Question

Arrange the following species in the decreasing order of their bond orders. NO \quad NO$^+$ \quad NO$^-$ (I) \quad (II) \quad (III)

• A. (I)>(II)>(III)
• B. (II)>(I)>(III)
• C. (II)>(III)>(I)
• D. (III)>(II)>(I)

Answer 1

The Correct Option is B

To determine the bond order of these diatomic species, we use Molecular Orbital Theory (MOT). 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. Total number of valence electrons for each species (considering only valence shell for simplicity, or total electrons if using full MO diagram): N (Atomic number 7): $1s^2 2s^2 2p^3$ (5 valence e$^-$) O (Atomic number 8): $1s^2 2s^2 2p^4$ (6 valence e$^-$) 1. NO (I): Total valence electrons = $5 (\text{N}) + 6 (\text{O}) = 11$. (Or total electrons = $7+8=15$). MO configuration for species with 14-16 total electrons (like N$_2$, O$_2$, NO) without $s-p$ mixing affecting order for $2p$ MOs: $(\sigma_{2s})(\sigma^*_{2s})(\sigma_{2p})(\pi_{2p})^4(\pi^*_{2p})^1$. (Using valence electron count of 11). Valence MO filling for 11 electrons: $\sigma_{2s}$ (2e, bonding) $\sigma^*_{2s}$ (2e, antibonding) $\sigma_{2p_z}$ (2e, bonding) $\pi_{2p_x}, \pi_{2p_y}$ (4e, bonding) $\pi^*_{2p_x}, \pi^*_{2p_y}$ (1e, antibonding) $N_b = 2(\sigma_{2s}) + 2(\sigma_{2p_z}) + 4(\pi_{2p}) = 8$. (Considering MOs from $2s, 2p$). $N_a = 2(\sigma^*_{2s}) + 1(\pi^*_{2p}) = 3$. Bond Order (NO) = $\frac{1}{2} (8 - 3) = \frac{5}{2} = 2.5$. 2. NO$^+$ (II): Total valence electrons = $5 (\text{N}) + 6 (\text{O}) - 1 (\text{for + charge}) = 10$. (Or total electrons = $15-1=14$). This is isoelectronic with N$_2$. Valence MO filling for 10 electrons: $\sigma_{2s}$ (2e) $\sigma^*_{2s}$ (2e) $\sigma_{2p_z}$ (2e) $\pi_{2p_x}, \pi_{2p_y}$ (4e) The $\pi^*_{2p}$ orbitals are empty. $N_b = 2+2+4 = 8$. $N_a = 2$. Bond Order (NO$^+$) = $\frac{1}{2} (8 - 2) = \frac{6}{2} = 3.0$. 3. NO$^-$ (III): Total valence electrons = $5 (\text{N}) + 6 (\text{O}) + 1 (\text{for - charge}) = 12$. (Or total electrons = $15+1=16$). This is isoelectronic with O$_2$. Valence MO filling for 12 electrons: $\sigma_{2s}$ (2e) $\sigma^*_{2s}$ (2e) $\sigma_{2p_z}$ (2e) $\pi_{2p_x}, \pi_{2p_y}$ (4e) $\pi^*_{2p_x}, \pi^*_{2p_y}$ (2e, antibonding) (one in $\pi^*_{2p_x}$, one in $\pi^*_{2p_y}$ by Hund's rule) $N_b = 2+2+4 = 8$. $N_a = 2+2 = 4$. Bond Order (NO$^-$) = $\frac{1}{2} (8 - 4) = \frac{4}{2} = 2.0$. Summary of bond orders: NO (I): 2.5 NO$^+$ (II): 3.0 NO$^-$ (III): 2.0 Decreasing order of bond orders: NO$^+$ (3.0)>NO (2.5)>NO$^-$ (2.0) So, (II)>(I)>(III). This matches option (b). \[ \boxed{\text{(II)>(I)>(III)}} \]