Question

# The common features among the species CN−,CO and NO+ are Bond order three and isoelectronic Bond order three and weak field ligands Bond order two and -acceptors Isoelectronic and weak field ligands

Solution

## The correct option is A Bond order three and isoelectronic First, we have to draw energy level diagram of CN−, CO and NO+ are and we will write the electronic configuration of all molecules. CN−⟶(σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px2≡π2py2)(π2pz2) CO⟶(σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px2≡π2py2)(π2pz2) NO+⟶(σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px2≡π2py2)(π2pz2) [Note:Because number of electrons > 15,that's why in 2p orbitals, π2px≡π2py will be filled first then σ2pz] Now,number of electrons in case of:              CN−=6+7+1=14              CO=6+8=14              NO+=7+8−1=14 Bond order of: CN−=12[BO−ABO]=12[10−4]=62=3 CO=12[BO−ABO]=12[10−4]=62=3 NO+=12[BO−ABO]=12[10−4]=62=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 π - donors ⟶ High field Ligands are all π - acceptors I−<Br−<S2−<SCN−<Cl−<NO3−<N3−<F−<OH−<C2O42−<H2O<NCS−<CH3CN <pyridine<NH3<ethylenediamine<(2,21−bipyridine<(1,10−phenanthroline)<NO2−<PPh3<CN−<CO ⇒ Therefore, we can see CN− & CO are not weak field Ligands.

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