Consider the following species, $CN^{+}$, $CN^{-1}$, $NO$, and $CN$. Which one of these will have the…

Consider the following species, $CN^{+}$, $CN^{-1}$, $NO$, and $CN$. Which one of these will have the highest bond order?
  1. CN+
  2. CN-
  3. NO
  4. CN

Solution

Ion/Species Total electron Bond order
NO 15 2.5
CN- 14 3
CN+ 12 2
CN 13 2.5

Number of $e^{-}$ in $NO=7+8=15$ $NO=\sigma_{1}s^{2}<\sigma_{*1}s^{2}<\sigma_{2}s^{2}<\sigma_{*2}s^{2}<\sigma_{2}P_{z}^{2}<\pi_{2}P_{y}^{2}=\pi_{2}Px^{2}<\pi_{*2}P_{y}^{1}=\pi_{*2}Px^{0}$ Bond order $=\frac{\text{{(number of electron in bonding molecular orbital - number of electron in anti bonding molecular orbital)}}}{2}$ $BO=\frac{6-1}{2}=2.5$ Number of $e^{-}$ in $CN=6+7=13$ $NO=\sigma_{1}s^{2}<\sigma_{*1}s^{2}<\sigma_{2}s^{2}<\sigma_{*2}s^{2}<\sigma_{2}P_{z}^{2}<\pi_{2}P_{y}^{2}=\pi_{2}Px^{2}<\pi_{*2}P_{y}^{1}=\pi_{*2}Px^{0}$ Bond order $=\frac{5}{2}=2.5$ Number of $e^{-}$ in $CN^{+}=6+7-1=12$ $NO=\sigma_{1}s^{2}<\sigma_{*1}s^{2}<\sigma_{2}s^{2}<\sigma_{*2}s^{2}<\sigma_{2}P_{z}^{2}<\pi_{2}P_{y}^{2}=\pi_{2}Px^{2}<\pi_{*2}P_{y}^{1}=\pi_{*2}Px^{0}$ Bond order $=\frac{4-0}{2}=2$ $CN^{-}$ number of $e^{-}$ in $CN^{-}=6+7+1=14$ $CN^{-}=\sigma_{1}s^{2}<\sigma_{*1}s^{2}<\sigma_{2}s^{2}<\sigma_{*2}s^{2}<\sigma_{2}P_{z}^{2}<\pi_{2}P_{y}^{2}=\pi_{2}Px^{2}<\pi_{*2}P_{y}^{1}=\pi_{*2}Px^{0}$ Bond order $=\frac{6-0}{2}=3$ $CN^{-}$ as highest bond order.

Asked in: NEET 2018

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