Do you know how the position of elements in a periodic table helps to predict their properties, why some elements react vigorously, while others are more stable? The answer to all these questions lies in NCERT Exemplar Class 11 Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in properties. This chapter is important for understanding the elements and the changes that occur while moving across the periodic table. In the periodic table, elements are arranged in order of their atomic number.
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The topics like periodicity, atomic mass, electronegativity, and magnetic properties of elements are all discussed in this chapter. The NCERT Exemplar solutions for class 11 are designed to offer a structured approach to understand these concepts in the easiest way. These NCERT Exemplar Solutions will help you develop a clear understanding of the topics through a series of solved exercises and conceptual explanations. The higher-order thinking skills (HOTS) questions are included in the article to improve your critical thinking. NCERT Solutions provide a valuable resource to enhance performance in board exams as well as in the competitive exams.
The NCERT Exemplar Class 11 Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in Properties MCQ questions are given below. This chapter is very important from exam point of view.
Question 1. Consider the isoelectronic species, $Na^{+} , Mg^{2+}, F^{-} and\; O^{2-}.$
The correct order of increasing length of their radii is _________.
(i) $F^{-} < O^{2-} <Mg^{2+}<Na^{+}$
(ii) $Mg^{2+}<Na^{+} <F^{-} < O^{2-}$
(iii) $O^{2-} <F^{-} < Na^{+} <Mg^{2+}$
(iv) $O^{2-} <F^{-} < Mg^{2+}<Na^{+}$
Answer:
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The answer is the option (ii) $Mg^{2+}<Na^{+} <F^{-} < O^{2-}$
Explanation: All of them are isoelectronic species because they have the same number of valence electrons and moreover they even have the same number of electrons on the whole, which is 10. So, their radii would be different because of their different nuclear charges. The greater the charge in the cation greater the force with which the nucleus pulls the electrons or the greater the force of attraction of electrons to the nucleus and thus will have a lesser radius. Whereas in the case of anions greater the charge lesser the force of attraction thus will have a larger radius, and in this very case the net repulsion between the electrons will outweigh the nuclear charge, and the ion will thus expand in size.
Question 2. Which of the following is not an actinoid?
(i) Curium (Z = 96)
(ii) Californium (Z = 98)
(iii) Uranium (Z = 92)
(iv)Terbium (Z = 65)
Answer:
The answer is option (iv) Terbium (Z = 65)
Explanation: Actinoids are the elements with Z= 90 to 103, and they are characterized by the outer electronic configuration $(n-2)f^{14}(n-1)d$ $ns^{2}$. In an actinoid the last electron enters the f-orbital, and that is not the case in Terbium, and hence it is not an actinoid, but it is a lanthanoid.
Question 3. The order of screening effect of electrons of s, p, d and f orbitals of a given shell of an atom on its outer shell electrons is:
(i) s >p >d >f
(ii) f> d > p> s
(iii) p< d < s > f
(iv) f> p > s> d
Answer:
The answer is the option (i) s >p >d >f
Explanation: The valence electrons experience a shielding or screening effect from the nucleus because of the intervening core electrons, wherein the effective nuclear charge experienced by a valence electron in an atom will be less than the actual charge on the nucleus. Like the 2s electron in lithium is shielded from the nucleus by the inner core of 1s electrons.
Question 4. The first ionisation enthalpies of Na, Mg, Al and Si are in the order:
(i) Na < Mg > Al < Si
(ii) Na > Mg > Al > Si
(iii) Na < Mg < Al < Si
(iv) Na > Mg > Al < Si
Answer:
The answer is the option (i) Na < Mg > Al < Si
Explanation: While moving across the periods, the increasing nuclear charge overpowers the shielding effect. Consequently, the valence electrons are held more and more tightly, and thus; the ionization enthalpy increases across the period. In the case of Mg, the valence electron is in the s-orbital which is a stable gas configuration, and the penetration effect of the s-subshell is more as compared to aluminum (Al) in which the outermost electron is in pthe -subshell.
Question 5. The electronic configuration of gadolinium (Atomic number 64) is
(i) $[Xe] \; 4f^{3} 5d^{5}6s^{2}$
(ii) $[Xe] \; 4f^{7} 5d^{2}6s^{1}$
(iii) $[Xe] \; 4f^{7} 5d^{1}6s^{2}$
(iv) $[Xe] \; 4f^{8} 5d^{6}6s^{2}$
Answer:
The answer is the option (iii) $[Xe] \; 4f^{7} 5d^{1}6s^{2}$
Explanation: Gadolinium is a lanthanoid, it is one of the f-block elements and has the outer electronic configuration as $(n - 2) f^{14} (n-1) d^{0-1}-1 ns^{2}$
Question 6. The statement that is not correct for periodic classification of elements is:
(i) The properties of elements are periodic functions of their atomic numbers.
(ii) Non-metallic elements are less in number than metallic elements.
(iii) For transition elements, the 3d-orbitals are filled with electrons after 3p-orbitals and before 4s-orbitals.
(iv) The first ionization enthalpies of elements generally increase with the increase in atomic number as we go along a period.
Answer:
The answer is option (iii) For transition elements; the 3d-orbitals are filled with electrons after 3p-orbitals and before 4s-orbitals.
Explanation: According to the Aufbau principle, the 3d-orbitals are filled after the 4s and before the 4p orbitals, as the energy of the 4s orbital is lesser than the 3d orbital.
Question 7. Among halogens, the correct order of amount of energy released in electron gain 3 (electron gain enthalpy) is:
(i) F > Cl > Br > I
(ii) F < Cl < Br < I
(iii) F < Cl > Br > I
(iv) F < Cl < Br < I
Answer:
The answer is the option (iii) F < Cl > Br > I
Explanation: Within the same group, electron gain enthalpy becomes less negative while going down in the group. However, the addition of an electron to the 2p orbital leads to greater repulsion than the addition of an electron to the larger 3p- orbital. Hence, in the following halogens, chlorine is the element with the highest negative electron gain enthalpy.
Question 8. The period number in the long form of the periodic table is equal to
(i) magnetic quantum number of any element of the period.
