
The pair of compounds which can form a coordinate bond is:
A.\[{\left( {{C_2}{H_5}} \right)_3}B\] and \[{\left( {C{H_3}} \right)_3}N\]
B.\[HCl\,and\,{H_2}\]
C.\[B{F_3}\] and \[N{H_3}\]
D.\[(1)\] and $(3)$ both
Answer
466.2k+ views
Hint: We have to know that the coordinate bond is a type of covalent bond and it is established from a single atom by sharing the pair of electrons. And each pair of electrons are contributed by the same atom. And this coordinate bond is also known as dipolar bond. After the sharing of electrons in a coordinate bond, the atom will become more stable. And the atom which shares the electron which acts as donor.
Complete answer:
\[{\left( {{C_2}{H_5}} \right)_3}B\] and \[{\left( {C{H_3}} \right)_3}N\] Pair of compounds which can form a coordinate bond. But these pairs of compounds are not the only ones forming coordinate bonds. Hence, option (A) is incorrect.
The pair of hydrochloric acid and hydrogen does not form a coordinate bond. Hence, the option (B) is incorrect.
\[B{F_3}\] and \[N{H_3}\] Pair of compounds which can form a coordinate bond. But \[{\left( {{C_2}{H_5}} \right)_3}B\] and \[{\left( {C{H_3}} \right)_3}N\] is also form the coordinate bond. Hence, option (C) is incorrect.
Among these options, \[{\left( {{C_2}{H_5}} \right)_3}B\] and \[{\left( {C{H_3}} \right)_3}N\] and \[B{F_3}\] and \[N{H_3}\] pairs of compounds which can form a coordinate bond. Because, here the triethyl borane and triethylamine are making a bond by the sharing of electrons which is present in nitrogen and boron. One empty p-orbital is present in boron. And it can be represented as,
And in the case of ammonia and boron trifluoride can represented as,
Therefore, a pair of compounds which can form a coordinate bond.
Hence, the option (D) is correct.
Note:
Among these given compounds, triethyl borane – triethylamine pair and boron trifluoride – ammonia pair will form the coordinate bond. Because, each pair of central atoms have unpaired electrons and there is a formation of covalent bond by sharing this pair of electrons. And the pairing of electrons is attached by both nuclei present in the central atom.
Complete answer:
\[{\left( {{C_2}{H_5}} \right)_3}B\] and \[{\left( {C{H_3}} \right)_3}N\] Pair of compounds which can form a coordinate bond. But these pairs of compounds are not the only ones forming coordinate bonds. Hence, option (A) is incorrect.
The pair of hydrochloric acid and hydrogen does not form a coordinate bond. Hence, the option (B) is incorrect.
\[B{F_3}\] and \[N{H_3}\] Pair of compounds which can form a coordinate bond. But \[{\left( {{C_2}{H_5}} \right)_3}B\] and \[{\left( {C{H_3}} \right)_3}N\] is also form the coordinate bond. Hence, option (C) is incorrect.
Among these options, \[{\left( {{C_2}{H_5}} \right)_3}B\] and \[{\left( {C{H_3}} \right)_3}N\] and \[B{F_3}\] and \[N{H_3}\] pairs of compounds which can form a coordinate bond. Because, here the triethyl borane and triethylamine are making a bond by the sharing of electrons which is present in nitrogen and boron. One empty p-orbital is present in boron. And it can be represented as,
And in the case of ammonia and boron trifluoride can represented as,
Therefore, a pair of compounds which can form a coordinate bond.
Hence, the option (D) is correct.
Note:
Among these given compounds, triethyl borane – triethylamine pair and boron trifluoride – ammonia pair will form the coordinate bond. Because, each pair of central atoms have unpaired electrons and there is a formation of covalent bond by sharing this pair of electrons. And the pairing of electrons is attached by both nuclei present in the central atom.
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