A) \[{{H}_{2}}S\]
B) \[{{H}_{2}}Te\]
C) \[{{H}_{2}}Se\]
D) \[{{H}_{2}}O\]
Correct Answer: D
Solution :
As the size of central metal atom increases, \[HXH\](here,\[X=0,S,Se,Te\]) bond sirength 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,\[{{H}_{2}}O\]is non-reducing among the VI group hydrides.
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