(ii) atomic number of any element of the period.
(iii) maximum Principal quantum number of any element of the period.
(iv) maximum Azimuthal quantum number of any element of the period.
Answer:
The answer is option (iii) maximum principal quantum number of any element of the period.
Explanation: The period indicates the value of n for the outermost or valence shell. Hence, the Period Number maximum n of any element, where n refers to a principal quantum number.
Question 9. The elements in which electrons are progressively filled in 4f-orbital are called
(i) Actinoids
(ii) Transition elements
(iii) Lanthanoids
(iv) Halogens
Answer:
The answer is the option (iii) Lanthanoids
Explanation: The sixth period (n = 6) contains 32 elements, and after that, the successive electrons enter the 6s, 4f, 5d, and 6p orbitals and in that order, filling up of ff orbital begins with cerium (Z=58) and ends at Lutetium(Z=71). This complete 4f inner transition series is called the lanthanoid series.
Question 10. Which of the following is the correct order of size of the given species:
(i) I > I- > I+
(ii) I+ > I- > I
(iii) I > I+ > I-
(iv) I- > I > I+
Answer:
The answer is the option (iv) I- > I > I+
Explanation: A cation is always smaller in size than its parent atom because, in the case of cations, there are fewer electrons in it than its parent atom, but the nuclear charge is the same in both; thus there is a greater pull towards the nucleus thereby reducing the size of the cation. The opposite happens in the case of an anion where the nuclear charge remains the same, but the number of electrons is increased by one or more than the parent atom, and this results in increased repulsions among electrons and thus causes a decrease in effective nuclear charge, and as a result, the size of the anion increases manifold.
Question 11. The formation of the oxide ion, $O^{2-}(g)$, from oxygen atom requires first an exothermic and then an endothermic step as shown below: Thus process of formation of $O^{2-}$ in gas phase is unfavourable even though $O^{2-}$ is isoelectronic with neon. It is due to the fact that,
$O(g)+e^{-}\rightarrow O^{-}(g);\Delta H^{\ominus }= - 141 kJ\; mol^{-1}$
$O^{-}(g)+e^{-}\rightarrow O^{2-}(g);\Delta H^{\ominus }= +780\; kJ\; mol^{-1}$
(i) Oxygen is more electronegative.
(ii) The addition of electrons in oxygen results in a larger size of the ion.
(iii) electron repulsion outweighs the stability gained by achieving noble gas configuration.
(iv) O– ion has a comparatively smaller size than the oxygen atom.
Answer:
The answer is that option (iii) electron repulsion outweighs the stability gained by achieving noble gas configuration.
Explanation: Energy is released whenever an electron is added to the O atom to form O- ion. Thus the first electron gain enthalpy of O is negative. Contrary to this whenever an electron is added to O- ion to form O2- ion, energy has to be given out in order to overcome the very strong electronic repulsions, and that is why the second gain enthalpy of O is positive.
Answer:
The answer is the option (iii) d-block
Explanation: Lanthanoids are formed by the filling of the 4f-orbitals, and they are the elements following lanthanum from 58Ce to 71Lu. Actinoids formed by filling of 5f orbitals are the elements the following actinium from 90Th to 103Lr. The characteristic outer electronic configuration is as follows:
$(n-2)f^{14}(n-1)d^{0-1}ns^{2}$
Question 12. Comprehension given below is followed by some multiple choice questions. Each question has one correct option. Choose the correct option. In the modern periodic table, elements are arranged in order of increasing atomic numbers which is related to the electronic configuration. Depending upon the type of orbitals receiving the last electron, the elements in the periodic table have been divided into four blocks, viz, s, p, d and f. The modern periodic table consists of 7 periods and 18 groups. Each period begins with the filling of a new energy shell. In accordance with the Aufbau principle, the seven periods (1 to 7) have 2, 8, 8, 18, 18, 32 and 32 elements respectively. The seventh period is still incomplete. To avoid the periodic table being too long, the two series of f-block elements, called lanthanoids and actinoids are placed at the bottom of the main body of the periodic table.
(b) The last element of the p-block in 6th period is represented by the outermost electronic configuration.
(i) $7s^{2}7p^{6}$
(ii) $5f^{14}6d^{10}7s^{2}7p^{0}$
(iii) $4f^{14}5d^{10}6s^{2}6p^{6}$
(iv) $4f^{14}5d^{10}6s^{2}6p^{4}$
Answer:
The answer is the option (iii) $4f^{14}5d^{10}6s^{2}6p^{6}$
Explanation: The sixth period, which is when n equals 6, contains 32 elements and the successive electrons enter the 6s, 4f, 5d, and 6p orbitals. The last element in the p-block in the 6th period is [86]Rn.
The answer is the option (iii) 126
Explanation: In the present setup of the long form of the periodic table, there are seven periods and four blocks. So, the maximum number of elements according to the Aufbau principle, which can be accommodated in the present setup of the periodic table is 118.
Question 12. Comprehension given below is followed by some multiple choice questions. Each question has one correct option. Choose the correct option. In the modern periodic table, elements are arranged in order of increasing atomic numbers which is related to the electronic configuration. Depending upon the type of orbitals receiving the last electron, the elements in the periodic table have been divided into four blocks, viz, s, p, d and f. The modern periodic table consists of 7 periods and 18 groups. Each period begins with the filling of a new energy shell. In accordance with the Aufbau principle, the seven periods (1 to 7) have 2, 8, 8, 18, 18, 32 and 32 elements respectively. The seventh period is still incomplete. To avoid the periodic table being too long, the two series of f-block elements, called lanthanoids and actinoids are placed at the bottom of the main body of the periodic table.
(d) The electronic configuration of the element which is just above the element with atomic number 43 in the same group is ________.
(i) $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{5} 4s^{2}$
(ii) $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{5} 4s^{3}4p^{6}$
(iii) $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{6} 4s^{2}$
(iv) $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{7} 4s^{2}$
Answer:
The answer is the option (i) $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{5} 4s^{2}$
Explanation: The atomic number of the element that lies just above the element with atomic number 43 is 25 (Mn) or manganese. Its electronic configuration is $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{5} 4s^{2}$.
