MP Board Class 11th Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in Properties

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MP Board Class 11th Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in Properties

MP Board Class 11 Chemistry Classification of Elements and Periodicity in Properties Textbook Questions and Answers

Question 1.
What is the basic theme of organisation in the periodic table?
Solution.
The basic theme of organisation of elements in the periodic table is to facilitate the study of the properties of all the elements and their compounds. On the basis of similarities in chemical properties, the various elements have been divided into different groups. This has made the study of elements simple because their properties are now studied in the form of 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 that?
Solution.
Mendeleev used atomic weight as the basis of classification of elements in the periodic table. He arranged the known elements in order of increasing atomic weights grouping together elements with similar properties. Mendeleev relied on the similarities in the properties of the compounds formed by the elements. 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.

He ignored the order of atomic weights and placed the elements with similar properties together for example, iodine with lower atomic weight than tellurium (Group VI) was placed in group VII along with fluorine, chlorine, etc., because of similarities in properties. Thus, Mendeleev did not stick strictly to arrangement of elements in the increasing order of atomic weight.

Question 3.
What is the basic difference in approach between Mendeleev’s Periodic Law and the Modern Periodic Law?
Solution.
Mendeleev’s Periodic Law states that the physical and chemical properties of the elements are a periodic function of their atomic weights whereas Modern Periodic Law states that the 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 Modem Periodic Law is the change in basis of arrangement of elements from atomic weight to atomic number.

Question 4.
On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Solution.
In the modern periodic table, each period starts with the filling of a new principal energy level. Thus, the sixth period begins with the filling of 6s orbital and continues till the filling of seventh energy level starts. According to Aufbau rule the subshells which follow 6s are 4f 5d, 6p, 7s. Therefore, in the 6th period, electrons can be filled in only 6s, 4f, 5d, and 6p-subshells. Now s-subshell has two, p-subshell has three, d-subshell has five and f-subshell has seven orbitals. Hence, in all there are 16 orbitals that can be filled in this period which, at the maximum, can accommodate, 32 electrons and therefore, sixth period has 32 elements.

Question 5.
In terms of period and group, where would you locate the element with Z = 114?
Solution.
The element with atomic number 114 would have the electronic configuration 86[Rn] 5f146d10, 7s2 6p2. Thus, the element belongs to: Period 7, group-14 and block p.

MP Board Class 11th Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in Properties

Question 6.
Write the atomic number of the element in the third period and seventeenth group of the periodic table.
Solution.
The elements of group-17 (halogens) have ns2 np5 valence shell electronic configuration. The element of group-17 belonging to third period would have valence shell electronic configuration 3s23p5. The complete electronic configuration of this element would be 1s2, 2s2, 2p6, 3s2, 3p5.
There are 17 electrons in it. Hence, the atomic number of the element in the third period and group-17 is 17.

Question 7.
Which element do you think would have been named by.
(i) Lawrence Berkeley Laboratory
(ii) Seaborg’s group?
Solution.
(i) Berkelium (Bk)
(ii) Seaborgium (Sg)

Question 8.
Why do elements in the same group have similar physical and chemical properties?
Solution.
Elements in the same group have similar valence shell electronic configuration and hence have similar physical and chemical properties.

Question 9.
What does atomic radius or ionic radius mean to you?
Solution.
Atomic radius gives us an idea about size of the atom. Atomic radius may be taken as the distance between the centre of the nucleus and the outermost shell of the atom. It can be measured either by X-ray or by spectroscopic methods. The ionic radius tells us about size of the ion. It is defined as the distance between centre of the nucleus and the point upto which the ion has influence in the ionic bond.

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

On the other hand, the atomic size decreases as we move from left to right in a period. This is due to the reason that within a period the valence shell remains 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 the outer electrons increases and hence the atomic size decreases.

