NCERT Solutions Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties: Download Questions & Answers PDF

A 3.1 The basic theme of organisation of elements in the periodic table is to simplify, systematize and classify the elements based on their similarities in the properties. This arrangement makes it easier for us to group the elements and makes it less confusing to present the periodic table.

A 3.2 Mendeleev arranged the elements horizontally, in periods and vertically, in groups in the increasing order of their atomic masses. This classification was done in such a way that the elements with similar properties fall in the same group.

A 3.3 Mendeleev’s periodic law classifies the elements based on their atomic weight. Modern periodic law classifies the elements based on their atomic number.

A 3.4 The sixth period means that the highest principle quantum number, = 6. The subsequent periods comprise of 8, 8, 18, 18 and 32 elements in the same order. The sixth period comprises of 32 elements where electrons enter the 6s, 4f, 5d and 6p orbitals. Number of orbitals present in 6s=1. 4f= 7, 5d= 5 and 6p= 3. The total number of orbitals= 16. A total of 32 electrons can be filled in these 32 orbitals; therefore, sixth period should have a maximum of 32 elements.

A 3.5 Elements with atomic numbers from Z = 87 to Z = 114 are present in the 7th period of the periodic table. So, this elements lies in the Period – 7 and Group -14 of Block-p.

A 3.6 The element is chlorine (Cl) with atomic number (Z) = 17.

A 3.7

• (i) Lawrencium (Lr) with atomic number (z) = 103
• (ii) Seaborgium (Sg) with atomic number (z) = 106.

A 3.8 The elements in a group have same number of valence electrons and hence have similar physical and chemical properties.

A 3.9

• Atomic radius: It is the distance from the centre of nucleus to the outer most shell of electrons in the atom of any element. It incorporates both the covalent, metallic radius or van der Waal’s radius depending on whether the element is a non-metal or a metal.
• Ionic radius: It is the distance between the centre of the nucleus of an ion up to the point where it exerts its influence on the electron cloud of a canton or anion.

A 3.10 Across a period, the atomic radii decrease from left to right due to increase in effective nuclear charge from left to right across a period Within a group, atomic radius increases down the group due to continuous increases in the number of electronic shells or orbit numbers in the structure of atoms of the elements down a group.

A 3.11 Species (atoms/ions) which have same number of electrons are called isoelectronic species. The isoelectronic species out of the given atoms/ions are:

• (i) Na⁺ is isoelectronic to F^- 
• (ii) K⁺ is isoelectronic to Ar
• (iii) Na⁺ is isoelectronic to Mg²⁺
• (iv) Sr²⁺ is isoelectronic to Rb⁺

A 3.12 (a) All of them have 10 electrons each and are isoelectronic in nature. (b) In isoelectronic species, higher the nuclear charge, smaller will be the atomic or ionic radius. Al³⁺< Mg²⁺< Na⁺< F^–< O^2-< N^3-

A 3.13 A cation is obtained by removing an electron from the outermost shell. The removal of electron(s) results in decrease of the size of the resulting ion than the parent atom because it has fewer electrons while its nuclearcharge remains the same. Anions are obtained by addition of electron(s) in the outermost shell. This results in increased repulsion among the electrons and a decrease in effective nuclear charge.

A 3.14 

• Significance of the term ‘isolated gaseous atom’. The atoms in the gaseous state are far separated in the sense that they do not have any mutual attractive and repulsive interactions. These are therefore regarded as isolated atoms. In this state, the value of ionization enthalpy and electron gain enthalpy are not influenced by the presence of the other atoms. It is not possible to express these when the atoms are in the liquid or solid state due to the presence of inter atomic forces.
• Significance of the term ‘ground state’. Ground state of the atom represents the normal – energy state of an atom. It means electrons in a particular atom are in the lowest energy state and they neither lose nor gain electron. Both ionisation enthalpy and electron gain enthalpy are generally expressed with respect to the ground state ofan atom only.

A 3.15  The ionisation enthalpy is for 1 mole atoms. Therefore, ground state energy of theatoms may be expressed as E_{ground state} = (– 2.18 x 10^-18 J) x(6.022 x 10²³ mol^-1)      = –1.312 x 10⁶ J mol^-1 Ionisation enthalpy =E_∞–E_{ground state}        = 0–(–1.312 x 10⁶mol^-1)
       = 1.312 x 10⁶ J mol^-1.

Among the second period elements, the actual ionization enthalpies are in the order: Li
Explain why
(i) Be has higher ∆_i H than B?
(ii) O has lower ∆_i H than N and F?                                                                 
Answer: (i) In the electronic configuration of Be (1s ² 2s ²) the outermost electron is present in 2s-orbital while in B (1s ² 2s ² 2p ¹) it is present in 2p-orbital. Since 2s – electrons are more strongly attracted by the nucleus than 2p-electrons, therefore, lesser amount of energy is required to knock out a 2p-electron than a 2s – electron. Therefore, Be has higher ∆i H than that of B.
(ii) The electronic configuration of oxygen is 1s ² 2s ² 2p ⁴. The 2p orbital contains 4 electrons out of which 2 are present in the same 2p-orbital. Due to this, the electron repulsion increases. N has stable half-filled configuration. F has higher effective nuclear charge. Hence, the ionization enthalpy of O is lower than that of N and F.

