Being systematic a $2$-special group is completely by two $\mathbb Z/2$-vector spaces $U=G/Z(G)$ and $V=Z(G)$ together with the induced square map $\Gamma^2U\to V$ which is non-degenerate in the following sense: We have an injective map $\Lambda^2U\to\Gamma^2U$ mapping $u\land v\mapsto\gamma_1(u)\gamma_1(v)$ so that the composite $\Lambda^2U\to\Gamma^2U\to V$ is non-degenerate. Every map $\Gamma^2U\to V$ occurs for some central extension of $V$ by $U$ and the extension gives a $2$-special group precisely when $\Lambda^2U\to\Gamma^2U\to V$ is non-degenerate.

For $V=(\mathbb Z/2)^2$ a map $\Lambda^2U\to V$ corresponds to a pair of alternating forms on $U$ and the non-degeneracy says exactly that the radicals of the two forms intersect in zero. Now, if $V$ is even-dimensional there is a single non-degenerate alternating form and if $V$ is odd-dimensional, then any $1$-dimensional subspace is the radical of some alternating form so that if $\dim V>1$ there are always two forms whose radicals intersect trivially. Picking such a map $\Lambda^2U\to V$ we simply may take any linear extension to a map $\Gamma^2U\to V$ to get a $2$-special group. This means that the answer to the question is no.

Being systematic a $2$-special group is specified completely by two $\mathbb Z/2$-vector spaces $U=G/Z(G)$ and $V=Z(G)$ together with the induced square map $\Gamma^2U\to V$ which is non-degenerate in the following sense: We have an injective map $\Lambda^2U\to\Gamma^2U$ mapping $u\land v\mapsto\gamma_1(u)\gamma_1(v)$ so that the composite $\Lambda^2U\to\Gamma^2U\to V$ is non-degenerate. Every map $\Gamma^2U\to V$ occurs for some central extension of $V$ by $U$ and the extension gives a $2$-special group precisely when $\Lambda^2U\to\Gamma^2U\to V$ is non-degenerate.

For $V=(\mathbb Z/2)^2$ a map $\Lambda^2U\to V$ corresponds to a pair of alternating forms on $U$ and the non-degeneracy says exactly that the radicals of the two forms intersect in zero. Now, if $V$ is even-dimensional there is a single non-degenerate alternating form and if $V$ is odd-dimensional, then any $1$-dimensional subspace is the radical of some alternating form so that if $\dim V>1$ there are always two forms whose radicals intersect trivially. Picking such a map $\Lambda^2U\to V$ we simply may take any linear extension to a map $\Gamma^2U\to V$ to get a $2$-special group. This means that the answer to the question is no.

Being systematic a $2$-special group is completely by two $\mathbb Z/2$-vector spaces $U=G/Z(G)$ and $V=Z(G)$ together with the induced square map $\Gamma^2U\to V$ which is non-degenerate in the following sense: We have an injective map $\Lambda^2U\to\Gamma^2U$ mapping $u\land v\mapsto\gamma_1(u)\gamma_1(v)$ so that the composite $\Lambda^2U\to\Gamma^2U\to V$ is non-degenerate. Every map $\Gamma^2U\to V$ occurs for some central extension of $V$ by $U$ and the extension gives a $2$-special group precisely when $\Lambda^2U\to\Gamma^2U\to V$ is non-degenerate.

For $V=(\mathbb Z/2)^2$ a map $\Lambda^2U\to V$ corresponds to a pair of alternating forms on $U$ and the non-degeneracy says exactly that the radicals of the two forms intersect in zero. Now, if $V$ is even-dimensional there is a single non-degenerate alternating form and if $V$ is odd-dimensional, then any $1$-dimensional subspace is the radical of some alternating form so that if $\dim V>1$ there are always two forms whose radicals intersect trivially. Picking such a map $\Lambda^2U\to V$ we simply may take any linear extension to a map $\Gamma^2U\to V$ to a $2$-special group.

Being systematic a $2$-special group is completely by two $\mathbb Z/2$-vector spaces $U=G/Z(G)$ and $V=Z(G)$ together with the induced square map $\Gamma^2U\to V$ which is non-degenerate in the following sense: We have an injective map $\Lambda^2U\to\Gamma^2U$ mapping $u\land v\mapsto\gamma_1(u)\gamma_1(v)$ so that the composite $\Lambda^2U\to\Gamma^2U\to V$ is non-degenerate. Every map $\Gamma^2U\to V$ occurs for some central extension of $V$ by $U$ and the extension gives a $2$-special group precisely when $\Lambda^2U\to\Gamma^2U\to V$ is non-degenerate.

For $V=(\mathbb Z/2)^2$ a map $\Lambda^2U\to V$ corresponds to a pair of alternating forms on $U$ and the non-degeneracy says exactly that the radicals of the two forms intersect in zero. Now, if $V$ is even-dimensional there is a single non-degenerate alternating form and if $V$ is odd-dimensional, then any $1$-dimensional subspace is the radical of some alternating form so that if $\dim V>1$ there are always two forms whose radicals intersect trivially. Picking such a map $\Lambda^2U\to V$ we simply may take any linear extension to a map $\Gamma^2U\to V$ to get a $2$-special group. This means that the answer to the question is no.