Periodic Table and Its Properties Solved Examples

SOLVED EXAMPLES Ex. 1 Following are the valence shell electronic configurations of some elements. (i) 3s2 3p5 (ii) 3d10 4s2 (iii) 2s2 3p6 4s1 (iv) 1s2 2s2 Find out the blocks to which they belong in the periodic table ? Ans. (i) p-block (ii) d-block (iii) s-block(iv) s-block Sol. The block of the elements depend on the type of sub-shell which receive the last electron. In case of (i) it enters in 3p-subshell, (ii)z it enters 3d-subshell, (iii) it enters 4s-subshell and (iv) it enters 2s-subshell. Ex. 2. A M2+ ion derived from a metal in the first transition metal series has four electrons in 3d subshell. What element might M be ? Ans. Chromium Sol.  Electron configuration of M2+ is : [Ar]18 4s0 3d4 Electron configuration of M is : [Ar]18 4s1 3d5 (and not 4s2 3d4) So total number of electrons = 24. Hence, metal M is chromium (Cr). Ex. 3 Find out the group of the element having the electronic configuration, 1s2 2s2 2p6 3s2 3p6 3d6 4s2 . Ans. As last electron enters in d-subshell, therefore this belongs to d-block. For d-block element the group number is equal to the number of valence shell electrons + number of electrons in (n-1) d-subshell. So, group number = 6 + 2 = 8. Ex. 4 Arrange the following ions in the increasing order of their size : Be2+ , Cl–, S2– , Na + , Mg2+ , Br– ? Ans. Be2+ < Mg2+ < Na + < Cl– < S2– < Br – Sol. Be2+ is smaller than Mg2+ as Be 2+ has one shell where as Mg2+ has two shells. Mg2+ and Na+ are isoelectronic species : Ionic radius  1/nuclear charge. Cl– and S2– are isoelectronic species : Ionic radius  1/nuclear charge. Cl– is smaller than Br – as Cl– has three shells where as Br– has four shells. Ex. 5 In Column-I, there are given electronic configurations of some elements. Match these with the correct metals given in Column-II : Column-I Column-II (A) ns2, np5 (B) (n – 1) d10 , ns1 (C) (n – 1) d , ns (D) (n – 1) d , ns , np 5 10 1 2 6 (p) Chromium (q) Copper (r) Krypton (s) Bromine Ans. (A)  (s) ; (B)  (q) ; (C) (p) ; (D) (r). Sol. (A) ns2np5 is general valence shell electron configuration of halogens. So this configuration belongs to bromine. (B) (n – 1) d1-10 ns1-2 ; This is electron configuration of d-block elements. As it contains (n – 1) d 10ns1 configuration it belongs to copper. (C) (n – 1) d1-10 ns1-2 ; This is electron configuration of d-block elements. As it contains (n – 1) d 5 ns1 configuration it belongs to chromium. (D) Noble gases has valence shell electron configuration ns2 np6, so it belongs to krypton. PERIODIC TABLE AND ITS PROPERTIES Ex. 6 Match the metals given in Column-II with their type given in Column-I : Column-I Column-II (A) Metalloid (p) Sulphur (B) Radioactive (q) Gold (C) Transition metal (r) Arsenic (D) Chalcogen (s) Uranium Ans. (A)  (r) ; (B)  (s) ; (C) (q) ; (D) (p) Sol. (A) Arsenic is a metalloid because it behaves as metal (forming cation, As3+ -AsCl3) as well as nonmetal (forming anion, As3– -AsH3). (B) Uranium is a radioactive element. (C) Those elements which in their neutral atoms or in most common oxidation state have partially filled d-orbitals are called as transition elements. Gold in its +3 oxidation state has electron configuration [Xe] 54, 5d86s0 . (D) 16th group elements like oxygen and sulphur are ore forming elements and therefore are called as chalcogens. Ex. 7 Match the metals given in Column-II with their type given in Column-I : Column-I Column-II (A) Representative element (p) Cerium (B) Lanthanide (q) Aluminium (C) Coinage metal (r) Thorium (D) Actinide (s) Gold Ans. (A)  (q) ; (B)  (p) ; (C) (s) ; (D) (r) Sol. (A) s-block and p-block elements are collectively called as representative elements. As in aluminium last electron enters in p-subshell ([Ne]103s23p1). (B) Lanthanide series follows lanthanum (atomic number 57) and starts from cerium (atomic number 58) to lutetium (atomic number 71), fourteen 4f- series elements. (C) Group 11- transition elements copper, silver & gold are known as coinage metals (used for making the coins). (D) Actinides series follows actinium (atomic number 89) and starts from thorium (atomic number 90) to lawrencium (atomic number 103), fourteen 5f- series elements. Ex. 8 The (IE1) and the (IE 2) in kJ mol–1 of a few elements designated by Roman numerals are shown below: I II III IE1 403 549 1142 IE2 2640 1060 2080 Which of the above