Which statement correctly explains why the boiling points of the Group 14 hydrides (CH4\text{CH}_4CH4, SiH4\text{SiH}_4SiH4, GeH4\text{GeH}_4GeH4, and SnH4\text{SnH}_4SnH4) increase down the group?
The covalent M-HM\text{-}HM-H bonds (M=C,Si,Ge,SnM = \text{C}, \text{Si}, \text{Ge}, \text{Sn}M=C,Si,Ge,Sn) become increasingly polar because the electronegativity of the central atom increases down the group, creating stronger permanent dipole-dipole attractions.
The strength of the covalent M-HM\text{-}HM-H bonds increases down the group, requiring more thermal energy to break these bonds in the liquid state.
The total number of electrons in the molecules increases down the group, resulting in a more polarisable electron cloud and stronger induced dipole-dipole (London) forces.
The molecular geometry transitions from tetrahedral in CH4\text{CH}_4CH4 to square planar in SnH4\text{SnH}_4SnH4, allowing tighter packing of the heavier molecules in the liquid state.
