Tiêu đề 3.PERIODIC CLASSIFICATION OF ELEMENTS ( 104 - 122 ).pdf ✅

PERIODIC TABLE 104 NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED - VOL - I NISHITH Multimedia India (Pvt.) Ltd., PERIODIC TABLE LEVEL-V SINGLE ANSWER QUESTIONS 1. Which of the following orders is incorrect with respect to the property indicated ? (A) NH PH AsH 3 3 3   -- acidic behaviour (B) Li Be B C    -- first ionisation energy (C) Al O MgO Na O K O 2 3 2 2    -- basic character (D) Li Na K Cs        -- ionic radius 2. The strength of an oxyacid ( ) E OH n ; where E is the central atom, depends upon (A) electronegativity of E but not on atomic size (B) atomic size of E but not on electronegativity of E (C) not on oxidation state of E in oxyacids (D) atomic size, electronegativity and oxidation state of E 3. Molecular sizes of ICl and Br2 are nearly same but boiling point of ICl is about 400C higher than Br2 . Because. (A) I-Cl is weaker than Br–Br bond (B) ionisation energy of Br atom is less than I atom (C) ICl is a polar where as Br2 is a non-polar molecule (D) ICl is non-polar where as Br2 is polar 4. In general, the configuration of lanthanides is 1 14 2 6 0 1 2 ( 2) ( 1) n f n s p d ns     . It has been observed that, with increase in atomic num- ber of lanthanides, there is a decrease in ionic radii from 0 La A (1.22 ) to 0 Lu A (0.99 ). The reason for decrease in ionic radii is an increase in (A) electronegative character (B) valency electrons and number of shells (C) atomic and ionic volumes (D) nuclear attraction for valence electrons lead- ing to inward shrinking of shell 5. The dipole moment of H O2 2 is more than that of H O2 but H O2 2 is not a good solvent because (A) it has a very high dielectric constant so that ionic compounds cannot be dissolved in it (B) it does not act as an oxidising agent (C) it acts as a reducing agent (D) it dissociates easily and acts as an oxidising agent in chemical reactions 6. An increase in both atomic and ionic radii with atomic number occurs in any group of the periodic table and in accordance with this the ionic radii of Ti(IV) and Zr(IV) ions are 0 0.68A and 0 0.74A respectively; but for Hf(IV) ion, the ionic radius is 0 0.75A , which is almost the same as that for Zr(IV) ion. This is due to (A) greater degree of covalency in compounds of 4 Hf  (B) lanthanide contraction (C) actinide contraction (D) difference in co-ordination number of Zn4+ and 4 Hf  in their compounds 7. Mercury is the only metal which is liquid at 0 0C. This is due to its (A) very high ionisation energy and weak metal- lic bond (B) low ionisation potential and high electrone- gativity (C) high atomic mass and small size (D) high electronegativity and low ionisation po- tential 8. The Radioactive element which was discov- ered by Madam Curie and the name of which is based on her country is (A) Uranium (B) Radium (C) Polonium (D) Neptunium 9. Which is the correct order of ionisation en- ergies? (A) F F Cl Cl      (B) F F Cl Cl      (C) F F Cl Cl      (D) F F Cl Cl     
NISHITH Multimedia India (Pvt.) Ltd., 105 JEE MAINS - CW - VOL - I NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED - VOL - I PERIODIC TABLE 10. Consider two changes 2 1 2 A A e E and A A e E ; ;          Where, E1 and E2 are the two energies re- quired to pull out two electrons. The cor- rect order of E1 and E2 would be (A) E E 1 2  (B) E E 1 2  (C) E E 1 2  (D) E E 1 2  11. Fluorine was discovered very late because of its high reactivity. It is always found in combined state in which it cannot be identi- fied easily. The fact that accompanied with the high reactivity of fluorine is (A) low ionization potential (B) the smallest size amongst halogen (C) low F–F bond-energy (D) the higher electronegativity 12. The atomic radii of fluorine and neon (in Ang- strom unit) respectively will be (A) 0.72, 1.60 (B) 1.60, 1.60 (C) 0.72, 0.72 (D) 0.82, 0.82 13. The ionic radii of 3 2 N O F Na , , ,     follows the order (A) 3 2 N O F Na        (B) 3 2 N Na O F        (C) 2 3 Na O N F        (D) 2 3 O F N Na        14. Aqueous solutions of two compounds M O H 1   and M O H 2   are prepared in two different beakers. If the electronega- tivity of M1 =3.4, M2 =1.2, O=3.5 and H=2.1, then the nature of the two solutions will be respectively. (A) acidic, basic (B) acidic, acidic (C) basic, acidic (D) basic, basic 15. The attraction that an atom exerts on a pair of electron that are being shared with an- other for forming single covalent bond is referred to as its (A) electron affinity (B) electronegativity (C) ionisation potential (D) valency 16. The correct order of the first ionisation po- tentials is (A) Ne Cl P S Al Mg      (B) Ne Cl P S Mg Al      (C) Ne Cl S P Mg Al      (D) Ne Cl S P Al Mg      17. Three elements X, Y and Z have atomic num- bers 19, 37 and 55 respectively. Then the correct statements(s) is/are (A) Their ionisation potential would increase with increasing atomic numbers (B) Y would have an ionisation potential between those of X and Z (C) Y would have tha highest ionisation potential (D) Z would have the highest ionisation potential 18. Which of the following pairs is chemically similar? (A) Zr–Hf (B) Cu–Ag (C) Fe–Au (D) Hf–La 19. The first ionisation energy of elements in- creases across the period. The first ionisation energy of nitrogen and oxygen will be respectively. (A) 13.6 eV, 14.6 eV (B) 14.6eV, 13.6eV (C) 13.6eV, 13.6 eV (D) 14.6eV, 14.6eV 20. Pauling's electronegativity values for ele- ments are useful in predicting (A) polarity of bonds in molecules (B) positions of elements in electrochemical se- ries (C) co-ordination number of elements (D) oxidation number of elements 21. The correct order of covalent, Van der Waals's and crystal radii is (A) covalent crystal Van der Wall r r r   (B) covalent Van der Wall crystal r r r   (C) crystal covalent Van der Wall r r r   (D) crystal Van der Wall covalent r r r  
PERIODIC TABLE 106 NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED - VOL - I NISHITH Multimedia India (Pvt.) Ltd., 22. The statement which is not correct for peri- odic classification of elements is (A) the properties of elements are a periodic func- tion of their atomic numbers (B) non-metallic elements are less in number than metallic elements (C) the first ionisation energies of elements along a period do not vary in a regular manner with increase in atomic number (D) generally electronegativity and electron af- finity increase across the period 23. Be and Mg have zero values of electron af- finity because (A) their 2s and 3s - orbitals are fully occupied (B) their first ionisation - energies are very small (C) their electronegativity are very high (D) their electro-affinity are very high in 2 Mg  and 2 Be  24. Nitrogen and phosphorous belong to the same group in the periodic table and yet the later forms 3 3 3 4 3 4 2 7 , ,( ) , H PO H PO HPO H P O n and the former forms only HNO3 , HNO2 and HNO4 i.e. former forms less number of oxy- acids. This is because (A) N is much more electronegative than P (B) N atom is smaller in size compared to P- atom (C) N atom does not have d-orbitals in its va- lence shell but P has (D) N has a lower affinity for H than P 25. Which one of the following arrangements is the incorrect representation of the property indicated with it? (A) Br Cl F Electronegativity   : (B) F Br Cl Electron affinity    : (C) 2 2 2 Br Cl F Bond energy   : (D) 2 2 2 Br Cl F   : Oxidising strength 26. The correct increasing order of electrone- gativity for 2 4 Sn Sn and Sn ,   species is (A) 2 4 Sn Sn Sn     (B) 4 2 Sn Sn Sn     (C) 2 4 Sn Sn Sn    (D) 2 4 Sn Sn Sn     27. The C–C single bond length is 1.54 A0 and that of Cl–Cl is 1.98A0 . If the electronega- tivity of Cl and C are 3.0 and 2.5 respec- tively, the C–Cl bond-length will be equal to (A) 3.12 A0 (B) 1.67A0 (C) 1.71A0 (D) 2.12A0 28. The radii of F, F– ,O and O2– follows the or- der of (A) 2 O F O F      (B) 2 O F O F      (C) 2 O F F O      (D) 2 F O F O      29. Periodic classification of elements based on atomic volume