Question

Regarding complex formation by $2^{nd}$ group metals, the correct statement(s) is/are:

• A. Beryllium forms complexes only with halide ligands.
• B. $M{g}^{2+}$ and $C{a}^{2+}$ can form complexes with EDTA, while $B{e}^{2+}$ cannot.
• C. $B{e}^{2+}$ forms complexes with oxalic acid, catechol to give tetrahedral complexes.
• D. In chlorophyll, 2 N-atoms of porphyrin ring are bonded to the magnesium atom in the complex.

Answer 1

Answer: (B), (C)

$B{e}^{2+}$ forms complexes with oxalic acid, catechol to give tetrahedral complexes.
$M{g}^{2+}$ and $C{a}^{2+}$ can form complexes with EDTA, while $B{e}^{2+}$ cannot.

1) Small sized $B{e}^{2+}$ has a strong tendency to form complexes with several ligands such as $F^{-},C{l}^{-},H_{2}O$, oxalic acid and catechol.
2) $M{g}^{2+}$ and $C{a}^{2+}$ ions form six coordination complexes with EDTA (a hexadentate ligand). However, due to lack of vacant d orbitals, $B{e}^{2+}$ cannot form hexadentate complexes. It forms four coordinated complexes.
3) In chlorophyll, 4 N-atoms of porphyrin are bonded to the magnesium atom in the complex. Below shown is the structure of chlorophyll.
Hence, options A and D are incorrect whereas, options B and C are correct.