A 3.17 Electronic configuration of Na and Mg are Na = 1s² 2s² 2p⁶ 3s¹
Mg = 1s² 2s² 2p⁶ 3s² First electron in both cases has to be removed from 3s-orbital but the nuclear charge of Na (+ 11) is lower than that of Mg (+ 12) therefore first ionization energy of sodium is lower than that of magnesium.
After the loss of first electron, the electronic configuration of Na⁺ = 1s² 2s² 2p⁶
Mg⁺ = 1s² 2s² 2p⁶ 3s¹ Here electron is to be removed from inert (neon) gas configuration which is very stable and hence removal of second electron from Na⁺ requires more energy in comparison to Mg.
Therefore, second ionization enthalpy of sodium is higher than that of magnesium.

A 3.18 Atomic size: With the increase in atomic size, the number of electron shells increase. Therefore, the force that binds the electrons with the nucleus decreases. The ionization enthalpy thus decreases with the increase in atomic size.
Screening or shielding effect of inner shell electron: With the addition of new shells, the number of inner electron shells which shield the valence electrons increases. As a result, the force of attraction of the nucleus for the valence electrons further decreases and hence the ionization enthalpy decreases.

B        Al       Ga       In        Tl
801    577     579     558       589

A 3.19 On moving down the group, the ionization enthalpy decreases. This is true for B and Al due to the bigger size of Al. The ionization enthalpy of Ga is unexpectedly higher than Al because Ga contains 10d electrons in inner shell whose shielding is less effective than that of s and p electrons. The outer electron is held fairly strongly by the nucleus. The ionization enthalpy increases slightly. A similar increase is observed from In to Tl due to presence of 14f electrons in the inner shell of Tl which have poor screening effect.

A 3.20 (i) The electronic configuration of O and F are:                                     O: 1s² 2s² 2p⁴ F: 1s² 2s² 2p⁵ F would readily accept an electron to attain the noble gas electronic configuration of Ne as compared to O. Hence, F has a more negative electron gain enthalpy. (ii) The electronic configuration of Cl and F are:                                     Cl: [Ne] 3s² 3p⁵ F: 1s² 2s² 2p⁵ Due to a larger size, Cl can accommodate an additional electron. F, on the other hand, has a smaller atomic size and therefore the accommodation of another electron increases the inter-electronic repulsion thereby, decreasing the stability. Therefore, Cl has a higher negative electron gain enthalpy.

A 3.21 When an electron is added to oxygen atom, energy is released to form a negative ion. This enthalpy change called the first electron gain enthalpy is thus negative. On adding another electron to the O⁻ ion, it experiences repulsion from the anion due to which the instability of the ion increases. Thus, the addition of the second electron requires energy due to which the second electron gain enthalpy is positive. O + e⁻ → O⁻     ΔH₁ = -141KJ O⁻ + e⁻ → O²⁻   ΔH₂ = 780 KJ

A 3.22 Electron gain enthalpy refers to tendency of an isolated gaseous atom to accept an additional electron to form a negative ion. Whereas electronegativity refers to tendency of the atom of an element to attract shared pair of electrons towards it in a covalent bond.

A 3.23 On Pauling scale, the electronegativity of nitrogen is 3.0 which indicates that it is quite electronegative. But it is not correct to say that the electronegativity of nitrogen in all the compounds is 3. It varies with the atoms that N is bonded with. For example, in NH₃, N has a different electronegativity than N in NO₂.

A 3.24 (a) Gain of an electron leads to the formation of an anion. The size of an anion will be larger than that of the parent atom because the addition of one or more electrons would result in increased repulsion among electrons and decrease in effective nuclear charge.

(b) Loss of an electron from an atom results in the formation of a cation. A cation is smaller than its parent atom because it has former electrons while its nuclear charge remains the same.

A 3.25 Isotopes have same number of electrons and protons, only the number of neutrons is different. The atomic number remains the same and only atomic mass differs. Hence the ionization energy of the isotopes of a chemical element remains the same.

A 3.26 

A 3.27 (a) Element belonging to nitrogen family (group 15) e.g., nitrogen.
(b) Element belonging to alkaline earth family (group 2) e.g., magnesium.
(c) Element belonging to oxygen family (group 16) e.g., oxygen

A 3.28 The elements of Group I have only one electron in their respective valence shells and thus have a strong tendency to lose this electron. The tendency to lose electrons in turn, depends upon the ionization enthalpy. Since the ionization enthalpy decreases down the group therefore, the reactivity of group 1 elements increases in the order Li < Na < K < Rb < Cs. In contrast, the elements of group 17 have seven electrons in their respective valence shells and thus have strong tendency to accept one more electron to make stable configuration. So for group 17, the electron gain enthalpy and electronegativity decreases down the group and thus the reactivity also decreases.