The answer is the option (ii) halogens
Explanation: The elements with atomic numbers 35, 53, and 85 are in the group before noble gases which are the halogens (group 17).
Question 13. Electronic configurations of four elements A, B, C and D, are given below :
(A) $1s^{2} 2s^{2} 2p^{6}$
(B) $1s^{2} 2s^{2} 2p^{4}$
(C) $1s^{2} 2s^{2} 2p^{6}3s^{1}$
(D) $1s^{2} 2s^{2} 2p^{5}$
Which of the following is the correct order of increasing tendency to gain electron?
(i) A < C < B < D
(ii) A < B < C < D
(iii) D < B < C < A
(iv) D < A < B < C
Answer:
The answer is the option (i) A < C < B < D
Explanation: Electron Gain Enthalpy: When an isolated gaseous atom or ion in its ground state adds an electron to form the corresponding anion, then the molar enthalpy change that occurs is known as Electron Gain Enthalpy.
The Classification of Elements and Periodicity in properties Chapter 3 NCERT Exemplar MCQ questions are given below. The chapter is vast yet easy to learn.
Question 14. Which of the following elements can show covalency greater than 4?
(i) Be
(ii) P
(iii) S
(iv) B
Answer:
The answer is the option (ii) and (iii)
Explanation: Both P and S have vacant d-orbitals in their atoms; that is why they are able to extend their covalency.
Question 15. Those elements impart color to the flame on heating in it, the atoms of which require low energy for the ionization (i.e., absorb energy in the visible region of the spectrum). The elements of which of the following groups will impart color to the flame?
(i) 2
(ii) 13
(iii) 1
(iv) 17
Answer:
The answer is the option (i) and (iii)
Explanation: Group I is the alkali metals group and Group II elements are the alkaline earth metals, they are reactive metals with low ionization enthalpies.
Question 16. Which of the following sequences contain atomic numbers of only representative elements?
(i) 3, 33, 53, 87
(ii) 2, 10, 22, 36
(iii) 7, 17, 25, 37, 48
(iv) 9, 35, 51, 88
Answer:
The answer is the option (i) and (iv)
Explanation: The P-block elements comprise the elements belonging to Groups 13 to 18 and these together with the s-block elements are called the Representative Elements or Main Group Elements and only options (i) and (iv) belong to s and p blocks.
Question 17. Which of the following elements will gain one electron more readily in comparison to other elements of their group?
(i) S (g)
(ii) Na (g)
(iii) O (g)
(iv) Cl (g)
Answer:
The answer is the option (i) and (iv)
Explanation: There are many elements where energy is released when an electron is added to the atom of those elements, thereby the electron gain enthalpy being negative for such elements. For example, the group 17 elements which are the halogens have extremely high negative electron gain enthalpies because they can attain stable noble gas electronic configurations by picking up an electron.
Question 18. Which of the following statements are correct?
(i) Helium has the highest first ionization enthalpy in the periodic table.
(ii) Chlorine has less negative electron gain enthalpy than fluorine.
(iii) Mercury and bromine are liquids at room temperature.
(iv) In any period, the atomic radius of alkali metal is the highest.
Answer:
The answer is the option (i), (iii) and (iv)
Explanation: Helium is the smallest noble gas, and that is the reason why it has the highest ionization enthalpy.
Question 19. Which of the following sets contain only isoelectronic ions?
(i) $Zn^{2+}, Ca^{2+}, Ga^{3+}, Al^{3+}$
(ii) $K^{+}, Ca^{2+}, Sc^{3+}, Cl^{-}$
(iii) $P^{-3}, S^{2-}, Cl^{-}, K^{+}$
(iv) $Ti^{4+}, Ar, Cr^{3+}, V^{5+}$
Answer:
The answer is the option (ii) and (iii)
Explanation: Isoelectronic species are those species that have the same number of electrons.
Question 20. In which of the following options order of arrangement does not agree with the variation of property indicated against it?
(i) Al3+< Mg2+< Na+ < F– (increasing ionic size)
(ii) B < C < N < O (increasing first ionisation enthalpy)
(iii) I < Br < Cl < F (increasing electron gain enthalpy)
(iv) Li < Na < K < Rb (increasing metallic radius)
Answer:
The answer is the option (ii) and (iii)
Explanation: The ionization enthalpy of N is always higher than that of O due to the greater stability of the half-filled electronic configuration of oxygen. So (ii) is incorrect. Similarly, in option (iii), the electron gain enthalpy of F is lower than that of Cl due to the small size of F. Hence option (iii) is also wrong.
Question 21. Which of the following have no unit?
(i) Electronegativity
(ii) Electron gain enthalpy
(iii) Ionisation enthalpy
(iv) Metallic character
Answer:
The answer is the option (i) and (iv)
Explanation: The electronegativity of an element is the tendency of an atom of that element to attract shared electrons towards itself, and unlike ionization enthalpy and electron gain enthalpy, it is not a measurable quantity.
Question 22. Ionic radii vary in
(i) inverse proportion to the effective nuclear charge.
(ii) inverse proportion to the square of effective nuclear charge.
(iii) direct proportion to the screening effect.
(iv) direct proportion to the square of screening effect.
Answer:
The answer is the option (i) and (iii)
Explanation: A cation is always smaller in size than its parent atom because, in the case of cations, there are fewer electrons in it than its parent atom, but the nuclear charge is the same in both; thus there is a greater pull towards the nucleus thereby reducing the size of the cation. The opposite happens in the case of an anion where the nuclear charge remains the same, but the number of electrons is increased by one or more than the parent atom, and this results in increased repulsions among electrons and thus causes a decrease in effective nuclear charge, and as a result, the size of the anion increases manifold.
Question 23. An element belongs to the 3rd period and group-13 of the periodic table. Which of the following properties will be shown by the element?
(i) Good conductor of electricity
(ii) Liquid, metallic
(iii) Solid, metallic
(iv) Solid, non-metallic
Answer:
The answer is the option (i) and (iii)
Explanation: Group-13 third period element is Aluminum which is a metal that is solid, metallic, and a good conductor of electricity.
