PSEB 11th Class Chemistry Important Questions Chapter 3 Classification of Elements and Periodicity in Properties

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 3 Classification of Elements and Periodicity in Properties Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 3 Classification of Elements and Periodicity in Properties

Very Short Answer Type Questions

Question 1.
Which properties of the elements depend on’ the electronic configuration of the atoms and which do not?
Answer:
Chemical and many physical properties of the elements depends on the electronic configuration of the atoms, whereas the nuclear properties do not.

Question 2.
Write the number designation of a group that has 2 electrons beyond a noble gas configuration.
Answer:
The number designation of a group that has 2 electrons beyond a noble gas configuration will be 2 which means it will belong to group 2 of periodic table.

Question 3.
Why is it more logical to call the atomic radius as the effective atomic radius?
Answer:
This is because the size of atom is very small and it has no sharp boundaries.

Question 4.
A boy has reported the radii of Cu, Cu⁺ and Cu²⁺ as 0.096 nm, 0.122 nm and Question072 nm respectively. However, it has been noticed that he interchanged the values by mistake. Assign the correct values to different species.
Answer:
Cu [0.122 nm], Cu⁺ [0.096 nm], Cu²⁺ [0.072 nm].
∵ Size ∝ \(\frac{1}{\text { positive charge }}\)

Question 5.
Atomic radii of fluorine is 72 pm where as atomic radii of neon is 160 pm. Why? [NCERT Exemplar]
Answer:
Atomic radius of F is expressed in terms of covalent radius while, atomic radius of neon is usually expressed in terms of van der Waals’ radius, van der Waals’ radius of an element is always larger than its covalent radius.
Therefore, atomic radius of F is smaller than atomic radius of Ne (F = 72 pm, Ne = 160 pm)

Question 6.
Arrange the following elements in order of decreasing electron gain enthalpy : B, C, N, O.
Answer:
N has positive electron gain enthalpy while all others have negative electron gain enthalpies. Since size decreases on moving from B → C → O, therefore, electron gain enthalpies become more and more negative from B → C → O. Thus, the overall decreasing order of electron gain enthalpies is N, B, C, O.

Question 7.
Which of the following atoms would most likely form an anion (i) Be, (ii) Al, (iii) Ga, (iv) I ?
Answer:
I, because of high electron gain enthalpy, it can accept an electron readily to form an anion F < Cl < Br > I.

Question 8.
Explain why chlorine can be converted into chloride ion more easily as compared to fluoride ion from fluorine.
Answer:
Electron gain enthalpy of Cl is more negative than that of F.

Question 9.
Among alkali metals which element do you expect to be least electronegative and why? [NCERT Exemplar]
Answer:
On moving down the group, electronegativity decreases because atomic size increases. Fr has the largest size, therefore it is least electronegative.

Question 10.
Arrange the following elements in the increasing order of non-metallic character. B, C, Si, N, F
Answer:
The given non-metals are arranged in the increasing order of non-metallic character as follows:

Short Answer Type Questions

Question 1.
What would be IUPAC names and symbols for elements with atomic numbers 122, 127, 135, 149 and 150? .
Answer:
The roots 2, 7, 5, 9 and 0 are referred as bi, sept, pent, enn and nil respectively. Therefore, their names and symbol are

Z (Atomic number)	Name	Symbol
122	Unbibium	Ubb
127	Unbiseptium	Ubs
135	Untripentium	Dtp
149	Unquadennium	Uqe
150	Unpentnilium	Upn

Question2.
All transition elements are d-block elements, but all d-block elements are not transition elements. Explain.
Answer:
Elements in which the last electron enters in the d-orbitals, are called d-block elements or transition elements. These elements have the general outer electronic configuration (n – 1)d^1-10ns^0-2 Zn, Cd and Hg having the electronic configuration, (n – l1)d¹⁰ns² do not show most of the properties of transition elements. The d-orbitals in these elements are completely filled in the ground state as well as in their common oxidation states. Therefore, they are not regarded as transition elements. Thus, on the basis of properties, all transition elements are d-block elements but on the basis of electronic configuration, all d-block elements are not transition elements.

Question 3.
Arrange the elements N, P, O and S in the order of
(i) increasing first ionisation enthalpy.
(ii) increasing non-metallic character.
Give reason for the arrangement assigned.
Answer:

	Group 15	Group 16
2nd period	N	0
3rd period	P	S

(i) Ionisation enthalpy of nitrogen (₇N = 1s², 2s², 2p³) is greater than oxygen (₈O = 1s² , 2s² , 2p⁴ ) due to extra stable half-filled 2p-orbitals. Similarly, ionisation enthalpy of phosphorus (₁₅P = 1s², 2s², 2p⁶, 3s², 3p³) is greater than sulphur (₁₆S = 1s², 2x², 2p⁶, 3s², 3p⁴).
On moving down the group, ionisation enthalpy decreases with increasing atomic size. So, the increasing order of first ionisation enthalpy is S < P < O < N.

(ii) Non-metallic character across a period (left to right) increases but on moving down the group it decreases. So, the increasing order of non-metallic character is P < S < N < 0.

Question 4.
What do you understand by exothermic reaction and endothermic reaction? Give one example of each type.
Answer:
Exothermic reactions : Reactions which are accompanied by evolution of heat are called exothermic reactions. The quantity of heat produced is shown either along with the products with a ‘+’ sign or in terms if ΔH with a sign, e.g.,

C(s) + O₂(g) → CO₂(g) + 393.5 kJ
H₂(g) + \(\frac{1}{2}\)O₂(g) → H₂O(l) ΔH = -285.8 kJ mol^-1

Endothermic reactions : Reactions which proceed with absorption of heat are called endothermic reactions. The quantity of heat absorbed is shown either alongwith the products with a sign or in terms of ΔH with a ‘-‘ sign, e.g.,

C(s) + H₂O(g) → CO(g) + H₂(g) -131.4 kJ
N₂(g) + 3H₂(g) → 2NH₃(g); ΔH = +92.4 kJ mol^-1

Question5.
How does the metallic and non-metallic character vary on moving from left to right in a period?
Answer:
As we move from left to right in a period, the number of valence electrons increases by one at each succeeding element but the number of shells remains same. Due to this, effective nuclear charge increases.

More is the effective nuclear charge, more is the attraction between nuclei and electron.
Hence, the tendency of the element to lose electrons decreases, this results in decrease in metallic character.
Furthermore, the tendency of an element to gain electrons increases with increase in effective nuclear charge, so non-metallic character increases on moving from left to right in a period.

Long Answer Type Questions

Question 1.
Write the drawbacks in Mendeleev’s Periodic Table that led to its modification.
Answer:
The main drawbacks of Mendeleev’s Periodic Table are:
(i) Some elements having similar properties were placed in different groups whereas some elements having dissimilar properties were placed in the same group. For example alkali metals such as Li, Na, K, etc., (I A group) are grouped together with coinage metals such as Cu, Ag, Au (I B group) though their properties are quite different. Chemically similar elements such as Cu (I B group) and Hg (II B group) have been placed in different groups.

(ii) Some elements with higher atomic weights are placed before the elements with lower atomic weights in order to maintain the similar chemical nature of elements. For example,

(iii) Isotopes did not find any place in the Periodic Table. However, according to Mendeleev’s classification, these should be placed at different places in the Periodic Table.
(All the above three defects were however removed when modern periodic law based on atomic number was given.)

(iv) Position of hydrogen in the Periodic Table is not fixed but is
controversial. ,
(v) Position of elements of group VIII could not be made clear which have been arranged in three triads without any justification.
(vi) It could not explain the even and odd series in IV, V and VI long periods.
(vii) Lanthanides and actinides which were discovered later on, have not been given proper positions in the main frame of Periodic Table.

Question 2.
p-block elements form acidic, basic and amphoteric oxides. Explain each property by giving two examples and also write the reactions of these oxides with water.
Answer:
In p-block, when we move from left to right in a period, the acidic character of the oxides increases due to increase in electronegativity, e.g.,

(i) 2nd period
B₂O₃ < CO₂ < N₂O₃ acidic character increases.

(ii) 3rd period
Al₂O₃ < SiO₂ < P₄O₁₀ < SO₃ < Cl₂O₇ acidic character increases.
On moving down the group, acidic character decreases and basic character increases, e.g.,

(a) Nature of oxides of 13 group elements

(b) Nature of oxides of 15 group elements

Among the oxides of same element, higher the oxidation state of the element, stronger is the acid. e.g., SO₃ is a stronger acid than SO₂. B₂O₃ is weakly acidic and on dissolution in water, it forms orthoboric acid. Orthoboric acid does not act as a protonic acid (it does not ionise) but acts as a weak Lewis acid.

Al₂O₃ is amphoteric in nature. It is insoluble in water but dissolves in alkalies and react with acids.

Tl₂O is as basic as NaOH due to its lower oxidation state (+1).
Tl₂O + 2HCl → 2TlCl + H₂O

P₄O₁₀ on reaction with water gives orthophosporic acid.

Cl₂O₇ is strongly acidic in nature and on dissolution in water, it gives perchloric acid.