Ncert Solution Chapter 3 Chemistry 11th is the solution of all the questions provided in the exercise of classification of elements and periodicity in properties. This is the 3rd chapter of 11th chemistry. This is a basic parts of chemistry. We can not do well better without the study of this chapter properly. We have to learn a lot of typical concepts in this chapter. The ideas about elements we learn in this chapter make interested in study of the properties and preparation of elements and their compounds.
The solution of all the questions in Ncert Solution Chapter 3 Chemistry 11th has been written in simple and proper way which will increase the comprehensive knowledge of the students in chemistry. This solution will help to score good marks not only in the board exams but also boost their knowledge for any entrance exams like Jee(mains), NEET, IIT advance etc. We hope that any student who approach this solution notes should take complete benefit after through study.
Ncert Solution Chapter 3 Chemistry 11th
Here we shall learn that what will be the correct answer of the questions. All these questions have been asked in the exercise of classification of elements and periodicity in properties. This is the 3rd chapter of chemistry 11th. This chapter is considered as the 1st chapter of inorganic chemistry. All the answers are accurate and in simple language. Any student can enhance their knowledge after reading this solution notes.
Ncert solutions of Classification of Elements and periodicity in properties
Important questions and their answers are followings:-
1. What is the basic theme of organization in the periodic table?
Ans:- The basic theme of organisation of elements in the periodic table is to simplify and systemaize the study of the properties of all the elements and millions of their compounds on the basis of similarities in chemical properties. The various elements have been divided into different groups. This has made the study simple because the properties of elements are now studied in form of groups rather than individually.
2. Which vital property did Mendeleev use to order the elements in the periodic table that he designed and did he adhere to that? (Ncert Solution Chapter 3 Chemistry 11th
Ans:- Mendeleev used atomic weights as the basis of classification of elements in the periodic table. He arranged the then known elements in order of increasing atomic weights grouping together elements with similar properties. He sincerely stuck to the basis leaving out blank spaces where elements were not known at that time. For example, the element gallium after aluminium and germanium after silicon were not known at that time. He called these elements as eka- aluminium and eka- silicon. Later these elements were discovered and their properties were found to be similar to those predicted by Mendeleev.
3. State difference between Mendeleev’s Approach for periodic law and the Modern approach for the periodic law.
Ans:- Mendeleev Periodic Law states that the physical and chemical properties of the elements are a periodic function of their atomic weights while Modern Periodic Law states that physical and chemical properties of elements are a periodic function of their atomic numbers. Thus, the basic difference in approach between Mendeleev’s Periodic Law and Modern Periodic Law is the change in basis of classification of elements from atomic weight to atomic number.
4. On the premise of the quantum numbers, verify that the 6th period of periodic table ought to have 32 components.
Ans:- In the modern periodic table, each period starts with the filling of a new principal energy level. Thus the sixth period begins with filling of principal quantum number, n = 6. Where n = 6, l = 0, 1, 2, 3. That means electrons have to be filled in 6s, 4f, 5d and 6p which have 16 orbitals. All these orbitals can be completely filled by 32 electrons. Hence, this period should have only 32 elements.
5. In groups and periods of periodic table where will you find the elements which is having Z =114?
Ans:- The filling of the 6th period ends at 86Rn. Thereafter, the filling of 7th period starts. Like in 6th period, in the 7th period, also the filling of four subshells as 7s, 7p, 6d and 5f occurs. But according to aufbau principle, their energies increase in the order: 7s < 5f < 6d < 7p. Therefore, after 86Rn, the next two elements with Z = 87 and 88 are s- block elements, the next next fourteen for Z = 90 to 103 are f- block elements, the next 10 for Z = 104 to 112 are d- block elements and the last six for Z = 113 to 118 are p- block elements.
Therefore, the element Z = 114 is the second p- block element. Thus the location of the element Z = 114 in the periodic table is period = 7th, Block = p- block and Group = 14.