A 3.29 (i) s-Block elements: ns^1-2 where n = 2-7.
(ii) p-Block elements: ns² np^1-6 where n = 2-6.
(iii) d-Block elements: (n-1)d^1-10 ns ^0-2  where n = 4-7.
(iv) f-Block elements: (n-2)f^0-14 (n-1)d^0-1 ns²where n = 6-7.

A 3.30 (i) For n = 3, the element belong to 3rd period, p-block element.
The electronic configuration is =1s² 2s² 2p⁶ 3s² 3p⁴. The element name is sulphur.
(ii) For n = 4, the element belongs to 4th period and since the valence shell has 4 electrons it belongs to group 4.
Electronic configuration= 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s², and the element name is Titanium (T_i).
(iii) For n = 6, the element belongs to 6th period. Last electron goes to the f-orbital, element is from f-block. It belongs to group = 3
The element is gadolinium (z = 64)
Complete electronic configuration =[Xe] 4 f⁷ 5d¹ 6s².

A 3.31 

(a) The element V has highest first ionization enthalpy (∆_iH₁) and positive electron gain enthalpy (∆_egH) and hence it is the least reactive element. Since inert gases have positive ∆_egH, therefore, the element-V must be an inert gas. The values of ∆_iH₁, ∆_iH₂ and ∆_egH match that of He.
(b) The element II which has the least first ionization enthalpy (∆_iH₁) and a low negative electron gain enthalpy (∆_egH) is the most reactive metal. The values of ∆_iH₁, ∆_iH₂ and ∆_egH match that of K (potassium).
(c) The element III which has high first ionization enthalpy (∆_iH₁) and a very high negative electron gain enthalpy (∆_egH) is the most reactive non-metal. The values of ∆_iH₁, ∆_iH₂ and ∆_egH match that of F (fluorine).
(d) The element IV has a high negative electron gain enthalpy (∆_egH) but not so high first ionization enthalpy (∆_egH). Therefore, it is the least reactive non-metal. The values of ∆_iH₁, ∆_iH₂ and ∆_egH match that of I (Iodine).
(e) The element VI has low first ionization enthalpy (∆_iH₁) but higher than that of alkali metals. Therefore, it appears that the element is an alkaline earth metal and hence will form binary halide of the formula MX₂ (where X = halogen). The values of ∆_i H₁, ∆_iH₂ and ∆_egH match that of Mg (magnesium).
(f) The element I has low first ionization (∆_iH₁) but a very high second ionization enthalpy (∆_iH₂), therefore, it must be an alkali metal. Since the metal forms a predominantly stable covalent halide of the formula MX (X = halogen), therefore, the alkali metal must be least reactive. The values of ∆_iH₁, ∆_iH₂ and ∆_egH match that of Li (lithium).

A 3.32 (a) LiO₂ (Lithium oxide) (b) Mg₃N₂ (Magnesium nitride) (c) AlI₃ (Aluminium iodide) (d) SiO₂ (Silicon dioxide) (e) Phosphorous pentafluoride (f) Z = 71 The element is Lutenium (Lu). Electronic configuration [Xe] 4 f7 5d1 6s2. With fluorine it will form a binary compound = LuF₃.

A 3.33 In the modern periodic table, each period begins with the filling of a new shell. Therefore, the period indicates the value of principal quantum number. Thus, option (c) is correct.

A 3.34 Statement (b) is incorrect.  

A 3.35 (c) Nuclear mass.

A 3.36 (a) Nuclear charge (Z).

A 3.37 (d) is incorrect.

3.38. Considering the elements B, Al, Mg and K, the correct order of their metallic character is: (a) B>Al>Mg>K (b) Al>Mg>B>K (c) Mg>Al>K>B (d) K>Mg>Al>B                                                                                                       

Answer: In a period, metallic character decreases as we move from left to right. Therefore, metallic character of K, Mg and Al decreases in the order: K > Mg > Al. However, within a group, the metallic character, increases from top to bottom. Thus, Al is more metallic than B. Therefore, the correct sequence of decreasing metallic character is: K > Mg > Al > B, i.e., option (d) is correct.

3.39. Considering the elements B, C, N, F and Si, the correct order of their non-metallic character is: (a) B>C>Si>N>F (b) Si>C>B>N>F (c) F>N>C>B>Si (d) F>N>C>Si>B                                                                                                       
Answer: In a period, the non-metallic character increases from left to right. Thus, among B, C, N and F, non-metallic character decreases in the order: F > N > C > B. However, within a group, non-metallic character decreases from top to bottom. Thus, C is more non-metallic than Si. Therefore, the correct sequence of decreasing non-metallic character is: F > N > C > B > Si, i.e., option (c) is correct.

3.40. Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidising property is:                                                           
(a) F > Cl> O > N (b) F > O > Cl> N (c) Cl> F > O > N (d) O > F > N > Cl
Answer: Within a period, the oxidising character increases from left to right. Therefore, among F, O and N, oxidising power decreases in the order: F > O > N. However, within a group, oxidising power decreases from top to bottom. Thus, F is a stronger oxidising agent than Cl. Further because O is more electronegative than Cl, therefore, O is a stronger oxidising agent than Cl. Thus, overall decreasing order of oxidising power is: F > O > Cl > N, i.e., option (b) is correct.