Some quick problems before my exam 2morrow

[ACLerok]

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

 

[himanshu121]

1,) 6 corresponding to l= 2,3,4 each with 2 spins

2.) possible sp^2-sp^2

3.) no of electrons

$$= \sum_0^{6} (2i+1) = 36$$

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

 

[ravenprp]

For the first question, the answer is C. 10. This is because the number of electrons allowed in an energy level is equal to 2n^2, where n is the principal quantum number. With n=5, there can be a maximum of 50 electrons, but since we are given other quantum numbers, we must take into account the exclusion principle and Hund's rule. Therefore, we must consider that l=0,1,2,3,4 and m_l=-2, and since m_s can have two values (+1/2 or -1/2), we can have 10 electrons in this scenario.

For the second question, the answer is D. sp^2 - sp^2. This is because the pi bond in CH2NF is formed between two sp^2 hybrid orbitals, one from carbon and one from nitrogen. The overlap of these two orbitals creates the pi bond.

For the third question, the answer is D. 49. This is because with n=5, there can be a maximum of 25 orbitals (2n^2). However, since we are only considering integer values of l, the allowed values are 0,1,2,3,4. This means there can be a maximum of 1+3+5+7+9=25 orbitals. However, since we are only considering l=0,1,2,3, there can be a maximum of 1+3+5+7=16 orbitals. Since we are considering n=5, there can be 5 different values of l, so the total number of orbitals is 5x16=80. However, since we are only considering integer values of l, we must divide by 2, giving us 80/2=40 orbitals. Since we are only considering m_l=-2, there can be a maximum of 2 electrons in each orbital, giving us 40x2=80 electrons. However, since we are only considering m_s=+1/2 or -1/2, we must divide by 2 again, giving us 80/2=40 electrons. Therefore, the maximum number of orbitals with n=5 is 40/2=20 orbitals.

For the fourth question, the answer is C. C2H4. This is because C2H4 has a double bond between the two carbon atoms