English
Karnataka Board PUCPUC Science 2nd PUC Class 12

Match the items of Columns I and II and mark the correct option. Column I Column II (A) HX2SOX4 (1) Highest electron gain enthalpy (B) CClX3NOX2 (2) Chalcogen (C) ClX2 (3) Tear gas (D) Sulphur (4)

Advertisements
Advertisements

Question

Match the items of Columns I and II and mark the correct option.

Column I Column II
(A) \[\ce{H2SO4}\] (1) Highest electron gain enthalpy
(B) \[\ce{CCl3NO2}\] (2) Chalcogen
(C) \[\ce{Cl2}\] (3) Tear gas
(D) Sulphur (4) Storage batteries

Options

  • A - (4), B - (3), C - (1), D - (2)

  • A - (3), B - (4), C - (1), D - (2)

  • A - (4), B - (1), C - (2), D - (3)

  • A - (2), B - (1), C - (3), D - (4)

MCQ
Advertisements

Solution

A - (4), B - (3), C - (1), D - (2)

Explanation:

Column I Column II
(A) \[\ce{H2SO4}\] (4) Storage batteries
(B) \[\ce{CCl3NO2}\] (3) Tear gas
(C) \[\ce{Cl2}\] (1) Highest electron gain enthalpy
(D) Sulphur (2) Chalcogen
shaalaa.com
  Is there an error in this question or solution?
Chapter 7: The p-block Elements - Multiple Choice Questions (Type - I) [Page 98]

APPEARS IN

NCERT Exemplar Chemistry Exemplar [English] Class 12
Chapter 7 The p-block Elements
Multiple Choice Questions (Type - I) | Q 61 | Page 98

RELATED QUESTIONS

a. Explain the trends in the following properties with reference to group 16:

1 Atomic radii and ionic radii

2 Density

3 ionisation enthalpy

4 Electronegativity

b. In the electolysis of AgNO3 solution 0.7g of Ag is deposited after a certain period of time. Calulate the quantity of electricity required in coulomb. (Molar mass of Ag is 107.9g mol-1)

 


Give reasons: SO2 is reducing while TeO2 is an oxidising agent.


Give reasons for the following : Oxygen has less electron gain enthalpy with negative sign than sulphur.


Why is H2O a liquid and H2S a gas?


The HNH angle value is higher than HPH, HAsH and HSbH angles. Why? [Hint: Can be explained on the basis of sp3 hybridisation in NH3 and only s−p bonding between hydrogen and other elements of the group].


Knowing the electron gain enthalpy values for \[\ce{O -> O-}\] and \[\ce{O -> O^{2-}}\] as −141 and 702 kJ mol−1 respectively, how can you account for the formation of a large number of oxides having O2− species and not O?

(Hint: Consider lattice energy factor in the formation of compounds).


Draw the structures of `H_3PO_2`

 


 Give reactions for the following: 
O – O single bond is weaker than S – S single bond. 


The boiling points of hydrides of group 16 are in the order:


Given below are two statements labelled as Assertion (A) and Reason (R).

Assertion (A): Electron gain enthalpy of oxygen is less than that of Flourine but greater than Nitrogen.

Reason (R): Ionisation enthalpies of the elements follow the order Nitrogen > Oxygen > Fluorine.

Select the most appropriate answer from the options given below:


Strong reducing behaviour of \[\ce{H3PO2}\] is due to ______.


Write a balanced chemical equation for the reaction showing catalytic oxidation of NH3 by atmospheric oxygen.


Out of \[\ce{H2O}\] and \[\ce{H2S}\], which one has higher bond angle and why?


The correct order of ΔiHs among the following elements is


Which of the following compound is a peroxide?


What is the basicity of \[\ce{H3PO4}\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×