MP Board Class 11th Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in Properties

Question 11.
What do you understand by isoelectronic species? Name the species that will be isoelectronic with each of the following atoms or ions.
(i) F-
(ii) Ar
(iii) Mg2+
(iv) Rb+
Solution.
The species which have the same number of electrons but different magnitude of the nuclear charge are called isoelectronic species.
(i) F- has 10 (9 + 1) electrons. The species nitride ion, N3-(7 + 3); oxide ion, O2- (8 + 2); neon, Ne (10 + 0); sodium ion, Na+ (11 – 1); magnesium ion, Mg2+ (12 – 2); aluminium ion, Al3+ (13 – 3), are isoelectronic with it.
(ii) Ar has 18 electrons. The species phosphide ion, P3- (15 + 3), sulphide ion; S2- (16 + 2); chloride ion, Cl (17 + 1), potassium ion, K+ (19 – 1), calcium ion, Ca2+ (20 – 2) are isoelectronic with it. .

(iii) Mg2+ has 10 (12 – 2) electrons. The species nitride ion, N3- (7 + 3); oxide ion, O2- (8 + 2) and sodium ion, Na+ (11 – 1) are isoelectronic with it.
(iv) Rb+ has 36 (37 – 1) electrons. The species bromide ion, Br (35 + 1), krypton, Kr (36 + 0) and strontium Sr2+ (38 – 2) each one of which has 36 electrons, are isoelectronic with it.

Question 12.
Consider the following species. N3-, O2-, F, Na+, Mg2+ and Al3+.
(a) What is common in them?
(b) Arrange them in order of increasing ionic radii.
Solution.
(a) Each one of these ions contains 10 electrons and hence these are isoelectronic ions.
(b) The ionic radii of isoelectronic ions decrease with the increase in the magnitude of the nuclear charge. Among the isoelectronic ions: N3-, O2-, F, Na+, Mg2+ and Al3+, nuclear charges increase in the order:
N3- < O2- < F < Na+ < Mg2+ < Al3+.
Therefore, their ionic radii decrease in the order:
N3- > O2- > F > Na+ > Mg2+ > Al3+.

Question 13.
Explain why cations are smaller and anions are larger in radii than their parent atoms.
Solution.
The ionic radius of a cation is always smaller than the parent atom because a cation is formed by loss of one or more electrons by the neutral atom. 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 an anion is formed by addition of one or more electrons to the neutral atom. 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.

MP Board Class 11th Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in Properties

Question 14.
What is the significance of the terms ‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy?
Solution.
Ionization enthalpy is the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom in ground state, so as to convert it into a gaseous cation. Electron gain enthalpy is the energy released when an isolated gaseous atom in the ground state accepts an extra electron to form the gaseous negative ion. The force with which an electron is attracted by the nucleus of an atom is appreciably affected by presence of other atom within its molecule or in the neighbourhood.

Therefore, for the purpose of determination of ionization enthalpy and electron gain enthalpy it is essential that these interatomic forces of attraction should be minimum. Since in the gaseous state, the atoms are widely separated, therefore, these interatomic forces are minimum. It is because of these reasons, that the term isolated gaseous atom has been included in the definition of ionization enthalpy and electron gain enthalpy.

The term ground state here means that the atom must be present in the most stable state, i.e., the ground state. The reason being that when the isolated gaseous atom is in the excited state, less amount of energy is required to remove the outermost electron. Similarly, less amount of energy is released when an electron is added to the atom in excited state. Therefore, for comparison purposes, the ionization enthalpies and electron gain enthalpies of gaseous atoms must be determined in their respective ground states.

Question 15.
Energy of an electron in the ground state of the hydrogen atom is – 2.18 × 10-18 J. Calculate the ionization enthalpy of atomic hydrogen in terms of kJ mol-1.
Solution.
The energy required to remove an electron in the ground state of hydrogen atom
= E – E1 = 0 – E1
= – (- 2.18 × 10-18 J) = 2.18 × 1018J.
∴ Ionization enthalpy .per mole of hydrogen atoms
= \(\frac{2.18 \times 10^{18} \times 6.02 \times 10^{23}}{1000}\) kJ mol-1
= 1312.36 kJ mol-1.

Question 16.
Among the second period elements, the actual ionization energies are in the order:Li<B<Be<C<0 Explain why
(i) Be has higher ΔiH than B
(ii) O has lower ΔiH than N and F?
Solution.
(i) 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, due to greater penetration, are more strongly attracted by the nucleus than 2p-electrons, therefore, more amount of energy is required to knock out a 2s-electron than a 2p-electron. Consequently, ΔiH of Be is higher than that ΔiH of B.
(ii) 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 0(ls2 2s2 2p2 2py 2p\) in which the 2p-orbitals are neither half-filled nor completely filled. Therefore, it is difficult to remove an electron from N than from O. As a result, ΔiH of O is less than that of N. Because of higher nuclear charge (+ 9) and smaller atomic size, the first ionization enthalpy of F is higher than that of O. Therefore, ionization enthalpy of O is less than that of N as well as that of F.

MP Board Class 11th Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in Properties

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.
MP Board Class 11th Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in Properties 1
The first ionization enthalpy of Mg is higher than that of Na because it has greater nuclear charge, smaller atomic size and stable electronic configuration. However, the second ionization enthalpy of sodium is much higher than that of magnesium because Na+ has a stable noble gas configuration. Moreover in Na+ the outermost shell is second whereas in Mg+ it is third.

Question 18.
What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down the group?
Solution.
Within the main group elements, the ionization enthalpy decreases regularly as we move down the group due to the following two factors.

  • Atomic size. On moving down the group, the atomic size increases progressively due to the addition of one new 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.
  • Screening effect. With the addition of new shells, the shielding effect or the screening effect increases. As a result, the force of attraction of the nucleus for the valence electrons further decreases and hence the ionization enthalpy decreases.

Question 19.
The first ionization enthalpy values (in kJ mol-1) of group 13 elements are:

B Al Ga In Ti
801 577 579 558 589

How will you explain this deviation from the general trend?
Solution.
On moving down the group 13 from B to Al, the ionization enthalpy decreases as expected due to an increase in atomic size and screening effect which outweigh the effect of increased nuclear charge. However, ΔiH1 of Ga is slightly higher than that of Al while that of T1 is higher than those of Al, Ga and In. These deviations can be explained as follows:

Al follows immediately after s-block elements while Ga and In follow after d-block elements and T1 after d- and f-block elements. This extra d- and f-electrons do not shield the valence shell-electrons from the nucleus effectively.
As a result, the valence electrons are more tightly held by the nucleus and hence larger amount of energy is needed for their removal. This explains why Ga has higher ionization 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 ΔiH1 of In is less than, that of Ga. Thereafter, the effect of increased nuclear charge predominates and hence the ΔiH1 of Tl is higher than that of In.

MP Board Class 11th Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in Properties

Question 20.
Which of the following pairs of elements would have a more negative electron gain enthalpy? Explain.
(i) O or F,
(ii) F or Cl.
Solution.
(i) F has more negative electron gain enthalpy due to its smaller size and greater effective nuclear charge.
(ii) Chlorine (Cl) has more negative electron gain enthalpy than fluorine (F). F has less negative electron gain enthalpy because in it the added electron goes to the smaller energy level (n = 2) and hence suffers significant repulsion from the electrons already present in this shell.

Question 21.
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 positive as explained below:
When an 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 electron being added.
O(g) + e (g) → O2-(g); ΔH = ΔegH2 = + ve
Therefore, the second electron gain enthalpy of oxygen is positive.

MP Board Class 11th Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in Properties

Question 22.
What is the basic difference between the terms electron gain enthalpy and electronegativity?
Solution.
Both electron gain enthalpy and electronegativity are measures of 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 electron to form a negative ion, electronegativity refers to the tendency of the atom to attract the shared pair of electrons towards itself in a molecule.

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 electronegativity of any given atom is not constant; it varies depending on the element to which it is bound and also on the state of hybridization of the atom of the element. Therefore, the statement, that the electronegativity of N on Pauling scale is 3.0 in all nitrogen compounds, is wrong.

Question 24.
Describe the theory associated with the radius of an atom as it (a) gains electron (b) loses electron.
Solution.
(a) Gain of electron. When a neutral atom gains one 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 the same as the parent atom. Since the same nuclear charge now attracts greater number of electrons, therefore, effective nuclear charge decreases and hence the electron cloud expands and the atomic radius increases.

