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

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(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

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$N_x > N_y$

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(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

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