number of electrons from left to right, and, therefore, have different valences. |
| 3.3. Assertion: Their synthesis and characterisation of new elements with very high atomic numbers require highly sophisticated costly equipment and laboratory. Reason: The new elements with very high atomic numbers are so unstable that only minute quantities, sometimes only a few atoms of them are obtained. |
| Ans. (a) |
| 3.4. Assertion: Metallic elements have strong tendency to gain electrons. Reason: Electronegativity is inversely related to the metallic properties of elements. |
| Ans: (d) Assertion is a wrong statement. Non-metallic elements have a strong tendency to gain electrons. Therefore, electronegativity is directly related to those non-metallic properties of elements. It can be further extended to say that the electronegativity is inversely related to the metallic properties of elements. Thus, the increase in electronegativities across a period is accompanied by an increase in non-metallic properties (or a decrease in metallic properties) of elements |
| 3.5. Assertion: Hydrogen is placed separately at the top of the Periodic Table. Reason: It is the lightest element on the periodic table with a weight of just 1.0079 amu. |
| Ans. (b) It has only ones-electron and hence can be placed in group 1(alkali metals). It can also gain an electron to achieve a noble gas arrangement and hence it can behave similarly to a group 17 (halogen family) element. Because it is a special case, we shall place hydrogen separately at the top of the Periodic Table |
| 3.1 Choose the incorrect statement (a) In a triad, the atomic mass of the middle element is the mean of the atomic masses of the first and third elements. (b) Electronegativity refers to the tendency of atom so share electrons with another atom. (c) Number of elements in each period is twice the number of atomic orbitals available in the energy level that is being filled. (d) According to Mendeleev, periodic properties of elements are a function of their atomic masses. |
| Ans: (b) Electronegativity refers to the tendency of an atom to share electrons with another atom. |
| 3.2. Mendeleev’s series are (a) Families (b) Groups (c) Periods (d) None |
| Ans. (c) |
| 3.3. Among the following elements, which is the highest electronegative element? (a) Carbon (b) Oxygen (c) Chlorine (d) Fluorine |
| Ans. (d) |
| 3.4. Diagonal relationships are shown by (a) He and Be (b) Mg and Al (c) Li and Mg (d) B and P |
| Ans. (c) |
| 3.5. Which one of the following is isoelectronic with Ne? (a) N3- (b) Mg2+ (c) Al3+ (d) All of the above |
| Ans. (d) |
| 3.1. State the Modern Periodic Law. |
| Ans. Modern Periodic Law states that physical and chemical properties of the elements are a periodic function of their atomic numbers. |
| 3.2. What is ionization enthalpy? Why is ionization enthalpy of nitrogen greater than that of oxygen? |
| Ans: Ionization enthalpy represents the energy required to remove an electron from an isolated gaseous atom in its ground state. Nitrogen has a greater ionization enthalpy than that of oxygen due to its exactly half-filled p-orbitals. |
| 3.3. Why are electron gain enthalpies of Be and Mg positive? |
| Ans: They have filled s-orbitals and hence do not tend to accept an additional electron. That is why energy is needed if an extra electron is to be added. Therefore, the electron gain enthalpies of Be and Mg are positive. |
| 3.4. What are representative elements? |
| Ans: The elements of group 1 (alkali metals), group 2 (alkaline earth metals) and group 13 to 17 constitute the representative elements. They are elements of the s-block and the p-block. |
| 3.5. Define electron gain enthalpy. What is its unit? |
| Ans: The energy which is released by an atom in gaining an electron from outside the atom or ion to form a negative ion (or an anion) is called electron gain enthalpy. The unit of electron gain enthalpy is kJ/ mol. In some cases, like in noble gases, atoms do not have any attraction to gain an electron. In that case, energy must be supplied. |
| 3.6. What are ‘Representative elements’? |
| Ans: The elements of the s-block and p-block are collectively called as representative or main group elements. These include elements of group I (alkali metals), group 2 (alkaline earth metals). |
| 3.7. Discuss briefly the various factors on which ionization enthalpy depends. |
