Question

Calculate the electronegativity differences of the following and determine if they are largely ionic, non-polar covalent, or polar covalent?

S8, CaCl2, SOCl2, NaF, CBr4, BrF, LiF and AsH3

Answer 1

bona du) Concepts if E.No of diffrence > 2 800, it will be balay a) Sq structure. s - s-sos bond s-s S S SS Electro negativity of S-S= E. Niet = 2.5-2.5 = 0 So bond a non-polare covalente b) Calle structure , Cance bond carce EON diff. of Carld = EN• of Ct - Eo No of Ca & 3-103 = 1.7 So. bond is palar covalente o as=0 C) Socle structuel. bond, S-o s-d. Pane diff. of S-O= B.No of O-E. Noofs = 3.5-2.5 =) so. bond is halore Covalent. Folo of S-Cd = Go No of Cl - CoNo ofs = 3 - 2.5=oer Ao bond is polare covolent.

MAA d) Not structuree Nauf bondo Na F. ko No difference of Na-f= h. N. off - 6. No of Na = 4-0.9=301 (32) 8 Soo bond is the lonico е) СВИ Br kond: C-B& Bello difference of C-B&= Ee No of Br - Eonloofc = 2.8-2.5 = 0.3 so bond is halare - Covalont. f) Brf Bruf Bo BV bond, Br-f. E. No difference of Br-fbond = Eolo off - CoNoof By = 4 - 2.8 = 102 is polare- Covalent, so bond

g) Ur estructuree lif. bond. Lif. BoNo diffrence of Li- f bond = F.Nooff - E.An of la = 4-123 so. bond as lonic. h) ds la structure As . H bond = As-H Eono diffrence of As-H= E No of As - Ello oft = 2.2-201 =0:1 so bond is polere covalent. Thank you, may it help your ..