Answer
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Hint: Bond length is the equilibrium distance between the nuclei of two bonded atoms in a molecule. Bond length is determined experimentally by X-ray diffraction or electron diffraction methods or spectroscopic methods.
Complete step-by-step answer:
- Valence electrons in an s orbital penetrate to the nucleus better than electrons in p orbitals, and as a result, they are more tightly bound to the nucleus and less able to participate in bond formation. A pair of such electrons is called an inert pair.
- These are the series of thionyl halides, ${ OsX }_{ 2 }$. They all have ${ sp }^{ 3 }$ hybridization of S atoms and possess a pyramidal structure. The bond angle is directly proportional to the size of the side atom while it is inversely proportional to the size of the central atom.
- As we know that down the group size of an atom increases. Due to the inert pair effect, when we go down the group,s shell electrons become reluctant to undergo hybridization and only pure p orbitals participate in bonding.
- Therefore, as p orbits are placed at ${ 90 }^{ \circ }$ to each other, then for F the bond angle between F-S-F bonds is very near. The bond angle increases due to lone pair-bond pair on halogen increases due to which repulsion occurs and the bond angle increases down the group.
Thus, ${ OsF }_{ 2 }$ has the smallest bond angle of the F-S-F bond.
Hence, the correct option is A.
Note: The possibility to make a mistake is that you may choose option D. But down the group as the size of an atom increases, bond angle increases as bond pair-lone pair repulsion increases.
Complete step-by-step answer:
- Valence electrons in an s orbital penetrate to the nucleus better than electrons in p orbitals, and as a result, they are more tightly bound to the nucleus and less able to participate in bond formation. A pair of such electrons is called an inert pair.
- These are the series of thionyl halides, ${ OsX }_{ 2 }$. They all have ${ sp }^{ 3 }$ hybridization of S atoms and possess a pyramidal structure. The bond angle is directly proportional to the size of the side atom while it is inversely proportional to the size of the central atom.
- As we know that down the group size of an atom increases. Due to the inert pair effect, when we go down the group,s shell electrons become reluctant to undergo hybridization and only pure p orbitals participate in bonding.
- Therefore, as p orbits are placed at ${ 90 }^{ \circ }$ to each other, then for F the bond angle between F-S-F bonds is very near. The bond angle increases due to lone pair-bond pair on halogen increases due to which repulsion occurs and the bond angle increases down the group.
Thus, ${ OsF }_{ 2 }$ has the smallest bond angle of the F-S-F bond.
Hence, the correct option is A.
Note: The possibility to make a mistake is that you may choose option D. But down the group as the size of an atom increases, bond angle increases as bond pair-lone pair repulsion increases.
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