Answer
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Hint: To answer this question you should first write the electronic configuration of both these atoms. This can give you the required information for finding the correct answer. You can use the elimination method if you want.
Complete step by step answer:
Let’s look at all the options one by one,
In options A and B, both of these molecules are nonpolar because in hydrogen and chlorine molecules both atoms are the same, which results in EN difference equal to zero. Hence, these options are incorrect.
In option C, this statement is completely wrong. Because these are not able to form H-bond due to their low electronegativity. Generally, the compounds containing fluorine, oxygen, and nitrogen are able to form a hydrogen bond.
In option D, this statement is true and we can understand it by the following explanation,
Coordinate bonds are formed by donation of lone pairs of electrons in vacant orbitals of other atoms.
The electronic configuration of hydrogen is 1${ s }^{ 1 }$. It uses this 1 electron to form bonds with another hydrogen atom to form a hydrogen molecule.
The electronic configuration of Chlorine is [Ne] 3${ s }^{ 2 }$3${ p }^{ 5 }$. Now in forming a molecule, it uses one electron while the four electrons or two lone pairs are available for donation to form coordination bonds.
So, the correct answer is “Option D”.
Note:We should know that Hydrogen bonding is a special type of dipole-dipole attraction between molecules, not a covalent bond to a hydrogen atom. It results from the attractive force between a hydrogen atom covalently bonded to a very electronegative atom such as an N, O, or F atom and another very electronegative atom.
Complete step by step answer:
Let’s look at all the options one by one,
In options A and B, both of these molecules are nonpolar because in hydrogen and chlorine molecules both atoms are the same, which results in EN difference equal to zero. Hence, these options are incorrect.
In option C, this statement is completely wrong. Because these are not able to form H-bond due to their low electronegativity. Generally, the compounds containing fluorine, oxygen, and nitrogen are able to form a hydrogen bond.
In option D, this statement is true and we can understand it by the following explanation,
Coordinate bonds are formed by donation of lone pairs of electrons in vacant orbitals of other atoms.
The electronic configuration of hydrogen is 1${ s }^{ 1 }$. It uses this 1 electron to form bonds with another hydrogen atom to form a hydrogen molecule.
The electronic configuration of Chlorine is [Ne] 3${ s }^{ 2 }$3${ p }^{ 5 }$. Now in forming a molecule, it uses one electron while the four electrons or two lone pairs are available for donation to form coordination bonds.
So, the correct answer is “Option D”.
Note:We should know that Hydrogen bonding 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 such as an N, O, or F atom and another very electronegative atom.
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