
The compound that has the largest H-M-H bond angle (M=N, O, S, C) is:
A. \[{\rm{C}}{{\rm{H}}_{\rm{4}}}\]
B. \[{{\rm{H}}_{\rm{2}}}{\rm{S}}\]
C. \[{\rm{N}}{{\rm{H}}_{\rm{3}}}\]
D. \[{{\rm{H}}_{\rm{2}}}{\rm{O}}\]
Answer
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Hint: The bond angle defines the average angle formed between orbitals possessing the bonding electrons around the central atom of a molecule. The bond angle is dependent on many factors such as hybridization, count of lone pairs around the central atom, etc.
Complete Step by Step Solution:
Let's first understand how the hybridization tells the bond angle of a molecule. The order of decreasing bond angle is sp>\[s{p^2} > s{p^3}\]>Drago. This order says that molecules that have sp hybridization possess the largest bond angle and Dargo compounds have the lowest bond angle.
Let's understand Drago's rule. This rule explains the bond angle of hydrides of the 15th and 16th group elements. And the bond angles of Drago compounds are the lowest.
Another factor that determines the bond angle is the count of lone pairs. The relation is,
Bond angle \[\alpha \dfrac{1}{{{\rm{Number}}\,{\rm{of}}\,{\rm{lone}}\,{\rm{pair}}}}\]
Let's discuss the options one by one.
The molecule\[{\rm{C}}{{\rm{H}}_{\rm{4}}}\]has four numbers of bond pairs, so it possesses \[s{p^3}\] hybridization. \[{{\rm{H}}_{\rm{2}}}{\rm{S}}\]is a Drago compound, so its bond angle is the smallest. The molecule \[{\rm{N}}{{\rm{H}}_{\rm{3}}}\]has one lone pair and three bond pairs. So, it has\[s{p^3}\]hybridization. The water molecule has two bond pairs and two lone pairs, so, its hybridization is \[s{p^3}\].
So, we find that, \[{\rm{C}}{{\rm{H}}_{\rm{4}}}\], \[{\rm{N}}{{\rm{H}}_{\rm{3}}}\]and \[{{\rm{H}}_{\rm{2}}}{\rm{O}}\]has same configuration and now we have compare their bond angle using the count of lone pairs.
\[{\rm{C}}{{\rm{H}}_{\rm{4}}}\]has no lone pair, \[{\rm{N}}{{\rm{H}}_{\rm{3}}}\]has one lone pair and \[{{\rm{H}}_{\rm{2}}}{\rm{O}}\] has two lone pairs. We know that bond angle is inversely proportional to the count of lone pairs. So, the largest bond angle is of \[{\rm{C}}{{\rm{H}}_{\rm{4}}}\]among the given molecules.
Hence, option A is right.
Note: The bond angle gives an idea of the location of the lone pairs and bond pairs also it gives an idea of the location of the surrounding atoms and the central atom. This helps us in determining the shape and geometry of the molecule.
Complete Step by Step Solution:
Let's first understand how the hybridization tells the bond angle of a molecule. The order of decreasing bond angle is sp>\[s{p^2} > s{p^3}\]>Drago. This order says that molecules that have sp hybridization possess the largest bond angle and Dargo compounds have the lowest bond angle.
Let's understand Drago's rule. This rule explains the bond angle of hydrides of the 15th and 16th group elements. And the bond angles of Drago compounds are the lowest.
Another factor that determines the bond angle is the count of lone pairs. The relation is,
Bond angle \[\alpha \dfrac{1}{{{\rm{Number}}\,{\rm{of}}\,{\rm{lone}}\,{\rm{pair}}}}\]
Let's discuss the options one by one.
The molecule\[{\rm{C}}{{\rm{H}}_{\rm{4}}}\]has four numbers of bond pairs, so it possesses \[s{p^3}\] hybridization. \[{{\rm{H}}_{\rm{2}}}{\rm{S}}\]is a Drago compound, so its bond angle is the smallest. The molecule \[{\rm{N}}{{\rm{H}}_{\rm{3}}}\]has one lone pair and three bond pairs. So, it has\[s{p^3}\]hybridization. The water molecule has two bond pairs and two lone pairs, so, its hybridization is \[s{p^3}\].
So, we find that, \[{\rm{C}}{{\rm{H}}_{\rm{4}}}\], \[{\rm{N}}{{\rm{H}}_{\rm{3}}}\]and \[{{\rm{H}}_{\rm{2}}}{\rm{O}}\]has same configuration and now we have compare their bond angle using the count of lone pairs.
\[{\rm{C}}{{\rm{H}}_{\rm{4}}}\]has no lone pair, \[{\rm{N}}{{\rm{H}}_{\rm{3}}}\]has one lone pair and \[{{\rm{H}}_{\rm{2}}}{\rm{O}}\] has two lone pairs. We know that bond angle is inversely proportional to the count of lone pairs. So, the largest bond angle is of \[{\rm{C}}{{\rm{H}}_{\rm{4}}}\]among the given molecules.
Hence, option A is right.
Note: The bond angle gives an idea of the location of the lone pairs and bond pairs also it gives an idea of the location of the surrounding atoms and the central atom. This helps us in determining the shape and geometry of the molecule.
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