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

Two hybrid orbitals have a bond angle of $120^o$. The percentage of $s-$ character in the hybrid orbital is nearly ............. $\%$

Write the bond angle and hybridization type of the water molecule.

In the compound $CH_3COCl$,which type of orbitals have been used by the carbonyl carbon in bond formation?

In which of the following pairs,both the species have the same hybridisation?
$I$. $SF_4, XeF_4$
$II$. $I_3^-, XeF_2$
$III$. $ICl_4^+, SiCl_4$
$IV$. $ClO_3^-, PO_4^{3-}$

Which bonds are formed by a carbon atom with $sp^{2}$-hybridisation?

Vedclass Test Series

Exam Paper Generator

Online Exam Module