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The Class 11 NCERT Exemplar Chemistry chapter 3 short answer type questions are given below. Periodic classification of elements is a vast yet easy to learn chapter.
Question 24. Explain why the electron gain enthalpy of fluorine is less negative than that of chlorine.
Answer:
The added electron in the fluorine atom goes to the second quantum level, and due to the very small size of fluorine, it experiences repulsion from other electrons much more in comparison to that in chlorine because, in chlorine, the electron is added to 3rd quantum level and chlorine, on the other hand, is not that small, so it experiences a much lower repulsion.
Question 25. All transition elements are d-block elements, but all d-block elements are not transition elements. Explain.
Answer:
Zn, Cd, and Hg which have the electronic configuration, (n-1) d10 ns2 do not show most of the properties of transition elements. Transition metals form a bridge between the chemically active metals of s-block elements and the less active elements of Groups 13 and 14.
Question 26. Identify the group and valency of the element having atomic number 119. Also predict the outermost electronic configuration and write the general formula of its oxide.
Answer:
Group: 1, Valency: 1
Outermost electronic configuration = 8s1
Formula of Oxide = M2O
Question 28. Among the elements B, Al, C and Si,
(i) which element has the highest first ionisation enthalpy?
(ii) which element has the most metallic character? Justify your answer in each case.
Answer:
(i) Moving along a period, ionization enthalpy increases, and moving down a group ionization enthalpy decreases. Therefore, C or Carbon has the highest first ionization enthalpy.
(ii) The metallic character increases while moving down the group and decreases along a period, so, Al has the most metallic character.
Question 29. Write four characteristic properties of p-block elements.
Answer:
(i) p-Block elements include both metals and non-metals.
(ii) p-Block elements, in most cases, form covalent bonds with other elements.
(iii) The ionization enthalpies are relatively higher than s-block elements.
(iv) Some of the p-block elements show variable or more than one oxidation state.
Question 30. Choose the correct order of atomic radii of fluorine and neon (in pm) out of the options given below and justify your answer.
(i) 72, 160
(ii) 160, 160
(iii) 72, 72
(iv) 160, 72
Answer:
The answer is the option (i) 72, 160
Explanation: The atomic size generally decreases across a period for the elements of the second period because within one period there are the same number of shells in all the elements within that period but the nuclear charge increases while going along the period and hence the electrons in the shells are attracted more towards the nucleus thereby reducing the atomic radii.
Question 31. Illustrate by taking examples of transition elements and non-transition elements that oxidation states of elements are largely based on electronic configuration.
Answer:
The oxidation state of an element is basically the charge acquired by its atom or appears to have when it is present in the combined state with other atoms. For transition elements, there are variable oxidation states due to the presence of both valence electrons and d-electrons. The most common oxidation states are +2 and +3. Titanium shows three oxidation states +2, +3, +4. In the case of non-transition elements, the value of the oxidation state is equal to a number of electrons in the outermost shell or 8 minus the number of electrons in the outermost shell.
Answer:
In the nitrogen atom, the three 2p-electrons reside in different atomic orbitals according to Hund's rule whereas, in the oxygen atom, two of the four 2p-electrons must occupy the same 2p-orbital resulting in an increased electron-electron repulsion. Oxygen has lower ionization enthalpy than nitrogen because by removing one electron from 2p-orbital, oxygen acquires a stable configuration, i.e., 2p3. On the other hand, in the case of nitrogen, it is not easy to remove one of the three 2p-electrons due to its stable configuration
Answer:
Anomalous Properties of Second Period Elements: The anomalous behavior is because of their small size, large charge/radius ratio, high electronegativity, and non-availability of d-orbitals in their valence shell.
example:
1. Compounds of lithium have a significant covalent character. whereas compounds of other alkali metals are predominantly ionic.
2. In p-block elements, the first member of each group has four orbitals, one 2s-orbital and three 2p-orbitals in their valence shell. So, these elements show a maximum covalency of four while other members of the same group or different groups show a maximum covalency beyond four due to the availability of vacant d-orbitals.
Question 34. p-Block elements form acidic, basic and amphoteric oxides. Explain each property by giving two examples and also write the reactions of these oxides with water.
Answer:
The oxides of p-block elements show acidic, basic, and amphoteric properties, due to the following factors:
Ionization enthalpy: The higher the ionization enthalpy of an element, the stronger will be the acid formed by that element. If the ionization enthalpy of an element is high, then its oxide will be acidic in nature, if low, then it will be basic in nature, and if intermediate, its oxide will be amphoteric in nature.
Electronegativity: The higher the electronegativity of the element, the more acidic is its oxide. For instance, $N_{2}O_{3}$ is more acidic than $B_{2}O_{3}$
Oxidation states: The higher the oxidation state of the elements; the stronger will be its acid. For instance, $SO_{3}$ is a stronger acid than $SO_{2}$
Reaction with water
$B_{2}O_{3}+3H_{2}O\rightleftharpoons 2H_{3}BO_{3}$
Boron Trioxide Orthoboric acid
$B(OH)_{3}+ H-OH\rightarrow [B(OH)_{4}]^{-}+H^{+}$
$Al_{2}O_{3}$ is amphoteric in nature. It is insoluble in water but dissolves in alkalies and reacts with acids.
$Al_{2}O_{3} +2NaOH\overset{\Delta }{\rightarrow}2NaAlO_{2}+H_{2}O$
Aluminium trioxide Sodium meta aluminate
$Al_{2}O_{3} +6HCl\overset{\Delta }{\rightarrow}2AlCl_{3}+3H_{2}O$
Aluminium Chloride
$Ti_{2}O$ is as basic as $NaOH$ due to its lower oxidation state (+1).
$Ti_{2}O+2HCl\rightarrow 2TiCl+H_{2}O$
Question 35. How would you explain the fact that first ionisation enthalpy of sodium is lower than that of magnesium, but its second ionisation enthalpy is higher than that of magnesium?
