# Classification of Elements and Periodicity in Properties Textbook Solutions

Question 1.
What is the basic theme of organisation in the periodic table?

Solution.
By 1865, number of identified elements was 63 and with such a large number of elements it was very difficult to study individually the chemistry of all these elements and their innumerable compounds individually.

To ease out this problem, scientists searched for a systematic way to organize their knowledge by classifying the elements. Here comes the basis of periodic table. Various elements have now been divided into different groups on the basis of similarities in chemical properties. This has made the study simpler as now the properties of elements are studied in groups rather than individually.

Question 2.
Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to it?

Solution.
Mendeleev arranged elements in horizontal rows and vertical columns of a table in order of their increasing atomic weights in such a way that the elements with similar properties occupied the same vertical column or group.

He realized that some of the elements did not fit in with his scheme of classification if the order of atomic weight was strictly followed. So, he did not stick to his criteria, he ignored the order of atomic weights, thinking that the atomic measurements might be incorrect, and placed the elements with similar properties together.

Question 3.
What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?

Solution.
According to Mendeleev’s periodic law, the physical and chemical properties of the elements are periodic functions of their atomic masses, but according to modern periodic law, the physical and chemical properties of the elements are periodic functions of their atomic numbers.

Question 4.
On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.

Solution.
In the sixth period, the orbitals to be filled are 6s, 4f, 5d and 6p. The complete filling of these orbitals require 2 + 14 + 10 + 6 = 32 electrons, hence, the sixth period of the periodic table should have 32 elements.

Question 5.
In terms of period and group where would you locate the element with Z= 114?

Solution.
The outermost electronic configuration of element (114Z) is
[Rn] 5f14 6d107s27p2. It has n = 7, so period → 7 It belongs to p-block so,
group number = 10 + 4 = 14.

Question 6.
Write the atomic number of the element present in the third period and seventeenth group of the periodic table.

Solution.
17Cl → It belongs to the third period. So, outermost shell is n = 3. Its configuration is [Ne] 3s23p5.
Therefore, its atomic number = 17.

Question 7.

Which element do you think would have been named by

• (i) Lawrence Berkeley Laboratory
• (ii) Seaborg’s group?

Solution.

• (i) Lawrencium  →  103Lr
• (ii) Seaborgium  →  106Sg

Question 8.
Why do elements in the same group have similar physical and chemical properties?

Solution.
In a group, the chemical properties of the elements remain nearly the same due to same valence shell configuration.

Question 9.

Solution.
Atomic radius: It is the distance between the centre of the nucleus and outermost shell where electrons are present.
Ionic radius : It is the distance between the nucleus and outermost shell of an ion.

Question 10.
How do atomic radius vary in a period and in a group? How do you explain the variation?

Solution.
Variation of atomic radii in a period :
As we move from left to right across a period, there is regular decrease in atomic radii of representative elements. This can be explained on the basis of effective nuclear charge which increases gradually in a period, i.c, electron cloud is attracted more strongly towards nucleus as the effective nuclear charge becomes more and more along the period. The increased force of attraction brings contraction in size.

Variation of atomic radii in a group : Atomic radii in a group increase as the atomic number increases. The increase in size is due to extra energy shell which outweighs the effect of increased nuclear charge.

Question 11.

What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.

• (i) F
• (ii) Ar
• (iii) Mg2+
• (iv) Rb+

Solution.

Isoelectronic species are those which have same number of electrons.

• (i) F  has 10 electrons. Therefore, the species N3+, O2-, Ne, Na+, Mg2+, etc., are isoelectronic with F .
• (ii) Ar has 18 electrons. Therefore, the species P3-, S2-, Cl, K+, Ca2+, etc., are isoelectronic with Ar
• (iii) Mg2+ has 10 electrons. Therefore, the species N3-, O2-, Ne, Na+, etc., are isoelectronic with Mg2+.
• (iv) Rb+ has 36 electrons. Therefore, the species Br, Kr, Sr2+, etc., are isoelectronic with Rb+

Question 12.

Consider the following species :

N3-, O2-, F, Na+, Mg2+ and Al3+

• (a) What is common in them?
• (b) Arrange them in the order of increasing ionic radii.

Solution.

• (a) All these are isoelectronic species as they are having same number of electrons i.c., 10.
• (b) As Z/e decreases, size increases so, order should be,

N3- > O2- > F  > Na+ > Mg2+ > Al3+.

Question 13.
Explain why cations are smaller and anions larger in radii than their parent atoms?

Solution.
A cation is smaller than its parent atom because it has fewer electrons while its nuclear charge remains the same. 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 the electrons and a decrease in effective nuclear charge.