(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 smaller number of electrons, therefore, effective nuclear charge increases and hence the atomic radius decreases.

MP Board Class 11th Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in Properties

Question 25.
Would you expect the first ionization enthalpies of two isotopes of the same element to be same or different? Justify your answer.
Solution.
Two isotopes of the same element have same number of electrons, same nuclear charge, same atomic radii. Therefore, they are expected to have same ionization enthalpies.

Question 26.
What are major differences between metals and non-metals?
Solution.
Elements which have a strong tendency to lose electrons to form cations are called metals while those which have a strong tendency to accept electrons to form anions are called non-metals.
Metals are highly electropositive whereas non-metals are highly electronegative. Metals are good reducing agents whereas non-metals are good oxidizing agents. Metals are generally solids whereas non-metals are generally gases.

Oxides of metals are basic or amphoteric whereas oxides of non-metals are acidic or neutral.
Metals are generally good conductors of heat and electricity whereas non-matals are generally bad conductors.

Question 27.
Use the periodic table to answer the following questions.
(a) Identify an element with five electrons in the outer sub-shell.
(6) Identify the element that would tend to lose two electrons.
(c) Identify the element that would tend to gain two electrons.
Solution.
(a) Fluorine (F). Its electronic configuration is 1s2, 2s2 2p5.
(b) Magnesium (Mg). Its electronic configuration is 1s2,2s2 2p6,3s2, by losing 2 electrons it attains noble gas configuration.
(c) Oxygen (O). Its electronic configuration is 1s2, 2s2 2p4, by gaining two electrons it attains stable noble gas configuration.
The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that of group 17 is F > Cl > Br > I. Explain.
Solution.
The elements of group-1 have a strong tendency to lose 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. On the other hand, the elements of group-17, have a strong tendency to accept electron.

The tendency to accept electrons, in turn, depends upon their electrode potential. Since the electrode potentials of group-17 elements decrease on descending the group, therefore, their reactivities decrease in the order: F > Cl > Br > I.

Question 29.
Write the general electronic configuration of s-, p-, d-, and f-block elements.
Solution.
s-Block elements: ns1-2
p-Block elements: ns2np1-6
d-Block elements: (n – 1) d1-10ns0-2.
f-Block elements: (n – 2) f1-14 (n – 1) d0-1 ns2.

Question 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.
Solution.
(i) n = 3 indicates that the element belongs to third period. Since the last electron enters the p-orbital, therefore, the given element is a p-block element. For p-block elements, group number = 10 + number of electrons in the valence shell.
∴ Group number of the element = 10 + 6 = 16.

(ii) n = 4 indicates that the element lies in the 4th period.
Since the d-orbital is incomplete, therefore, it is d-block element.
The group number of the element = number of (n – 1) d-electrons + number of (n) s-electrons
= 2 + 2 = 4.

(iii) n = 6 means that the element lies in the sixth 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.

MP Board Class 11th Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in Properties

Question 31.
The first (ΔiH1) and the second (ΔiH2) ionization enthalpies (in kJ mol-1) and the (Δeg H) electron gain enthalpy (in kJ mol-1) of a few elements are given below:

Element ΔiH1 ΔiH2 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:
(а) 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 MX, (X = halogen).
(f) the metal which can form predominantly stable covalent halide of the formula MX (X = halogen)?
Solution.
(a) The element V has highest first ionization enthalpy and positive electron gain enthalpy and hence it is likely to be the least reactive element.
(b) The element II which has the least first ionization enthalpy and a low negative electron gain enthalpy is the most reactive metal.
(c) The element III which has high first ionization enthalpy and a very high negative electron gain enthalpy is likely to be the most reactive non-metal.
(d) The element IV has a high negative electron gain enthalpy but not so high first ionization enthalpy. Therefore, it is the least reactive non-metal.
(e) The element VI has low values for first and second ionization enthalpies. Therefore, it appears that the element is an alkaline earth metal and hence will form binary halide of the formula MX2.
(f) The element I has low first ionization enthalpy and a very high second ionization enthalpy, therefore, it must be an alkali metal with smaller atomic number and is likely to form a predominantly stable covalent halide of the formula MX2.