6. What is the atomic number of element keeping in mind both the cases given below;
1. Element is in 3rd period of periodic table.
2. Element is in 17th group of periodic table.
Ans:- In the third period, the filling up of only 3s- and 3p- orbitals occurs. Therefore, in this period there are only two s- and six p- block elements. Since third period starts with Z = 11 and ends with Z = 18, therefore, elements with Z = 11 and Z = 12 are s- block elements. The next six elements with Z = 13 to 18 are p- block elements and belong to 13, 14, 15, 16, 17 and 18. Therefore, the element which will be in the third period and seventeenth group will have Z = 10 + 7 = 17.
7. Which element are named by
(a) Seaborg’s group
(b) Lawrence Berkeley Laboratory?
Ans:- (a) Seaborgium (Z = 106) (b) Lawrencium (Z = 103) and Berkelium (Z = 97).
8. Elements present in the same group are having similar chemical and physical properties. Why is it so?
Ans:- Elements present in the same group have similar electronic configuration and hence have similar physical and chemical properties.
9. What do you understand by the term ‘Ionic radius’ and ‘atomic radius’?
Ans:- Atomic radius literally means size of the atom. It is defined as one half the distance between the nuclei of two covalently bonded atoms of the same element in a molecule. For example, the internuclear distance between two chlorine atoms in chlorine molecule is 198 pm. Therefore, the covalent radius of chlorine atom is 198/2 = 99 pm.
In case of metal, atomic radius is called metallic radius. It can be defined as one half the distance between the two adjacent atoms in the crystal lattice. For example, the metallic radius of Cu is 128 pm.
Ionic radius means size of the ion. An ion can be cation or anion. It may be defined as the effective distance from the centre of the nucleus of the ion upto which it exerts its influence on its electronic cloud. The size of a cation is always smaller than that of the parent atom. On the other hand, the size of the anion is larger than the parent atom because the addition of one or more electrons decreases the effective nuclear charge.
For example, the size of the Ionic radius of Na+ is 95 pm while the atomic radius of Na is 186 pm. On the other hand, Ionic radius of fluoride ion is 136 pm whereas the atomic radius of Fluorine is only 72 pm.
10. Explain why there is variation in atomic radius in a group and period?
Ans:- Within a group, the atomic radius increases down the group. This is because a new energy shell is added at each succeeding element while the number of electrons in the valence shell remains to be the same. In other words, the electrons in the valence shell of each succeeding element lie farther and farther away from the nucleus. As a result, the force of attraction of the nucleus for the valence electrons decreases and hence the atomic size increases.
In contrast, the atomic size decreases as we move from left to right in a period. This is because that within a period the outer electrons remain in the same shell but the nuclear charge increases by one unit at each succeeding element. Due to this increased nuclear charge, the attraction of the nucleus for outer electrons increases and hence the atomic size decreases.
11 .Explain what is isoelectronic species? Give names of the species which will be isoelectronic species with each ion or atom given below.
1. Ar
2. Rb+
3. F−
4. Mg+
Ans:- Ions of different elements which have the same number of electrons but different magnitude of the nuclear charge are called isoelectronic species.
1. F– has 10 (9 + 1) electrons. Therefore, the species N3-, O2-, Ne, Na+, Mg2+ etc. each one have 10 electrons. Hence, they are isoelectronic of F–.
2. Ar has 18 electrons. Therefore, the species P3-, S2-, Cl–, K+, Ca2+ etc. each one have 18 electrons. Hence, they are isoelectronic of Ar.
3. Mg2+ has 10 (12 – 2) electrons. Therefore, the species N3-, O2-, Ne, Na+, Al3+ etc. each one have 10 electrons. Hence, they are isoelectronic of Mg2+
4. Rb+ has 36 (37 – 1) electrons. Therefore, the species Br–, Kr, Sr2+ etc. each one has 36 electrons. Hence, they are isoelectronic of Rb+.
