Important Questions of Periodic Classification of Elements Class 11
Important Questions of Periodic Classification of Elements Class 11
Que 1. An element is present in the third period of the p-block. It has 5 electrons in its outermost shell. Predict its group. How many unpaired electrons does it have?
Ans 1. It belongs to the 15th group (P). It has 3 unpaired electrons.
Que 2. An element X with Z = 112 has been recently discovered. Predict its electronic configuration and suggest the group in which it is present.
Ans 2. [Rn] 5f14 6d10 7s2. It belongs to the 12th group.
Que 3. The electronic configuration of an element is 1s2 2s2 2p6 3s2 3p5. Name the period and the group to which it belongs.
Ans 3. Third-period Group 17.
Que 4. Arrange Cl, Cl–, and Cl+ ions in order of increasing size.
Ans 4. C+ < Cl < C–.
Que 5. Arrange the following in increasing order of size.
N3-, Na+, F–, O2-, Mg2+
Ans 5. Mg2 + < Na+ < F– < O2- < N3-
Important Questions of Periodic Classification of Elements Class 11
Que 6. Give the formula of one species positively charged and one negatively charged that will be isoelectronic with Ne.
Ans 6. Na+, F–.
Que 7. Argon has atomic number 18 and belongs to the 3rd period and 18th group. Predict the group and period for the element having atomic number 19.
Ans 7. Group I, Period 4th.
Que 8. Do elements with high Ionisation Enthalpy have high Electron Affinity?
Ans 8. Normally is true that the elements with haying high value of Ionisation Enthalpy have a high value of E affinity. But however, there are marked exceptions. It is seen that elements, with stable electronic configurations, have very high values of I-Energies as it is difficult to remove electrons as is the case with 15th and 18th group elements but in such case, electrons cannot be added easily so that is why elements of 15th group have almost zero Electron Affinity and elements of 18th group have got zero Electron Affinity whereas their Ionization energy values are very high.
Que 9. What is a periodic classification of elements?
Ans 9. By periodic classification of the elements, we mean the arrangement of the elements in such a way that the elements with similar physical and chemical properties are grouped together, and for this various scientists made contributions but the contributions made by Mendeleev are of great significance and he gave a periodic table which called as Mendeleev’s Periodic ‘Table which was older and replaced by the long form of the periodic table.
Que 10. Distinguish between s and p block elements.
Ans 10. They can be distinguished as follows:
s – block elements:
i). They have got the general configuration of the valence shell, ns1-2.
ii). They are all metals.
iii). Their compounds are mostly ionic.
iv). They are generally strong reducing agents.
v). They mostly impart characteristic color to the flame.
vi). They have low ionization energies.
vii). They show fixed oxidation states,
p – block elements:
i). The valence shell electronic configuration of p block elements in ns2 p1-6.
ii) They are mostly non-metals.
iii). Their compounds are mostly covalent.
iv). They are generally strong oxidizing agents.
v). Mostly they do not impart color to the flame.
vi). They have got a comparatively higher value of I.E.
vii). They show variable oxidation states.
Important Questions of Periodic Classification of Elements Class 11
Que 11. Explain why ionization enthalpies decrease down a group of the Periodic Table.
Ans 11. The decrease in ionization enthalpies down any group is because of the following factors:
(i) There is an increase in the number of the main energy shells moving from one element to another in a group.
(ii) There is also an increase in the magnitude of the screening effect due to the gradual increase in the number of inner electrons.
Que 12. Why does the first ionization enthalpy increase as we move from left to right across a given period of the Periodic Table?
Ans 12. The value of ionization enthalpy increases with the increase in atomic number across the period. This is due to the fact that in moving across the period from left to right.,
(i) Nuclear charge increases regularly by one unit.
(ii) The progressive addition of electrons occurs at the same level.
(iii) Atomic size decreases.
This is due to the gradual increase in nuclear charge and a simultaneous decrease in atomic size the electrons are more and more tightly bound to the nucleus. This results in a gradual increase in ionization energy across the period.
Que 13. How do atomic radii vary across a period with an atomic number in the periodic table?
Ans 13. Variation of Atomic radii across a period: atomic radii decrease with the increase in the atomic number in a period. For example, atomic radii decrease from lithium to fluorine in the second period.
In moving from left to right across the period, the nuclear charge increases progressively by one unit but the additional electron goes” to the same principal shell. As a result, the electron cloud is pulled closer to the nucleus by increased effective nuclear charge. This causes a decrease in atomic size.
