NCERT Solutions for Class 11 Chemistry Chapter 3

The NCERT Solutions for Class 11 Chemistry helps students gain in-depth understanding of the topics discussed in the chapter. Students will be able to cover all the sections thoroughly. These solutions are extremely helpful in preparing last minute notes for revision and help recall all the crucial aspects of the chapter. The NCERT Solutions for class 11 Chemistry has been provided below to assist students in their upcoming examinations.

The NCERT Solutions for Class 11 Chemistry Chapter 3 Structure of Atom Classification of Elements and Periodicity in Properties is an extremely crucial segment for class 11 syllabus. Our subject experts at SimplyAcad have explained each solution in an elaborate manner, so all the doubts and confusions regarding the chapter are cleared.

Access the NCERT Solutions for Class 11 Chemistry Chapter 3: Structure of Atom Classification of Elements and Periodicity in Properties

Question 1. Consider the following isoelectronic species, Na+, Mg2+, F–and O2–. The correct order of increasing the length of their given radii is _________.

(i) F– < O2- < Mg2+ < Na+

(ii) Mg2+ < Na+< F–< O2-

(iii) O2-< F– < Na+< Mg2+

(iv) O2- < F– < Mg2+ < Na+

Answer 1: Option (ii) is the correct answer.

Question 2. Which among the following is not an actinoid?

(i) Curium (Z = 96)

(ii) Californium (Z = 98)

(iii) Uranium (Z = 92)

(iv) Terbium (Z = 65)

Answer 2: Option (iv) is the correct answer.

Question 3. The order of screening effect of the electrons of s, p, d and f orbitals for the given shell of an atom on their outer shell electrons will:

(i) s > p > d > f

(ii) f > p > s > d

(iii) p < d < s > f

(iv) f > d > p > s

Answer 3: Option (i) is the correct answer.

Question 4. The first ionisation enthalpies of the Na, Mg, Al and Si are in the order as follows:

(i) Na < Mg > Al < Si

(ii) Na > Mg > Al > Si

(iii) Na < Mg < Al < Si

(iv) Na > Mg > Al < Si

Answer 4: Option (i) is the correct answer.

Question 5. The electronic configuration of the given gadolinium (Atomic number 64) is

(i) [Xe] 4f3 5d5 6s2

(ii) [Xe] 4f7 5d2 6s1

(iii) [Xe] 4f7 5d1 6s2

(iv) [Xe] 4f8 5d6 6s2

Answer 5: Option (iii) is the correct answer.

Question 6. The statement which is not correct for the following periodic classification of elements is:

(i) The properties of elements are periodic functions of their given atomic numbers.

(ii) The Non-metallic elements are less in number than the metallic elements.

(iii) For the given transition elements, the 3d-orbitals fill with the electrons after the 3p-orbitals and before the 4s-orbitals.

(iv) The first ionisation enthalpies of the given elements generally increase with an increase in the atomic number as we go along the period.

Answer 6:

Option (iii) is the correct answer.

Question 7. Among halogens, correct order of the amount of the energy released in the given electron gain (electron gain enthalpy) will:

(i) F < Cl < Br < I

(iv) F > Cl > Br > I

(iii) F < Cl > Br > I

(ii) F < Cl < Br < I

Answer 7: Option (iii) is the correct answer.

Question 8. The period number at the long form of the given periodic table is equal to

(i) the magnetic quantum number of any given element of the period.

(ii) an atomic number of any given element of the period.

(iii) maximum Principal quantum number of any given period element.

(iv) maximum Azimuthal quantum number of any given period element.

Answer 8: Option (iii) is the correct answer.

Question 9. The elements in which the given electrons progressively fill in the 4f-orbitals call

(i) actinoids

(ii) transition elements

(iii) lanthanoids

(iv) halogens

Answer 9: Option (iii) is the correct answer.

Question 10. Which among the following is the correct order of the size of the given species:

(i) I > I– > I+

(ii) I+> I– > I

(iii) I > I+ > I-

(iv) I– > I > I+

Answer 10:Option (iv) is the correct answer.