| Answer: Ionisation enthalpy depends on the following factors: 1. Atomic size: The ionization enthalpy decreases with an increase in atomic size because with the increase in the atomic size, the number of electron shells increases. Therefore, the force that binds the electrons to the nucleus decreases. 2. Nuclear charge: The ionization enthalpy increases with the increase in the magnitude of the nuclear charge. This is because as the magnitude of the positive charge on the nucleus of an atom increases, the attraction with the electrons also increases. 3. Screening or shielding effect: Greater the magnitude of the screening effect less will be the value of ionization enthalpy. |
| 3.8. What are Eka-Aluminium and Eka-Silicon? |
| Ans: Elements gallium and germanium were unknown at the time Mendeleev published his Periodic Table. He left the gap under aluminium and a gap under silicon, and called these elements Eka-Aluminium and Eka-Silicon. Mendeleev predicted not only the existence of gallium and germanium but also described some of their general physical properties. |
| 3.9. What are transition elements? Write their important characteristics. |
| Ans: The d-block elements are known as transition elements. They have electronic configuration= (n – 1) d1-10 ns1-2 Characteristics of d-block elements are: They show variable oxidation states. Their compounds are generally paramagnetic. Most of the transition elements form coloured compounds. They are all metals with high melting and boiling points. These elements are used in the formation of alloys. |
| 3.10. What is screening or shielding effect? How does it influence the ionization enthalpy? |
| Ans. In a multi-electron atom, the electrons present in the inner shells shield the electrons in the valence shell from the attraction of the nucleus. They act as a screen between the nucleus and these electrons. This is known as the shedding effect or the screening effect. As the screening effect increases, the effective nuclear charge decreases. Consequently, the force of attraction by the nucleus for the valence shell electrons decreases, and hence the ionization enthalpy decreases. |
Classification of Elements and Periodicity in Properties Questions and Answers
Commonly asked questions
3.20. Which of the following pairs of elements would have a more negative electron gain enthalpy? (i) O or F (ii) F or Cl.
Solution:
Enthalpy in simple terms means chemical energy, the Electron Gain Enthalpy is equal to the energy released or absorbed when an atom gains an electron to form a anion (in gaseous state).
The Negative ion enthalpy is measured when net amount of energy decreases. This happens when atom releases some amount of energy in an Exothermic process.
(i) O or F:
The electronic configuration of O and F are:
O: 1s2 2s2 2p4
F: 1s2 2s2 2p5
Fluorine (F) has a more negative electron gain enthalpy than oxygen (O) because of its smaller size and its valency.
Due to smaller size, all electrons feels stronger electrostatic attraction force, hence it needs more energy to release the electron. Moreover fluorine needs only one electron to complete its outer shell and become stable unlike oxygen which needs two.
(ii) F or Cl:
The electronic configuration of Cl and F are:
Cl: [Ne] 3s2 3p5
F: 1s2 2s2 2p5
As per the NCERT "Value of electron gain enthalpy of Cl is more negative than that of F, though F has higher electronegativity. This is because F atom is small, so there is more repulsion in its 2p orbital."
Since Fluorine is very small, its electrons are packed. When an extra electron tries to enter, there’s too much repulsion due to lack of space. But chlorine is slightly bigger, so there’s more space for the new electron. So less repulsion and more energy released when Cl gains an electron. That’s why, chlorine releases more energy, so its electron gain enthalpy is more negative than fluorine.
3.16. Among the second period elements, the actual ionization enthalpies are in the order: Li
Explain why
(i) Be has higher ?i H than B?
(ii) O has lower ?i H than N and F?
3.16. (i) In the electronic configuration of Be (1s2 2s2) the outermost electron is present in 2s-orbital while in B (1s2 2s2 2p1) it is present in 2p-orbital. Since 2s – electrons are more strongly attracted by the nucleus than 2p-electrons, therefore, lesser amount of energy is required to knock out a 2p-electron than a 2s – electron.
