In the qualitative analysis scheme for metal ions, how are the analytical group iii cations separated from the cations of analytical groups iv and v? a. by addition of hcl, forming insoluble metal chlorides b. by addition of h2s in basic solution, forming insoluble metal sulfides or hydroxides c. by addition of h2s in acidic solution, forming insoluble metal sulfides d. by addition of (nh4)2co3 or (nh4)3po4, forming insoluble metal carbonates or phosphates e. by addition of h2so4, forming insoluble metal sulfates

they both have a y in them

your answer is c) plasma

~*"auba14"*~

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.