Question

Lucky 13. Refer to the following molecular formulas by their letter designations A-H when answering. If none of the letters designating a structure match the criteria, write “none”. Each answer could be one or more letters (or “none”). “Orbital geometry” refers to central atom in AXn type. NaCl KF SF6 XeF4 SO2 SO3 BF3 NO A B C D E F G H (a) (5 pts) Which one(s) have a high melting point? (b) (5 pts) Which one(s) have octahedral orbital geometry? (d) (5 pts) Which one(s) have no dipole moment? (e) (5 pts) Which one(s) are a free radical?

Answer 1

The given molecules me shown below- Nad KF SF 6 X e FA SOL SO3 BF3 NO B C D E A 6 H a) The ronic compounds have high melting points, because Tonic bonds are strongest among all bonds. This is due to the fact that lonic bonds are formed due to the conlombi attraction force behoeen cation and anion. So very nigh energy required to to spe separate them Among the given compounds, Nach and KF are ionic. So Mace, and KF has high melting point 6) The molecules with octahedral geometry having central atom connected to six ligands, grouporafous placed in six corner of octahedrat octahedron. The octahedron have have total eight faces, so it is called octahedral shape Among the given molecules, only SF6 has detaled fal orbibil geometry as central safon connected to six fators in an octahedron. So Sfo has octahedral orbital geometry. d) The moleants which are non polar has no net dipote o electronegativity differences beheen atonus forms molentes, or the resultant dipole moment of bonds cancelled each other, the molecente became non polar, i.e. having zero dipole moment. K-F haveanet dipole moment dipolemoment Both Nach and KF is Lonic so the electronegativity difference between cation and anion is very high. So both have dipole moment. moment When there is no St S- 8+ 8- Na-a haveanet

U SFG S Here all the sof bonds one polar due to efechonegativity difference between s and F. But here all the bond dipole is equal and at opposite direction. So resultant dipole moment is zero. So SF6 have no dipolen moment F . F Xefa of bonds F F te Xefa i square plane ner. the also the resultant dipole moment cancelled each other. So xefa does not have any dipole moment T T + + SOL : 8 Here 50, molemte is bent clue Co to presence of lone pair on sulphant g- satom. So it has a net dipole moment as shown the the structure. netdipole moment 7 503 In 503, the resultant dipole moment Son 행 11 ň zero 8- B7€ an 8+ NO = o + BF3 In BF3, the resultant dipole moment is zero. th Due to small electronegativity difference between Nando, no has a dipole moment Therefore among the given molenks SF6, Xefa, SO3, and BF3 do not have any dipole moment. e free radical is an unchanged atom or molecule which have an unpaired valence election the among the given molecules, only No contains an unpaired election on Natom. Sono is a free radical unpaired exeepron :N=. Free radical