Question

Three Centre two-electron bond is present in

• A. ${\left({\mathrm{BeCl}}_2\right)}_2$
• B. ${\mathrm{Li}}_2{\mathrm{C}}_2$
• C. ${\mathrm{LiAlH}}_4$
• D. ${\left({\mathrm{BeH}}_2\right)}_2$

Answer 1

The correct option is D

${\left({\mathrm{BeH}}_2\right)}_2$

Explanation for correct option:

(D) ${\left({\mathrm{BeH}}_2\right)}_2$

• An electron-deficient chemical bond in which three atoms share two electrons is known as a three-centre two-electron bond.
• Three atomic orbitals combine to generate three molecular orbitals: a bonding orbital, a non-bonding orbital, and an anti-bonding orbital.
• Instead of being distributed evenly among all three atoms as is the case in many common bonds of this type, the bonding orbital is moved towards two of the three atoms.
• The atomic no. of Beryllium $\left(\mathrm{Be}\right)$ atoms is $4$ and its electronic configuration is$1s^{2}2s^2$ - There are $2$ electrons in the valence shell.
• Among these $2$ electrons, one electron forms a covalent bond with one hydrogen atom and another electron is used to form a banana bond with another hydrogen atom.
• In the given compound ${\left(\text{Be}{\text{H}}_{\text{2}}\right)}_{\text{2}}$, the total no. of Beryllium atoms is $2$ and the total no. of hydrogen atoms is $4$; in which $2$ hydrogen forms a covalent bond & another $2$ take part in the formation of a banana bond.
• One Banana bond is formed by the $2$ electrons between the $3$ atoms but there is no uniform distribution of electrons among all $3$ atoms, it is more towards the $2$ atoms only.
• The structure of${\left(\text{Be}{\text{H}}_{\text{2}}\right)}_{\text{2}}$is shown as:

Explanation for incorrect option:

(A) ${\left({\mathrm{BeCl}}_2\right)}_2$

• ${\left({\mathrm{BeCl}}_2\right)}_2$ is a three-bond four-electron centre. So, this is an incorrect option.

(B) ${\mathrm{Li}}_2{\mathrm{C}}_2$

• ${\mathrm{Li}}_2{\mathrm{C}}_2$ does not have a three bonds two-electron system. So, this is an incorrect option.

(C) ${\mathrm{LiAlH}}_4$

• ${\mathrm{LiAlH}}_4$ does not have a three bonds two-electron system. So, this is an incorrect option.

Hence, the correct option is (D) ${\left({\mathrm{BeH}}_2\right)}_2$