Question

What is the orientation of the two unhybridized p orbitals on be with respect to the two be−f bonds?

Answer 1

Each sp2 hybrid orbital on a carbon atom contains one electron. Figure 9.21 shows how the four CH  bonds are formed by overlap of sp2 hybrid orbitals on C with the 1s orbitals on each H atom. We use eight electrons to form these four electron-pair bonds. The CC  bond is formed by the overlap of two sp2 hybrid orbitals, one on each carbon atom, and requires two more electrons. The C2H4 molecule has a total of 12 valence electrons, 10 of which form the one CC and the four CH  bonds.

Answer 2

Answer: The two unhybridized orbitals are perpendicular to each other as well as to the hybridized orbitals.

As it is given that Be forms two Be-F bonds that means the molecule is $BeF_2$.

VSEPR Theory is used to predict the geometry of the molecules from the number of electrons pairs that surround the central atom.

Formula used :$Number of electrons =\frac{1}{2}[V+N-C+A]$

where, V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

In the given molecule, Be is the central atom and F is the monovalent atom.

${\text{Number of electrons}} =\frac{1}{2}[2+2+0=0]=2$

The number of electrons is 2 that means the hybridization will be sp and geometry of the molecule will be linear. Thus two p orbitals will be feft unhybridized and will be arranged perpendicularly to each other as well as to the hybrid orbitals.