Question

In a change from BrCl3 + BrCl5, the hybrid state of
Br change from
A
sp2 to sp3
B
sp3d to sp3d?2
C
sp3 to sp3d
D
sp3 to dsp2​

Answer 1

Answer:

option- C sp 3to sp3d

Explanation:

As hybridisation is equal to sum of sigma bonds + number of lone pairs. Hence in PCl

3

we have 3 sigma bonds and one lone pair so total 4, hence hybrdisation is sp

3

.

Hybridization of PCl

5

is sp

3

d

The Hybrid state of p changes from sp

3

to sp

3

d

Answer 2

In a change from BrCl3 to BrCl5, the hybrid state of  Br changes from B . sp³d to sp³d².

• In BrCl₃ (Bromine trichloride) each Cl forms a single bond with Br.
• Hence Br has formed a total of 3 covalent bonds.
• Br has a total of 7 valence electrons. After forming 3 bonds, it has 4 valence electrons that are present as 2 lone pairs.
• Thus bromine has a total of 10 valence electrons (it can have an expanded octet).
• Total orbital involves = 3 for bonds + 2 for lp = 5. Hence the hybridization is sp³d.
• In BrCl₅, Br forms 5 sigma bonds and has single lone pair.
• Hence the total orbitals involved = 5 bonds + 1 lp = 6. So the hybridization is sp³d².