हिंदी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 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) - Chemistry

Advertisements
Advertisements

प्रश्न

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

विकल्प

  • 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

उत्तर

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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: The p-block Elements - Multiple Choice Questions (Type - I) [पृष्ठ ९८]

APPEARS IN

एनसीईआरटी एक्झांप्लर Chemistry [English] Class 12
अध्याय 7 The p-block Elements
Multiple Choice Questions (Type - I) | Q 61 | पृष्ठ ९८

संबंधित प्रश्न

Account for the following: Oxygen shows catenation behavior less than sulphur.


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


Give reasons for the following : H2Te is the strongest reducing agent amongst all the hydrides of Group 16 elements.


Which of the following does not react with oxygen directly?

Zn, Ti, Pt, Fe


Why does NH3 form hydrogen bond but PH3 does not?


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].


Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.


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).


Why are halogens strong oxidising agents?


Give reasons Thermal stability decreases from H2O to H2Te.


Explain the following properties of group 16 elements :
1) Electro negativity
2) Melting and boiling points
3) Metallic character
4) Allotropy


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


The formation of \[\ce{O^+_2[PtF6]^-}\] is the basis for the formation of first xenon compound. This is because ____________.


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 ______.


The correct order of ΔiHs among the following elements is


Which of the following compound is a peroxide?


These are physical properties of an elements.

  1. Sublimation enthalpy
  2. Ionisation enthalpy
  3. Hydration enthalpy
  4. Electron gain enthalpy

The total number of above properties that affect the reduction potential is ______. (Integer answer)


Given below are two statements:

Statement I: The boiling point of hydrides of Group 16 elements follows the order:

H2O > H2Te > H2Se > H2S

Statement II: On the basis of molecular mass, H2O is expected to have a lower boiling point than the other members of the group but due to the presence of extensive H-bonding in H2O, it has a higher boiling point.

In the light of the above statements, choose the correct answer from the options given below:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×