Question 14.
What is the significance of the terms – ‘isolated gaseous atom’and’ground state’while defining the ionization enthalpy and electron gain enthalpy?
[Hint: Requirements for comparison purposes.]
Solution.
In the definition of ionization enthalpy and electron gain enthalpy, isolated gaseous atom is required for comparison purposes. Ionization energy is the minimum amount of energy required to remove most loosely bound electron from an isolated atom in the gaseous state of an element so as to convert it into gaseous monovalent positive ion.

Electron gain enthalpy is the energy change accompanying the process of adding an electron to a gaseous isolated atom to convert it into a negative ion, i.e., a monovalent anion.

Both the above mentioned processes are carried out on an isolated gaseous atom, which in turn is obtained from either the excitation of a ground state atom (in case the element is monoatomic) or atomisation of polyatomic elements.

The force with which an electron is attracted by the nucleus is appreciably affected by presence of other atoms in the neighbourhood. Since in the gaseous state the atoms are widely separated, therefore these interatomic forces are minimum.
The term ground state means that the atom must be present in the most stable state.

Question 15.
Energy of an electron in the ground state of the hydrogen atom is -2.18 x 10-18 J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol-1.
[Hint: Apply the idea of mole concept to derive the answer]
Solution.
Energy of the electron in the ground state of H-atom, E1 = -2.18 x 10-18 J
Ionisation energy = E – En
Ionisation enthalpy per mole of atomic hydrogen = (E – E1)NA
= [0 – (- 2.18 x 10-18)] x 6.023 x 1023
= 2.18 x 6.023 x 105 J/mol = 13.13 x 105 j/mol
= 1.313 x 106 J/mol

Question 16.

Among the second period elements the actual ionization enthalpies are in the order Li <B <Be< C< 0 < N< F < Ne. Explain why?

• (i) Be has higher ∆iH than B
• (ii) O has lower ∆iH than N and F?

Solution.

(i) An s-electron is attracted to the nucleus more than a p-electron. In beryllium, the electron removed during the ionization is an s-electron whereas the electron removed during ionization of boron is a p-electron.

The penetration of a 2s-electron to the nucleus is more than that of a 2p-electron; hence the 2 p electron of boron is more shielded from the nucleus by the inner core of electrons than the 2s electrons of beryllium.

Therefore, it is easier to remove the 2p-electron from boron as compared to the removal of a 2s-electron from beryllium. Thus, boron has a smaller first ionization enthalpy than beryllium.

(ii) O has lower ionisation energy than N because N (1s2 2s2 ${ 2p }_{ x }^{ 1 }\quad { 2p }_{ y }^{ 1 }\quad { 2p }_{ z }^{ 1 }$) has extra stable electronic configuration whereas O (1s2 2s2 ${ 2p }_{ x }^{ 1 }\quad { 2p }_{ y }^{ 1 }\quad { 2p }_{ z }^{ 1 }$) does not. O has lower ionisation energy than F because O has larger size than F.

Question 17.

How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?

Solution.

The electronic configurations of Na and Mg are :

• Na : 1s2 2s2 2p6 3s1
• Na+ : 1s2 2s2 2p6
• Mg : 1s2 2s2 2p6 3s2
• Mg+ : 1s2 2s2 2p6 3s1
• Mg2+ : 1s2 2s2 2p6

The 1st ionization enthalpy of Na is lesser than that of Mg because Mg has an extra stable configuration and smaller size, so, a larger amount of energy would be required to remove an electron from the 3s orbital, which has a pair of electrons.

The 2nd ionization enthalpy of Na is more than that of Mg because Na+ has an extra stable configuration (complete octet), whereas Mg+ does not have an extra stable configuration.

Question 18.
What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?

Solution.
We have to consider two factors :
(i) the attraction of electrons towards the nucleus, and

(ii) the repulsion of electrons from each other. The effective nuclear charge experienced by a valence electron in an atom will be less than the actual charge on the nucleus because of shielding or screening of the valence electron from the nucleus by the intervening core electrons.

As we descend the 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. The increase in shielding outweighs the increasing nuclear charge and the removal of the outermost electron requires less energy down a group.

Question 19.

The first ionization enthalpy values (in kJ mol-1) of group 13 elements are :

B         Al      Ga    In       Tl

801    577    579   558   589

How would you explain this deviation from the general trend?

Solution.

• (i) Al has lower ionization enthalpy than B due to larger size.
• (ii) Ga has slightly higher ionization enthalpy than Al due to ineffective shielding by 3d electrons.
• (iii) In has lower ionization enthalpy than Ga due to larger size.
• (iv) Tl has higher ionization enthalpy than In due to ineffective shielding by 4f electrons.

Question 20.

Which of the following pairs of elements would have a more negative electron gain enthalpy?

• (i) O or F
• (ii) F or Cl

Solution.

(i) O or F : F has more negative electron gain enthalpy than O due to smaller size, higher nuclear charge and greater possibility of attaining the nearest stable noble gas configuration by gaining one electron.

