Is the answer d?
Answer is A because in NCERT, it is given that ionic bonds will be formed more easily when ionization energy is low and electron gain enthaly is high. So when least ease of formation is considered one with highest ionization energy should be the answer i.e. BeCl2 .
Out of four cations given beryllium has highest ionisation enthalpy because it has two valence electrons which is considered to be stable
My answer was based on Fajan's Rule. When bigger the size of the cation more is the ionic character.
correct answer my approch by applying hemismith equation d is correct..
Yes by fajan it should be d @sonu_2020 bro reply what is answer
Yes @Ankith_2020 answer is D
But how to check by taking into consideration the lattice enthalpy and ionization enthalpy of cations
@sonu_2020 these are simple s block elements and hence only see the size of cation, as size of cation increases ionic character increases .
Because according to Fajan’s rule smaller cation will have more polarizing power and hence are more covalent .
Then answer should be A @Shwetanshu_2018
I told according to ncert answer is A
@sonu_2020 that’s what i have explained in post#9 that lager cation will have more ionic characters and hence smallest cation(i.e Be2+ or BeCl2 ) will have least ease of forming ionic compound
@Samarth_2020 @Sneha_2021 @Ankith_2020
All of you have explained correctly that larger cation will have more ionic nature .
But, i guess you all have also done that same silly mistake that we tend to do during exam . In question they have asked about least ionic compound and hence all of your explanations are correct but answer will be least ionic one i.e smallest cation i.e option A
But the answer given in the book is option "D" .
Be has small radius, so it has high polarising power hence more covalent and least ionic