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) Lithium is an element of 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 formed 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 8 – 5 = 3. Thus, the formula of the compound formed would be Mg3N2.
(c) Aluminium is 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 formed would be AlI3.
(d) Silicon is a group-14 element with a valence of 4 while oxygen is a group-16 element with a valence of 2. Hence, the formula of the compound formed is SiO2.
(e) Phosphorus 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 formed would be PF3 or PF5.
(f) Element with atomic number 71 is a lanthanoid called lutetium (Lu). Its common valence is 3. Fluorine is a group-17 element with a valence of 1. Therefore, the formula of the compound formed would be LuF3.

Question 33.
In the modern periodic table, the period indicates the value of:
(a) atomic number
(b) mass number
(c) principal quantum number
(d) azimuthal quantum number.
Solution.
In the modern periodic table, the period number is same as the value of principal quantum number. Thus, option (c) is correct.

Question 34.
Which of the following statements related to the modern periodic table is incorrect?
(а) The p-block has six columns because a maximum of 6 electrons can occupy all the orbitals in p-subshell.
(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) A block indicates value of azimuthal quantum number (1) for the last subshell that received electrons in building up the electronic configuration.
Solution.
Statement (b) is incorrect. The d-block has 10 columns because a maximum of 10 electrons can occupy all the orbitals is a d-subshell.

MP Board Class 11th Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in Properties

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.
Nuclear mass does not affect the valence shell. Thus, option (c) is the wrong answer.

Question 36.
The size of isoelectronic species-F, Ne and Na+ is affected by
(а) nuclear charge (Z)
(б) valence principal quantum number (n)
(c) electron-electron interaction in the outer orbital
(d) none of factors because their size is the same.
Solution.
The size of the isoelectronic ions depends upon the nuclear charge (Z). As the nuclear charge increases the ionic size decreases. Therefore, statement (a) is correct.

Question 37.
Which 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 electrons 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.
Statement (d) is incorrect because an electron with lower value of n is more tightly held as it is closer to the nucleus.

MP Board Class 11th Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in Properties

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.
Across a period, metallic character decreases as we move from left to right. However, within a group, the metallic character, increases from top to bottom. Therefore, K is expected to be most metallic and B the least. Hence, the choice (d) is correct.

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.
Across a period, the non-metallic character increases from left to right. Therefore, 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, and the choice (c) is correct.

MP Board Class 11th Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in Properties

Question 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
Solution.
Across 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. Thus, overall decreasing order of oxidising power is: F > O > Cl > N and the choice (b) is correct.

MP Board Class 11 Chemistry Classification of Elements and Periodicity in Properties Important Questions and Answers

Question 1.
State Mendeleev’s periodic law.
Answer:
It states, “The physical and chemical properties of the elements are periodic functions of their atomic masses.”

Question 2.
What is the present-day name of Eka-silicon?
Answer:
Germanium.

Question 3.
How many groups are there in modern periodic table?
Answer:
Eighteen.

Question 4.
How many groups are there in the p-block of the periodic table?
Answer:
Six (groups 13, 14, 15, 16, 17 and 18).

Question 5.
Name the first element of the third period.
Answer:
Sodium (Na).

MP Board Class 11th Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in Properties

Question 6.
What are d-block elements?
Answer:
The elements in which the incoming electron enters the d-orbitals are known as d-block elements. They are also known as transition elements.

Question 7.
What are lanthanoides?
Answer:
The elements of 4 f-series are called lanthanoides.

Question 8.
What is the general electronic configuration of d-block elements?
Answer:
(n – l)d1 -10 ns1-2

Question 9.
What are main group elements?
Answer:
s- and p-block elements together are called main group elements.

Question 10.
By what name are the elements of group 17 known?
Answer:
Halogens.

Question 11.
What are the units of ionization enthalpy?
Answer:
Units of ionization enthalpy are kJ mol-1.

MP Board Class 11th Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in Properties

Question 12.
Explain the term electron gain enthalpy.
Answer:
It is the enthalpy released when an electron is added to a neutral gaseous atom (X) to convert it into a negative ion.
X(g) + e- → X(g) + ΔHeg.
where ΔHeg is electron gain enthalpy.

