Solved Question Bank On Periodic Table

July 30, 2017 | Author: faizan123khan | Category: Periodic Table, Electron Configuration, Ion, Fluorine, Chlorine
Share Embed Donate


Short Description

Solved Question Bank On Periodic Table IIT JEE , class 11...

Description

[Title]

Page 1

Section (A): 1. Why the fourth period contains 18 electrons and not 32? 2. Write group electronic configuration of the elements having the atomic number as given below and predict the period group number and block to which they belong. Atomic number : 9, 14, 20, 27 3. whth reference to periodic table indicate (a) An element that is in group III A and 3rd period. (b) Second transition element of fourth period. (c) The group which accommodates lanthanides and actinides. (d) The elements of 15th group which have metallic as well as non – metallic behaviour. 4. A particular atom (not ion) having atomic number between 22 to 30 has magnetic moment equal to 1.73 B.M then find the atomic number of the element which is just below it in the periodic table. 5. Explain the following: (i) There are only 14 lanthanides and only 14 actinides in periodic table (ii) Why argon (at Mass 39.94) has been placed before potassium (at mass 39.10) in the periodic table. Section (B): Atomic And Ionic Radius: 6. Arrange the following in increasing order as directed (A)

size

(B)

– atomic size

7. The atomic radii of palladium and platinum are nearly same why? 8. Pick out among the following,

species having smallest ionic

radius 9. Arrange the isoelectronic species (a) Increasing effective nuclear charge

and

in order of their

(b) Increasing ionic radius

(c) Increasing ionization energy 10. Why the atomic radius of neon is greater than fluorine? Section (C): Ionization Energy: 11. Pick out among the following

- species which is least stable why?

12. The ionization energies of the coinage metals falls in the order 13. The 1st ionization energy of potassium is less than that of Cu but the reverse is true for 2nd ionization energy. EDUDIGM 1B Panditya Road, Kolkata 29

www.edudigm.in

40034819

[Title]

Page 2

14. Compare qualitatively the first and second ionization potential of copper and zinc.. Explain the observation. 15. Why the 1st I.E of nitrogen is higher than oxygen and opposite is true for second I.E energy? 16. The sums of first and second ionization energies and those of third and fourth ionization energies (in Kj

of nickel and platinum are:

Ni

2.49

8.80

Pt

2.66

6.70

Based on this information, write the most common oxidation state of Ni ns Pt Section (D): Electron Gain Enthalpy (Electron Affinity ) 17. Of the two elements given in each of the following sets choose the element having a more negative electron gain enthalpy. (i) N and O

(ii) F and Cl

18. The formation of

from F(g) is exothermic where as that of

(g) from O is

endothermic, why? 19. Be and N have extremely low value of EA against the trend. Explain. Section (E): Electronegativity: 20. Calculate the electronegativity of chlorine from the bond energy of kcal

, F – F bond (38 Kcal

) and Cl – Cl bond (58 Kcal

bond (61 ) and

electronegativity of fluorine is 4.0 21. The ionization potentials of atoms A and B are 400 kcal electron affinities of these atoms are 80.0 and 85.0 kcal

respectively. The respectively. Prove that

which of the atoms have higher electrone- gativity. 22. For the gaseous reaction

was calculate to be 19 kcal under

conditions where the cations and anions were prevented by electrostatic separation from cabining with each other. The ionization energy of K is 4.3 eV. What is the E. A of F. Answer: 1. In fourth period 4f and 4d orbitals have higher enegies (even more than 5s).

EDUDIGM 1B Panditya Road, Kolkata 29

www.edudigm.in

40034819

[Title]

Page 3

2. Atomic Number

Period

Group no.

Block

9

2nd

17th

p – block

14

3rd

14th

p – block

20

4th

2nd

s – block

27

4th

9th

d – block

3. (a) Al

(b) Ti

3rd or III B (d) As & Sb.

4. 47 5. (i) In lanthanides and actinides, the differentiating electron enters to (n – 2)f – subshell. (ii) In modern periodic table, elements have been place in order of their increasing atomic numbers. The atomic number of argon is 18 and that of potassium is 19. Thus argon has been placed before potassium. 6. (A)

; (B)

7. Due to lanthanide contraction. 8. 9. (a)

(b)

(c) 10 Vander waas’s radius is considered in case of neon not covalent radius 11. 12. In all the 3 cases an s – electron in the unpaired state is to be removed. In the case of Cu a 4s electron is to be removed which is closer to the nucleus than the 5s electron of Ag. So I.P. decreasing from Cu to Ag. However from Ag to Au the 14 f electrons are added which provide very poor shielding effect. The nuclear charge is thus enhanced and therefore the outer electron of Au is more tightly held and so the IP is high. 13 The configuration of K is [Ar]4’s while that of Cu is [

]3

4

So the answer lies

in the presence of 10 d electrons. A the electrons have got very poor screening effect so the nuclear charge is not properly screened. Therefore the effective nuclear charge is high and so the outer most electron is tightly held and high energy is needed to remove the electron. 2nd ionization energy of potassium is higher because the 2nd electron from is to be removed from stable inert gas configuration.

EDUDIGM 1B Panditya Road, Kolkata 29

www.edudigm.in

40034819

[Title]

Page 4

14. Cu

744

1961

Zn

906

1736

of copper is less than that of zinc, because removed of electron takes palce from 4 (attaining a more stable configuration 3 completely filled 4

where as in case of zinc its from

(attaining the configuration 4

of copper is higher than zinc, because the removal of IInd electron from stable configuration

requires higher energy.

15. Due to the filled p – orbital of nitrogen 1st ionization energy is higher than oxygen. After removal of one electron oxygen becomes half filled and second ionization energy becomes more as compared to nitrogen 16.

2

4 since

of Ni is less than its

and reverse

is the case in Pt. 17. (i) O,

(ii) Cl

18. Addition of second electron is opposed by electrostatic repulsion of 19. In Be the extra electron is to be added in 2p-orbital because 2s orbital is completely filled and in N, it is to added to a half filled 2p orbital. Since half – filled and full – filled orbitals are more stable, reluctance in accept electron is found. 20.

20 *

+

3 22

21. Electronegativity of A = 3.84; electronegativity of B = 3.08. therefore A has higher electronegativity. 22.

(

)(

43

)(

)(

)

0 2

0 2 35

EDUDIGM 1B Panditya Road, Kolkata 29

www.edudigm.in

40034819

View more...

Comments

Copyright ©2017 KUPDF Inc.