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प्रश्न
Why are halogens strong oxidising agents?
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उत्तर १
The general electronic configuration of halogens is np5, where n = 2-6. Thus, halogens need only one more electron to complete their octet and to attain the stable noble gas configuration. Also, halogens are highly electronegative with low dissociation energies and high negative electron gain enthalpies. Therefore, they have a high tendency to gain an electron. Hence, they act as strong oxidizing agents.
उत्तर २
Halogens are strong oxidising agents because:
- They have a strong tendency to gain one electron to achieve a noble gas configuration.
- This is due to their high electronegativity and high negative electron gain enthalpy.
- Halogens readily gain electrons, undergoing reduction, which makes them strong oxidisers.
- The oxidising power decreases from fluorine to iodine due to decreasing standard reduction potential.
- Halogens with lower atomic numbers can displace halide ions of halogens with higher atomic numbers from their salts.
संबंधित प्रश्न
Give reasons for the following : Oxygen has less electron gain enthalpy with negative sign than sulphur.
Write the order of thermal stability of the hydrides of Group 16 elements.
Why does NH3 form hydrogen bond but PH3 does not?
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).
Arrange the following in the order of property indicated for the given set:
F2, Cl2, Br2, I2 - increasing bond dissociation enthalpy.
Explain the following properties of group 16 elements :
1) Electro negativity
2) Melting and boiling points
3) Metallic character
4) Allotropy
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 ____________.
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:
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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)
What is the basicity of \[\ce{H3PO4}\]?
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:
