Class 11 Chemistry Classification of Elements and Periodicity in Properties Question and Answers
Question 1. What is the basic theme of organization in the periodic table?
Answer: The basic theme of organization in the periodic table is to study different physical and chemical properties of all the elements and their compounds simply and systematically.
Elements belonging to the same group have similar physical and chemical properties.
So, if the physical and chemical property of any one element of a group is known, then it is possible to predict the physical and chemical properties of the remaining elements of that group.
Therefore, it is not important to keep in mind the physical and chemical properties of all elements in the periodic table.
Question 2. Which property did Medeleev use to classify the elements in his periodic table? Did he stick to that?
Answer: Mendeleev classified the elements based on their increasing atomic weights in the periodic table. He arranged almost 63 elements in order of their increasing atomic weights placing together elements with similar properties in a vertical column
He observed that while classifying elements in the periodic table according to increasing atomic weight, certain elements had different properties than those elements belonging to the same group.
For such cases, Mendeleev prioritized the properties of the element over its atomic weight.
So, he placed an element with a higher atomic weight before an element with a lower atomic weight.
For example, iodine [I (126.91)] with a lower atomic weight than tellurium [Te (127.61)] is placed after tellurium in group VII along with elements like fluorine, chlorine, etc., due to similarities in properties with these elements.
Thus, Mendeleev did not stick to his idea of classifying elements only according to the increasing atomic weights.
Question 3. What is the basic difference in approach between Mendeleev’s Periodic Law & the Modern Periodic Law?
Answer: According to Mendeleev’s periodic law, the physical and chemical properties of elements are a periodic function of their atomic weights.
On the other hand, the modern periodic table states that the physical and chemical properties of the elements are a periodic function of their atomic numbers.
Thus the basic difference in approach between Medeleev’s periodic law and modern periodic law is the change in the basis of classification of elements from atomic weight to atomic number.
Based on quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
In the modern periodic table, each period begins with the filling of a new principle energy level. Therefore, the sixth period starts with the filling of the principal quantum number, n = 6.
In the sixthperiod elements, the electron first enters the 6s -orbital, and then from left to right across a period the electrons enter the 4/, 5d, and 6p orbitals of the elements.
Filling of electrons in orbitals case of6th period continues till new principal energy level of quantum number, n = 7 begins, i.e., for elements of the sixth period, electrons fill up the 6s, 4/, 5d, and 6p orbitals total number of orbitals in this case =1 + 7 + 5 + 3 = 16. Since each orbital can accommodate a maximum of two electrons, there can be 16 X 2 or 32 elements in the sixth period.
Question 5. In terms of period and group where would you locate the element with Z = 114?
Answer: it is known that the difference between atomic numbers of the successive members of any group is 8,8,18,18 and 32 (from top to bottom). So, the element with atomic number 114 will lie just below the element with atomic number (114- 32) = 82.
The element with atomic number 82 is lead (Pb), which is a member of the 6th period belonging to group number 14 (p -block element). Thus the element with atomic number 114 takes its position in the 7th period and group- 14 (p -block element) ofthe periodic table.
Question 6. Write the atomic number of the element present in the third period & seventeenth group of the periodic table.
Answer: The general electronic configuration of the valence shell of the elements of group-17 (halogens) is ns2np5.
For the third period, n = 3. Therefore, the electronic configuration of the valence shell of the element ofthird period and group-17 Is 3ia3p1 and the complete electronic configuration of this element Is ls22s22p63s23p5.
There are a total of 17 electrons in this element. Thus, the element In the third period and seventeenth group of the periodic table has atomic number = 17.
Question 7. Which clement do you think would have been named by—Lawrence Berkeley Laboratory and Seaborg’s group?
Answer: Lawrencium, Lr and Berkellum, UK. Seaborgium, Sg.
Question 8. Why do elements In the same group have similar physical and chemical properties?
Answer: Elements belonging to the same group have similar valence shell electronic configurations so they have similar physical and chemical properties.
Question 9. What do you mean by isoelectronic species? Name a species that will be isoelectronic with each of the given atoms or ions: F- Ar, Mg2+, Rb+.
Answer: Isoelectronic ions are ions of different elements that have the same number of electrons but different magnitudes of nuclear charge.
There are (9 + 1) or 10 electrons in F-. Isoelectronic species of F- are nitride (N3-) ion [7 + 3], oxide (O2-) ion [8 + 2], neon (Ne) atom [10], sodium (Na+) ion [11-1], magnesium (Mg2+) ion [12-2], aluminum (Al3+) ion [13-3].
There are 18 electrons in Ar. Isoelectronic species of Ar phosphide (P3-) ion [15 + 3], sulfide (S2-) ion [16 + 2]. chloride (Cl ion [17 + 1], potassium (K+) ion [19 -1], calcium (Ca2+) ion [20-2].
There are (12-2) = 10 electrons in Mg2t. Isoelectronic species of Mg2+ are nitride (N3-) ton [7 + 3], oxide (O2-) ion [8 + 2], fluoride (F-)ion [9+1] sodium (Na+) ion [11-1].
