Ncert Exercises & Solutions

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 > Br₂>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.

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