Therefore, Be has higher? I H than that of B.
(ii) The electronic configuration of oxygen is 1s2 2s2 2p4. The 2p orbital contains 4 electrons out of which 2 are present in the same 2p-orbital. Due to this, the electron repulsion increases. N has stable half-filled configuration. F has higher effective nuclear charge. Hence, the ionization enthalpy of O is lower than that of N and F.
3.17. How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
3.17. Electronic configuration of Na and Mg are
Na = 1s2 2s2 2p6 3s1
Mg = 1s2 2s2 2p6 3s2
First electron in both cases has to be removed from 3s-orbital but the nuclear charge of Na (+ 11) is lower than that of Mg (+ 12) therefore first ionization energy of sodium is lower than that of magnesium.
After the loss of first electron, the electronic configuration of
Na+ = 1s2 2s2 2p6
Mg+ = 1s2 2s2 2p6 3s1
Here electron is to be removed from inert (neon) gas configuration which is very stable and hence removal of second electron from Na+ requires more energy in comparison to Mg.
Therefore, second ionization enthalpy of sodium is higher than that of magnesium.
3.18. What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down the group?
Atomic size: With the increase in atomic size, the number of electron shells increase. Therefore, the force that binds the electrons with the nucleus decreases. The ionization enthalpy thus decreases with the increase in atomic size.
Screening or shielding effect of inner shell electron: With the addition of new shells, the number of inner electron shells which shield the valence electrons increases. As a result, the force of attraction of the nucleus for the valence electrons further decreases and hence the ionization enthalpy decreases.
3.19. The first ionization enthalpy values (in kJ mol -1) of group 13 elements are:
B Al Ga In Tl
801 577 579 558 589
How would you explain this deviation from the general trend?
3.19. On moving down the group, the ionization enthalpy decreases. This is true for B and Al due to the bigger size of Al.
The ionization enthalpy of Ga is unexpectedly higher than Al because Ga contains 10d electrons in inner shell whose shielding is less effective than that of s and p electrons.
The outer electron is held fairly strongly by the nucleus. The ionization enthalpy increases slightly. A similar increase is observed from In to Tl due to presence of 14f electrons in the inner shell of Tl which have poor screening effect.
3.1. What is the basic theme of organisation in the periodic table?
3.1. The basic theme of organisation of elements in the periodic table is to simplify, systematize and classify the elements based on their similarities in properties. This arrangement makes it easier for us to group the elements and makes it less confusing to present the periodic table.
3.2. Which important property did Mendeleev use to classify the elements in this periodic table and did he stick to that?
3.2. Mendeleev arranged the elements horizontally, in periods and vertically, in groups in the increasing order of their atomic masses. This classification was done in such a way that the elements with similar properties fall in the same group.
3.3. What is the basic difference in approach between Mendeleev’s Periodic Law and the Modern Periodic Law?
3.3. Mendeleev's periodic law classifies the elements based on their atomic weight. Modern periodic law classifies the elements based on their atomic number.
3.5. In terms of period and group where will you locate the element with z = 114?
3.5. Elements with atomic numbers from Z = 87 to Z = 114 are present in the 7th period of the periodic table. So, this elements lies in the Period – 7 and Group -14 of Block-p.
3.14. What is the significance of the terms – isolated gaseous atom and ground state while defining the ionization enthalpy and electron gain enthalpy? [Hint:Requirements for comparison purposes]
3.14. Significance of the term 'isolated gaseous atom'. The atoms in the gaseous state are far separated in the sense that they do not have any mutual attractive and repulsive interactions. These are therefore regarded as isolated atoms. In this state, the value of ionization enthalpy and electron gain enthalpy are not influenced by the presence of the other atoms. It is not possible to express these when the atoms are in the liquid or solid state due to the presence of inter atomic forces.