12. Consider the accompanying species: N3−,O2−,F−,Na+, Mg2+, and Al3+
(i) What is similar in them?
(ii) Arrange them in the according to their increasing order of ionic radii.
Ans:- (i) Each one of these ions contains 10 electrons and hence all are isoelectronic ions.
(ii) The ionic radii of isoelectronic ions decrease with increase in the magnitude of the nuclear charge. All these ions have 10 electrons but their nuclear charge increase in the order : N3−(+7), O2−(+8), F− (+9), Na+ (+11), Mg2+(+12), and Al3+(+13). Therefore their ionic radii decrease in the order N3− > O2−> F−> Na+ > Mg2+> Al3+
13. Cation are having smaller radii then that of their parent atom and anion are having larger radii than their parent atom. Why?
Ans:- the Ionic radius of a cation is always smaller than the parent atom because the loss of one or more electrons increases the effective nuclear charge. As a result, the force of attraction of nucleus for the electrons increases and hence the ionic radii decrease. In contrast, the ionic radius of an anion is always larger than its parent atom because the addition of one or more electrons decreases the effective nuclear charge. As a result, the force of attraction of the nucleus for the electrons decreases and hence the ionic radii increase.
14. State significance of following terms:
1. “isolated gaseous atom”
2. “ground state”
in the definition of ionization enthalpy and electron gain enthalpy?
Ans:- 1. Ionization enthalpy is the minimum amount of energy required to remove the most loosely bonded electron from an isolated gaseous atom so as to convert it into a gaseous cation. The force with which an electron is attracted by the nucleus of an atom is appreciably affected by presence of other atoms within its molecule or in the neighborhood. Therefore, for the purpose of determination of ionization enthalpy, it is essential that these interatomic forces are minimum.
Further since it is not possible to isolate a single atom for the purpose of determination of its Ionisation enthalpy, therefore, the inter atomic distances are further reduced by carrying out the measurement at a low pressure of the gaseous atom. It is because of these reasons, that the term isolated gaseous atom has been included in the definition of ionization enthalpy.
2. Electron gain enthalpy is the energy released when an isolated gaseous atom in ground state accepts an extra electron to form the gaseous negative ion. The team isolated gaseous atom has already been explained above. The term ground state here means that the atom must be present in the most stable state or ground state. The reason being that when the isolated gaseous atom is in the exited state, lesser amount of energy will be released when it gets converted into gaseous atom after accepting an electron.
Therefore, for comparison purposes, the election gain enthalpies of gaseous atoms must be determined in their respective most stable state or ground state.
15. Determine ionization enthalpy of Hydrogen atom in Jmol−1.
Electron of hydrogen is having −2.18∗10−18J in ground state.
Ans:- Energy of the electron in the ground state = −2.18∗10−18J and energy of the electron at affinity = 0
The energy required to remove an electron in the ground state of hydrogen atom = 0 – (its energy in the ground state) = – (−2.18 × 10−18J) = 2.18 × 10−18J ∴ Ionisation enthalpy per mole of hydrogen atom = 2.18 × 10−18J × 6.022 × 1023 /1000 kJ = 1312.36 kJ mol–1 = 1312.36 × 103 J mol-1
16. For some elements of the 2nd period the arrangement according to their ionization enthalpy is given as follows
Li<B<C<O<N<F<Ne
Explain Why?
1. ΔiH for O is lower than ΔiH of N and F.
2. ΔiH for Be is higher than ΔiH than B?
Ans:-1. The electronic configuration of N (1s2 2s2 2px1 2py1 2pz1) in which 2p- orbitals are exactly half filled is more stable than the electronic configuration of O ((1s2 2s2 2px2 2py1 2pz1) in which 2p- orbitals are neither exactly half filled nor completely filled. Therefore, it is difficult to remove an electron from N than from O. As a result ΔiH of N is higher than that of O.
