In boron monofluoride, boron will need 5 bonds to attain octet rule while fluorine will require 1 bond to attain octet rule.
The valency electron of an atom determines the kind of bonding that occurs between atom of elements.
Therefore,
The bonding between Boron and fluorine to form boron monofluoride are as follows:
An electronically neutral boron atom has 5 electrons. Using electronic configuration, the outer electrons are 3 (Using the model 2 8 8 2). It will require 5 electrons to attain the octet rule.
On the other hand an electronically neutral fluorine atom has 9 electrons. The outer electrons are 7 electrons. Naturally, it will require 1 electron to attain octet rule.
In the case of boron monofluoride, 1 atom of boron combines with 1 atom of fluorine to form the compound. Therefore, the total valency electrons of the diatomic compound is 10 electrons altogether.
Generally, In boron monofluoride, boron will need 5 bonds to attain octet rule while fluorine will require 1 bond to attain octet rule.
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Boron needs four bonds in order to satisfy the octet rule while fluorine needs only one bond in order to satisfy the octet rule.
Boron is an element in group 13 of the periodic table. This means that boron has three valance electrons. The octet rule states that an atom must have eight valence electrons in order to attain stability.
This implies that boron must have a total of four bonds in order to satisfy the octet rule. On the other hand, fluorine which has seven valence electrons only needs one bond in order to satisfy the octet rule.
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