Decreasing order of relative nucleophilicity of the following nucleophiles in protic solvent is -
$⊖ S H, A C ⊖ O, P h ⊖ O, ⊖ O H, H_{2} O$

⊖ S H > ⊖ O H > H_{2} O > A C ⊖ O > P h ⊖ O

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⊖ S H > ⊖ O H > P h ⊖ O > A C ⊖ O > H_{2} O

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⊖ S H > P h ⊖ O > ⊖ O H > H_{2} O > A C ⊖ O

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⊖ O H > ⊖ S H > P h ⊖ O > A C ⊖ O > H_{2} O

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Solution

The correct option is B $⊖ S H > ⊖ O H > P h ⊖ O > A C ⊖ O > H_{2} O$
Neutral species have less nucleophilicity than negatively charged species. So, $H_{2} O$ has least nucleophilicity.
In these two compounds, negative charge is in decolalisation. So, less electron density will be available for bonding, hence nucleophilicity will be less.
In $P h O^{⊝}$ , negative charge will come on $^{'} C^{'}$ while in $A C O^{⊝},$ negative charge will come on $^{'} O^{'}$ which is more stable.

Nucleophilicity is directly proportional to size atom in a group.
So, $⊖ S H > ⊖ O H$

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