Question

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

Hint: The ${2^{nd}}$ group metal of the periodic table forms different metal complexes by joining with different types of ligand. The coordination complex formed depends on the type of ligand. Example EDTA is a hexadentate ligand.

Complete step by step answer:
The size of beryllium cation is very small due to high polarization characteristics which help beryllium to form covalent bond and binds with several halides ligands like ${F^ - }$, $C{l^ - }$ and ligand apart from halide like ${H_2}O$, oxalic acid and catechol. So the statement: Beryllium forms complexes only with halide ligand is incorrect.
The $M{g^{2 + }}$ cation and $C{a^{2 + }}$ cation forms a six coordination complex with EDTA which is a hexadentate ligand. But in $B{e^{2 + }}$, the d-orbital is vacant due to which it cannot form six coordination complexes rather than form a four coordination complex. So the statement: $M{g^{2 + }}$ and $C{a^{2 + }}$ can form complexes with EDTA, while $B{e^{2 + }}$ cannot is correct.
As in $B{e^{2 + }}$, the d-orbital is vacant due to which it cannot form six coordination complex therefore, it forms a four coordination complex. Thus, it forms a tetrahedral complex with oxalic acid. Therefore, this option is correct.
In chlorophyll molecules, the 4-N atom of the porphyrin ring is joined with the magnesium atom in the complex. Thus, the given statement: In chlorophyll, 2 N-atoms of porphyrin ring are bonded to the magnesium atom in the complex is incorrect.
Therefore, the correct option is B and C.

Note:
Because of higher positive charge of group 2 elements, the alkaline earth element forms complexes with Lewis base as compared to alkali metal. The tendency to form complex by group 2 element decreases as the atomic radii of metal ions increases.