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Br being a halogen has 7 electronsin in its valence shell or orbitals and is electronegative. However, in Phosphorus (P), total valence electron is 5. And all of them are bonded with the electronegative halogens.
Now, each electron enters a specific orbital to occupy it. The first onegoes to sand next 3 goes to px, py and pz orbitals respectively. The last one left enters dx orbital.
Since, 1 electron enters s, 3 in pand 1 in d, hence the hybridisation is s1p3d1, usually written as sp3d.
PBr5 exists only in gaseous state unlike PCl5 which exists both in solid and gaseous state.
In solid state it should ionize into PBr6-and PBr4+. Since P cannot hold six big Br atoms with its lone pairs around it and hence PBr6-doesn't exist and so does the gaseous state of PBr5.
So it's hybridization is sp3d.
Central atom is P. Z=15
Electronic configuration (ground state):- [Ne] 3s2 3p3
Excited state:- 3s1 3p3 3d1
Now these five orbitals (one s, three p and one d) hybridize to give you sp3d hybridization.
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