General chemistry Q

[299678]

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3) Which of the following is the electron configuration of an element in an excited state?
A) 1s22s22p63s23p64s23d8
B) 1s22s22p63s23p64s13d10
C) 1s22s22p63s23p64s24d1
D) 1s22s22p63s23p5

Answer is C, I am wondering why B can't be answer? since one electron on S orbitals moved to 3 d orbital.

11) Which of the following is diamagnetic?
A) Ca+
B) Zn
C) Zn2+
D) Cu+

The answer is d) and I also agree with the answer, but why not for B)?
ZN also 4S23D10 which is diamagnetic

13) Which of the following is most likely to decay by beta emission?
A) calcium-40
B) oxygen-16
C) carbon-14
D) potassium-39

the correct answer is C)
I am confusing between c and d) D also lose their electron

Please explain me in detail to support your answer. Thank you

 

[UndergradGuy7]

3) Which of the following is the electron configuration of an element in an excited state?
A) 1s22s22p63s23p64s23d8
B) 1s22s22p63s23p64s13d10
C) 1s22s22p63s23p64s24d1
D) 1s22s22p63s23p5

Answer is C, I am wondering why B can't be answer? since one electron on S orbitals moved to 3 d orbital.

11) Which of the following is diamagnetic?
A) Ca+
B) Zn
C) Zn2+
D) Cu+

The answer is d) and I also agree with the answer, but why not for B)?
ZN also 4S23D10 which is diamagnetic

13) Which of the following is most likely to decay by beta emission?
A) calcium-40
B) oxygen-16
C) carbon-14
D) potassium-39

the correct answer is C)
I am confusing between c and d) D also lose their electron

Please explain me in detail to support your answer. Thank you

For 1, there are some exceptions like Copper, Cr, Mo, Ag, Au. They are more stable if they have a full d shell, so electrons are moved from the 4s to the 3d. It is better to have a full 3d shell.

Answer choice, b is the normal configuration of copper. One electron from 4s2 is moved to the 3d9 so the new configuration is 4s1 3d10
Notice choice C, moved all the way to 4d1. So it skipped 3d, 5s etc.

For 2, B, C, and D are all diamagnetic. For D because as above the configuration of copper is 4s1 3d10. If you remove electrons you remove it from the 4s first. So for Cu+ it becomes 4s0 3d10 so there are no unpaired electrons. Zn is 4s2 3d10 so it also has no unpaired electrons. The 4s is full and so is the 3d. Diamagnetic means no unpaired electrons. I guess Zn is also diamagnetic? If it was Zn+ it would be paramagnetic. Zn2+ is also diamagnetic. The only one that is different is Ca+ it is paramagnetic. Maybe the question wanted to know paramagnetic? A typo?

3, not really sure if my reasoning is what they are looking for. But looking at the periodic table. A, B, and D the mass number given after their names is the number given in the periodic table. So they are the most common and stable forms. C is the only one that is not the most common. Carbon is normally carbon-12. So carbon-14 is unstable.

Also from chad's video. b- decay happens to the one with the highest N/Z ratio (neutrons/atomic number). I didn't try calculating them for each one, but check and see if carbon-14 has the highest. For example carbon 14 has 6 atomic number and 14-6=8 neutrons so 8/6.

 

[299678]

Thank you undergrad guy. It really helps
for the last question, on the chad video, he said under atomic number below 20, the ration of N/Z closed to 1 will be more like to be radioactive, I calculated each one, but
ca - 1, Oxygen - 1, Carbon = 2.3, K = 1, so carbon should be least radioactive? or I just misunderstood chad in opposite way?

 

[lvtran]

Don't dwell on this concept, I'd focus more on nuclear decays and half life.

 

[UndergradGuy7]

Thank you undergrad guy. It really helps
for the last question, on the chad video, he said under atomic number below 20, the ration of N/Z closed to 1 will be more like to be radioactive, I calculated each one, but
ca - 1, Oxygen - 1, Carbon = 2.3, K = 1, so carbon should be least radioactive? or I just misunderstood chad in opposite way?

I don't know about the video, but I looked at the word document he has with the video. In the word document there is a table and it says b- decay.. who? N/Z is too high. So the higher the N/Z ratio the more likely for it to do b-.