Further the electronic configuration of F is 1s2 2s2 2px2 2py2 2pz1. Because of higher nuclear charge (+9), the first Ionisation enthalpy of F is higher than that of O. Further, the effect of increased nuclear charge outweighs the effect of stability due to exactly half filled orbitals, therefore, the ΔiH of N and O are lower than that of F.
2. The Ionisation enthalpy, among other things, depends upon the type of electron to be removed from the same principal shell. In case 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 2s electron. Consequently, ΔiH for Be is higher than ΔiH than B.
17. Explain why the 1st ionization enthalpy of magnesium is higher than 1st ionization enthalpy of sodium but the 2nd ionization enthalpy of magnesium is lower than 2nd ionization enthalpy of sodium?
Ans:- The electronic configuration of Na and Mg are 1s2 2s2 2p6 3s1 and 1s2 2s2 2p6 3s2 respectively. Thus, the first electron in both the cases has to be removed from the 3s- orbital but the nuclear charge of Na(+11) is lower than that of Mg(+12), therefore, the first ionisation energy of sodium is lower than that of magnesium.
After the loss of first electron, the electronic configuration of Na+ is 1s2 2s2 2p6 . Here, the election is to be removed from inert gas configuration which is very stable and hence removal of second electron from sodium is very difficult. However, in case of Magnesium, after the loss of first electron, the electronic configuration of Mg2+ is 1s2 2s2 2p6 3s1 Here, the electron is to be removed from 3s orbital which is much easier than to remove an electron from inert gas configuration. Therefore, the second ionisation enthalpy of sodium is higher than that of magnesium.
18. State the factors because of which in elements of main group the ionization enthalpy decreases when we move down the group.
Ans:- Within the main group of elements, the ionisation enthalpy decreases regularly as we move down the group due to the following two factors.
(i) Atomic size. On moving down the group, the atomic size increases generally due to the addition of one new principal energy shell at each succeeding element. As a result, the distance of the valence electrons from the nucleus increases. Consequently the force of attraction of the nucleus for the valence electrons decreases and hence the ionisation enthalpy decreases.
(ii) Screening effects. With the addition of new shells, the number of inner electrons shells which shield the valence electrons increases. In other words, the shielding effect increases and the force of attraction of the nucleus for the valence electrons further decreases and hence the ionisation enthalpy decreases.
19 For the elements of group 13 the values of 1st ionization enthalpy is given below:
| B | Al | Ga | In | Tl |
| 801 | 577 | 579 | 558 | 589 |
Explain the ‘deviation from the general trend’?
Ans:- On moving down the group 13 from B to Al, the ionisation enthalpy decreases as expected due to an increase in atomic size and screening effect which outweigh the effect of increased nuclear charge. However, ΔiH of Ga is only slightly higher than that of Al while that of Tl is much higher than those of Al, Ga and In due to following reasons.
Al follows immediately after s-block elements while Ga and In follow after d- block elements Tl d- and f- block elements. These extra d- and f- electrons do not shield the outer shell electrons from the nucleus and hence larger amount of energy is needed for their removal. This explains why Ga has higher ionisation enthalpy than Al.
Further on moving down the group from Ga to In, the increased shielding effect due to the presence of additional 4d electrons outweighs the effect of increased nuclear charge and hence the first ionisation enthalpy of In is lower than that of Ga.
Thereafter, the effect of increased nuclear charge of Tl outweighs the shielding effect due to the presence of 4f and 5d electrons and hence the first ionisation enthalpy of Tl is higher than that of In.
20. Find out which of the pair given below will have high negative electron affinity?
(a) F or Cl (b) O or F
Ans:- (a) In general, the electron gain enthalpy becomes less negative on moving down the group. But the electron gain enthalpy of chlorine is more negative than that of Fluorine. The reason for this deviation is the small size of F atom. Due to its small size, the electron – electron repulsion in 2p subshell are larger and hence the incoming electron is not accepted with the same ease as in the case of chlorine.