Que 14. The 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
(ii) [Ar]3d10 4s2
(iii) [Ar]3d10 4s2 4p6 5s2
(iv) [Arl 3d10 4s2 4p6 5s1
Ans 14. (i) [Ar]4s2 is Calcium metal with At. no. = 20.
(ii) [Ar]3d10 4s2 is Zinc metal with At. no. = 30.
(iii) [Ar]3d10 4s2 4p6 5s2 is Strontium metal with At. no. = 38.
(iv) [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, the metallic character increases from
Zn < Ca < Sr < Rb or (ii) < (i) < (iii) < (iv)
Que 15. Explain the important general characteristics of groups in the modem periodic table in brief.
Ans 15. The elements of a group show the following important similar characteristics.
(i) 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 Earth) Here ns2 outer configuration, and halogens (group 17) have ns2 np5 configuration (where n is the outermost shell).
(ii) 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, and I are all non-metals, and their- reactivity goes on decreasing from top to bottom.
Important Questions of Periodic Classification of Elements Class 11
Que 16. Explain the electronic configuration in periods in the periodic table.
Ans 16. 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 gives 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.
Que 17. Explain the variation of valence in the periodic table.
Ans 17. Variation of valence in a group as well as across a period in the periodic table occurs as follows:
(i) 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 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.
(ii) 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 increase, and then decreases.
Que 18. Write the atomic number of the element present in the third period and a seventeenth group of the periodic table.
Ans 18. There are two elements in the 1 𝑠𝑡 period and eight elements in the 2 𝑛𝑑 period., The third period starts with the element Z = 11. Now, there are eight elements in the third period. Thus, the 3 𝑟𝑑 period ends with the element with Z = 18 i.e., the element in the 18𝑡ℎ group of the third period has Z = 18. Hence, the element in the 17𝑡ℎ group of the third period has atomic number Z = 17.
Que 19. Consider the following species: 𝑁3−, 𝑂2−, 𝐹−, N𝑎+, M𝑔2+ and A𝑙3+
(a) What is common in them?
(b) Arrange them in the order of increasing ionic radii.
Ans 19. (a) Each of the given species (ions) has the same number of electrons (10 electrons). Hence, the given species are isoelectronic.
(b) The ionic radii of isoelectronic species increase with a decrease in the magnitudes of nuclear charge.
The arrangement of the given species in order of their increasing nuclear charge is as follows: 𝑁3− < 𝑂2− < 𝐹− < N𝑎+ < M𝑔2+ < A𝑙3+
Nuclear charge = +7 +8 +9 +11 +12 +13
Therefore, the arrangement of the given species in order of their increasing ionic radii is as follows: A𝑙3+ < M𝑔2+ < N𝑎+ < 𝐹− < 𝑂2− < 𝑁3−
Que 20. Among the second-period elements, the actual ionization enthalpies are in the order Li < B < Be < C < O < N < F < Ne. Explain why
(i) Be has higher ∆𝑖H than B
(ii) O has lower ∆𝑖H than N and F?
Ans 20. (i) During the process of ionization, the electron to be removed from the beryllium atom is a 2𝑠−electron, whereas the electron to be removed from the boron atom is a 2𝑝−electron. Now, 2𝑠−electrons are more strongly attached to the nucleus than 2𝑝−electrons. Therefore, more energy is required to remove a 2𝑠−electron of beryllium than that required to remove a 2𝑝−electron of boron. Hence, beryllium has higher ∆𝑖H than boron.
(ii) In nitrogen, the three 2𝑝−electrons of nitrogen occupy three different atomic orbitals. However, in oxygen, two of the four 2𝑝−electrons of oxygen occupy the same 2𝑝−orbital. This results in increased electron-electron repulsion in the oxygen atom. As a result, the energy required to remove the fourth 2𝑝−electron from oxygen is less as compared to the energy required to remove one of the three 2𝑝−electrons from nitrogen. Hence, oxygen has lower ∆𝑖H than nitrogen. Fluorine contains one electron and one proton more than oxygen. As the electron is being added to the same shell, the increase in nuclear attraction (due to the addition of a proton) is more than the increase in electronic repulsion (due to the addition of an electron). Therefore, the valence electrons in fluorine atoms experience a more effective nuclear charge than that experienced by the electrons present in oxygen. As a result, more energy is required to remove an electron from a fluorine atom than that required to remove an electron from an oxygen atom. Hence, oxygen has lower ∆𝑖H than fluorine.
