Bromination of 2-methylpropane gives almost exclusively 2-bromo-2-methylpropane, whereas chlorination gives a mixture. The best explanation is that

1. Ahydrogen abstraction by $\mathrm{Br}\cdot$ is endothermic, so the transition state is late and resembles the alkyl radical, which magnifies differences in radical stability
2. B$\mathrm{Br}\cdot$ is more reactive than $\mathrm{Cl}\cdot$, so it finds the tertiary hydrogen more easily
3. Cthe $\mathrm{C}-\mathrm{Br}$ bond is stronger than the $\mathrm{C}-\mathrm{Cl}$ bond
4. D$\mathrm{Br}{}_2$ is more easily homolysed than $\mathrm{Cl}{}_2$ in light