Thread View: can.schoolnet.chem.sr
1 messages
1 total messages
Started by anonymous
Fri, 12 Nov 1999 00:00
tough problems!
Author: anonymous
Date: Fri, 12 Nov 1999 00:00
Date: Fri, 12 Nov 1999 00:00
59 lines
3003 bytes
3003 bytes
Hello all, I'm a college student from Canada and I would really appreciate some help in doing the following chemistry problems. Any solution you may propose is very welcome. Thanks. #1. Consider the following ionization energies for the aluminium: Al(g)--> Al+(g) + e- I1 = 580kJ/mol Al+(g)-->Al+2(g) + e- I2 = 1815kJ/mol Al+2(g)-->Al+3(g) + e- I3 = 2740kJ/mol Al+3(g)-->Al+4(g) + e- I4 = 11600 kJ/mol (a) Account for the trend in the values of ionization energies (b) Explain the large increase between I3 and I4. (c) Look up the definition of electron affinity and determine which of the four ions has the greatest electron affinity. Explain. (d) List the four aluminium ions given above in the order of increasing size, and explain your ordering. #2. From the informations below, identify element X. 1. The wavelenght of the radio waves sent by an FM station broadcasting at 97.1 MHz is 30 million times greater then the wavelenght correspomding to the energy difference between a particular excited state of the hydrogen atom and the ground state. 2. Let V represent the principal quantum number of the valence shell of element X. If an electron in the hydrogen atoms falls from energy level V to the inner energy level corresponding to the excited state mentionned in part 1, the wavelenght of light emmited is the same as the wavelenght of an electron moving at a speed of 570m/s. 3. The number of unpaired electrons for element X is the same as the maximum number of electrons has the maximum number of electrons inan atom that can have the quantum number designations n=2, m1 = -1 and ms = -1/2. 4. Let A represent the principal quantum number for the electron in an excitted He+ ion in which the single electron has the same energy as the electron in the ground state of a hydrogen atom. This value of A represents the angular momemtum quantum number for the orbital containing the unpaired electron(s) for element X. #3. Determine the Lewis structure for N2O. Taking into account the following bond lenghts : N-N 167 pm N=O 115 pm N=N 120 pm N-O 147 pm N=N 110 pm Rationalize the observations that NN bond lenght in the N2O is 112 pm and that the NO bond lenght is 119 pm. Assign formal charges to all possible structures. Can you eliminate any structures on the basis of a formal charge ?
Thread Navigation
This is a paginated view of messages in the thread with full content displayed inline.
Messages are displayed in chronological order, with the original post highlighted in green.
Use pagination controls to navigate through all messages in large threads.
Back to All Threads