elements is likely to be a (a) non-metal (b) alkali metal (c) alkaline earth metal ? Ans. (a) non-metal(III) – Due to highest ionisation energy, (IE1) and (IE2). (b) alkali metal (I) – Due to lowest ionisation energy, (IE1) and there is quite high jump in (IE2) due to inert gas configuration. (c) alkaline earth metal (II) – There is little difference in (IE1) and (IE2) and the value of (IE1) is slightly greater than(I) due to stable configuration(ns2). Ex. 9 Ionisation energy and electron affinity of fluorine are respectively 17.42 and 3.45 eV. Calculate electronegativity of fluorine atom. Sol. According to Mulliken’s electronegativity (M) = = Ionisation energy + Electron affinity 2 17.42 + 3.45 = 10.435 2 Therefore, electronegativity on Pauling’s scale (P) = 10.435 = 3.726 2.8 Ans. P = 3.726 Ex. 10 Why the electron gain enthalpy values of alkaline earth metals are lower (i.e. less negative) or positive ? Sol. The general valence shell electron configuration of alkaline earth metals is ns2 (stable configuration). The extra electron must enter np subshell, which is effectively shielding by the two ns electrons and the inner electrons. Consequently, the alkaline earth metals have little or no tendency to pick up an extra electron Ex. 11 Match the particulars given in Column-I with the process/metal / species given in Column-II. Column-I Column-II (A) Isoelectronic species (p) A+(g) + energy  A ++(g) + e– (g) (B) Half filled orbital (q) Ar, K+, Ca++ (C) Second ionisation energy (r) Lutetium (D) Inner transition element (s) Antimony Ans. (A)  (q) ; (B)  (s) ; (C) (p) ; (D) (r) Sol. (A) Species having same number of electrons but different nuclear charge are called isoelectronic species. Ar, K + & Ca++ have same number of electrons i.e. 18 but 18, 19 & 20 number of protons respectively. (B) np3, (n–1) d5 and (n–2) f7 represent half filled orbitals. Antimony has ([Kr]36 4d105s25p3). (C) The energy required to remove an electron from an univalent cation(g) is called second ionisation energy. (D) 4f and 5f- series elements are called inner transition elements because they have three outer most shells incomplete. Ex. 12 Ans. The Column-I has certain details about the elements of s-, p- and d-block elements. Match those with the group number of the elements listed in Column-II. Column-I Column-II (element / elements) (group number) (A) An element whose fourth shell contains two p-electrons (p) 8th group (B) An element whose valence shell contains one unpaired p-electron (q) 12th group (C) An element which receives last electron in (n – 1) d-subshell (r) 14th group (D) An element with the ground-state electron configuration [Ar]4s23d10 (s) 17th group (A)  (r) ; (B)  (s) ; (C) (p, q) ; (D) (q). PERIODIC TABLE AND ITS PROPERTIES Sol. (A) [Ar]3d104s24p2 : = 10 + 4 = 14. Fourth shell contains two electron in 4p-sub shell i.e., 4p2. Therefore, group number (B) Halogens (i.e. group number 17) have valence shell electronic configuration ns2np5 and there is one unpaired electron in p-subshell i.e., (C) The element in which last electron enters in d-subshell belongs to d-block. For d-block elements the group number = number of electrons in valence shell + number of electrons in (n – 1) d-subshell. Group number 8. Valence shell electronic configuration is ns2(n – 1)d6. Therefore, group number = 2 + 6 = 8. Like wise, group 12 is ns2(n – 1)d10. Therefore, group number = 2 + 10 = 12. So in group 8 and 12 last electron enters in d-subshell. (D) For electronic configuration. [Ar]4s23d10 the group number = 2 + 10 = 12. Ex. 13 Match the type of elements / characteristic of the elements listed in Column-I with the correct element listed in Column-II. Column-I (A) Highest (B) 1st Column-II ionisation energy (p) Technitium Highest electronegativity (q) Lithium (C) Synthetic element (r) Helium (D) Strongest reducing agent (s) Fluorine Ans. (A)  (r) ; (B)  (s) ; (C) (p) ; (D) (q). Sol. (A) Helium has highest 1st ionisation energy amongst all the elements of periodic table because of ns2 valence electron configuration and its small size of atom. (B) Fluorine has highest electronegativity i.e. 4.0 on Pauling scale on account of its small size. (C) Technitium is a man made element. (D) Lithium is a strongest reducing agent because of its highest negative value of Eº due to its higher hydration energy on account of its small size of atom.