(b) Both O and F lie in 2nd period. As we move from O to F, the atomic size decreases and the nuclear charge increases. Both these factors tend to increase the attraction of the nucleus for the incoming electron and hence electron gain enthalpy becomes more negative. Further, gain of one electron by F gives F– ion which has stable inert gas configuration while the gain of one electron by O gives O– ion which gives not stable inert gas configuration. As a result the energy released is much higher in going from F → F– than in going from O → O–. In other words, electron gain enthalpy of F is much more negative than that of O.
21. What is electron gain enthalpy of O(oxygen) atoms?
1.Positive
2.More negative
3.Less negative
Justify the answer.
Ans:- When an electron is added to O atom to form O– ion, energy is released. Thus, first electron gain enthalpy of O is negative. O(g) + e —– O– (g) ; ΔegH = – 141 KJmol-1
But when another electron is added to O– to form O2– ion, energy is required to overcome the strong electrostatic repulsion between the negatively charged O– ion and the second electron being added. Thus the second electron gain enthalpy of oxygen is positive. As O– (g) + e– (g) → O2– (g); ΔegH = + 780 KJmol-1
22. State the difference between the terms electron affinity and electronegativity.
Ans:- Both electron gain enthalpy and electronegativity refer to the tendency of the atom of an element to attract electrons. Whereas electron gain enthalpy refers to the tendency of an isolated gaseous atom to accept an additional electron to form a negative ion, electronegativity refers to the tendency of the atom of an element to attract the shared pair of electrons towards it in a covalent bond.
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:- The electronegativity of any given atom is not constant. Therefore, the statement that the electronegativity of N atom on Pauling scale is 3.0 in all nitrogen compounds is wrong. Actually electronegativity varies with the state of hybridisation and oxidation state of the element. The electronegativity increases as the percentage of s- character of a hybrid orbital increases or the oxidation state of the element increases.
24.Describe the theory associated with the radius of an atom as it
(a) gains an electron
(b) loses an electron
Ans:- (a) Gains of electrons. When a neutral atom gains an electron to form an anion, its radius increases. The reason being that the number of electrons in the anion increases while its nuclear charge remains same as the parent atom. Since the same nuclear charge now attracts greater number of electrons, therefore, the force of attraction of the nucleus on the electrons of all the shells decreases and hence the electron cloud expands.
(b) Loss of electrons. When a neutral atom loses one electron to form a cation, its atomic radius decreases. The reason being that the number of electrons in the cation decreases while its nuclear charge remains the same as the parent atom. Since the same nuclear charge now attracts lesser number of electrons. Therefore, the force of attraction of the nucleus on the electrons of all the shells increases and hence the size of cation decreases.
25. Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
Ans:- Ionisation enthalpy, among other things, depends upon the electronic configuration and nuclear charge. Since the isotopes of an element have the same electronic configuration and nuclear charge, they are expected to have same ionisation enthalpy.
26. What are the major differences between metals and non-metals?
Ans:- Elements which have a strong tendency to loss electrons to form cations are called metals while those which have a strong tendency to accept electrons to form anions are called non- metals. Thus metals are strong reducing agents, they have low ionisation enthalpy, have less negative electron gain enthalpy, low electronegativity, form basic oxides and Ionic compounds.
Non- metals, on the other hand, are strong oxidising agents, they have high ionisation enthalpy, have high negative electron gain enthalpy, high electronegativity, form acidic oxides and covalent compounds.
27. Use the periodic table to answer the following questions.
(a) Identify an element with five electrons in the outer subshell.
(b) Identify an element that would tend to lose two electrons.
(c) Identify an element that would tend to gain two electrons.
(d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature.
Ans:- (a) The general electronic configuration of the elements having five electrons in the outer subshell is ns2 np3. This electronic configuration is of elements of group 17 as of halogens and they are F, Cl, Br, I, At.
