Question

# Which of the following will qualify as Lewis base?\begin{align} & \text{A}\text{. BC}{{\text{l}}_{3}} \\ & \text{B}\text{. C}{{\text{H}}_{4}} \\ & \text{C}\text{. C}{{\text{l}}_{2}} \\ & \text{D}\text{. N}{{\text{H}}_{3}} \\ \end{align}

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Hint: Lewis base is those molecules which have complete octet also they can have more than 8 electrons in their outermost shell. They tend to donate a pair of electrons.

-As we all know that nitrogen has five electrons in the outermost shell so it can make a maximum of five bonds.
-But in ammonia, nitrogen has made only three bonds and a pair of the electrons remains unused.
-So, Nitrogen has one lone pair and also it has more than eight electrons.
-Hence, ammonia will be a Lewis base.
So, option D is the correct answer.
-In boron chloride, boron has three electrons in the outermost shell and makes three bonds.
-So, a total of bond pairs electrons is present in the outermost octet of the boron which proves that it is an electron deficient molecule.
So, option A. is an incorrect answer because it is a lewis acid.
-Similarly, chlorine and methane are also electron-deficient species and they tend to accept a pair of electrons.
-So, they are also considered as Lewis acid.
Hence, option B and C are also incorrect answers.

Therefore, option A. is the correct one.

Note: Lewis acid has an incomplete outermost shell and they can accept a pair of the electrons to become a stable molecule. For example, $\text{B}{{\text{F}}_{3}}\text{, Al}{{\text{F}}_{3}}$. In this boron, fluoride has a total of 6 electrons in the outermost shell so it is a Lewis acid.