Arrange the halogens , in order of their increasing reactivity with alkanes.

1. ${\mathrm{I}}_{2}<{\mathrm{Br}}_{2}<{\mathrm{Cl}}_{2}<{\mathrm{F}}_{2}$

2. ${\mathrm{Br}}_{2}<{\mathrm{Cl}}_{2}<{\mathrm{F}}_{2}<{\mathrm{I}}_{2}$

3. ${\mathrm{F}}_{2}<{\mathrm{Cl}}_{2}<{\mathrm{Br}}_{2}<{\mathrm{I}}_{2}$

4. ${\mathrm{Br}}_{2}<{\mathrm{I}}_{2}<{\mathrm{Cl}}_{2}<{\mathrm{F}}_{2}$

HINT- High electronegativity means high reactivity with alkanes
EXPLAINATION (1)
Alkane react with ${\mathrm{F}}_{2}$ is vigorously and with ${\mathrm{I}}_{2}$ the reaction is too slow that it requires a catalyst. It is because of high electronegativity of fluorine. Reactivity decreases with decrease in electronegativity and electronegativity decreases down the group.