Class 11 Chemistry Chapter 1 Ncert Solutions Study Rankers
NCERT Solutions Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Prop…– Here are all the NCERT solutions for Class 11 Chemistry Chapter 3. This solution contains questions, answers, images, explanations of the complete chapter 1 titled Classification of Elements and Periodicity in Prop… taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Chemistry, then you must come across chapter 3 Classification of Elements and Periodicity in Prop…. After you have studied the lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Prop… in one place.
NCERT Solutions Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties
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| Class | 11 |
| Subject | Chemistry |
| Book | Chemistry Part I |
| Chapter Number | 3 |
| Chapter Name | Classification of Elements and Periodicity in Properties |
NCERT Solutions Class 11 Chemistry chapter 3 Classification of Elements and Periodicity in Properties
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Question & Answer
Q.1: What is the basic theme of organisation in the periodic table? Ans : The basic theme of organisation of elements in the periodic table is to classify the elements in periods and groups according to their properties. This arrangement makes the study of elements and their compounds simple and svsternatic. In the periodic table, elements with similar properties are placed in the same group.
Q.2: Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that? Ans : Mendeleev arranged the elements in his periodic table ordered by atomic weight or mass. He arranged the elements in periods and groups in order of their increasing atomic weight He placed the elements with similar properties in the same group. However, he did not stick to this arrangement for long. He found out that if the elements were arranged strictly in order of their increasing atomic weights, then some elements did not fit within this scheme of classification. Therefore, he ignored the order of atomic weights in some cases. For example, the atomic weight of iodine is lower than that of tellurium. Still Mendeleev placed tellurium (in Group VI) before iodine (in Group V II) simply because iodine's properties are so similar to fluorine, chlorine, and bromine.
Q.3: What is the basic difference in approach between the Mendeleev's Periodic Law and the Modern Periodic Law? Ans : Mendeleev's Periodic Law states that the physical and chemical properties of elements are periodic functions of their atomic weights, On the other hand, the Modern periodic Law states that the physical and chemical properties of elements are periodic functions of their atomic numbers.
Q.4: On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements. Ans : In the periodic table of the elements, a period indicates the value of the principal quantum number (n) for the outermost shells. Each period begins With principal quantum number The value of n for the sixth period is 6. For n = 6, azimuthal quantum number (l) can have values of 0, 1, 2, 3, 4. According to Aufbau's principle, electrons are added to different orbitals in order of their increasing energies. The energy of the 6d subshell is even higher than that of the 7s subshell. In the 6th period, electrons can be filled in only 6s, 4f, 5d, and 6p subshells. Now, 6s has one orbital, 4f has seven orbitals, 5d has five orbitals, and 6p has three orbitals, Therefore, there are a total of sixteen (1 + 7 + 5 + 3 = 15) orbitals available. According to Pauli's exclusion principle, each orbital can accommodate a maximum of 2 electrons. Thus, 16 orbitals can accommodate a maximum of 32 electrons. Hence, the sixth period of the periodic table should have 32 elements.
Q.5: In terms of period and group where would you locate the element with Z =114? Ans : Elements With atomic numbers from Z = 87 to Z = 114 are present in the 7th period of the periodic table. Thus, the element with Z = 114 is present in the 7th period of the periodic table. In the 7th period, first two elements with Z = 87 and Z = 88 are s-block elements, the next 14 elements excluding Z = 89 i.e., those with Z = 90 - 103 are f - block elements, ten elements Kith Z = 89 and Z = 104 - 112 are d block elements, and the elements Z = 113 -118 are p — block elements. Therefore, the element with Z = 114 is the second p — block element in the 7th period. Thus, the element with Z = 114 is present in the 7th period and 4th group of the periodic table.
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Class 11 Chemistry Chapter 1 Ncert Solutions Study Rankers
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