Which element has the fewest number of electrons in its valence shell?OLiCoCaON Rank the electronegativity of potassium (K), copper (Cu), carbon (C) and chlorine (CI) in ascending order (order of increasing electronegativity).(hint: rank low to high)O K. Cu, C, CIOC, CI, K, CuO Cu, C, CI, KO Cu, C, K, CIOK, Cu, Cl, C

Answer 1) Valence shell electrons : are those electrons which are present in the highest principle shell (n) except case of transition metals. In case of transition metals : number of valence electrons are present in ns and (n-1) d electrons. This is because in transition metals d orbital are incomplete and that's why these can be used for their bonding. Number of valence electrons of Li = 1 (because, Li = [He] 2s1 ) Co = 9 (because, Co = [Ar] 3d7 4s2 ) Ca = 2 (because, Ca = [Ar] 2s2 ) N = 5 (because, N = [He] 2s2 2p3 ) O = 6 (because, O = [He] 2s2 2p4 ) I = 7 (because, I = [Kr] 4d10 5s2 5p5 ) Hence, correct option is Li. Answer 2) • As we move from left to right in a periodic table electronegativity increase. That's why : Cl = highest electronegativity because it belongs to 17 group of periodic table. C = higher electronegativity but lower than Cl because it belongs to 14 group of periodic table. Cu = low electronegativity than C and Cl but higher than K because it belongs to 11 group of periodic table. K = least electronegativity because it belongs to 1st group of periodic table. That's why correct option is : k < Cu < C < Cl.