Which ion has the MOST unpaired electrons?

[MedChad]

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Q: Which ion has the MOST unpaired electrons?

a. Cr 3+
b. Mn 2+
c. Cu +
d. Ni 2+

I must be over thinking this, but I can get it down between a and b, but don't understand how to conclude that the correct answer is b. Help would be great!!

 

[chiddler]

Q: Which ion has the MOST unpaired electrons?

a. Cr 3+
b. Mn 2+
c. Cu +
d. Ni 2+

I must be over thinking this, but I can get it down between a and b, but don't understand how to conclude that the correct answer is b. Help would be great!!

If chromium loses three electrons, then it leaves only 3 unpaired orbitals (4s1 3d5 --> 4s0 3d3)

Manganese loses its 4s2 electrons leaving its 3d5 shell intact. This means 5 unpaired! No competition.

 

[MedPR]

Q: Which ion has the MOST unpaired electrons?

a. Cr 3+
b. Mn 2+
c. Cu +
d. Ni 2+

I must be over thinking this, but I can get it down between a and b, but don't understand how to conclude that the correct answer is b. Help would be great!!

Look at the periodic table and think about the electron configurations. Remember that you lose 4s electrons before 3d electrons.

 

[SaintJude]

This question is testing whether you understand whether electrons are removed from 4s or 3d first.

The choices are all transition metals with d shells. A d subshell has 5 orbitals. Then each orbital can carry 2 electrons. According to the Hund's rule, every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied.

A. 3 unpaired electrons. Cr is [Ar] 4s^1 3d^5. Then removing 1 electron from the 4s subshell and 2 electrons from 3d subshell, you should get Cr3+ : [Ar]3d^3 --
B. Mn2+: [Ar]3d^5-- 5 unpaired electrons --You have to remove the electrons from the highest shell number first, which is 4s
C. Cu+: no unpaired electrons! Cu: is 4s^1 3d^10. So Cu+ will be: [Ar]3d^10. You removed the electron from the 4s shell first.
D. Ni2+: 2 unpaired electrons. Remember, [Ar] 3d8 4s2 for Ni and [Ar] 3d8 for Ni+2 because the electrons are removed from the 4sh shell first.

 

[Erythropoietin]

draw out electron configuration for each, 4s is lost before 3d

 

[osprey099]

Cr3+ means that you lose the 1 electron in the 4s orbital, then 2 electrons from the 3d orbital leaving you with 3 unpaired electrons. Cr electron configuration is 4s1 3d5 because having a half filled 3d orbital is more stable than a filled 4s orbital.

Mn2+ means that you lose both electrons in the 4s orbital, leaving you with 5 unpaired electrons.

Cu+ means that you lose the only electron in the 4s orbital leaving you with a filled 3d orbital so you have no unpaired electrons. Cu electron configuration is 4s1 3d10 for the similar reason as described for Cr.

Ni2+ means that you lose both electrons from the 4s orbital leaving you with 8 electrons in the 3d orbital so you have 2 unpaired electrons.

Therefore, the answer is Mn2+

 

[MedChad]

Thanks for all the great responses, I just had to take a break and look at it now and it makes perfect sense after reading your replies. The transition metals cause confusion for me sometimes just because they don't follow the normal trends of the periodic table.

 

[ilovemedi]

So removing the s before the d is only for TRANSITION metals ONLY?