What is the molecular geometry of $ PB{{r}_{5}} $ ? Draw its VSEPR and Lewis structure.
Answer
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Hint :Existence of species in what form depends on the given condition in which it is placed and whether it is stable in that condition or not. Now let us see what happens to hybridization of $ PB{{r}_{5}} $ in solid state. Lewis structure is defined as the position of atoms and the bonds nature which have been connected in the molecule.
Complete Step By Step Answer:
Lewis structure helps us to determine the bond pair and the lone pair of each atom present in the molecule. In solid state splits into stable tetrahedral structure as $ PB{{r}_{5}}\underset{{}}{\leftrightarrows}{{[PB{{r}_{4}}]}^{+}}+{{[Br]}^{-}} $ . The shape of $ {{[PB{{r}_{4}}]}^{+}} $ is tetrahedral and its hybridisation is $ s{{p}^{3}} $ . Therefore, in solid state hybridization of $ P $ in $ PB{{r}_{5}} $ is $ s{{p}^{3}} $ .
Let's start with its Lewis structure. $ P $ has $ 5 $ valence electrons and $ Br $ has $ 7 $ valence electrons, which means that a total of $ ~\left[ 5+\left( 5\times 7 \right) \right]=40~ $ electrons must be accounted for.
$ :\underset{\scriptscriptstyle\centerdot\centerdot}{\ddot{B}}r-(:\underset{\scriptscriptstyle\centerdot\centerdot}{\ddot{B}}r)(:\underset{\scriptscriptstyle\centerdot\centerdot}{\ddot{B}}r)P(\underset{\scriptscriptstyle\centerdot\centerdot}{\ddot{B}}r:)-\underset{\scriptscriptstyle\centerdot\centerdot}{\ddot{B}}r: $
$ P $ bonds with the five Br atoms account for $ 5\times 2=10 $ electrons, while the three lone pairs on each Br atom complete the rest of the electrons $ 5\times 6=30~ $ electrons.
Since, $ P $ forms bonds with $ 5 $ other atoms and has no lone pairs, it has a steric number and a coordination number equal to $ 5. $ According to VSEPR Theory; this corresponds to a trigonal bipyramidal molecular geometry.
Note :
Remember that the splitting of $ PB{{r}_{5}} $ is different because Bromine atoms are large and six atoms of bromine cannot be easily accommodated around smaller Potassium atoms. . The formal charge is not the real charge but a theoretical charge which helps to determine the lowest energy structure.
Complete Step By Step Answer:
Lewis structure helps us to determine the bond pair and the lone pair of each atom present in the molecule. In solid state splits into stable tetrahedral structure as $ PB{{r}_{5}}\underset{{}}{\leftrightarrows}{{[PB{{r}_{4}}]}^{+}}+{{[Br]}^{-}} $ . The shape of $ {{[PB{{r}_{4}}]}^{+}} $ is tetrahedral and its hybridisation is $ s{{p}^{3}} $ . Therefore, in solid state hybridization of $ P $ in $ PB{{r}_{5}} $ is $ s{{p}^{3}} $ .
Let's start with its Lewis structure. $ P $ has $ 5 $ valence electrons and $ Br $ has $ 7 $ valence electrons, which means that a total of $ ~\left[ 5+\left( 5\times 7 \right) \right]=40~ $ electrons must be accounted for.
$ :\underset{\scriptscriptstyle\centerdot\centerdot}{\ddot{B}}r-(:\underset{\scriptscriptstyle\centerdot\centerdot}{\ddot{B}}r)(:\underset{\scriptscriptstyle\centerdot\centerdot}{\ddot{B}}r)P(\underset{\scriptscriptstyle\centerdot\centerdot}{\ddot{B}}r:)-\underset{\scriptscriptstyle\centerdot\centerdot}{\ddot{B}}r: $
$ P $ bonds with the five Br atoms account for $ 5\times 2=10 $ electrons, while the three lone pairs on each Br atom complete the rest of the electrons $ 5\times 6=30~ $ electrons.
Since, $ P $ forms bonds with $ 5 $ other atoms and has no lone pairs, it has a steric number and a coordination number equal to $ 5. $ According to VSEPR Theory; this corresponds to a trigonal bipyramidal molecular geometry.
Note :
Remember that the splitting of $ PB{{r}_{5}} $ is different because Bromine atoms are large and six atoms of bromine cannot be easily accommodated around smaller Potassium atoms. . The formal charge is not the real charge but a theoretical charge which helps to determine the lowest energy structure.
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