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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

Out of HX2O and HX2S, which one has higher bond angle and why?

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प्रश्न

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

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उत्तर

Oxygen is more electronegative than sulphur, the bond angle of \[\ce{H2O}\] is greater, and the body pair electron of an \[\ce{OH}\] bond will be closer to oxygen, causing bond-pair bond-pair repulsion between bond pairs of two \[\ce{OH}\] bonds.

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अध्याय 7: The p-block Elements - Multiple Choice Questions (Type - I) [पृष्ठ ९६]

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एनसीईआरटी एक्झांप्लर Chemistry Exemplar [English] Class 12
अध्याय 7 The p-block Elements
Multiple Choice Questions (Type - I) | Q 45 | पृष्ठ ९६

संबंधित प्रश्न

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


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


Account for the following : There is large difference between the melting and boiling points of oxygen and sulphur.


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


List the important sources of sulphur.


Write the order of thermal stability of the hydrides of Group 16 elements.


Which of the following does not react with oxygen directly?

Zn, Ti, Pt, Fe


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.


Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.


Arrange the following in the order of property indicated for the given set:

F2, Cl2, Br2, I2 - increasing bond dissociation enthalpy.


Give reasons Thermal stability decreases from H2O to H2Te.


Arrange the following in order of the property indicated set.
HF, HCl, HBr, HI - decreasing bond enthalpy.


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


Which of the following statement is incorrect?


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


In forming (i) \[\ce{N2 -> N^{+}2}\] and (ii) \[\ce{O2 -> O^{+}2}\]; the electrons respectively are removed from:


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:


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