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Hybridisation Involving d Orbitals

Hybridization...

Question

Hybridization of N in methylamine is :

A

s p^{3}

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B

s p^{2}

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C

s p

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D

d s p^{2}

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Solution

The correct option is A $s p^{3}$
$s p^{3}$ hybridization like $N H_{3}$

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Similar questions

why Benzylamine is more basic than N-Methylamine

p K b

of aniline is more than that of methylamine. Reason: In aniline, the lone pair of electrons on the

N

atom is delocalized over the benzene ring. As a result, the electron density of nitrogen decreases. In contrast, in

{C H}_{3} {N H}_{2} (+ I)

{C H}_{3}

increases the electron density on the

N

atom. Therefore, aniline is a weaker base than methylamine and hence its

p K_{b}

value is higher than that of methylamine.

K_{b}

values for the aqueous methylamine, ammonia, urea and pyridine are:

5.6 \times 10^{- 4}

1.8 \times 10^{- 5}

1.5 \times 10^{- 14}

1.7 \times 10^{- 9}

respectively.
Which is the correct order of basic strength?

Q. Silver Chloride is soluble in Methylamine due to the formation of

Q. Account for the following:
(i)

p K_{b}

of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.

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Hybridisation Involving d Orbitals

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