  A \emph{frequency square} is a matrix in which each row and column is a
permutation of the same multiset of symbols. We consider only {\em binary}
frequency squares of order $n$ with $n/2$ zeroes and $n/2$ ones in each row and
column. Two such frequency squares are \emph{orthogonal} if, when superimposed,
each of the 4 possible ordered pairs of entries occurs equally often. In this
context we say that a $k$-MOFS$(n)$ is a set of $k$ binary frequency squares of
order $n$ in which each pair of squares is orthogonal.
  A $k$-MOFS$(n)$ must satisfy $k\le(n-1)^2$, and any MOFS achieving this bound
are said to be \emph{complete}. For any $n$ for which there exists a Hadamard
matrix of order $n$ we show that there exists at least $2^{n^2/4-O(n\log n)}$
isomorphism classes of complete MOFS$(n)$. For $2<n\equiv2\pmod4$ we show that
there exists a $17$-MOFS$(n)$ but no complete MOFS$(n)$.
  A $k$-maxMOFS$(n)$ is a $k$-MOFS$(n)$ that is not contained in any
$(k+1)$-MOFS$(n)$. By computer enumeration, we establish that there exists a
$k$-maxMOFS$(6)$ if and only if $k\in\{1,17\}$ or $5\le k\le 15$. We show that
up to isomorphism there is a unique $1$-maxMOFS$(n)$ if $n\equiv2\pmod4$,
whereas no $1$-maxMOFS$(n)$ exists for $n\equiv0\pmod4$. We also prove that
there exists a $5$-maxMOFS$(n)$ for each order $n\equiv 2\pmod{4}$ where $n\geq
6$.
