3.091 Fall Term 2002 Homework #4 Solutions

Transcription

1 3.091 all Term 2002 omework #4 olutions 5-5. We imply that sodium is a better electron donor than lithium. Evidence for this can be found in the lower value of AVEE which for these two elements is equivalent to the value of the first ionization energy. or sodium the value is ev; for lithium ev The very high third ionization energy for magnesium (7733 kj) The most important factor is the electron configuration and the number of electrons that must be gained or lost to achieve a filled outer shell or octet electron configuration The product of the reaction of strontium metal with phosphorus should have the formula r 3 P (a) Zn 2 (b) Al 3 (c) n 2 or n 4 (d) Mg 2 (e) (Bi 3 or Bi (a) 2 6 (b) 2 4 (c) 2 2 (d) 2 2

2 3.091 omework #4 page In the molecule N 2 5, there are 40 valence electrons available. A structure that contains 2 N-N 3 requires 42 valence electrons. The correct Lewis structure is: N N Exceptions to the Lewis octet rule are encountered when the central atom has fewer than or more than eight valence electrons. (a) 2 2 is not an exception to the octet rule. (b) Be 2 Be Be is surrounded by four valence electrons. This is an exception to the octet rule. (c) 4 ulfur is surrounded by ten valence electrons. This is an exception to the octet rule. (d) 3 3 is not an exception to the octet rule. An alternate structure that minimizes the formal charge is an exception.

3 3.091 omework #4 page The resonance structures of N are: N N N (a) EN δ = G N B EN + EN 4.19 δ = = = 0 (b) 4.19 δ = = = 0.29 (c) 4.19 δ = = = (a) P 3 P Number of bonding domains = 3; Number of nonbonding domains = 1 Geometry = trigonal pyramidal (b) Ga 3 Ga Number of bonding domains = 3; Number of nonbonding domains = 0 Geometry = trigonal planar

4 3.091 omework #4 page 4 (c) I 3 I Number of bonding domains = 3; Number of nonbonding domains = 2 Geometry = T-shaped (d) Xe 3 + Xe Number of bonding domains = 3; Number of nonbonding domains = 2 Geometry = T-shaped (a) P P P

5 3.091 omework #4 page 5 (b) 4 (c) Xe 4 Xe Xe (d) Mn 4 Mn Mn

6 3.091 omework #4 page (a) n 2 n Number of bonding domains = 2; Number of nonbonding domains = 1 Geometry = bent (b) n 3 n Number of bonding domains = 3; Number of nonbonding domains = 1 Geometry = trigonal pyramidal (c) n 4 n (d) n 6 n Number of bonding domains = 6; Number of nonbonding domains = 0 Geometry = octahedral

7 3.091 omework #4 page All are polar The Lewis structures of thionyl chloride ( 2 ) and sulfuryl chloride ( 2 2 ) are: thionyl chloride sulfuryl chloride Thionyl chloride has a dipole moment along the - bond. Therefore, it is a polar molecule. There are two possible structures that could be drawn for sulfuryl chloride, and both structures have a tetrahedral arrangement. ince there are two types of polar bonds in the molecule, - and -, the individual bond dipoles do not cancel. ulfuryl chloride is a polar molecule. 4-A5. 4-A6. (a) 4 sp 3 (b) 2 sp 2 (c) - 2 sp 2 (a) 4 sp 3 d (b) Br 3 sp 3 (c) Xe 3+ sp 3 d (d) 2 sp 2 Additional questions: 1. E lattice = Mq 1 q 2 4πε o r o 1 1 n and r o = r s + + r olve first for r o r o = Mq 1 q 2 N Av 4πε o E lattice 1 1 n = ( ) = m = 3.50 Å = r s + + r r s + = = 1.69 Å 1 1 4π

8 3.091 omework #4 page 8 2. (a) hlorine will be liberated if the - bond breaks, compute its strength and show that ultraviolet photons have enough energy to break the bond. E = E E ( χ χ ) 2 = ( ) 2 = / = J/bond photon will break this bond if E ph > E bond critical λ is λ = hc = = m E bond which lies in the u.v. part of the electromagnetic spectrum. (b) Draw the Lewis structure of reon 12 and indicate the polarities of each bond within this compound. δ- δ+ or δ- δ- δ- (c) - χ = 0.61 ~9% ionic character - χ = 1.43 ~40% ionic character