(b) The elements which have tendency to loss two electrons must have two electrons in the valence shell. Therefore, their general electronic configuration should be ns2. This configuration is the characteristics of Group- 2 elements and they are Mg, Ca, Sr and Ba.
(c) The elements which have tendency to accept two electrons must have four electrons in the valence shell. Therefore, their general electronic configuration is ns2 np4. This configuration is the characteristics of Group- 16 elements and they are O, S, Se, Te and Po.
(d) A metal which is liquid at room temperature is mercury. It is a transition metal and belongs to group 12. A non- metal which is a gas at room temperature is nitrogen (group 15), oxygen (group 16), Fluorine, chlorine (group 17) and inert gases (group 18). A non- metal which is liquid at room temperature is bromine (group 17).
28. The increasing order of reactivity among group 1 elements is Li < Na < K < Rb <Cs whereas that among group 17 elements is F > CI > Br > I. Explain.
Ans:- The elements of group 1 have only one electron in their respective valence shells and thus have a strong tendency to loss this electron. This tendency to loss electrons, in turn, depends upon the Ionisation enthalpy. Since the Ionisation enthalpy decreases down the group, therefore, the reactivity of group 1 elements increases as Li < Na < K < Rb < Cs.
In contrast, the elements of group 17 ,have 7 electrons in their respective valence shells and thus have a strong tendency to accept one more electron. This tendency to accept electron depends upon their electrode potentials. Since the electrode potentials of group 17 elements decrease in the order: F (+ 2.87 V), > Cl (+1.36 V) > Br (+ 1.08 V) > I(+ 0.53 V). Therefore, their reactivities also decrease in the same order: F > Cl > Br > I.
29. Write the general outer electronic configuration of s-, p-, d- and f- block elements.
Ans:- (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 ns0 – 2 where n = 4 – 7. (iv) f – block elements: (n -2)f1-14 (n-1)d 0-2 ns2 where n = 6-7.
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.
Ans:- (i) n = 3 suggests that the element belongs to third period since the last electron enters the p- orbital, therefore, the given element is a p- block element. Further, since the valence shell contains 6 electrons, therefore, group number of the element = 10 + 6 = 16. The complete electronic configuration of the element is 1s2 2s2 2p6 3s2 3p4 and the element is sulphur.
(ii) n = 4 suggests that the element lies in the 4th period. Since the d- orbitals are incomplete, therefore, it is d- block element. The group number of the element = no. of d- electrons + no. of s- electrons = 2 + 2 = 4. Thus, the element lies in group 4 and 4th period. The complete electronic configuration of the element is 1s2 2s2 2p6 3s2 3p6 3d2 4s2 and the element is Titanium.
(iii) n = 6 suggests that the element lies in 6th period. Since, the last electron goes to the f- orbital, therefore, the element is a f- block element. All f- block elements lie in group 3. The complete electronic configuration of the element is [Xe] 4f7 5d1 6s2. The atomic number of the element 54 + 7 + 1 + 2 = 64 and the element is Gd (Gadolinium).
31. The first (ΔiH1) and the second (ΔiH) ionization enthalpies (in kJ mol–1) and the (ΔegH) electron gain enthalpy (in kJ mol–1) of a few elements are given below:
| Elements | ΔiH | ΔiH | 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)?
Ans:- (a) The element V has highest first Ionisation enthalpy (ΔiH) and positive electron gain enthalpy (ΔegH) hence it is the least reactive element. Since inert gases have positive electron gain enthalpy. Therefore, the element V must be an inert gas and this is He.
(b) The element ll which has the least first Ionisation enthalpy (ΔiH1) and a low negative electron gain enthalpy (ΔegH) is the most reactive metal and this is K.
(c) The element lll which has high first Ionisation enthalpy (ΔiH1) and a very high electron gain enthalpy (ΔegH) is the most reactive non metal and this is F.
