NCERT MOST IMPORTANT QUESTIONS CLASS – 11 | CHEMISTRY IMPORTANT QUESTIONS PART-1 | CHAPTER -3| CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES | EDUGROWN |

In This Post we are  providing Chapter-3 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

Question 1.
Calculate the energy required to convert all atoms of magnesium to magnesium ions present in 24 mg of magnesium vapors. IE₁ and IE₂ of Mg are 737.76 and 1450.73 kJ mol-1 respectively.

Answer:
Mg (g) + IE₁ → Mg+ (g) + e^– (g)
IE₁ = 737.76 kJ mol^-1

Mg+ (g) + IE₂ → Mg²⁺ + e^– (g)
IE₂ = 1450.73 kJ mol^-1

Total amount of energy needed to convert Mg (g) into ,
Mg²⁺ ion = IE₁ + IE₂

Now 24 mg of Mg = 241000 g = 241000×24 mol [1 Mole of Mg = 24 g]
= 10^-3 mole.

Question 2.
The IE₁ and IE₂ of Mg. (g) are 740 and 1450 kJ mol^-1. Calculate the percentage of Mg⁺ (g) and Mg²⁺ (g) if 1 g of Mg (g)
absorbs 50 kJ of energy.

Answer:
No. of moies of Mg vapours present in I g = 124 = 0.0147

Energy absorbed to convèrt Mg (g) to Mg⁺ (g) = 0.0417 × 740
=30.83 kJ

Energy left unused = 50 – 30.83
= 19.17 kJ
Now 19.17 kJ will be used to e Mg⁺ (g) to Mg²⁺ (g) .
∴ No. of moles of Mg+ (g) converted to Mg2+ (g) = 19.171450 = 0.132

%age of Mg⁺ (g) = 0.02850.0417 × 100 = 68.35
and %age of Mg²⁺ (g) = 100 – 68.35 = 31.65

Question 3.
Which of the elements Na, Mg, Si, P would have the greatest difference between the first and the second ionization enthalpies. Briefly explain your answer

Answer:
Among Na, Mg, Si, P, Na is an alkali metal. It has only one electron in the valence shell. Therefore, its IE₁ is low: However, after the removal of the first electron, it acquires a Neon gas configuration i.e., Na+ (1s², 2s² 2p⁶). Therefore its IE₂ is expected to be very high. Consequently, the difference in IE₁ and IE₂ comes to be greatest in the case of Na.

Question 4.
The IE₂ of Mg is higher than that of Na. On the other hand, the IE₂ of Na is much higher than that of Mg. Explain,

Answer:
The first electron in both cases has to be removed from the 3s orbital, but the nuclear charge of Na is less than that of Mg. After the removal of the first electron from Na, the electronic configuration of Na⁺ is 1s², 2s² 2p⁶, i.e., that of noble gas which is very stable and the removal of the 2nd electron is very difficult. In the case of Mg after the removal of the first electron, the electronic configuration of Mg⁺ is 1s², 2s² 2p⁶ 3s. The 2nd electron to be removed is again from 3s orbital which is easier.

Question 5.
The amount of energy released when 1 × 10¹⁰ atoms of chlorine in vapor state are converted to Cl^– ions according to the equation.
Cl (g) + e^– → Cl^– (g) is 57.86 × 10^-10 J
Calculate the electron gain enthalpy of the chlorine atom in terms of kJ mol^-1 and eV pet atom.
Answer:
The electron gain enthalpy of chlorine

Question 6.
Electronic configuration of the four elements are given below:
Arrange these elements in increasing order of their metallic character. Give reasons for your answer.

(i) [Ar]4s²
Answer:
[Ar]4s² is Calcium metal with At. no. = 20.

(ii) [Ar]3d¹⁰ 4s²
Answer:
[Ar]3d¹⁰ 4s² is Zinc metal with At. no. = 30.

(iii) [Ar]3d¹⁰ 4s² 4p⁶ 5s²
Answer:
[Ar]3d¹⁰ 4s² 4p⁶ 5s² is Strontium metal with At. no. = 38.

(iv) [Arl 3d¹⁰ 4s² 4p⁶ 5s¹
Answer:
[Ar] 3d¹⁰ 4s² 4p⁶, 5s¹ is*Rubidium metal with At. no. = 37.

Alkali metals are the most metallic, followed by alkaline earth metals and transition metals. Among alkali metals – Rubidium (37) is the most metallic. Among alkaline earth metals (Ca, Sr) Sr (Strontium) is more metallic than Calcium (Ca) as the metallic character increases from top to bottom in a group. Zinc – the transition metal is the least metallic. Thus metallic character increases from
Zn < Ca < Sr < Rb or (ii) < (i) < (iii) < (iv)

Question 7.
The formulation of F (g) from F (g) is exothermic whereas that of O^2- (g) from O (g) is endothermic. Explain.

