Saloni G.

a) Acc to fajans rule, if the size of cation is small and anion is large then cation distorts the electron cloud of the anion making it polarised and more covalent. in this case Li⁺ (cation) is small in size and out of  I^- and F^-, flourine is small anion and iodine is large anion. so as LiI will show polarisation it would be covalent.

b) lattice enthalpy is the enrgy to break out bonds to release atoms. as Li is small and F itself is small so the bond will be small and thus more energy will be required to break the bond and thus the enthapy would be higher.

a) alkali metals have only 1 electron in their last valence shell so thats why they show +1 oxidation state by attaining the noble gas configuration.

b) if the electrons are single in the last shell they possess very high energy so as the metal gets heated up they excite to the higher energy level by attaining some energy and when they re-emit the absorbed energy the color is imparted.

c) Li⁺ is very small in size. and its in general that a smaller anion is stablized by smaller cation. so being oxide O^2- and peroxide 0₂^2- the smaller and the larger anion respectively. therefore Li⁺ will form oxide not superoxide.

d) reducing agents are those which reduces others but oxidise themselves. oxidation is loosing of electron so because Li readily forms Li⁺ on loosing one electron thus is a reducing agent.