Question 13.
What is meant by atomic radius?
Answer:
Atomic radius refers to either covalent radius or metallic radius depending on non-metallic or metallic nature of the element.

Question 14.
Select the one having smallest ionic radius Li+, Na+, K+.
Answer:
Li+.

Question 15.
Why are s-block elements most reactive?
Answer:
The ionization enthalpy values of s-block elements are quite low and so they are most reactive (reactivity depends on the case with which an element can lose on electron).

Question 16.
Name the element with highest value of electron affinity.
Answer:
Chlorine (Cl).

Question 17.
How many elements are there in first transition series?
Answer:
Ten.

MP Board Class 11th Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in Properties

Question 18.
What is the relationship between electron gain enthalpy and electron affinity.
Answer:
Electron affinity is the negative of electron gain enthalpy.
Electron affinity is defined at absolute zero and therefore at any temperature, heat capacities of reactants and products have to be taken into consideration. Then, electron gain enthalpy (AEGH) is given by
ΔEGH = -Ae(Electron affinity) – \(\frac{1}{2}\) RT

Question 19.
What is the long form of the periodic table?
Answer:
The arrangement of elements on the basis of their electronic configuration is called long form of the periodic table. It consists of four blocks (i.e., s-block, p-block, d-block and f-block) of elements.

Question 20.
What is meant by periodicity of the elements?
Answer:
Periodicity is the repetition of the properties of elements (physical and chemical) in a particular group (vertical row) when they are arranged in the periodic table on the basis of their order of increasing atomic number.
The properties of the elements placed in vertical row are repeated after certain regular interval of atomic numbers 2, 8, 8, 18, 18, 32. These numbers are often called Magic Numbers.

Question 21.
The elements Z = 107 and Z = 109 have been made recently. Element Z = 108 has not yet been made. Indicate the groups in which you will place these elements.
Answer:
Element (Z = 107) in group 7.
Element (Z = 108) in group 8.
Element (Z = 109′ in group 9.

Question 22.
Noble gases have zero electron affinity. Why?
Answer:
Noble gases have complete outermost shell, i.e., have 8 electrons in outermost shell, so they show no tendency to accept an extra electron. Hence they have zero electron affinity.

Question 23.
Account for the low values of electron affinities of nitrogen and phosphorus.
Answer:
These elements have exactly half filled p-orbitals. This provides an extra stability to p-orbitals in nitrogen and phosphorus atoms
N (Z = 7): 1s2, 2s2 2px1 2py1 2pz1
P (Z = 15): 1s2, 2s2, 2p6, 3s2, 3p3
The electrons are not added with ease since they have a weak tendency to accept additional electrons. Hence, they have low values of electron affinities.

MP Board Class 11th Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in Properties

Question 24.
What are f-block elements? By which alternative name are they known and why?
Answer:
Elements in the form of two parallel rows i.e., from Ce-g to Lu71 and from Th90 to Lr103 placed at the bottom of periodic table are known as f-block elements since in these elements the last electron enters in the antipenultimate subshell (i.e., 3rd from the outmost). Their outer electronic configuration varies from (n – 2) f1 (n – 1) d0-1 ns2 to (n – 2) f14(n – 1) d1 ns2 where n is the outermost shell, f-block consists of two series of 14 elements each known as
(4f) Ce58 – Lu71 = Lanthanodes/Lanthanons
(5f) Th90 – Lr103 = Actinodes/Actinons.
f-block elements are also called inner transition elements because they form a transition series within transition elements. The atoms of these elements have their three outermost shells incomplete.

Question 25.
The electrons affinity of chlorine is 349 kJ mol-1. How much energy in kJ is released when 1 g of chlorine is converted completely to Cl- ions in gaseous state?
Answer:
Cl(g) + e→ Cl-(g) + 349 kJ mol-1 (E.A.)
Gram atomic mass of Cl- = 35.5 g
Thus, energy released by 35.5 g of chlorine on being converted to Cl- ion = 349 kJ.
Hence, energy released when 1 g of chlorine is converted to Cl ion = \(\frac{349}{35.5}\) × 1 = 9.83 KJ.