Question 11. The formation of the given oxide ion, O2– (g), from the oxygen atom, requires first an exothermic and then an endothermic step as stated below:

O (g) + e– → O–(g) ;thus, ∆ HV = – 141 kJ mol–1

O– (g) + e– → O2– (g) ;thus, ∆ HV = + 780 kJ mol–1

Thus, the formation process of O2– in the given gas phase is unfavourable even though O2– is isoelectronic with the neon. It is because of this fact that

(i) oxygen is more electronegative.

(ii) addition of the electrons in oxygen results in the larger size of the ion.

(iii) electron repulsion outweighs stability gained by achieving the noble gas configuration.

(iv) O– ion is smaller than a given oxygen atom.

Answer 11:Option (iii) is the correct answer.

Question 12. Electronic configurations of the following four elements, A, B, C and D, have given below :

(A) 1s2 2s2 2p6

(B) 1s2 2s2 2p4

(C) 1s2 2s2 2p6 3s1

(D) 1s2 2s2 2p5

Which among the following is the correct order of increasing the tendency to gain the electrons :

(i) A < C < B < D

(ii) D < A < B < C

(iii) D < B < C < A

(iv) A < B < C < D

Answer 12:Option (i) is the correct answer.

Question 13. Which among the following elements can show the covalency greater than the 4?

(i) B

(ii) P

(iii) S

(iv) Be

Answer 13:Option (ii) and (iii) are the correct answers.

Question 14. Those elements which impart colour to the flame on heating in it are the atoms that require low energy for the ionisation (i.e., absorb energy in the visible region of the spectrum). The elements of which among the following groups will impart colour to the flame?

(i) 2

(ii) 13

(iii) 1

(iv) 17

Answer 14:Option (i) and (iii) are the correct answers.

Question 15. Which among the following sequences contains 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 15:Option (i) and (iv) are the correct answers.

Question 16. Which among the following elements will gain one electron more readily in comparison to the other elements of the given group?

(i) S (g)

(ii) Na (g)

(iii) O (g)

(iv) Cl (g)

Answer 16:Option (i) and (iv) are the correct answers.

Question 17. Explain why the electron gain enthalpy of the elemental fluorine is less negative than the elemental chlorine. .

Answer 17:Fluorine has a smaller size as compared to chlorine. As a result, the attraction outside the shell to gain the electrons is less. Moreover, they possess inter-electronic repulsions in the 2p orbitals, resulting in less negative electron gain enthalpy.

Question 18. All the transition elements are d-block elements, but all the d-block elements are not the transition elements. Explain.

Answer 18:The elements having their outermost shell filled with the d electrons are called the d block elements. All the d blocks are not the transition elements as it is important to have an incompletely filled d orbital of the given element like calcium and zinc etc.

Question 19. Identify the group and valency of the following element having atomic number 119. Also, predict the outermost electronic configuration for it and write the general formula of the oxide.

Answer19: There are 118 elements found in the seven periods of the modern periodic table. Thus, the element with the atomic number 119 will lie in the 8th period of the first group and have the outermost electronic configuration of 8s1. It belongs to group 1 and has the valency one. The formula of the oxide would be M2O.

Question 20. Ionisation enthalpies of a few elements of the second period are given below: Ionisation enthalpy/ kcal mol–1: 520, 899, 801, 1086, 1402, 1314, 1681, 2080.

Here, Match the correct enthalpy by the given elements and complete the graph in the Fig. Also, write symbols of the elements with the given atomic number.

Answer 20:

The ionisation enthalpy of the given elements varies across the period and group. The ionisation enthalpy decreases down the group and increases as we move from left to right in the period.

Question 21. Among the given elements B, Al, C and Si,

(i) which of the following elements has the highest first ionisation enthalpy?

(ii) which element contains the most metallic character? Explain your answer in each case.

Answer 21: (i) Carbon possesses the highest ionisation enthalpy. It increases from left to right along the period and also decreases as we go down the group.

(ii) Aluminium possesses the most metallic character. On moving down, the metallic character increases and decreases across the period from left to right.

Question 22. Write the four characteristic properties for the p-block elements.