Significance of the term 'ground state'. Ground state of the atom represents the normal – energy state of an atom. It means electrons in a particular atom are in the lowest energy state and they neither lose nor gain electron. Both ionisation enthalpy and electron gain enthalpy are generally expressed with respect to the ground state of an atom only.
3.15. Energy of an electron in the ground state of the hydrogen atom is- 2.18 x 10-18 J.Calculate the ionization enthalpy of atomic hydrogen in terms of J mol-1. [Hint: Apply the idea of mole concept to derive the answer]
3.15. The ionisation enthalpy is for 1 mole atoms.
Therefore, ground state energy of theatoms may be expressed as
Eground state = (– 2.18 x 10-18 J) x (6.022 x 1023 mol-1)
= –1.312 x 106 J mol-1
Ionisation enthalpy =E? –Eground state
= 0– (–1.312 x 106mol-1)
= 1.312 x 106 J mol-1.
3.46. Choose the incorrect statement
(a) In a triad, the atomic mass of the middle element is the mean of the atomic masses of the first and third elements.
(b) Electronegativity refers to the tendency of atom so share electrons with another atom.
(c) Number of elements in each period is twice the number of atomic orbitals available in the energy level that is being filled.
(d) According to Mendeleev, periodic properties of elements are a function of their atomic masses.
3.46. (b) Electronegativity refers to the tendency of an atom to share electrons with another atom.
3.53. Why are electron gain enthalpies of Be and Mg positive?
3.53. They have filled s-orbitals and hence do not tend to accept an additional electron. That is why energy is needed if an extra electron is to be added. Therefore, the electron gain enthalpies of Be and Mg are positive.
3.58. What are Eka-Aluminium and Eka-Silicon?
3.58. Elements gallium and germanium were unknown at the time Mendeleev published his Periodic Table. He left the gap under aluminium and a gap under silicon, and called these elements Eka-Aluminium and Eka-Silicon. Mendeleev predicted not only the existence of gallium and germanium but also described some of their general physical properties.
3.59. What are transition elements? Write their important characteristics.
3.59. The d-block elements are known as transition elements.
They have electronic configuration= (n – 1) d1-10 ns1-2
Characteristics of d-block elements are:
They show variable oxidation states.
Their compounds are generally paramagnetic.
Most of the transition elements form coloured compounds.
They are all metals with high melting and boiling points.
These elements are used in the formation of alloys.
3.60. What is screening or shielding effect? How does it influence the ionization enthalpy?
3.60. In a multi-electron atom, the electrons present in the inner shells shield the electrons in the valence shell from the attraction of the nucleus. They act as a screen between the nucleus and these electrons. This is known as the shedding effect or the screening effect.
As the screening effect increases, the effective nuclear charge decreases. Consequently, the force of attraction by the nucleus for the valence shell electrons decreases, and hence the ionization enthalpy decreases.
3.4. On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
3.4. The sixth period means that the highest principal quantum number = 6. The subsequent periods comprise of 8, 18, 18 and 32 elements in the same order. The sixth period comprises of 32 elements where electrons enter the 6s, 4f, 5d and 6p orbitals. Number of orbitals present in 6s=1. 4f= 7, 5d= 5 and 6p= 3. The total number of orbitals is 16. A total of 32 electrons can be filled in these 32 orbitals; therefore, the sixth period should have a maximum of 32 elements.
3.6. Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
3.6. The element is chlorine (Cl) with atomic number (Z) = 17.
3.7. Which element do you think would have been named by
(i) Lawrence Berkeley Laboratory
(ii) Seaborg’s group?
3.7. (i) Lawrencium (Lr) with atomic number (z) = 103
(ii) Seaborgium (Sg) with atomic number (z) = 106.
3.8. Why do elements in the same group have similar physical and chemical properties?
3.8. The elements in a group have same number of valence electrons and hence have similar physical and chemical properties.
