Question

The shape of $BeC{l_2}$ molecules is linear.
A.True
B.False

Answer 1

Hint: This question can be solved from the knowledge of the VSEPR theory or the Valence Shell Electron Pair Repulsion Theory and the hybridization of the central atom in the molecule. The theory predicts that for compounds with molecular formula $X{Y_m}$the number of electron pairs on the central atom determines the arrangement of these lone pairs around the central atom.

Complete step by step solution:
Beryllium belongs to group 12 of the periodic table, has atomic number 4 and has an electronic configuration: $1{s^2}2{s^2}$. Hence the $2p$ orbital of \[Be\] remains vacant. As it has two electrons in the valence shell, so \[Be\] can form two covalent bonds.
According to the VSEPR theory, there is electrostatic repulsion between any two electron pairs and thus they tend to push the atomic orbitals as far as possible. Also the strength of this repulsion is higher for lone pairs of electrons, compared to the bond pairs.
So, in \[Be\], there being two bond pairs, the atomic orbitals will stay most far apart from each other at ${180^ \circ }$ angle. Hence, the molecule should have a linear structure.

So, the answer is (A), the statement is true.

Note: Alternatively, the hybridization of the central atom also predicts the shape of the molecule. According to The formula of hybridization:
$H = \dfrac{{V + X - C + A}}{2}$
Where, H is the hybridization of the atom, X is the no. of number of monovalent atoms around the central atom, C is the positive charge on the cationic species if the compound is cationic while A is the anionic charge if the compound is anionic.
Therefore for $BeC{l_2}$ , $H = \dfrac{{2 + 2 - 0 + 0}}{2}$
$ \Rightarrow H = \dfrac{4}{2}$ = 2. For, H = 2, the hybridization on the central atom is $sp$ and the structure is “linear.”