A) \[{{H}_{2}}S\]
B) \[~{{H}_{2}}Te\]
C) \[{{H}_{2}}Se\]
D) \[~{{H}_{2}}\text{O}\]
Correct Answer: D
Solution :
As the size of central metal atom increases, H?X?H (here,\[X=O,S,Se,Te\]) bond strength decreases and thus, the reducing power increases. The order of reducing power is \[{{H}_{2}}O<{{H}_{2}}S<{{H}_{2}}Se<{{H}_{2}}Te\] Hence,\[{{\text{H}}_{\text{2}}}\text{O}\]is non-reducing among the VI group hydrides.
You need to login to perform this action.
You will be redirected in
3 sec
