Some quick problems before my exam 2morrow

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The discussion focuses on specific chemistry problems related to quantum numbers and molecular bonding, particularly in preparation for a chemistry exam. Key questions include determining the number of electrons allowed with quantum number n=5, identifying the orbital overlap for the pi bond in CH2NF, calculating the number of allowed orbitals for n=5, and comparing C-C bond lengths in various molecules. The correct answers are 10 electrons for n=5, sp2-sp2 for CH2NF, 36 orbitals for n=5, and C2H2 having the shortest C-C bond length.

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Some quick questions before my Chem exam tomorrow. Here are some problems that were on a practice exam but I don't have the answers to them. I am stuck on all of these so i was wondering if you guys/gals can help me out with these. If you can't do all then one is fine.

how many electrons are allowed with n=5, l=?, m_l=-2, m_s=?
A. 3
B. 6
C. 10
D. 8
E. 25

The pi bond in CH2NF is formed with the following orbital overlap.
A. sp^2 - 2py
B. 2px - 2py
C. 2py - sp^3
D. sp^2 - sp^2
E. 2py - 2py

If the allowed values of the quantum number l were all integers from 0 to n, how many orbitals are allowed with n=5?
A. 64
B. 36
C. 83
D. 49
E. 25

In which of the following molecules is the C to C the shortest?
A. C2H6
B. C2H2
C. C2H4
D. H3CCOOH
E. H3CCN
 
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1,) 6 corresponding to l= 2,3,4 each with 2 spins

2.) possible sp^2-sp^2

3.) no of electrons

[tex]= \sum_0^{6} (2i+1) = 36[/tex]

4.) triple bond will have shortest C-C distance hence C2H2
 
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