Answer 22: 1. They show the variable oxidation states. The trend of the reducing character increases on moving down the group, and oxidising character increases as we move along the period.

1. They have a higher ionisation enthalpy than the s-block elements.
2. They usually form the given covalent compounds.
3. Both metals and non-metals can be found in this group, but the non-metals are slightly more in number.

Question 23. Illustrate by taking examples of the following transition and non-transition elements for which the oxidation states of the given elements are mostly based on the electronic configuration.

Answer 23: Ti possesses an atomic number of 22 and electronic configuration [Ar]3d24s2 and has three oxidation states at +2,+3 and +4 in the various compounds like TiO29(+4), Ti2O3(+3) and TiO(+2). The non-transition elements like the p-block elements have variable oxidation states like the phosphorus. It has -3,+3 and +5.

Question 24. Nitrogen possesses the positive electron gain enthalpy, whereas the oxygen possesses the negative. But oxygen has lower ionisation enthalpy than nitrogen. Explain.

Answer 24:The ionisation enthalpy of oxygen is lower than that of nitrogen as when we remove one electron from the oxygen then it easily donates it to attain half-filled stability; however, in the case of nitrogen, it is difficult to remove one electron because it already has half-filled stability and it will become unstable after that.

Question 25. The first member of each group of the representative elements, s and p-block elements, shows abnormal behaviour. Illustrate with the help of two examples.

Answer 25: Lithium and beryllium are the examples. Li is the first group element. It has different properties as well as forms the covalent compounds as well as nitrides. Beryllium is the first element of the second group. It has various anomalies like it forms the covalent compound with coordination number four, unlike other elements with the coordination number 6.

Question 38. How will you explain that the first ionisation enthalpy of sodium is quite lower than that of magnesium; however, its second ionisation enthalpy is quite higher than that of magnesium?

Answer 38: Sodium attains the stable configuration if it loses on an electron from the outermost shell. That is why its first ionisation enthalpy is less than the magnesium. However, in the case of the second ionisation, magnesium has one electron in its outermost shell to attain stability; it loses the electron easily compared to sodium, which is already stable.

Question 26. What is the basic theme for the organisation in the modern periodic table?

Answer 26. The basic theme of organisation in the modern periodic table is to classify the given elements in periods and groups according to its properties. Hence, the course of action makes the study of elements and their compounds simple and systematic. In the periodic table, elements with similar properties are placed in the same group.

Question 27. How does the metallic and non-metallic character vary as we move from left to right in the period?

Answer 27:

Metallic character decreases as we move along left to right across the period, and non-metallic character increases as you can find an increase in the ionisation enthalpy and electron gain enthalpy along the period.

Question 28. The radius of the Na+ cation is less than that of the Na atom. Give reasons for your answer.

Answer 28:

Sodium atoms can lose one electron to form sodium cation. After the formation of the cation, the effective nuclear charge on the ion increases on the left electrons, resulting in the decrease of the radius.

Question 29. Among alkali metals which element can be least electronegative and why?

Answer 29:

Caesium is the least electronegative alkali metal because electronegativity decreases as we move from top to bottom due to increase in size.. Caesium is the group 1 element and lies down the group because it has the largest size due to the decrease in the effective nuclear charge.

Question 30. What do you understand about the term exothermic reaction and endothermic reaction? Give one example of each type.

Answer 30: Exothermic reaction- The reaction where heat evolved is called an exothermic reaction.

for example,

Cao + CO2→ CaCO3 ΔH=-178 kJmol-1

Endothermic reaction- The reaction where heat is absorbed is called an endothermic reaction.

for example,

2NH3 → 3N2+ H2 ΔH=918kJmol-1

Question 31. Arrange the following elements N, P, O and S in the order of-

(i) increasing first ionisation enthalpy.

(ii) increasing non-metallic character.

Give an appropriate reason for the arrangement assigned.

Answer 31:

(i) S< P< O< N is the accurate, increasing order of the first ionisation enthalpy.

On going down the group, the ionisation enthalpy decreases, and as we move along the period, then it increases; however, in the case of oxygen and nitrogen, because of the half-filled stability of 2p orbitals of nitrogen, it has the higher ionisation enthalpy than oxygen.

(ii) P<S<N<O is the accurate, increasing order of non-metallic character.

Moving down the group, we will see non-metallic character decrease as the effective nuclear charge on the outermost shell decreases, which helps to gain an electron. The effective nuclear charge increases moving along the period, increasing the non-metallic character.

Question 32. p-Block elements can form acidic, basic and amphoteric oxides. Explain each given property by giving two examples and write the reactions of these oxides with water.

Answer 32:

ACIDIC OXIDES

SO2 and B2O3 are the acidic oxides in the p block elements.

The required reaction of SO2 with water

SO2 + H2O → H2SO3

The required reaction of B2O3 with water

B2O3 +3 H2O → 2H3BO3

Acidic oxides are those oxides that can form acids after reacting with water.

BASIC OXIDES

Cao, BaO, and Ti2O form the basic oxides

The required reaction of Ti2O with water.

BaO + H2O → Ba(OH)2

REACTION OF CaO WITH WATER

Cao + H2O → Ca(OH)2

Basic oxides are those oxides that can form bases after reacting with the water.

AMPHOTERIC OXIDES

Zinc oxides, as well as aluminium oxides, are the two given amphoteric oxides.

Reaction of ZnO with the water:

ZnO + 2H2O + 2NaOH → Na3Zn[OH]4 + H2

ZnO +2HCl → ZnCl2 + H2O

Reaction of Al2O3 with the water:

Al2O3 (s) + 6 NaOH(aq) + 3H2O(l) → Na3 [Al(OH)6] (aq)

Al2O3 (s) + 6HCl(aq) + 9H2O(l) → 2[Al(H2O)]3+(aq) + 6Cl–

Question 33. Explain the deviation in the ionisation enthalpy of some given elements from the general trend using Fig.

Answer 33:

The ionisation enthalpy of given elements varies across periods and groups. The ionisation enthalpy decreases down the group and increases as we go on from left to right in the period.

• Effective nuclear charge for the outermost electrons.
• Electron-electron repulsion force.
• Stability of the element because of half-filled and filled orbitals are some of the parameters affected.

Question 34. Explain the following:

(a) Electronegativity of the given elements increases as we move from left to right in the periodic table.

(b) Ionisation enthalpy decreases in a group from top to bottom?

Answer 34:

(a) As we move left to right in the period, the size of the atoms decreases because of the increase in the effective nuclear charges on the outermost electron. As a result, the electronegativity of the elements increases as we move along left to right in the periodic table.

(b) we move down the group, then the atomic size increases, which results by the increase in the distance of the electrons in the outermost shell. As a result, the effective nuclear charge decreases. That results in the decrease of the ionisation enthalpy.

Question 35. Which important properties did Mendeleev use to classify the elements in his periodic table, and did he stick to that?

Answer 35. Mendeleev organised the components in his periodic table according to the order of their atomic weight. He organised the components in groups and periods according to the increment order of atomic weight. He placed the elements with equal properties in the same group.

So, he did not stick to this arrangement for long. He discovered that if the elements were organised according to their increasing atomic weights, thus some of the elements didn’t fit in with his classification scheme.

Thus, he ignored the order of atomic weights in some cases. For example, the atomic mass of the iodine is lower than the atomic mass of the tellurium.

Still, Mendeleev set tellurium (in the Group 6) ahead of iodine (in Group 7) along with fluorine, chlorine, and bromine because of similarities in properties.

Question 36. What is the basic difference in the required approach between Mendeleev’s Periodic Law and the Modern Periodic Law?

Answer 36.

Mendeleev’s Periodic law	Modern Periodic Law
Mendeleev’s Periodic Law states that elements’ physical and chemical properties are periodic functions of their atomic weights.	Modern Periodic Law states that elements’ physical and chemical properties are periodic functions of their atomic numbers.

Question 37. Based on quantum numbers, justify that the sixth period of the periodic table should have 32 elements.

