Question

Find the number of species in which $d\pi-p\pi$ bonds are present?

$SiF_6^{2-}$, $C_2H_2$, $CO_2$, $ClO_3^-$, $PO_4^{3-}$, $SO_3$, $NO_3^-$

Answer 1

3

Answer 2

To identify species with $d\pi-p\pi$ bonds, we look for two main criteria:

1. A central atom from the third period or beyond, which has available vacant d-orbitals.
2. A surrounding atom (usually oxygen or fluorine) that has lone pairs in p-orbitals and forms pi bonds with the central atom. The pi bond is formed by the overlap of the p-orbital of the surrounding atom and a d-orbital of the central atom. This type of bonding often helps in minimizing formal charges on the central atom by allowing it to expand its octet.

Let's analyze each given species:

1. $SiF_6^{2-}$:

   • Central atom: Silicon (Si) is in the 3rd period and has vacant 3d orbitals.
   • Surrounding atom: Fluorine (F) is in the 2nd period and has lone pairs in 2p orbitals.
   • Bonding: In $SiF_6^{2-}$, silicon forms six sigma bonds with fluorine atoms. The hybridization of Si is $sp^3d^2$. There are no pi bonds in this species. The hypervalency of silicon is explained by the involvement of d-orbitals in sigma bonding (hybridization), not pi bonding.
   • Conclusion: No $d\pi-p\pi$ bonds.

2. $C_2H_2$ (Acetylene):

   • Central atom: Carbon (C) is in the 2nd period and does not have vacant d-orbitals in its valence shell.
   • Bonding: The carbon-carbon triple bond consists of one sigma and two pi bonds. These pi bonds are formed by the overlap of 2p orbitals of carbon atoms ($p\pi-p\pi$ bonding).
   • Conclusion: No $d\pi-p\pi$ bonds.

3. $CO_2$ (Carbon Dioxide):

   • Central atom: Carbon (C) is in the 2nd period and does not have vacant d-orbitals.
   • Bonding: The carbon-oxygen double bonds consist of one sigma and one pi bond each. These pi bonds are formed by the overlap of 2p orbitals of carbon and oxygen ($p\pi-p\pi$ bonding).
   • Conclusion: No $d\pi-p\pi$ bonds.

4. $ClO_3^-$ (Chlorate ion):

   • Central atom: Chlorine (Cl) is in the 3rd period and has vacant 3d orbitals.
   • Surrounding atom: Oxygen (O) is in the 2nd period and has lone pairs in 2p orbitals.
   • Bonding: The most stable Lewis structure for $ClO_3^-$ involves one Cl-O single bond and two Cl=O double bonds (due to resonance) to minimize formal charges on chlorine. These pi bonds are formed by the overlap of oxygen's 2p orbitals and chlorine's vacant 3d orbitals.
   • Conclusion: Yes, $d\pi-p\pi$ bonds are present.

5. $PO_4^{3-}$ (Phosphate ion):

   • Central atom: Phosphorus (P) is in the 3rd period and has vacant 3d orbitals.
   • Surrounding atom: Oxygen (O) is in the 2nd period and has lone pairs in 2p orbitals.
   • Bonding: The most stable Lewis structure for $PO_4^{3-}$ involves one P=O double bond and three P-O single bonds (due to resonance) to minimize formal charges on phosphorus. This pi bond is formed by the overlap of oxygen's 2p orbitals and phosphorus's vacant 3d orbitals.
   • Conclusion: Yes, $d\pi-p\pi$ bonds are present.

6. $SO_3$ (Sulfur Trioxide):

   • Central atom: Sulfur (S) is in the 3rd period and has vacant 3d orbitals.
   • Surrounding atom: Oxygen (O) is in the 2nd period and has lone pairs in 2p orbitals.
   • Bonding: The most stable Lewis structure for $SO_3$ involves three S=O double bonds (due to resonance) to minimize formal charges on sulfur. These pi bonds are formed by the overlap of oxygen's 2p orbitals and sulfur's vacant 3d orbitals.
   • Conclusion: Yes, $d\pi-p\pi$ bonds are present.

7. $NO_3^-$ (Nitrate ion):

   • Central atom: Nitrogen (N) is in the 2nd period and does not have vacant d-orbitals.
   • Bonding: The nitrate ion exhibits resonance with one N=O double bond and two N-O single bonds. The pi bond is formed by the overlap of 2p orbitals of nitrogen and oxygen ($p\pi-p\pi$ bonding). Nitrogen cannot expand its octet.
   • Conclusion: No $d\pi-p\pi$ bonds.

Based on the analysis, the species containing $d\pi-p\pi$ bonds are $ClO_3^-$, $PO_4^{3-}$, and $SO_3$. The number of such species is 3.