The correct basicity order of various atoms is :

1 < 2 < 3 < 4

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2 < 3 < 4 < 1

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3 < 2 < 1 < 4

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3 < 2 < 4 < 1

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Solution

The correct option is B 2 < 3 < 4 < 1
Basicity order is: N $s p^{3}$ (localised lone pair) > N $s p^{2}$ (localised lone pair) > N $s p^{2}$ (Delocalised lone pair).

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Similar questions

Q. Arrange in decreasing order of basicity:

Q. The sum of the series

1 + \frac{1}{1!} . \frac{1}{4} + \frac{1.3}{2!} . {(\frac{1}{4})}^{2} + \frac{1.3.5}{3!} . {(\frac{1}{4})}^{3} + . . . .

\infty

Q. Arrange the following in the order of decreasing basic character:
1. p-Toluidine
2. N,N-Dimethyl-p-toludine
3. p-Nitroaniline
4. Aniline

Q. Show that

\frac{4^{3} - 1}{3} = 4^{2} + 4 + 1

Evaluate :
$(i) \frac{2}{3} - \frac{4}{5} (i i) \frac{- 4}{9} - \frac{2}{- 3} (i i i) - 1 - \frac{4}{9} (i v) \frac{- 2}{7} - \frac{3}{- 14} (v) \frac{- 5}{18} - \frac{- 2}{9} (v i) \frac{5}{21} - \frac{- 13}{42}$

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The correct basicity order of various atoms is :

The correct option is B 2 < 3 < 4 < 1Basicity order is: N sp3 (localised lone pair) > N sp2 (localised lone pair) > N sp2 (Delocalised lone pair).

The correct option is B 2 < 3 < 4 < 1
Basicity order is: N $s p^{3}$ (localised lone pair) > N $s p^{2}$ (localised lone pair) > N $s p^{2}$ (Delocalised lone pair).