Answer 37. In the periodic table containing elements, a period shows the value of a principal quantum number (n) for the outermost shells. Every period starts with filling the principal quantum number (n). The value for n for the 6th period is equal to 6. then, for n = 6, the azimuthal quantum number (l) could have “0, 1, 2, 3, 4” values.

According to Aufbau’s principle, electrons have been added to the various orbitals in order of their increasing energies. Here, the 6d subshell has much higher energy than the energy of the 7s subshell.

In the sixth period, the electrons could fill in only 6s, 4f, 5d, and 6p subshells. 6s has only one orbital, 4f has seven orbitals, 5d has five orbitals, also 6p has three orbitals. Therefore, there are a sum of 16 (1 + 7 + 5 + 3 = 16) orbitals available. As per Pauli’s exclusion principle, each orbitals can accommodate a maximum of 2 electrons.

Hence, sixteen orbitals can accommodate a maximum of 32 electrons.

So, the 6th period of the periodic table should have 32 elements.

Question 38. In terms of the period and group, where would you locate the element with Z =114?

Answer 38. Elements of the atomic numbers from Z = 87 to the Z = 114 are present in the seventh period of the periodic table. Hence, the element with Z = 114 is present in the seventh period for the periodic table.

At the seventh period, first 2 elements with Z = 87 and Z= 88 are s-block elements and the next 14 elements except Z = 89 i.e., those with Z = 90 to Z = 103 are f – block elements, as well as next 10 elements with Z = 89 as well as Z = 104 to Z = 112 are d-block elements as well as the elements with Z = 113 to Z = 118 are the p-block elements. Thus, the element with Z = 114 is the second p-block element in the seventh period. So, the element with Z = 114 is present in the periodic table’s seventh period and fourteenth group.

Therefore,

Period = 7th, Group = 14 and Block = p-block.

Question 39. In the modern periodic table, elements are arranged in order of their increasing atomic number related to the electronic configuration. Depending upon the type of the orbitals that receive the last electron, the periodic table elements are categorised into four blocks, viz, s, p, d and f. The modern periodic table has 7 periods and 18 groups. Each period starts with the filling of the new energy shell. By the Aufbau principle, the seven periods (1 to 7) contain 2, 8, 8, 18, 18, 32 and 32 elements, respectively. The seventh period is still incomplete. To avoid today’s periodic table being too long, the two series of the f-block elements, called lanthanoids and actinoids, are placed at the bottom of the main body of today’s periodic table.

(a) The element having the atomic number 57 belongs to

(i) s-block

(ii) p-block

(iii) d-block

(iv) f-block

(b) The outermost electronic configuration represents the last element of the existing p-block in the 6th period.

(i) 4f14 5d10 6s2 6p4

(ii) 5 f 14 6d10 7s2 7p0

(iii) 4f 14 5d10 6s2 6p6

(iv) 7s2 7p6

(c) Which of the following elements whose atomic numbers given below cannot be accommodated in the present set-up of the long form of the periodic table?

(i) 102

(ii) 118

(iii) 126

(iv) 107

(d) The electronic configuration of the given element that is just above the element with the atomic number 43 in the same group will ________.

(i) 1s2 2s2 2p6 3s2 3p6 3d5 4s2

(ii) 1s2 2s2 2p6 3s2 3p6 3d6 4s2

(iii) 1s2 2s2 2p6 3s2 3p6 3d7 4s2

(iv) 1s2 2s2 2p6 3s2 3p6 3d5 4s3 4p6

(e) The elements having atomic numbers 35, 53 and 85 will all ________.

(i) light metals

(ii) halogens

(iii) heavy metals

(iv) noble gases

Answer 39:

(a) d-block,

(b) 4f 14 5d10 6s2 6p6

(c) 126

(d) 1s2 2s2 2p6 3s2 3p6 3d5 4s2

(e) halogens

Question 40. What is the atomic number of elements keeping in mind both the cases given below;

1. Element is in the 3rd period of the periodic table.
2. Element is in the 17th group of the periodic table.

Answer 40. In the third period, we provided that element. The element’s highest principal quantum number (n) into which the last electron enters is called the period number. As a result, n=3 for the 3rd period. In addition, the element could be found in the seventeenth group. The elements for the seventeenth group have the following general configuration: ns2np5.

