Explanation:
It is known that an electron pair represents a domain. Hybridization of BF_{3}BF3 will be calculated as follows.
Hybridization = \frac{1}{2}[\text{Valence electrons of central atom + no. of bonded atoms + negative charge - positive charge}]21[Valence electrons of central atom + no. of bonded atoms + negative charge - positive charge]
Valence electrons of boron are 3 and no. of bonded atoms in a BF_{3}BF3 molecule is 3.
Therefore, calculate hybridization of BF_{3}BF3 as follows.
Hybridization = \frac{1}{2}[\text{Valence electrons of central atom + no. of bonded atoms + negative charge - positive charge}]21[Valence electrons of central atom + no. of bonded atoms + negative charge - positive charge]
= \frac{1}{2}[3 + 3 + 0]21[3+3+0]
= 3
Hence, hybridization of BF_{3}BF3 is sp^{2}sp2 , that is, trigonal planar.
Thus, we can conclude that the given image represents that it is a trigonal planar molecule with three domains.
Answer:
electron pair => 1 covalent bond => same as H:H
Explanation:
The bar notation is generally accepted as a pair of electrons depicting a 'single' chemical bond. If two bars are shown (e.g; C=C) then two chemical bonds are represented. If three bars are shown (e.g; N≡N) then three chemical bonds are represented. This is the highest number of bonds that can exist between two elements. Hope this helps. :-)
