Question

The common features among the species $C{N}^{-},COandN{O}^{+}$ are

• A. Bond order three and isoelectronic
• B. Bond order three and weak field ligands
• C. Bond order two and -acceptors
• D. Isoelectronic and weak field ligands

Answer 1

The correct option is A

Bond order three and isoelectronic

First, we have to draw energy level diagram of $C{N}^{-}$, $CO$ and $N{O}^{+}$ are and we will write the electronic configuration of all molecules.

$C{N}^{-}⟶\left(\sigma 1s{)}^2\right({\sigma }^{\ast }1s{)}^2\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2(\pi 2p_{x^2}\equiv \pi 2p_{y^2})\left(\pi 2p_{z^2}\right)$

$CO⟶\left(\sigma 1s{)}^2\right({\sigma }^{\ast }1s{)}^2\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2(\pi 2p_{x^2}\equiv \pi 2p_{y^2})\left(\pi 2p_{z^2}\right)$

$N{O}^{+}⟶\left(\sigma 1s{)}^2\right({\sigma }^{\ast }1s{)}^2\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2(\pi 2p_{x^2}\equiv \pi 2p_{y^2})\left(\pi 2p_{z^2}\right)$

[Note:Because number of electrons > 15,that's why in 2p orbitals, $\pi 2p_x\equiv \pi 2p_y$ will be filled first then $\sigma 2p_z$]

Now,number of electrons in case of:

$C{N}^{-}=6+7+1=14$

$CO=6+8=14$

$N{O}^{+}=7+8-1=14$

Bond order of:

$C{N}^{-}=\frac{1}{2}[BO-ABO]=\frac{1}{2}[10-4]=\frac{6}{2}=3$

$CO=\frac{1}{2}[BO-ABO]=\frac{1}{2}[10-4]=\frac{6}{2}=3$

$N{O}^{+}=\frac{1}{2}[BO-ABO]=\frac{1}{2}[10-4]=\frac{6}{2}=3$

Therefore, All three are isoelectronic and bond order = 3

Option (b) and (d) can be discarded because, all three are not weak field ligands.

$⟶$ Low field Ligands are all $\pi$ - donors

$⟶$ High field Ligands are all $\pi$ - acceptors

$I^{-}<B{r}^{-}<S^{2-}<SC{N}^{-}<C{l}^{-}<{N{O}_3}^{-}<{N_3}^{-}<F^{-}<O{H}^{-}<{C_2{O}_4}^{2-}<H_{2}O<NC{S}^{-}<C{H}_{3}CN$
$<pyridine<N{H}_3<ethylenediamine<(2,2^1-bipyridine<(1,10-phenanthroline)<{N{O}_2}^{-}<PP{h}_3<C{N}^{-}<CO$

$\Rightarrow$ Therefore, we can see $C{N}^{-}$ & $CO$ are not weak field Ligands.