No it is not in 2NF because candidate key is $PQS$ and $PQ → R$ and $S → T$ are both partial keys.

To convert into 2NF find $PQ^+$ and $S^+$

So R(P, Q, R, S, T) will decompose into $R_1(P,Q,S), R_2(P,Q,R), R_3(S,T)$

Please mention the source of this question in title and if it is a self doubt mention it in title and ask question in question description.

To convert into 2NF find $PQ^+$ and $S^+$

So R(P, Q, R, S, T) will decompose into $R_1(P,Q,S), R_2(P,Q,R), R_3(S,T)$

Please mention the source of this question in title and if it is a self doubt mention it in title and ask question in question description.