Answer:
The first ionization enthalpy of magnesium is higher than that of Na due to the higher nuclear charge and slightly smaller atomic radius of Mg than Na. After the loss of the first electron, Na+ formed has the electronic configuration of neon (2,8) which means it becomes highly stable because of achieving the nearest noble gas configuration, and its quite clear that the higher the stability more energy is required to remove an electron; thus more is the ionization enthalpy. The higher stability of the completely filled noble gas configuration leads to a very high second ionization enthalpy for sodium. On the other hand, Mg+ formed after losing the first electron still has one more electron in its outermost (3s) orbital. As a result, the second ionization enthalpy of magnesium is much smaller than that of sodium.
Question 36. What do you understand by exothermic reaction and endothermic reaction? Give one example of each type.
Answer:
Exothermic Reaction: $\Delta H$ is negative for reactions in which heat evolves during the reaction and are called exothermic. Example: $CaO+CO_{2}\rightarrow CaCO_{3}$
This reaction leads to the release of -178.3 kJ mol-1 of energy and is an exothermic reaction as $\Delta H$ is negative.
Endothermic Reaction: $\Delta H$ is positive in an endothermic reaction which absorbs heat from the surroundings
Example: $2NH_{3}\rightarrow N_{2}+3H_{2}$
This reaction requires the energy of +91.8 kJ mol-1 and is an endothermic one because $\Delta H$ is positive.
Question 37. Arrange the elements N, P, O, and S in the order of-
(i) increasing first ionization enthalpy.
(ii) increasing non-metallic character.
Give a reason for the arrangement assigned.
Answer:
(i) S < P < N < O
(ii) P < S < N < O
The ionization enthalpy increases on moving from left to right in a period, and similarly, the Non-metallic character increases while moving from left to right across the period.
Question 38. Explain the deviation in ionization enthalpy of some elements from the general trend by using Fig. 3.2.
Answer:
First ionization enthalpy increases while moving along a period because the nuclear charge increases while moving along the period. There is the deviation of ionization enthalpy of some elements from the general trend as shown in the figure. The first ionization enthalpy of B is lower than that of Be due to the filled s-orbital in Be and in the case of nitrogen, the first ionization enthalpy is higher than that of O due to half-filled stability.
Question 39. Explain the following:
(a) Electronegativity of elements increases on moving from left to right in the periodic table.
(b) Ionisation enthalpy decrease in a group from top to bottom?
Answer:
(a) Electronegativity of an element is the tendency of an atom of that element to attract shared electrons towards itself, and unlike ionization enthalpy and electron gain enthalpy, it is not a measurable quantity. The increase in electronegativities across a period is due to a decrease in the size of an atom and an increase in nuclear charge.
(b) As we go down a group, the outermost electron increasingly becomes far from the nucleus, and there is an increase in atomic size. Therefore, ionization enthalpy decreases down the group.
Question 40. How does the metallic and non-metallic character vary on moving from left to right in a period?
Answer:
Metallic character: The tendency of an element to lose electrons and thereby form positive ions is called electropositive or metallic character. The elements with lower ionization energies have a higher tendency to lose electrons; thus, they are electropositive or metallic in their behavior.
Periodicity: In a period, the electropositive or metallic character decreases from left to right in a period. In a group, the electropositive or metallic character increases from top to bottom in a group.
Non-metallic character: The tendency of an element to accept electrons to form an anion is called its non-metallic or electronegative character.
Periodicity: (i) In period: The electronegative or non-metallic character increases while going from left to right in a period.
(ii) In group: The electronegative or non-metallic character decreases while going from top to bottom in a group.
Question 41. The radius of Na+ cation is less than that of Na atom. Give reason.
Answer:
The radius of Na+ cation is less than that of the Na atom because of the decrease of one shell, also there is a greater pull towards the nucleus thereby reducing the size of the cation.
Question 42. Among alkali metals which element do you expect to be least electronegative and why?
Answer:
Electronegativity decreases in a group from top to bottom, and thereby cesium is the least electronegative element.
Class 11 chemistry Chapter 3 Classification of Elements and Periodicity in properties important questions are discussed below. These are generally asked in exams to test your knowledge. These exemplar solutions is quite helpful for competitive exams.
Question 43. Match the correct atomic radius with the element.
|
Element |
Atomic radius (pm) |
|
Be |
74 |
|
C |
88 |
|
O |
111 |
|
B |
77 |
|
N |
66 |
Answer:
Be = 111, O = 66, C = 77, B = 88, N = 74.
Question 44. Match the correct ionisation enthalpies and electron gain enthalpies of the following elements.
|
Elements |
$\Delta H_{1}$ |
$\Delta H_{2}$ |
$\Delta_{eg} H$ |
|
(i) Most reactive non-metal |
A.419 |
3051 |
-48 |
|
(ii) Most reactive metal |
B.1681 |
3374 |
-328 |
|
(iii) Least reactive element |
C.738 |
1451 |
-40 |
|
(iv) Metal forming binary halide |
D.2372 |
5251 |
+48 |
Answer:
Most reactive non metal = B, Most reactive metal = A, Least reactive element = D, Metal forming binary halide = C
|
Column (I) Electronic configuration |
Column (II) Electron gain enthalpy / kJ mol -1 |
|
(i)$1s^{2}2s^{2}2p^{6}$ |
(A) -53 |
|
(ii)$1s^{2}2s^{2}sp^{6}3s^{1}$ |
(B) -328 |
|
(iii)$1s^{2}2s^{2}2p^{5}$ |
(C) -141 |
|
(iv)$1s^{2}2s^{2}2p^{4}$ |
(D) +48 |
Answer:(i) $\rightarrow$ (d); (ii) $\rightarrow$ (a); (iii) $\rightarrow$ (b); (iv) $\rightarrow$ (c)
The Assertion and Reason type questions of Chapter 3 test a students conceptual clarity and logical reasoning skills. These Periodic classification of elements Class 11 NCERT Exemplar solutions help you understand the reason behind periodic trends and element classification.
The assertion is a correct statement and the reason is a wrong statement.
Assertion and reason both are correct statements and reason is a correct explanation of assertion.
Assertion and reason both are wrong statements.
