Home | | Chemistry 11th std | Evaluate Yourself: Chemistry Periodic Classification of Elements

Solved Example Problems - Evaluate Yourself: Chemistry Periodic Classification of Elements | 11th Chemistry : UNIT 3 : Periodic Classification of Elements

Chapter: 11th Chemistry : UNIT 3 : Periodic Classification of Elements

Evaluate Yourself: Chemistry Periodic Classification of Elements

Chemistry : Periodic Classification of Elements : Evaluate Yourself

Evaluate Yourself

1. What is the basic difference in approach between Mendeleev's periodic table and modern periodic table ?

 

2. The element with atomic number 120 has not been discovered so far. What would be the IUPAC name and the symbol for this element? Predict the possible electronic configuration of this element.

Answers:

Atomic number : 120

IUPAC temporary symbol : Unbinilium

IUPAC temporary symbol : Ubn                                                                   

Possible electronic configuration :[Og] 8s2

 

3. Predict the position of the element in periodic table satisfying the electronic configuration (n-1)d2, ns2 where n=5

Answers:

Electronic Configuration : (n – 1)d2 ns2

for n = 5, the electronic configuration is,

1s22s2 2p6 3s2 3p6 4s2 3d10 4p6 4d2

5s2Atomic number : 40

4th group 5th period (d block element)

= Zirconium

 

4. Using Slater's rule calculate the effective nuclear charge on a 3p electron in aluminium and chlorine. Explain how these results relate to the atomic radii of the two atoms.

Answers:

Electronic Configuration of Aluminium


Effective nuclear charge = Z – S

= 13 – 9.5

(Zeff)Al   = 3.5

Electronic Configuration of chlorine


 

5. A student reported the ionic radii of isoelectronic species X3+, Y2+ and Z- as 136 pm, 64 pm and 49 pm respectively. Is that order correct? Comment.

Answers:

X3+, Y2+, Z– are isoelectronic.

Effective  nuclear  charge  is  in the order  (Zeff )Z-  < (Zeff )Y2 +  < (Zeff )X3+ and

Hence, ionic radii should be in the order

rZ -  > rY2 +  > rX3+

The correct values are,


 

6. The first ionisation energy (IE1) and second ionisation energy (IE2) of elements X, Y and Z are given below.

Element   :  IE1 (kJ mol-1) : IE2 (kJ mol-1)

X :  2370 :  5250

Y  : 522 :  7298

Z :  1680  : 3381

Which one of the above elements is the most reactive metal, the least reactive metal and a noble gas?

Answers:

Noble gases : Ioniation energy ranging from 2372 KJmol–1 to 1037 kJ mol–1.

For element X, the IE1, value is in the range of noble gas, moreover for this element both IE1 and IE 2 are higher and hence X is the noble gas.

For Y, the first ionisation energy is low and second ionisation energy is very high and hence Y is most reactive metal.

For Z, both IE1 and IE2 are higher and hence it is least reactive.

 

7. The electron gain enthalpy of chlorine is 348 kJ mol-1. How much energy in kJ is released when 17.5 g of chlorine is completely converted into Cl- ions in the gaseous state?

Answers:

Cl(g) + e– → Cl–(g) ∆H = 348 kJ mol–1 

For one mole (35.5g) 348 kJ is released.


 

Tags : Solved Example Problems , 11th Chemistry : UNIT 3 : Periodic Classification of Elements
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
11th Chemistry : UNIT 3 : Periodic Classification of Elements : Evaluate Yourself: Chemistry Periodic Classification of Elements | Solved Example Problems


Privacy Policy, Terms and Conditions, DMCA Policy and Compliant

Copyright © 2018-2023 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.