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A table has fields $F_{1}, F_{2}, F_{3}, F_{4}, F_{5}$ with the following functional dependencies:
$F_{1} \rightarrow F_{2}, F_{2} \rightarrow F_{3},F_{3} \rightarrow F_{4},F_{4} \rightarrow F_{5},F_{1}F_{2} \rightarrow F_{4},F_{1}F_{2} \rightarrow F_{5}.$ The highest normalization form this table is in is _________

  1. $\text{1 NF}$
  2. $\text{2 NF}$
  3. $\text{3 NF}$
  4. $\text{BCNF}$
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Using the closure set of attributes we get the primary key as $F_{1}.$

Since, there is only one candidate key and it has only one attribute we cannot have a partial functional dependency. So, the table must be in $2\;NF.$

We have $F_2 \to F_3,$ where $F_2$ is not a prime attribute and hence this violates $3\;NF$ requirement. So, correct answer is B.
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