Assertion is a wrong statement and the reason is the correct statement
Answer:
The answer is option (ii) Assertion, and reason both are correct statements, and the reason is a correct explanation of assertion.
Explanation: Ionisation enthalpy depends upon the nuclear force that attracts electrons and on the repulsion of electrons from one another.
Assertion and reason both are correct statements, but reason is not correct explanation for assertion.
The assertion is a correct statement, but the reason is a wrong statement.
Assertion and reason both are correct statements and reason is the correct explanation for the assertion.
Assertion and reason both are wrong statements.
Answer:
The answer is option (iii) Assertion and reason both are correct statements and reason is the correct explanation for the assertion.
Explanation: During ionization, the electron removed in the case of beryllium is from the s-orbital, and the electron removed from the boron atom is from the p-orbital, and the penetration of 2s electron to the nucleus is more than that of 2p electron hence 2p electron of boron is more shielded from the nucleus than the 2s electron.
Assertion and reason both are correct statements, but reason is not correct explanation for assertion.
Assertion and reason both are correct statements and reason is the correct explanation for assertion.
Assertion and reason both are wrong statements.
The assertion is a wrong statement, but the reason is a correct statement.
Answer:
The answer is option (iv) Assertion is a wrong statement, but the reason is a correct statement.
Explanation: Electron gain enthalpy does not always become less negative as we go down a group in the Modern periodic table.
Ex. $\Delta H_{e g}(2 p$ series $)<\Delta H_{e g}(3 p$ series $)$
The Long Answer Type questions of Chapter 3 test a students conceptual clarity and logical reasoning skills. These Periodic classification of elements Class 11 NCERT Exemplar solutions help you understand the reason behind periodic trends and element classification.
Question 49. Discuss the factors affecting electron gain enthalpy and the trend in its variation in the periodic table.
Answer:
Factors Affecting Electron Gain Enthalpy:
Nuclear charge: As nuclear charge increases the gain enthalpy becomes more negative.
Size of the atom: As nuclear charge increases, the gain enthalpy becomes less negative.
Electronic configuration: Electron gain enthalpy is less negative for elements that have a stable, half-filled, or completely filled electronic configuration.
Trends in variation in the periodic table:
In period: The electron gain enthalpy increases from left to right in a period.
In group: The electron gain enthalpy decreases from top to bottom in a group.
Question 50. Define ionisation enthalpy. Discuss the factors affecting ionisation enthalpy of the elements and its trends in the periodic table.
Answer:
Ionisation Enthalpy: It is the change in the molar enthalpy when an electron is removed from a gaseous phase atom in its ground state.
Factors on which Ionisation Enthalpy Depends:
(i) Size of the atom: The larger the atomic size, smaller is the value of ionisation enthalpy. In a larger atom, the outer electrons are far away from the nucleus and thus force of attraction with which they are attracted by the nucleus is less and hence can be easily removed.
(ii) Screening effect: Higher the screening effect, the lesser is the value of ionisation enthalpy as the screening effect reduces the force of attraction towards nucleus and hence the outer electrons can be easily removed.
(iii) Nuclear charge: As the nuclear charge increases among atoms having a same number of energy shells, the ionization enthalpy increases because the force of attraction towards the nucleus increases.
(iv) Half filled and fully filled orbitals: The atoms having half filled and fully filled orbitals are comparatively more stable, hence more energy is required to remove the electron from such atoms. The ionization enthalpy is rather higher than the expected value in case of such an atom.
(v) Shape of orbital: The s-orbital is more close to nucleus than the p-orbital of the same orbit. Thus, it is easier to remove electron from a p-orbital in comparison to s-orbital. In general, the ionisation enthalpy follows the following order
(s>p> d>f) orbitals of the same orbit.
Variation of ionisation enthalpy in the periodic table
ln general, the ionisation energy decreases down the group due to increase in atomic size. Whereas, the ionisation energy increases across the period from left to right, again due to decrease in atomic size from left to right.
Question 51. Justify the given statement with suitable examples— "the Properties of the elements are a periodic function of their atomic numbers".
Answer:
The physical and chemical properties of elements are periodic functions of the atomic numbers means that chemical and physical properties repeat after a regular interval. Elements having similar outer electronic configurations in their atoms are arranged in vertical columns, referred to as groups or families.
Question 52. Write down the outermost electronic configuration of alkali metals. How will you justify their placement in group 1 of the periodic table?
Answer:
All the elments of Alkali metals have the similar outer electronic configuration, i.e. ns1 where n refers to the number of principle shell. These electronic configuration are given below.
$_{3}Li\; \; \; \; \; \; 1s^{2}2s^{1}or\left [ He \right ]2s^{1}$
$_{11}Na\; \; \; \; \; \; 1s^{2}2s^{2}2p^{6}3s^{1}or\left [ Ne \right ]3s^{1}$
$_{19}K\; \; \; \; \; \; 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}or\left [ Ar \right ]4s^{1}$
$_{37}Rb\; \; \; \; \; \; 1s^{2}2s^{2}2p^{6}3s^{1}3p^{6}3d^{10}4s^{2}4p^{6}5s^{1}or\left [ Kr \right ]5s^{1}$
$_{55}Cs\; \; \; \; \; \; 1s^{2}2s^{2}2p^{6}3s^{1}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^{6}6s^{1}or\left [ Xe \right ]6s^{1}$
$_{87}Fr\; \; \; \; \; \; \left [ Rn \right ]7s^{1}$
Hence, placement of all the elements in group 1 of the periodic table because of similarity in electronic configuration and all the elements have similar properties.
Question 53. Write the drawbacks in Mendeleev's periodic table that led to its modification.
Answer:
Following are the drawbacks in Mendeleev's periodic table: -
Hydrogen's position-
Hydrogen resembled the properties of both alkali metals(like lithium) and also that of halogens(like iodine). Hence, the position of hydrogen(whether hydrogen is to be placed with halogens or alkali metals) was not specified.
Position for Isotopes-
Mendeleev's periodic table was based on arranging elements in increasing order of atomic masses. But isotopes were not included in his periodic table.
