If N_x is the number of bonding orbitals of a | vedclass

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

Question

(A) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as: $	ext{Bond Order} = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the nu

Vedclass · Accepted Answer

$N_x > N_y$

Vedclass · Answer

(A) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as: $	ext{Bond Order} = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of electrons in bonding molecular orbitals and $N_a$ is the number of electrons in antibonding molecular orbitals.$A$ molecule is considered stable if the bond order is greater than zero,which implies that the number of electrons in bonding orbitals must be greater than the number of electrons in antibonding orbitals $(N_b > N_a)$.In the context of the given variables,$N_x > N_y$ ensures a positive bond order,indicating stability.

If $N_x$ is the number of bonding orbitals of an atom and $N_y$ is the number of antibonding orbitals,then the molecule/atom will be stable if

Similar Questions

Group the following molecules/ions according to their bond order: $O_2^{2-}, Li_2, O_2^{2+}, F_2, N_2, He_2^{2+}$.

Which of the following pairs of species have the same bond order?

During the change of $O_{2}$ to $O_{2}^{-}$,the incoming electron goes to the orbital:

Observe the following molecules: $C_2, N_2, O_2, F_2$. Which one of the following statements is correct for the above molecules?

The correct bond order in the following species is

Vedclass Test Series

Exam Paper Generator

Online Exam Module