Consider the following table:

Accoridng to given information of correct order is/are:

Options

A. `x_(1) gt x_(2) gt x_(3) gt x_(4)`
B. `y_(1) gt y_(2) gt y_(3) gt y_(4)`
C. `z_(1) gt z_(4)gt z_(3) gt z_(2)`
D. `x_(1) gt x_(4) gt x_(3) gt x_(2)`

Answer

A, B, C

Explanation

As trend for decreasing bond enegy in the given hydrides is `H_(2)O gt H_(2)S gt H_(2)Se gt H_(2) Te`. Hence their enthalpy of formation aso varies in the same trend.
As extent of delocalization of lone pair on central atom into vaccat d-orbital increased downward hence bond angle decreases from `H_(2)O` to `H_(2)Te`.
Order of `B.P.H_(2)O gt H_(2)Te gt H_(2)Se gtH_(2)S`
Boiling point of `H_(2)Te`, `H_(2)Se` and `H_(2)S` is decided by van der waals' forces i.e., higher van der waals' forces among molecules more will be B.P. But B.P. of `H_(2)O` is decided by H-bond. strength of H-bond in force among molecules more will be B.P. but B.P. of `H_(2)O` B.P. of `H_(2)O` is decided by H-bond. strength of H-bond in `H_(2)O` is greater than van der waals' forces of attraction among any other give molecular species.