Certain elements were arranged in reverse order. Elements having higher atomic mass were placed in front(or before) the elements with less atomic mass. Example - Cobalt and Nickel
Cobalt being more heavier than Nickel ,was placed before Nickel , in the Mendeleev's periodic table.
Several gaps were left in the periodic table as he believed that several elements are yet to be discovered. For instance – Gallium was not discovered at that time.
The position of VIII Group is inappropriate.
Question 54. In what manner is the long form of the periodic table better than Mendeleev's periodic table? Explain with examples.
Answer:
Modern Periodic Table is better than Mendeleev's periodic table because:
This table is based on the atomic number, which makes it more accurate.
There is a correlation between the position of the elements and their electron configurations clearly.
Complete separation of metals and non-metals has been achieved.
It eliminates the even and odd series of IV, V, VI and VII periods of Mendeleev's table.
Position of VIII Group is more appropriate than before.
Question 55. Discuss and compare the trend in ionisation enthalpy of the elements of group1 with those of group 17 elements.
Answer:
Ionisation Enthalpy is defined as the change in the molar enthalpy when an atom loses an electron.
Factors Affecting Ionisation enthalpy:
Nuclear charge: As nuclear charge increases the Ionisation Enthalpy increases.
Size of the atom: As nuclear charge increases the Ionisation Enthalpy becomes less.
When one moves from lithium to fluorine across the second period, successive electrons are added to orbitals in the same principal quantum number, and the shielding of the nuclear charge by the inner core of electrons does not increase very much. Thus, across a period, increasing nuclear charge outweighs the shielding. Consequently, the outermost electrons are held more and more tightly, and the ionisation enthalpy increases across a period.
As one goes down a group, the outermost electron being increasingly farther from the nucleus, there is an increased shielding of the nuclear charge by the electrons in the inner levels. So ionisation enthalpy decreases down the group.
HOTS questions of Classification of Elements and Periodicity in properties Chapter 3 NCERT Exemplar are designed to enhance analytical thinking and application-based understanding.
Question 1: The comprehension given below is followed by some multiple-choice questions. Each question has one correct option. Choose the correct option. In the modern periodic table, elements are arranged in order of increasing atomic numbers which is related to the electronic configuration. Depending upon the type of orbitals receiving the last electron, the elements in the periodic table have been divided into four blocks, viz, s, p, d, and f. The modern periodic table consists of 7 periods and 18 groups. Each period begins with the filling of a new energy shell. By the Aufbau principle, the seven periods (1 to 7) have 2, 8, 8, 18, 18, 32, and 32 elements respectively. The seventh period is still incomplete. To avoid the periodic table being too long, the two series of f-block elements, called lanthanoids and actinoids are placed at the bottom of the main body of the periodic table.
The elements with atomic numbers 35, 53, and 85 are all ________.
(1). noble gases
(2). Halogen
(3). Heavy metals
(4). light metals
Answer:
The answer is halogens.
Explanation: The elements with atomic numbers 35, 53, and 85 are in the group before noble gases which are the halogens (group 17).
Hence, the answer is the option (2).
Question 2: The electron affinity value are negative for :
A. $\mathrm{Be} \rightarrow \mathrm{Be}^{-}$
B. $\mathrm{N} \rightarrow \mathrm{N}^{-}$
C. $\mathrm{O} \rightarrow \mathrm{O}^{2-}$
D. $\mathrm{Na} \rightarrow \mathrm{Na}^{-}$
E. $\mathrm{Al} \rightarrow \mathrm{Al}^{-}$
Choose the most appropriate answer from the options given below :
Answer:
(A) $\mathrm{Be}+\mathrm{e}^{-} \rightarrow \mathrm{Be}^{-}, \quad$ E.A $=$-ive
(B) $\mathrm{N}+\mathrm{e} \rightarrow \mathrm{N}^{-} \quad$ E.A $=-$ ive
(C) $\mathrm{O}^{+} \mathrm{e}^{-} \rightarrow \mathrm{O}^{-}$
$\mathrm{O}^{-}+\mathrm{e}^{-} \rightarrow \mathrm{O}^{-2} \quad$ E.A $=-$ ive
(D) $\mathrm{Na}+\mathrm{e}^{-} \rightarrow \mathrm{Na}^{-} \quad$ E.A $=+$ ive
(E) $\mathrm{A} \ell+\mathrm{e}^{-} \rightarrow \mathrm{A} \ell^{-} \quad \mathrm{E} \cdot \mathrm{A}=+\mathrm{ive}$
$\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ only
Hence, the answer is option (4)
Question 3: Given below are two statements :
Statement (I) : The first ionisation enthalpy of group 14 elements is higher than the corresponding elements of group 13.
Statement (II) : Melting points and boiling points of group 13 elements are in general much higher than those the corresponding elements of group 14.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1) Statement I is correct but Statement II is incorrect
(2) Statement I is incorrect but Statement II is correct
(3) Both Statement I and Statement II are incorrect
(4) Both Statement I and Statement II are correct
Answer:
Statement 1 is correct since left to right, I.E. increases in general in the periodic table.
Statement 2 is incorrect since the M.P. of group 14 elements is more than that of group 13 elements.
Hence, the correct answer is option (1).
Question 4: Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy?
(1) $\quad \mathrm{CaSO}_4$
(2) $\quad \mathrm{BeSO}_4$
(3) $\quad \mathrm{BaSO}_4$
(4) $\quad \mathrm{SrSO}_4$
Answer:
As we move down the group, size of metal increases. Be has lower size while $\mathrm{SO}_4^{2-}$ has bigger size, that's why $\mathrm{BeSO}_4$ breaks easily and lattice energy becomes smaller but due to lower size of $B e$, water molecules are gathered around and hence hydration energy increases. On the other hand, rest of the metals, i.e., $C a, B a, S r$ have bigger size and that's why lattice energy is greater than hydration energy.
Hence, the correct answer is option (2).