There are (37-1)- 36 electrons In Kb 1 . Isoelectronic species of Rb+ are bromide (Br-) Ion [35 + 1], krypton (Kr) atom [36], strontium (Sr2′) Ion [30-2].
Question 12. Consider the given species: N3-, 02-, I1-, Nn+, Mg2+ and Al3+. What Is Common In them? Arrange them in the order of increasing ionic radii.
Answer: Each of the given ions has 10 electrons. Hence, they are all isoelectronic species. Ionic radii isoelectronic ions decrease with an increase in the magnitude of the nuclear charge.
The order of increasing nuclear charge of the given isoelectronic ions is N3- < O2- <F- < Na+ < Mg2+ < Al3+.
Therefore, the order of increasing ionic radii is: Al3+ < Mg2+ < Na+ < F- < O2- < N3-.
Question 13. 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]
Answer: The ionization energy of an element is defined as the minimum amount of energy required to remove the most loosely bound electron from the valence shell of an isolated gaseous atom existing in its ground state to form a cation in the gaseous state.
Electron-gain enthalpy is defined as the enthalpy change involved when an electron is added to an isolated gaseous atom in its lowest energy state (ground state) to form a gaseous ion carrying a unit negative charge.
The force with which an electron gets attracted by the nucleus of an atom is influenced by the presence of other atoms in the molecule or the neighborhood. Thus, to determine the ionization enthalpy, the interatomic forces should be minimal.
Interatomic forces are minimal in the case of the gaseous state as the atoms are far apart from each other.
Consequently, the value of ionization enthalpy is less affected by the surroundings. Similarly, for electron affinity, the interatomic forces of attraction should be minimal for the corresponding atom. Tor Tills reason, the term ‘Isolated gaseous atom’ Is used while defining Ionisation enthalpy and electron-gain enthalpy.
Ground state means the state at which the atom exists In Its most stable state. If the atom Is In the excited state, then the amount of heat applied to remove an electron or the amount of heat liberated due to the addition of an electron is low.
So, for comparison, the term ‘ground state’ Is used while defining ionization enthalpy and electron-gain enthalpy.
Question 15. The energy of an electron in the ground state of the Hatom is -2.18 X 10-18J. Calculate the ionization enthalpy of atomic hydrogen in terms of J . mol-1. [Hint: Apply the idea of the mole concept.]
Answer: Amount of energy required to remove an electron from a hydrogen atom at ground state =Eog-El = 0-El = -(-2.18 X 10-18)J =2.18 X 10-8 J
Ionization enthalpy of atomic hydrogen per mole = 2.18 X 10“18 X 6.022 X 1023 = 1312.8 X 103 J.mol-1.
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?
Answer: image-
Evident that in both atoms, the valence electrons enter the 3s orbital. However, the nuclear charge of the Mg atom (+12) is greater than that of the Na atom (+11).
Again, the 3s orbital of the Mg atom being filled is more stable than half-filled.
3s -orbital of Na atom. Thus, the first ionization enthalpy of sodium is lower than that of magnesium.
On the other hand, the removal of one electron from the valence shell of the Na atom leads to the formation of the Na+ ion whose electronic configuration is highly stable (similar to inert gas, neon).
So high amount of energy is required to remove the second electron because it disturbs the stable electronic configuration. However, the electronic configuration of Mg+ is not as stable as that of Na+, but electronic configuration of Mg2+ is more stable as it is similar to the electronic configuration of the inert gas, neon.
So, less amount of energy is required to remove an electron from Mg+. Thus, the second ionization enthalpy of sodium is higher than that of magnesium.
Question 18. What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
Answer: The two factors are: atomic size and screening effect.
Question 19.First ionisation enthalpy values (in kjmol-1) of group-13 elements are: B = 801, A1 = 577, Ga = 579, In = 558 and T1 = 589. How would you explain this deviation from the general trend?
Answer: On moving down group-13 from B to Al, ionization enthalpy decreases due to an increase in atomic size and shielding effect which jointly overcome the effect of an increase in nuclear charge. However, ionization enthalpy increases slightly on moving from Al to Ga (2 kj.mol-1).
This is because due to poor shielding of valence electrons by 3d -electrons effective nuclear charge on Ga is slightly more than Al.
On moving from Ga to In, the shielding effect of all the inner electrons overcomes the effect of the increase in nuclear charge. Thus, the ionization enthalpy of In is lower than Ga.
Again, on moving from Into T1, there is a further increase in nuclear charge which overcomes the shielding effect of all electrons present in the inner shells including those of 4f- and 5d -orbitals. So, the ionization enthalpy of T1 is higher than In.
Question 20. Which of the given pairs would have a more negative electron-gain enthalpy: O or F F or Cl?
Answer: 0 and F both belong to the second period. As one moves from O to F, atomic size decreases and nuclear charge increases. Due to these factors, the incoming electron enters the valence shell, and the amount of energy liberated in the case of F is more than that of O.
Again, Fatom (ls22s22p5) accepts one electron to form F- ion (lsz2s22p6) which has a stable configuration similar to neon. However, O-atom when converted to O- does not attain any stable configuration.
Thus energy released is much higher in going from F to F- than in going from 0 to 0-. So, the electron-gain enthalpy of is much more negative than that of O
