W, X are monovalent ions (W^+X^-) and Y, Z are trivalent ions (Y^3+ Z^3-). The i
ID: 1023386 • Letter: W
Question
W, X are monovalent ions (W^+X^-) and Y, Z are trivalent ions (Y^3+ Z^3-). The interatomic energies for both ion sets are given as E_attractive = - A(Ze)^2/r & E_repulsive = B/r^9 where A. and B are constants that are the same for both ion pairs, Z is the valence, and e is the fundamental charge of an electron. What would be the ratio of the equilibrium bond distances ro for WX compared to YZ? Diamond is elemental carbon. What would be the 4 quantum numbers n. 1, m_1, m_s for its highest energy electron? In order to reduce r_0 by 10%, energy must be increased. True or False and WHY?Explanation / Answer
1. The total energy of bonding in terms of an isolated ion pair is sumof the attractive and repulsive terms: Eattractive and Erepulsive. At equlibirum the attractive and repuslive forces are equal. At this equlibirum distance (r0) the bond distances depend on charge of the system. Assuming A and B are constants the equlibirum distance depend on Z . Hence ratio of the equlibirum bond distances depend are 1x1 (charge on the cation and anion) in WX system and 3x3 in YZ system. The ratio of the equilibrum bond distances is 1/9.
2. The electronic configuration of carbon is 1s22s22p2, i.e. with four valence electrons spread in the s and p orbitals. In order to create covalent bonds in diamond, the s orbital mixes with the three p orbitals to form sp3 hybridization. Each of the SP3 hybridised orbital has one electron to overlap with other carbon atoms to form four covalent bond. The four quantum numbers of of the electrons in hybridised carbon (all are degenerate) are n=2; l=0, 1; ml= 0, -1,0, +1 and ms = 1/2, 1/2,1/2, 1/2.
3. True. The equlbrium, r0 is the minmium energy system. Any reduction in the distance increases the repulsion and hence the total energy of the system.
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