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dlinn [17]
4 years ago
7

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

Biology
2 answers:
olganol [36]4 years ago
5 0
Each sp2<span> hybrid orbital on a carbon atom contains one electron. Figure 9.21 shows how the four C</span><span>H </span><span> bonds are formed by overlap of </span>sp2<span> hybrid orbitals on C with the 1</span>s<span> orbitals on each H atom. We use eight electrons to form these four electron-pair bonds. The C</span><span>C </span><span> bond is formed by the overlap of two </span>sp2<span> hybrid orbitals, one on each carbon atom, and requires two more electrons. The C</span>2H4<span> molecule has a total of 12 valence electrons, 10 of which form the one C</span>C and the four C<span>H </span><span> bonds.</span>
inna [77]4 years ago
3 0

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.



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