Important Questions of Periodic Classification of Elements Class 11
Que 21. Explain the meaning of a positive electron gain enthalpy.
Ans 21. Positive electron gain enthalpy means that energy is required rather than released during the addition of an electron to a neutral gaseous atom to form a gaseous ion.
Que 22. Which is smaller in size – Li or F and why?
Ans 22. F is smaller in size than Li because the effective nuclear charge increases moving from left to right across the period, consequently, the size of the atom decreases.
Que 23. Why zero group elements are inert?
Ans 23. Zero group elements are inert because they have completely filled valence shells. Neither do they have the tendency to lose electrons nor to gain or share electrons with other elements.
Que 24. Why argon (atomic mass 39.94) has been placed before potassium (atomic mass 39.10) in the periodic table?
Ans 24. In the modern periodic table, elements have been placed in order of their increasing atomic numbers, not on the basis of atomic masses. Since the atomic number of argon is 18 and that of potassium is 19, argon is placed before potassium.
Que 25. Out of the metallic radius and covalent radius of an element, which is larger and why?
Ans 25. The metallic radius of an atom is always larger than the covalent radius of the element because a metallic bond is a weak bond in comparison to a covalent bond. The atoms held by covalent bonds are closer to each other.
Important Questions of Periodic Classification of Elements Class 11
Que 26. Among the following compounds CsI, CsF, LiF, and NaF, which one has the highest cation-to-anion size ratio?
Ans 26. Amongst alkali metals cation i.e., Cs, Na, and Li, the largest size of cation is of Cs+. Among the halides i.e., F and I, the smallest size is of F–. Therefore, the ratio of Cs+ and F– is the highest.
Que 27. Why lanthanoids and actinoids are placed in separate rows at the bottom of the periodic table?
Ans 27. These are separately placed at the bottom of the periodic table for convenience. If they are placed within the body of the periodic table, the periodic table will become very long and cumbersome.
Que 28. Arrange the isoelectronic species O2–, F–, Na+ and Mg2+ in order of their,
(a) Increasing effective nuclear charge
(b) Increasing ionic radius
(c) Increasing ionization enthalpy. Give reasons for their arrangement also.
Ans 28. (a) As the positive charge increases on the atom. The effective nuclear charge increases, thus increasing the force of attraction of the nucleus on the valence electrons. The order of species becomes O2– < F– < Na+ < Mg2+
(b) As the effective nuclear charge increases, the size of the ion decreases accordingly due to the increased force of attraction by the nucleus on the valence electrons. So the size of the ions can be arranged accordingly in increasing order as follows: Mg2+ < Na+ < F– < O2–
(c) The increased effective nuclear charge increases the hold of the nucleus on electrons means a higher amount of energy will be required to remove electrons from the valence shell. So, the order of ionization enthalpy becomes as O2– < F– < Na+ < Mg2+.
Que 29. (i) What is the basic difference between the terms electron gain enthalpy and electronegativity?
(ii) What do you understand by shielding effect or screening effect?
Ans 29. (i) Electron gain enthalpy is the property of an isolated atom of an element in a gaseous state whereas electronegativity is the property of an atom of the element when it is bonded to some other atom. The value of electron gain enthalpy can be experimentally determined while that of electronegativity cannot be determined experimentally.
(ii) In multi-electron atoms, the electrons present in the outermost shell do not experience the complete nuclear charge because of the repulsive interaction of the intervening electrons. Thus, the outermost electrons are shielded or screened from the nucleus by the inner electrons. This is known as the screening effect. If the number of electrons in the inner shells is large, the screening effect will be large.
Que 30. For the following sets of elements, select the element which has the property noted below: (i) Largest van der Waals radius: O, N, Cl, H
(ii) Most positive electron gain enthalpy: Mg, Ne, Ar, N
(iii) Largest screening effect: Na, Cs, Mg, Al
Ans 30. (i) Largest van der Waal’s radius is of Cl because, with the increase in size, the van der Waal’s radii increase.
(ii) Ne has the most positive electron gain enthalpy due to high effective nuclear charge.
(iii) Cs has the largest screening effect because the screening effect is more pronounced whenever there is a full shell of electrons between the outermost shell and nucleus.
Important Questions of Periodic Classification of Elements Class 11
