Yeah transition elements are pretty good at coming up with surprises.
lets examine qn 1
1) S is a transition element. The 3d sub-shell of S in the compound K(S(C2O4)2(NH3)2) contains 3 electrons. how many unpaired electrons does S contain when it is in elemental state? TJC prelim 2008
A – 3
B – 4
C – 5
D – 6
alright btw S is not sulphur
that is the first trick we gotta see through 🙂
Next, K is suspiciously in the compound… no transition element complex is going to have K in it. what K is doing there, is to form an ionic compound with (S(C2O4)2(NH3)2) –
the 2nd trick is to note that (S(C2O4)2(NH3)2) comes with a minus charge
the 3rd trick is easier to see, now element S cannot possibly be negatively charged as transition element ions are all positive… so the negative charge in (S(C2O4)2(NH3)2) comes from C2O4 which has a charge of 2-
4th trick is to process the infomation from trick 3, if the 2 C2O4s in (S(C2O4)2(NH3)2) each as a charge of -2, and overall (S(C2O4)2(NH3)2) has a charge of -1, by comparison, S has a charge of +3
Trick 5, now that you know that S has a charge of +3, and S in (S(C2O4)2(NH3)2) has 3 electrons in its 3d subshell, all we need to do is to add 3 more electrons to S to know its electronic configuration at elemental state.
Remember that when you add electrons to transition metal ions… fill up the 4s orbital first!
Trick 6. By following above, you will get something (3d)4(4s)2 so your answer will be 4 unpaired electrons. BUT BUT BUT BUT BUT the correct answer is (3d)5(4s)1 because all the 3d orbitals singly filled is OMGly stable, think of chromium 🙂