Question 5: The successive 5 ionisation energies of an element are $800,2427,3658,25024$ and $32824 \mathrm{~kJ} / \mathrm{mol}$, respectively. By using the above values predict the group in which the above element is present:
(1) Group 2
(2) Group 13
(3) Group 4
(4) Group 14
Answer:
The $\mathrm{IE}_4$ is suddenly very high therefore element must have 3 valence $\mathrm{e}^{-}(\mathrm{s})$ and it belong to group 13
Hence, the correct answer is option (2).
Question 6: Given below are two statements :
Statement I : The metallic radius of $\mathrm{Na}$ is $1.86 \mathrm{~A}^{\circ}$ and the ionic radius of $\mathrm{Na}^{+}$is lesser than $1.86 \mathrm{~A}^{\circ}$.
Statement II: Ions are always smaller in size than the corresponding elements.
In the light of the above statements, choose the correct answer from the options given below :
(1) Statement I is correct but Statement II is false
(2) Both Statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is incorrect but Statement II is true
Answer:
$\mathrm{r}_{\mathrm{Na}}>\mathrm{r}_{\mathrm{Na}^{+}}$
So, Statement (I) is correct but the size of anions is greater than the size of neutral atoms.
So statement (II) is incorrect.
Hence, the answer is the option (1).
Question 7: Given below are two statements:
Statement I : The correct order of first ionization enthalpy values of Li, Na, F and Cl is Na < Li < Cl < F.
Statement II : The correct order of negative electron gain enthalpy values of Li, Na, F and Cl is Na < Li < F < Cl
In the light of the above statements, choose the correct answer from the options given below :
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is false but Statement II is true
(4) Statement I is true but Statement II is false
Answer:
1. Ionization energy data for given elements -
| Element | Na | Li | Cl | F |
| I.E1 in KJ/mol | 496 | 520 | 1256 | 1681 |
2. Electron gain enthalpy energy data for given elements -
| Element | Na | Li | F | Cl |
| $\Delta_{\text {eg }} H$ in KJ/mol | - 53 | - 60 | - 328 | - 349 |
Hence, the answer is the option (1).
NCERT Exemplar Solutions for Class 11 Chemistry Chapter 3 forms the basis of chemistry. The approaches given below are helpful and provide step-by-step approach to solve questions.
1. Before solving questions, of Chapter 3 it is important to have proper knowledge of basic concepts like atomic structure, electronic configuration and quantum numbers.
2. Learn to discover the elements placement in the Periodic Table according to their atomic numbers, groups and periods. Try to understand the meaning of the s, p, d, and f blocks and the difference between metals, non-metals, and metalloids.
3. Examine trends such as metallic/non-metallic character, ionisation enthalpy, electron gain enthalpy, electronegativity and atomic/ionic radius.
Also, learn to recognise the causes of each trend, including the rise and fall in the atomic radius as you move through the periodic table. Use periodic trends to solve problems on reactivity, bond formation, basic/acidic character of oxides, etc. It is advisable to relate the electron configuration with the position of elements in the periodic table and to know why the elements are divided into different blocks, like s-block or p-block.
4. Learn to compare two or more elements based on trends and write logical explanations. Focus on anomalies and exceptions. Students can refer to Classification of Elements and Periodicity in properties notes.
5. Practice in-text, exercise and exemplar questions regularly. Try to attempt assertion-reason, match-the-following and previous year trend-based MCQs from competitive exams.
Practice previous year board and NEET/JEE questions.
The class 11 chemistry chapter 3 Classification of Elements and Periodicity in properties question answer helps students to understand the fundamental principles of the periodic table. The advantages of using these NCERT Exemplar Class 11 Chemistry solutions are given below:
NCERT Exemplar Class 11 Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in properties covers key concepts related to the classification of elements and periodic trends. Understanding these concepts are essential for building a strong foundation in Classification of Elements and Periodicity in Properties.
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:
The S-block Elements
The P-block Elements
The D-block Elements (Transition Elements)
The F-block Elements (Inner-transition Elements)
Metals, Nonmetals and Metalloids
Periodic Trends in Properties of Elements
Trends in Physical Properties
Periodic Trends in Chemical Properties
Periodic Trends and Chemical Reactivity.
Students can follow the links given below to access all NCERT Exemplar Solutions for Class 11.
Students can follow the links given below to access all NCERT Solutions for Class 11 Chemistry
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Given below are links for Class 11 NCERT Solutions
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Frequently Asked Questions (FAQs)
We need to classify elements Because the classification of elements organizes the elements based on similarities in their properties which makes it easier to predict their behaviour and reactions.
The periodic table is organized by increasing atomic number, which results in a repeating pattern of physical and chemical properties known as periodicity.
Mendeleev's Periodic Table was based on grouping elements with similar chemical properties and increasing atomic numbers.
Elements in the periodic table are primarily classified into three categories: metals, non-metals, and metalloids.
The Periodic Table is divided into four blocks:
The main reason for classifying elements is to organise them based on shared properties and behaviours, making it easier for scientists to study and understand their relationships.
Class 11 NCERT Exemplar Chemistry chapter 3 is based on the topic Classification of Elements and Periodicity in Properties.This chapter focuses on how elements are classified into groups and periods in the periodic table according to their properties.
You can find NCERT Exemplar Solutions for Class 11 Chemistry on the official NCERT website, or on trusted educational platforms such as Careers360. These solutions are also available in PDF format for free download.
NCERT Exemplar Solutions cover all Class 11 and 12 Chemistry topics, including basic concepts of chemistry, atomic structure, chemical bonding, thermodynamics, equilibrium, organic chemistry, and environmental chemistry.
Class 11 chemistry chapter 3 Classification of Elements and Periodicity in properties question answer are based on topics like modern periodic law, electronic configuration, periodic trends, and classification of elements.
A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
| Option 1)
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Option 2)
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| Option 3)
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Option 4)
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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
| Option 1)
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Option 2)
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| Option 3)
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Option 4)
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A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
| Option 1)
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Option 2)
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| Option 3)
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Option 4)
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In the reaction,
| Option 1)
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Option 2)
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| Option 3)
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Option 4)
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How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?
| Option 1)
0.02 |
Option 2)
3.125 × 10-2 |
| Option 3)
1.25 × 10-2 |
Option 4)
2.5 × 10-2 |
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