Advertisements
Advertisements
प्रश्न
Out of \[\ce{H2O}\] and \[\ce{H2S}\], which one has higher bond angle and why?
Advertisements
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.
APPEARS IN
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.
Why is H2O a liquid and H2S a gas?
Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.
Arrange the following in order of the property indicated set.
HF, HCl, HBr, HI - decreasing bond enthalpy.
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 ____________.
Which of the following statement is incorrect?
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 |
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:
Which of the following statements are correct?
(i) \[\ce{CaF2 + H2SO4 -> CaSO4 + 2HF}\]
(ii) \[\ce{2HI + H2SO4 -> I2 + SO2 + 2H2O}\]
(iii) \[\ce{Cu + 2H2SO4 -> CuSO4 + SO2 + 2H2O}\]
(iv) \[\ce{Nacl + H2SO4 -> NaHSO4 + HCl}\]
In forming (i) \[\ce{N2 -> N^{+}2}\] and (ii) \[\ce{O2 -> O^{+}2}\]; the electrons respectively are removed from:
These are physical properties of an elements.
- Sublimation enthalpy
- Ionisation enthalpy
- Hydration enthalpy
- Electron gain enthalpy
The total number of above properties that affect the reduction potential is ______. (Integer answer)
______ is a gaseous element of group 16.
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:
