The hybridisations of N,C and O shown in the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Hybridisation is determined from the steric number (number of atoms bonded to the central atom $+$ the number of lone pairs). The number of hybrid

Vedclass · Accepted Answer

$sp^2, sp, sp^2$

Vedclass · Answer

(A) Hybridisation is determined from the steric number (number of atoms bonded to the central atom $+$ the number of lone pairs). The number of hybrid orbitals must be equal to the steric number.From the Lewis structure of the isocyanate group $(R-N=C=O)$:$I$. For the $N$-atom: It is bonded to $2$ atoms ($R$ and $C$) and has $1$ lone pair. Steric number $= 2 + 1 = 3$. Therefore,hybridisation is $sp^2$.$II$. For the $C$-atom: It is bonded to $2$ atoms ($N$ and $O$) and has $0$ lone pairs. Steric number $= 2 + 0 = 2$. Therefore,hybridisation is $sp$.$III$. For the $O$-atom: It is bonded to $1$ atom $(C)$ and has $2$ lone pairs. Steric number $= 1 + 2 = 3$. Therefore,hybridisation is $sp^2$.Thus,the hybridisations are $sp^2, sp, sp^2$.

The hybridisations of $N$,$C$ and $O$ shown in the following compound are:

Similar Questions

Of the following compounds,the one having a linear structure is

The hybridization states of the chlorine atom in $ClO_4^-$,$ClO_3^-$,and $ClO_2^-$ are respectively .....

The hybridizations of $C_2$ and $C_3$ in the molecule $H_3C-CH=C=CH-CH_3$ are respectively:

Which hybrid orbitals are used by carbon atom in the following molecules?
$(a)$ $CH_3-CH_3$
$(b)$ $CH_3-CH=CH_2$
$(c)$ $CH_3-CH_2-OH$
$(d)$ $CH_3-CHO$
$(e)$ $CH_3COOH$

The type of hybrid orbitals used by the chlorine atom in $ClO_2^-$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module