3.9. What does atomic radius and ionic radius really mean to you?
3.9. Atomic radius: It is the distance from the centre of nucleus to the outer most shell of electrons in the atom of any element. It incorporates both the covalent, metallic radius or van there Waal's radius depending on whether the element is a non-metal or a metal.
Ionic radius: It is the distance between the centre of the nucleus of an ion up to the point where it exerts its influence on the electron cloud of a canton or anion.
3.10. How do atomic radii vary in a period and in a group? How do you explain the variation?
3.10. Across a period, the atomic radii decrease from left to right due to increase in effective nuclear charge from left to right across a period
Within a group, atomic radius increases down the group due to continuous increases in the number of electronic shells or orbit numbers in the structure of atoms of the elements down a group.
3.11. What do you understand by isoelectronic species? Name a species that will be iso electronic with each of the following atoms or ions.
(i) F–(ii) Ar (iii) Mg2+(iv) Rb+
3.11. Species (atoms/ions) which have same number of electrons are called isoelectronic species. The isoelectronic species out of the given atoms/ions are:
(i) Na+ is isoelectronic to F-
(ii) K+ is isoelectronic to Ar
(iii) Na+ is isoelectronic to Mg2+
(iv) Sr2+ is isoelectronic to Rb+
3.12. Consider the following species:
N3-, O2-, F–, Na+, Mg2+, Al3+
(a) What is common in them?
(b) Arrange them in order of increasing ionic radii?
3.12. (a) All of them have 10 electrons each and are isoelectronic in nature.
(b) In isoelectronic species, higher the nuclear charge, smaller will be the atomic or ionic radius.
Al3+< Mg2+< Na+< F–< O2-< N3-
3.13. Explain why cations are smaller and anions larger in radii than their parent atoms?
3.13. A cation is obtained by removing an electron from the outermost shell. The removal of electron (s) results in decrease of the size of the resulting ion than the parent atom because it has fewer electrons while its nuclearcharge remains the same.
Anions are obtained by addition of electron (s) in the outermost shell. This results in increased repulsion among the electrons and a decrease in effective nuclear charge.
3.21. Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.
3.21. When an electron is added to oxygen atom, energy is released to form a negative ion. This enthalpy change called the first electron gain enthalpy is thus negative. On adding another electron to the O? ion, it experiences repulsion from the anion due to which the instability of the ion increases. Thus, the addition of the second electron requires energy due to which the second electron gain enthalpy is positive.
O + e? ? O? ? H1 = -141KJ
O? + e? ? O2? ? H2 = 780 KJ
3.22. What is basic difference between the terms electron gain enthalpy and electro negativity?
3.22. Electron gain enthalpy refers to tendency of an isolated gaseous atom to accept an additional electron to form a negative ion. Whereas electronegativity refers to tendency of the atom of an element to attract shared pair of electrons towards it in a covalent bond.
3.23. How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?
3.23. On Pauling scale, the electronegativity of nitrogen is 3.0 which indicates that it is quite electronegative. But it is not correct to say that the electronegativity of nitrogen in all the compounds is 3. It varies with the atoms that N is bonded with. For example, in NH3, N has a different electronegativity than N in NO2
3.24. Describe the theory associated with the radius of an atom as it:
(a) Gains an electron (b) Loses an electron?
3.24. (a) Gain of an electron leads to the formation of an anion. The size of an anion will be larger than that of the parent atom because the addition of one or more electrons would result in increased repulsion among electrons and decrease in effective nuclear charge.
(b) Loss of an electron from an atom results in the formation of a cation. A cation is smaller than its parent atom because it has former electrons while its nuclear charge remains the same.
3.25. Would you expect the first ionization enthalpies of two isotopes of the same element to be the same or different? Justify your answer.
3.25. Isotopes have same number of electrons and protons, only the number of neutrons is different. The atomic number remains the same and only atomic mass differs. Hence the ionization energy of the isotopes of a chemical element remains the same.
3.26. What are major differences between metals and non-metals?
3.26.
