Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising powers of F2 and Cl2.
The oxidising powers of both the members of halogen family are expressed in terms of their electron accepting tendency and can be compared as their standard reduction potential values.
F2 + 2e– → 2F–; E° = 2-87 V, Cl2 + 2e– → 2Cl– ; E° = 1-36 V
Since the E° of fluorine is more than that of chlorine, it is a stronger oxidising agent.
Explanation : Three factors contribute towards the oxidation potentials of both the halogens. These are :
(i) Bond dissociation enthalpy: Bond dissociation enthalpy of F2 (158 kJ mol-1) is less compared to that of Cl2 (242·6 kJ mol-1).
(ii) Electron gain enthalpy: The negative electron gain enthalpy of F (- 332·6 kJ mol-1) is slightly less than of Cl (-348·5 kJ mol-1).
(iii) Hydration enthalpy: The hydration enthalpy of F- ion (515 kJ mol-1) is much higher than that of Cl- ion (381 kJ mol-1) due to its smaller size.
From the available data, we may conclude that lesser bond dissociation enthalpy and higher hydration enthalpy compensate lower negative electron gain enthalpy of fluorine as compared to chlorine. Consequently, F2 is a more powerful oxidising agent than Cl2.
Write balanced equations for the following:
(i) NaCl is heated witlrsulphuric acid in the presence of MnO2
(ii) Chlorine gas is passed into a solution of Nal in water.
Which of the following statements are true?
(a) Only type of interactions between particles of noble gases are due to weak dispersion forces.
(b) Ionisation enthalpy of.molecular oxygen is very close to that of xenon.
(c) Hydrolysis of XeF6 is a redox reaction.
(d) Xenon fluorides are not reactive.
What happens when sulp’hur dioxide is passed through an aqueous solution of Fe(III) salt?
Match the formulas of oxides given in Column I with the type of oxide given in Column II and mark the correct option.
How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions . involved.
What are the oxidation states of phosphorus in the following: –
(i) H3PO3 (ii)PCl3
(iii) Ca3P2(iv)Na3PO4
(v) POF3
If chlorine gas is passed through hot NaOH solution, two changes are observed in the oxidation number of chlorine during the reaction. These are —— and ——-
Why does nitrogen show catenation properties less than phosphorus ? (C.B.S.E. Foreign 2009)
Which of the following acid forms three series of salts?
(a) H2PO2 (b) H3BO3 (C)H3PO4(d)H3PO3
Which of the following is correct for P4 molecule of white phosphorus?
(a) It has 6 lone pairs of electrons (b) It has six P – P single bonds
(c) It has three P – P single bonds (d) It has four lone pairs of electrons,
Explain why does the stability of oxoacids of chlorine increase in the order given below:
HClO < HClO2 < HClO3 < HClO4
Give an example to show the effect of concentration of nitric acid on the formation of oxidation product.
How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions . involved.
Why is BiH3 the strongest reducing agent amongst all the hydrides of Group 15 elements?
What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO2 ?
What happens when sulp'hur dioxide is passed through an aqueous solution of Fe(III) salt?
In qualitative analysis when H2S is passed through an aqueous solution of salt acidified with dil. HCl, a black precipitate is obtained. On boiling the precipitate with dil. HNO3, it forms a solution of blue colour. Addition of excess of aqueous solution of ammonia to this solution gives
Which of the following statements are correct?
(a) All three N – O bond lengths in HNO3 are equal.
(b) All P – Cl bond lengths in PCl5 molecule in gaseous state are equal.
(c) P4 molecule in white phosphorus have angular strain therefore white phosphorus is very reactive.
(d) PCl5 is ionic in solid state in which cation is tetrahedral and anion is octahedral.
In PCl5, phosphorus is in sp3d hybridised state but all its five bonds are not equivalent. Justify your answer with reason.
Knowing the electron gain enthalpy values of O—>O– and O—>O2- 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–?