Answer:
F (g) + e^– (g) → F (g); ΔH = Negative
Energy is released when an extra electron from outside is added to a neutral isolated gaseous atom of an element.
To convert, O (g) to O^2- (g) two steps are required
(i) O (g) + e^– (g) → O^– (g); ΔH₁ = – 141 kJ mol^-1
(ii) O^– (g) + e^– (g) → O^2- (g); ΔH₂ = + 780 kJ mol^-1

Hence the over all processes endothermic (+ 780 – 141 = + 639 kJ mol^-1) whereas F (g) to F^– (g) is exothermic.

Question 8.
Explain the important general characteristics of groups in the modem periodic table in brief.

Answer:
The elements of a group show the following important similar characteristics.
(0 Electronic configuration. All elements in a particular group have similar outer electronic configuration e.g., all elements of group I’, i.e., alkali metals have ns1 configuration in their valency shell. Similarly, group 2 elements (alkaline Earths) haye ns² outer configuration and halogens (group 17) have ns² np⁵ configuration (where n is the outermost shell).

(it) Valency. The valency of an element depends upon the number of electrons in the outermost shell. So elements of a group show the same valency, e.g., elements of group 1 show + 1 valency and group 2 show + 2 valencies i.e. valency i.e., NaCl > MgCl₂ etc.

(iii) Chemical properties. The chemical properties of the elements are related to the number of electrons in the outermost shell of their atoms. Hence all elements belonging to the same group show similar chemical properties. But the degree of reactivity varies gradually from top to bottom in a group. For example, in group 1 all the elements are highly reactive metals but the degree of reactivity increases from Li to Cs. Similarly, elements of group 17, i.e., halogens: F, Cl, Br, I are all non-metals and they’re- reactivity goes on decreasing from top to bottom.

Question 9.
Explain the electronic configuration in periods in the periodic table.

Answer:
Each successive period in the periodic table is associated with the filling Up of the next higher principal energy level (n – 1, n – 2, etc.). It can be readily seen that the number of elements in each period is twice the number of atomic orbitals available in the energy level that is being filled. The first period starts with the filling of the lowest level (1s) and has thus the two elements – hydrogen (1s¹) and helium (1s²) when the first shell (K) is completed. The second period starts with lithium and the third electron enters the 2s orbital.

The next element, beryllium has four electrons and has the electronic configuration 1s² 2s². Starting from the next element boron, the 2p orbitals are filled with electrons when the L shell is completed’ at neon (2s² 2p⁶). Thus there are 8 elements in the second period. The third period (n = 3) being at sodium, and the added electron enters a 3s orbital. Successive filling of 3s and 3p orbitals give rise to the third period of 8 elements from sodium to argon.

The fourth period (n = 4) starts at potassium with the filling up of 4p of 4s orbital. Before the 4p orbital is filled, the filling up of 3d orbitals becomes energetically favorable and we come across the so-called 3d transition series of elements. The fourth period ends at krypton with the filling up of the 4p orbitals. Altogether we have 18 elements in the fourth period. The fifth period (n = 5) beginning with rubidium is similar to the fourth period and contains the 4d transition series starting at yttrium (Z = 39).

This period ends at xenon with the filling up of the 5p orbitals. The sixth period (n = 6) contains 32 elements and successive electrons enter 6s, 4/, 5d, and 6p orbitals, in that order. Filling up of the 4/ orbitals being with cerium, (Z = 58) and ends at lutetium (Z = 71) to give the 4/-inner transition series which is called the lanthanide series. The seventh period (n = 7) is similar to the sixth period with the successive filling up of the 7s, 5f, 6d, and 7p orbitals and includes most of the man-made radioactive elements.

This period will end at the element with atomic number 118 which would belong to the noble gas family. Filling up of the 5f orbitals after actinium (Z = 89) gives the 5f-inner transition series known as the actinide series. The 4f and 5f transition series of elements are placed separately in the periodic table to maintain its structure and to preserve the principle of classification by keeping elements with similar properties in a single column.

Question 10.
Explain the variation of valence in the periodic table.

Answer:
Variation of valence in a group as well as across a period in the periodic table occurs as follows:
1. In a group: All elements in a group show the same valency. For example, all alkali metals (group 1) show a valency of 1+. Alkaline earth metals (group 2) show a valency of 2+.

However, the heavier elements of p-block elements (except noble gases) show two valences: one equal to the number of valence electrons or 8-No. of valence electron# and the other two less. For example, thallium (Tl) belongs to group 13. It shows valence of 3+ and 1+.

Lead (Pb) belongs to group 14. If shows valance of 4+ and 2+.
Antimony (Sb) and Bismuth (Bi) belong to group 15. They show valence of 5+ and 3+ being more stable.

This happens due to the non-participation of tie two s-electrons present in the valence shell of these elements. This non-participation of one pair of s-electrons in bonding is called the inert-pair effect.

2. In a period: The number of the valence electrons increases – in going from left to right in a period of the periodic table. Therefore the valency of the elements in a period first increases, and then decreases.