curve was given by (A) Newland (B) Lother Mayer (C) Dobereiner (D) Mendeleev 30. Match the coloumn Scientis Periodic Table (I) Duma (A) Octave rule (II) Newland (B) Atomic volume curve (III) Lother Meyer (C) Homologous se- ries (IV) Doberiner (D) Triad rule The correct option is (A) I-A, II-B, III-C, IV-D (B) I-C, II-A, III-B, IV-D (C) I-C, II-A, III-D, IV-B (D) I-D, II-B, III-A, IV-C 31. The transition elements (d-block elements) show variable oxidation states because (A) of the presence of ns, np and nd electrons (B) the energy difference between (n-1)d and ns electrons is very less, thus (n-1)d electrons also behave like valence electrons. (C) of the presence of ns and nd orbitals (D) of the presence of electrons in np and nd orbitals
NISHITH Multimedia India (Pvt.) Ltd., 107 JEE MAINS - CW - VOL - I NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED - VOL - I PERIODIC TABLE 32. According to Born-Haber's cycle, the en- thalpy of formation of ionic compounds can be determined. The formation of NaCl in- volves following steps (I) Energy Sublimation ( ) ( ) S Na S Na S  ( I I ) Ionisation potential ( ) ( ) I Na g Na g e     (III) 2 Bond-dissociation energy ( ) 2 ( ) D Cl g Cl g  ( IV) Electron-affinity ( ) ( ) E Cl g e Cl g      (V) Lattice energy ( ) ( ) ( ) U Cl g Na g NaCl S      The enthalpy of formation of NaCl will be (A) 2 f D       H S I E U (B) 2 f D       H S I E U (C) 2 f E        H S I D U (D) 2 f E        H S I D U 33. The correct order of electron affinity of B, C, N and O is (A) O C N B    (B) B N C O    (C) O C B N    (D) O B C N    34. The electonegativity of oxygen and an ele- ment A on Pauling's scale are 3.5 and 1.1 respectively. The nature of oxide of A will be (A) acid or neutral (B) basic (C) amphoteric (D) given information is not sufficient 35. Mendeleev's periodic table is based on atomic masses of elements. It was the first successful attempt to classify all the known elements (63) that time. One of the most importance advantage of this classification was that Mandeleev predicted the physical and chemical properties of three elements : Eka-boron, Eka-silicon and Eka-aluminium. These elements were discovered as (A) germanium, scandium, gallium (B) scandium, germanium, gallium (C) iron, sulphur and germanium (D) iron, sulphur and scandium 36. If the ionic radii of each K+ and F– are 1.34A0 , then the atomic radii of K and F will be re- spectively (A) 1.34 A0 ,1.34 A0 (B) 0.72 A0 , 1.96A0 (C) 1.96A0 , 0.72A0 (D) 1.96A0 , 1.34A0 37. It has been found that the atoms combines to form bonds in order to keep 8 valence electrons attained stability. Hypothetically, if stability is attained with 7 electrons in- stead of 8 electrons, the formula of the stable nitride ion would be (A) + N (B) 2 N  (C) 2 N  (D) 3 N  38. Which of the following statements is not true about the long form of modern periodic table? (A) it reflects the sequence of filling of electrons in order to sub-energy levels s, p, d and f (B) it helps to predict the stable valency states of the elements (C) it reflects trends in physical and chemical properties of the elements (D) it helps to predict the relative ionicity of the bond between any two elements 39. If the atomic number of an element is 33. It will be placed in the periodic table in the A) 1 st group B) 13th group C) 15th group D) 17th group 40. Amongst the following elements ( whose electronic configurations are given below) the one having the highest ionisation energy is A)   2 1 Ne s p 3 3 B)   2 3 Ne s p 3 3 C)   2 2 Ne s p 3 3 D)   10 2 3 Ne d s p 3 4 4 41. 'A' Chloride of an element gives neutral solution in water. In the periodic table, the element 'A' belongs to group A) 13 B) 15 C)1 D) 16 42. An element with atomic number 20 will placed in which period of the periodic table A) 4 B) 3 C) 2 D) 1 43. The electron affinities of N,O,S and Cl are A) O N Cl S    B) O S Cl N    C) O Cl N S    D) N O S Cl    44. The sizes of given species are such that A) I I I     B) I I I     C) I I I     D) I I I    