
Is $ CaC{l_2} $ a covalent bond?
Answer
493.5k+ views
Hint: The attractive force which holds various constituents like atoms, ions etc. together in different chemical species is called a chemical bond. This force holds the atom together in the molecules. Bonding is nature’s way of lowering the energy to attain stability.
Complete answer:
To answer the question first we will understand what all types of bonds are. Having learnt that, we will be able to tell whether $ CaC{l_2} $ is a covalent bond or not, and if not then what type of chemical bond it forms.
There are primarily four types of bonds. These are-
Ionic bond- It involves the transfer of electrons from one atom or molecule to another. An atom loses an electron and this electron is gained by another atom involved in bonding. This results in positively and negatively charged ions.
Covalent bond- When two atoms share an electron pair then it results in covalent bond formation. Each combining atom contributes at least one electron to the shared pair.
Polar covalent bonding- In polar covalent bonds electrons are shared unequally, the more electronegative atom pulls the electron pair closer to itself and this results in development of polar and non-polar charges.
Hydrogen bonding- It is a special type of dipole-dipole attraction between molecules, not a covalent bond to a hydrogen atom. It results from the attractive force between a hydrogen atom covalently bonded to a very electronegative atom.
It is clear from the definition that $ CaC{l_2} $ cannot have hydrogen bonding or polar covalent bonding. lf we look at the structure of $ CaC{l_2} $ chemical bonding the $ Calcium $ gives up electron to each $ Chlorine $ atom and thus becomes $ C{a^{2 + }} $ and $ C{l^ - } $ . As we have learned, when electrons are transferred from one atom to another anion and cations are formed, which is a type of ionic bonding. Therefore $ CaC{l_2} $ is not a covalent bond but an ionic bond.
Note:
As earlier we learned about chemical bonds. Therefore it can be inferred that chemical bonds are reliant on the strength of the chemical bonds between its constituents , the stronger the bonding between the constituents the more stable the resulting compound will be.
Complete answer:
To answer the question first we will understand what all types of bonds are. Having learnt that, we will be able to tell whether $ CaC{l_2} $ is a covalent bond or not, and if not then what type of chemical bond it forms.
There are primarily four types of bonds. These are-
Ionic bond- It involves the transfer of electrons from one atom or molecule to another. An atom loses an electron and this electron is gained by another atom involved in bonding. This results in positively and negatively charged ions.
Covalent bond- When two atoms share an electron pair then it results in covalent bond formation. Each combining atom contributes at least one electron to the shared pair.
Polar covalent bonding- In polar covalent bonds electrons are shared unequally, the more electronegative atom pulls the electron pair closer to itself and this results in development of polar and non-polar charges.
Hydrogen bonding- It is a special type of dipole-dipole attraction between molecules, not a covalent bond to a hydrogen atom. It results from the attractive force between a hydrogen atom covalently bonded to a very electronegative atom.
It is clear from the definition that $ CaC{l_2} $ cannot have hydrogen bonding or polar covalent bonding. lf we look at the structure of $ CaC{l_2} $ chemical bonding the $ Calcium $ gives up electron to each $ Chlorine $ atom and thus becomes $ C{a^{2 + }} $ and $ C{l^ - } $ . As we have learned, when electrons are transferred from one atom to another anion and cations are formed, which is a type of ionic bonding. Therefore $ CaC{l_2} $ is not a covalent bond but an ionic bond.
Note:
As earlier we learned about chemical bonds. Therefore it can be inferred that chemical bonds are reliant on the strength of the chemical bonds between its constituents , the stronger the bonding between the constituents the more stable the resulting compound will be.
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