(ii) F or Cl : Cl has more negative electron gain enthalpy because in F the incoming electron is added to the smaller n = 2 quantum level and suffers significant repulsion from the other electrons present in this level.

In Cl, the added electron goes to n = 3 quantum level and occupies a larger region of space and electron-electron repulsion experienced is far less.

Question 2.
Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.

Solution.
The second electron gain enthalpy of O is +ve. This is because energy has to be supplied to convert O(g) to O2-(g) in order to overcome the repulsive forces.

Question 22.
What is the basic difference between the terms electron gain enthalpy and electronegativity?

Solution.

Question 23.
How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?

Solution.
The statement is not correct because electronegativity of an element varies with the state of hybridisation and oxidation state of the element.

Question 24.

Describe the theory associated with the radius of an atom as it

• (a) gains an electron
• (b) loses an electron.

Solution.

The distance between the nucleus and the outermost shell of an ion is known as ionic radius.

(a) The gain of an electron leads to the formation of an anion. The radius of the anion is larger than the atomic radius of its parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge which acts on more electrons so, each electron is held less tightly and thereby the electron cloud expands.

(b) The removal of an electron from an atom results in the formation of a cation. A cation is smaller than its parent atom because it has fewer electrons while its nuclear charge remains the same and since it is now acting on lesser number of electrons and pulls them closer, the ion is smaller.

Question 25.
Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.

Solution.
Two isotopes of the same element have the same first ionization enthalpies because of same effective nuclear charges.
e.g., $_{ 17 }^{ 35 }{ Cl }$ and $_{ 17 }^{ 37 }{ Cl }$ have same ionization enthalpies.

Question 26.
What are the major differences between metals and non-metals?

Solution.

Question 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.

Solution.

• (a) Fluorine
• (b) Magnesium
• (c) Oxygen
• (d) Group 17 (Halogens) :
• F, Cl – Non metals and gases
• Br – Non metal and liquid
• I – Shows metallic lustre.

Question 28.
The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that among group 17 elements is F > Cl > Br > I. Explain.

Solution.
The trend Li < Na < K < Rb < Cs is observed for chemical reactivity because upon descending the group the ionization energy of alkali metals decreases i.e., it is easy for them to lose an electron from their valence shell and attain the nearest stable noble gas configuration. The trend F > Cl > Br > I is observed for chemical reactivity in halogens because the standard reduction potential decreases as we descend the group. F being the most electronegative readily accepts an e~ to complete its octet and hence the trend.

Question 29.
Write the general outer electronic configuration of s-, p-, d- and f- block elements.

Solution.
General outer electronic configuration :

Question 30.

Assign the position of the element having outer electronic configuration

• (i) ns2 np4 for n = 3
• (ii) (n – 1)d2ns2 for n = 4, and
• (iii) (n-2)f 7(n-1)d1ns2 for n = 6, in the periodic table.

Solution.

Question 31.

The first (∆iH1) and the second (∆iH1) ionization enthalpies (in kJ mol-1) and the (∆H1) electron gain enthalpy (in kJ mol-1) of a few elements are given below:

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)?

Solution.

(a) V : The element V has highest first ionization enthalpy and positive electron gain enthalpy hence, it is least reactive.

(b) II : The element II has the least first ionization enthalpy hence, it is most reactive metal.

(c) III; The element III has very high negative electron gain enthalpy hence, it is most reactive non-metal.

(d) IV : Element IV has high negative electron gain enthalpy but ionization energy is not that high hence, it is least reactive non-metal.

(e) VI : The first and second ionization energies of element VI indicate that it can form a stable binary halide.

(f) I : The element 1 has very low value of first ionization energy but very high second ionization energy. Hence, it will form a stable covalent halide of the formula MX.

Question 32.

Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.

• (a) Lithium and oxygen
• (b) Magnesium and nitrogen
• (c) Aluminium and iodine
• (d) Silicon and oxygen
• (e) Phosphorus and fluorine
• (f) Element 71 and fluorine

Solution.

• (a) Li2O
• (b) Mg3N2
• (c) AlI3
• (d) SiO2
• (e) PF5
• (f) LuF3

Question 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.

Solution.

(c) Principal quantum number

Question 34.

Which of the following statements related to the modern periodic table is incorrect?

(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-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 (/) for the last subshell that received electrons in building up the electronic configuration.

Solution.

(b): The d-block has 10 columns, because a maximum of 10 electrons can occupy all the orbitals in a d-subshell.

Question 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.

Solution.

(c): Nuclear mass

Question 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.

Solution.

(a): Nuclear charge (Z).

Question 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.
Solution.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.

Question 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

Solution.

(d) K > Mg > Al > B

Question 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

Solution.

(c) F>N>C>B>Si

Question 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) CI>F>0>N
• (d) O > F > N > Cl

Solution.

(b) F > O > Cl > N