(d) The element IV has a high negative electron gain enthalpy (ΔegH) but not so high first Ionisation enthalpy (ΔiH1). Therefore, it is the least reactive non metal and it is Iodine.
(e) The element VI has low first Ionisation enthalpy (ΔiH1) but higher than that of alkali metals. Therefore, it appears that the element is an alkaline earth metal and hence will form binary halides of the formula MX2 and this metal appears Mg.
(f) The element l has low first Ionisation enthalpy (ΔiH1) but a very high second Ionisation enthalpy therefore, it must be an alkali metal which can form a predominantly stable covalent halide of the formula MX (X = halogen) but the alkali metal must be least reactive then this element should be Li.
32. Predict the formula 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) Phosphorus and fluorine (f) Element 71 and fluorine
Ans:- (a) Lithium is an alkali metal (group 1). It has only one electron in the valence shell, therefore, its valence is 1. Oxygen is a group 16 element with a valence of 2. Therefore, formula of the compound would be Li2O.
(b) Magnesium is an alkaline earth metal (group 2) and hence has a valence of 2. Nitrogen is a group 15 element with a valence of 3. Thus, the formula of the compound would be Mg3N2.
(c) Aluminium is a group 13 element with a valence of 3 while Iodine is a halogen (group 17) with a valence of 1. Therefore, the formula of the compound would be AlI3
(d) Silicon is a group 14 element with a valence of 4 while Oxygen is a group 16 with a valence of 2. Therefore, the formula of the compound will be SiO2.
(e) Phosphorous is a group 15 element with a valence of 3 or 5 while Fluorine is a group 17 element with a valence of 1. Hence the formula of the compound would be PF3 or PF5.
(f) Element with atomic number 71 is a Lanthanoid called lutetium (Lu). Its common valence is 3. Fluorine has its valence 1. Therefore the formula of the compound will be LuF3.
33. In the modern periodic table, the period indicates the value of:
(a) Atomic number
(b) Atomic mass
(c) Principal quantum number
(d) Azimuthal quantum number.
Ans:- In the Modern periodic table, each period begins with the filling of a new shell. Therefore, the periods indicate the value of principal quantum number. Hence the option (c) is correct.
(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell.
(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.
Ans:- Statement (b) is incorrect while other statements are correct. The correct statement (b) is : the d- block has 10 columns, because a maximum of 10 electrons can occupy all the d- orbitals in a d- subshell.
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.
Ans:- Nuclear mass does not affect the valence shell because nucleus consists of protons and neutrons. Whereas protons i.e., nuclear charge affects the valence shell but neutrons do not. Thus option (c) is wrong.
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
Ans:- The size of the isoelectronic ions depends upon the nuclear charge (Z). As the nuclear charge increases the size decreases. Therefore, the correct option will be (a) nuclear charge (Z).
37. Which one 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 electron 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.
Ans:- Statement (d) is incorrect. The correct statement is: Removal of electron from orbitals bearing lower n value is difficult than from orbital having higher n value. All other statements are correct.
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
Ans:- 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. Hence option (d) is correct.
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
Ans:- 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. Hence the correct option is (c).
40. Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is:
(a) F > Cl > O > N (b) F > O > Cl > N
(c) Cl > F > O > N (d) O > F > N > Cl
Ans:- 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 stronger oxidising agent than Cl. Thus overall decreasing order of oxidising power is: F > O > Cl > N. Hence the correct option is (b).
Conclusion of Ncert Solution Chapter 3 Chemistry 11th
As we know that Classification of Elements and Periodicity of Properties is the 3rd chapter of chemistry class 11th. We have included all the questions provided in the exercise of this chapter. There are 40 questions to solve. We have solved all the questions in simple manner with suitable reasons and proper examples. We hope that this NCERT Notes will be very helpful during your preparation for examinations like NEET, Jee(mains), IIT advance and other entrance exams. We have a kind request to you all that please share this article among your friends and favourites as far as possible. Thank you.