Which of the following options are not in accordance with the property mentioned against them?

a. \(\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2\) Oxidising power
b. \(\mathrm{MI}>\mathrm{MBr}>\mathrm{MCl}>\mathrm{MF}\)
(M represents the Metal)
Ionic character of metal halide
c. \(\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2\) Bond dissociation enthalpy
d. \(\mathrm{HI}<\mathrm{HBr}<\mathrm{HCl}<\mathrm{HF}\) Hydrogen-halogen bond strength

1. (a, b)
2. (b, c)
3. (c, d)
4. (a, d)

 
 



Hint: Cl2 has Higher bond dissociation energy than F2

a.

F2 >Cl2 > Br2 > I2  As ability to gain electron increases oxidising property increases. Here, F is the most electronegative element having the highest value of SRP hence it has the highest oxidizing power. 

The reduction potential values are as follows:

F2 = 2.87 V; Cl2 = 1.36 V; Br2 =1.09 V; I2 = 0.54

Thus,  statement a is correct.

b.
As the electronegativity difference between metal and halogen increases ionic character increases.

The given order is as follows:
 
MI > MBr > MCl > MF
This is the incorrect order of ionic character of metal halide.
The correct order can be written as follows:

MI < MBr < MCI < MF

Thus, b is incorrect
c. 

The given order of bond dissociation energy, that is, F2 >Cl2 > Br2 > I2 is incorrect.

The correct order is I2 > F2 > Br>Cl2 due to electronic repulsion among lone pairs in F2 molecule.

The bond dissociation enthalpy is as follows:

F2 = 158.8 kJ mol-1; Cl2 =  242.6 kJ mol-1; Br2 = 192.8 kJ mol-1, I2= 151.1  kJ mol-1.

d.

Bond energy decreases down the group because halogen size increases and the hydrogen and halogen bond strength decreases. 
The correct order is as follows:

H–F > H–Cl > H–Br > H–I.

The bond dissociation energy value is as follows:

H-F = 574 kJ mol-1; H-Cl = 432 kJ mol-1; H-Br = 363 kJ mol-1, H-I = 295 kJ mol-1

The d is correct.