Metals | Non-metals |
1. They have strong tendency to lose electrons to form cations. | 1. They have strong tendency to accept electrons to form anions. |
2. Metals are strong reducing agents. | 2. They are strong oxidising agents. |
3. Metals have low ionization enthalpies. | 3. Non-metals have high ionization enthalpies. |
4. They form basic oxides and ionic compounds. | 4. They form acidic oxides and covalent compounds. |
3.27. Use periodic table to answer the following questions:
(a) Identify the element with five electrons in the outer sub shell.
(b) Identify the element that would tend to lose two electrons.
(c) Identify the element that would tend to gain two electrons.
3.27. (a) Element belonging to nitrogen family (group 15) e.g., nitrogen.
(b) Element belonging to alkaline earth family (group 2) e.g., magnesium.
(c) Element belonging to oxygen family (group 16) e.g., oxygen.
3.28. The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that of group 17 is F > Cl > Br > I. Explain?
3.28. The elements of Group I have only one electron in their respective valence shells and thus have a strong tendency to lose this electron. The tendency to lose electrons in turn, depends upon the ionization enthalpy. Since the ionization enthalpy decreases down the group therefore, the reactivity of group 1 elements increases in the order Li < Na < K < Rb < Cs. In contrast, the elements of group 17 have seven electrons in their respective valence shells and thus have strong tendency to accept one more electron to make stable configuration. So for group 17, the electron gain enthalpy and electronegativity decreases down the group and thus the reactivity also decreases.
3.29. Write the general electronic configuration of s-, p-, d-, and f-block elements?
3.29. (i) s-Block elements: ns1-2 where n = 2-7.
(ii) p-Block elements: ns2 np1-6 where n = 2-6.
(iii) d-Block elements: (n-1)d1-10 ns 0-2 where n = 4-7.
(iv) f-Block elements: (n-2)f0-14 (n-1)d0-1 ns2where n = 6-7.
3.30. Assign the position of the element having outer electronic configuration,
(i) ns2 np4 for n = 3
(ii) (n – 1) d2 ns2 for n = 4 and
(iii) (n – 2) f7 (n – 1) d1 ns2 for n = 6 in the periodic table?
3.30. (i) For n = 3, the element belong to 3rd period, p-block element.
The electronic configuration is =1s2 2s2 2p6 3s2 3p4. The element name is sulphur.
(ii) For n = 4, the element belongs to 4th period and since the valence shell has 4 electrons it belongs to group 4.
Electronic configuration= 1s2 2s2 2p6 3s2 3p6 3d2 4s2, and the element name is Titanium (Ti).
(iii) For n = 6, the element belongs to 6th period. Last electron goes to the f-orbital, element is from f-block. It belongs to group = 3
The element is gadolinium (z = 64)
Complete electronic configuration = [Xe] 4 f7 5d1 6s2.
3.32. Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements:
(a) Lithium and oxygen
(b) Magnesium and nitrogen
(c) Aluminium and iodine
(d) Silicon and oxygen
(e) Phosphorous and fluorine
(f) Element 71 and fluorine.
3.32. (a) LiO2 (Lithium oxide)
(b) Mg3N2 (Magnesium nitride)
(c) AlI3 (Aluminium iodide)
(d) SiO2 (Silicon dioxide)
(e) Phosphorous pentafluoride (f) Z = 71
(f) The element is Lutenium (Lu). Electronic configuration [Xe] 4 f7 5d1 6s2.
With fluorine it will form a binary compound = LuF3.
3.33. In the modern periodic table, the period indicates the value of
(a) Atomic number
(b) Mass number
(c) Principal quantum number
(d) Azimuthal quantum number?
3.33. In the modern periodic table, each period begins with the filling of a new shell. Therefore, the period indicates the value of principal quantum number. Thus, option (c) is correct.
3.34. Which of the following statements related to the modem periodic table is incorrect?
(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-subshell.
(b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.
(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
(d) The block indicates value of azimuthal quantum number (l) for the last subshell that received electrons in building up the electronic configuration.
3.34. Statement (b) is incorrect.
3.35. Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?
(a) Valence principal quantum number (n)
(b) Nuclear charge (Z)
(c) Nuclear mass
(d) Number of core electrons
3.35. (c) Nuclear mass.
3.36. The size of isoelectronic species-F–, Ne and Na+ is affected by
(a) Nuclear charge (Z)
(b) Valence principal quantum number (n)
(c) Electron-electron interaction in the outer orbitals
(d) None of the factors because their size is the same
3.36. (a) Nuclear charge (Z).
3.37. Which of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electrons from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
3.37. (d) is incorrect.
3.38. Considering the elements B, Al, Mg and K, the correct order of their metallic character is:
(a) B>Al>Mg>K
(b) Al>Mg>B>K
(c) Mg>Al>K>B
(d) K>Mg>Al>B
3.38. In a period, metallic character decreases as we move from left to right. Therefore, metallic character of K, Mg and Al decreases in the order: K > Mg > Al. However, within a group, the metallic character, increases from top to bottom. Thus, Al is more metallic than B. Therefore, the correct sequence of decreasing metallic character is: K > Mg > Al > B, i.e., option (d) is correct
3.39. Considering the elements B, C, N, F and Si, the correct order of their non-metallic character is:
(a) B>C>Si>N>F
(b) Si>C>B>N>F
(c) F>N>C>B>Si
(d) F>N>C>Si>B
3.39. In a period, the non-metallic character increases from left to right. Thus, among B, C, N and F, non-metallic character decreases in the order: F > N > C > B. However, within a group, non-metallic character decreases from top to bottom. Thus, C is more non-metallic than Si. Therefore, the correct sequence of decreasing non-metallic character is: F > N > C > B > Si, i.e., option (c) is correct.
3.40. Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidising property is:
(a) F > Cl> O > N
(b) F > O > Cl> N
(c) Cl> F > O > N
(d) O > F > N > Cl
3.40. Within a period, the oxidising character increases from left to right. Therefore, among F, O and N, oxidising power decreases in the order: F > O > N. However, within a group, oxidising power decreases from top to bottom. Thus, F is a stronger oxidising agent than Cl. Further because O is more electronegative than Cl, therefore, O is a stronger oxidising agent than Cl. Thus, overall decreasing order of oxidising power is: F > O > Cl > N, i.e., option (b) is correct.
3.42. Assertion: Elements of the same group exhibit similar chemical properties.
Reason: Elements of the same group have similar valence shell electronic configuration.
3.42. (a) Elements of the same group have similar valence shell electronic configuration and, therefore, exhibit similar chemical properties. However, the elements of the same period have an incrementally increasing number of electrons from left to right, and, therefore, have different valences.
3.44. Assertion: Metallic elements have strong tendency to gain electrons.
Reason: Electronegativity is inversely related to the metallic properties of elements.
3.44. (d) Assertion is a wrong statement. Non-metallic elements have a strong tendency to gain electrons. Therefore, electronegativity is directly related to those non-metallic properties of elements. It can be further extended to say that the electronegativity is inversely related to the metallic properties of elements.
Thus, the increase in electronegativities across a period is accompanied by an increase in non-metallic properties (or a decrease in metallic properties) of elements
3.45. Assertion: Hydrogen is placed separately at the top of the Periodic Table.
Reason: It is the lightest element on the periodic table with a weight of just 1.0079 amu.
3.45. (b) It has only ones-electron and hence can be placed in group 1 (alkali metals). It can also gain an electron to achieve a noble gas arrangement and hence it can behave similarly to a group 17 (halogen family) element. Because it is a special case, we shall place hydrogen separately at the top of the Periodic Table
3.51. State the Modern Periodic Law.
3.51. Modern Periodic Law states that physical and chemical properties of the elements are a periodic function of their atomic numbers.
3.52. What is ionization enthalpy? Why is ionization enthalpy of nitrogen greater than that of oxygen?
3.52. Ionization enthalpy represents the energy required to remove an electron from an isolated gaseous atom in its ground state. Nitrogen has a greater ionization enthalpy than that of oxygen due to its exactly half-filled p-orbitals.
3.54. What are representative elements?
3.54. The elements of group 1 (alkali metals), group 2 (alkaline earth metals) and group 13 to 17 constitute the representative elements. They are elements of the s-block and the p-block.
3.55. Define electron gain enthalpy. What is its unit?
3.55. The energy which is released by an atom in gaining an electron from outside the atom or ion to form a negative ion (or an anion) is called electron gain enthalpy.
The unit of electron gain enthalpy is kJ/ mol. In some cases, like in noble gases, atoms do not have any attraction to gain an electron. In that case, energy must be supplied.
3.56. What are ‘Representative elements’?
3.56. The elements of the s-block and p-block are collectively called as representative or main group elements. These include elements of group I (alkali metals), group 2 (alkaline earth metals).
3.57. Discuss briefly the various factors on which ionization enthalpy depends.
3.57. Ionisation enthalpy depends on the following factors:
1. Atomic size: The ionization enthalpy decreases with an increase in atomic size because with the increase in the atomic size, the number of electron shells increases. Therefore, the force that binds the electrons to the nucleus decreases.
2. Nuclear charge: The ionization enthalpy increases with the increase in the magnitude of the nuclear charge. This is because as the magnitude of the positive charge on the nucleus of an atom increases, the attraction with the electrons also increases.
3. Screening or shielding effect: Greater the magnitude of the screening effect less will be the value of ionization enthalpy.
3.31. The first (ΔiH1) and the second (ΔiH2) ionization enthalpies (in kJ mol–1) and the (ΔegH) electron gain enthalpy (in kJ mol–1) of a few elements are given below:
Elements ΔH1 ΔH2 ΔegH
I 520 7300 –60
II 419 3051 –48
III 1681 3374 –328
IV 1008 1846 –295
V 2372 5251 +48
VI 738 1451 –40
Which of the above elements is likely to be:
(a) The least reactive element
(b) The most reactive metal
(c) The most reactive non-metal
(d) The least reactive non-metal
(e) The metal which can form a stable binary halide of the formula MX2 (X = halogen)
(f) The metal which can form a predominantly stable covalent halide of the formula MX (X = halogen)?
(a) The element V has the highest first ionization enthalpy (? iH1) and positive electron gain enthalpy (? egH) and hence it is the least reactive element. Since inert gases have positive? egH, therefore, the element-V must be an inert gas. The values of? iH1, ? iH2 and? egH match that of He.
(b) The element II which has the least first ionization enthalpy (? iH1) and a low negative electron gain enthalpy (? egH) is the most reactive metal. The values of ? iH1, ? iH2 and? egH match that of K (potassium).
(c) The element III which has high first ionization enthalpy (? iH1) and a very high negative electron gain enthalpy (? egH) is the most reactive non-metal. The values of? iH1, ? iH2 and? egH match that of F (fluorine).
(d) The element IV has a high negative electron gain enthalpy (? egH) but not so high first ionization enthalpy (? egH). Therefore, it is the least reactive non-metal. The values of? iH1, ? iH2 and? egH match that of I (Iodine).
(e) The element VI has low first ionization enthalpy (? iH1) but higher than that of alkali metals. Therefore, it appears that the element is an alkaline earth metal and hence will form binary halide of the formula MX2 (where X = halogen). The values of? i H1, ? iH2 and? egH match that of Mg (magnesium).
(f) The element I has low first ionization (? iH1) but a very high second ionization enthalpy (? iH2), therefore, it must be an alkali metal. Since the metal forms a predominantly stable covalent halide of the formula MX (X = halogen), therefore, the alkali metal must be least reactive. The values of? iH1, ? iH2 and? egH match that of Li (lithium).
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