1) A Cobalt (II) ion has the electron configuration ____? and is ____?

Question

The answer choices were:
[Ar] 3d5 4s2 diamagnetic
[Ar] 3d7 paramagnetic
[Ar] 3d5 4s2 para
[Ar] 3d7 dia
None of these

I chose [Ar] 3d5 4s2 diamagnetic

2) Choose the corect statement concerning ionization energies.

- The IEof Na is greater than the IE of Cl
- The IE of S is greater than the IE of O
- The IE of Se is greater than the IE of Br
- Of the elements Li, Be, and Na, Li has the smallest IE
- Of the elements K, Ca, and Sr, Ca has the greatest IE.

I chose that the IE of S is greater than the IE of O

Are these correct?

Answer 1

1). No, you don't have it right.
Co electron configuration is [Ar]3d74s2.
To make the Co(II) ion, we remove the two 4s electrons so it now is [Ar]3d7. Since there are five d orbitals, placing 7 electrons can't be done evenly so it must have some unpaired electrons which makes it paramagnetic.
(Lat me give you the DrBob222 rule for the 3d transition elements. For the +2 ion of ALL of the 3d elements, the number of electrons in the 3d orbitals is the same as the second number of the atomic number. Neat, huh? That means Sc is 1, Ti is 2, V is 3, Cr is 4, Mn is 5, Fe is 6, Co is 7, Ni is 8, Cu is 9, and Zn is 10(not zero))

2.
O is 13.6, S is 10.4. Look at choice 5.

Answer 2

for chose 5 Ca is next to K and above Sr so wouldnt Sr be greater?

Answer 3

K = 4.3
Ca = 6.1
Sr = 5.7

Answer 4

ok where do you get those numbers from?