Two precursor alkenes
H₃C CH₃
I I
H₂C=C-CH-CH₃ 2,3-dimethyl-1-butene
H₃C CH₃
I I
H₃C-CH=CH-CH₃ 2,3-dimethyl-2-butene
alkane
H₃C CH₃
I I
H₃C-CH-CH-CH₃ 2,3-dimethylbutane
H₃C CH₃ H₃C CH₃
I I I I
H₂C=C-CH-CH₃ + H₂ → H₃C-CH-CH-CH₃
H₃C CH₃ H₃C CH₃
I I I I
H₂C-C=CH-CH₃ + H₂ → H₃C-CH-CH-CH₃
Answer:
order = SrS > SrCl2 > RbCl > CsBr
Explanation:
Comparison of the melting points of compounds is basically dependent on the charge on their cation and anion, the more the charges on the cation and anion, the stronger and greater the force of attraction and as such the melting point will be relatively higher as well.
The ionic radii is also another factor to be considered, the more the distance between ions, the lesser the bond strength and the lesser the melting point.
from the options, in terms of ionic radii SrS > SrCl2 and RbCl > CsBr
also both SrS and SrCl2 have more charges on their ions compared to RbCl and CsBr and as such the arrangement of the highest melting point will be in the order SrS > SrCl2 > RbCl > CsBr.