As a result, the needed element’s overall electronics configuration is 3s23p5 (because n=3).

The element’s complete electrical configuration will now be: 1s22s22p63s23p5.

then add up the total number of electrons in the element’s ground state: 1+2+6+2+5=17 electrons.

The element’s atomic number equals its ground state’s total number of electrons.

As determined above, the element has 17 electrons in its ground state; its atomic number is 17. Chlorine is an atomic number 17 element (Cl).

Question 41. Discuss those factors affecting electron gain enthalpy as well as the trend in its variation in the periodic table.

Answer 41. factors affecting electron gain enthalpy as well as the trend in its variation in the periodic table are:

1)ATOMIC SIZEAs, we go down in the group, the electron gain enthalpy decreases as the distance of the nucleus from outermost shell increases, which decreases its tendency to gain electrons and electron gain enthalpy becomes less negative.

2)EFFECTIVE NUCLEAR CHARGEAs we go from the left to right in a period, the effective nuclear charge increases, as well as when we move down the group, it decreases, that results in the attraction of the electrons from the outermost shell

3) ELECTRONIC CONFIGURATIONThe tendency to gain electrons depends upon the stability of the element. Elements having complete or half-filled stable orbitals have a low tendency to gain electrons; thus, they have very low electron gain enthalpy.TRENDSAcross a period, electron gain enthalpy becomes more negative. Down the group, the electron gain enthalpy becomes less negative.

Question 42. Define the ionisation enthalpy. Discuss the factors affecting ionisation enthalpy of elements and their trends in the periodic table.

Answer 42. Ionisation enthalpy is the energy required by the isolated and gaseous atom in its ground state to remove an electron. The effective nuclear charge is due to the screening effect; inner core electrons shield the valence electrons. eThe effective nuclear charge is less than the actual charge on the atom. Penetrated orbital: It is difficult to remove an electron from the orbitals closer to the nucleus and penetrate towards the nucleus. The order of the penetration has been given by s>p>d>f Stability of the orbitals: Half-filled and filled orbital have a high ionisation enthalpy as well as they don’t want to lose their stability. Across a period, ionisation enthalpy increases along with the period. Down the group, the ionisation enthalpy decreases.

Question 43. Assertion (A): ionisation enthalpy increases from the left to right in a period.

Reason (R): When the successive electrons added to the orbital in the same principal quantum level, the shielding effect of the inner core of electrons does not increase much to compensate for increased attraction of the electron to the nucleus.

(i) The assertion is the correct statement, and the reason is the wrong statement.

(ii) Assertion and reason both were correct statements, and the reason is the correct explanation of assertion.

(iii) Assertion and reason both were wrong statements.

(iv) The assertion is the wrong statement, and the reason is the correct statement.

Answer 43. (ii) Assertion and reason both were correct statements, and the reason is the correct explanation of assertion.

Question 44. Assertion (A): Boron has a smaller first ionisation enthalpy than beryllium.

Reason (R): The penetration of the 2s electron to the nucleus is more than the 2p electron; therefore, the 2p electron is more shielded by the inner core of the electron than 2s electrons.

(i) Assertion and reason both were correct statements, but the reason is not the correct explanation for the assertion.

(ii) The assertion was a correct statement, but the reason was the wrong statement.

(iii) Assertion and reason both were correct statements, and the reason is the correct explanation for the assertion.

(iv) Assertion and reason both were wrong statements.

Answer 44. (iii) Assertion and reason both were correct statements, and the reason is the correct explanation for the assertion.

Question 45. Assertion (A): Electron gain enthalpy becomes less negative as we go down a group.

Reason (R): The size of the atom increases on going down the group, and the added electron would be farther from the nucleus.

(i) Assertion and reason both were correct statements, but the reason is not the correct explanation for the assertion.

(ii) Assertion and reason both were correct statements, and the reason is the correct explanation for the assertion.

(iii) Assertion and reason both were wrong statements.

(iv) The assertion was the wrong statement, but the reason was the correct statement.

Answer 45. (iv) The assertion was the wrong statement, but the reason was the correct statement.