Problem

Why are $\mathrm{Mn}^{2+}$ compounds more stable …

Problem 1 Easy Difficulty

Write down the electronic configuration of:
(i) $\mathrm{Cr}^{3+}$
(iii) $\mathrm{Cu}^{+}$
(v) $\mathrm{Co}^{2}+$
(vii) $\mathrm{Mn}^{2+}$
(ii) $\mathrm{Pm}^{3+}$
(iv) $\mathrm{Ce}^{4+}$
(vi) $\mathrm{Lu}^{2+}$
(viii) $\mathrm{Th}^{4+}$

Answer

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Chemistry 102

NCERT Class 12 Part 1 - Chemistry

Chapter 8

The d-and f-Block Elements

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Video Transcript

When we're talking about transition metal ions and their electron configurations, we start with the ground state and then remove electrons from the outer move shell going inward. So if we start with vanadium, we have are gone now vanadium in the ground state will be three d three for as to that's neutral vanadium. So if we go to Vanadium by waas, we've taken away those five outer electrons so right, that is three D zero for s zero and we continue that way we look at chromium three plus starting with are gone in the neutral state Crow mamma's three d 34 s to. So instead we have three D to four as zero looking at banging knees. Manganese in the ground state is three d five. We were asked to. So we have Megan ease two. Plus, we take those two s electrons away for iron three plus, As always, we're starting with are gone for this first row iron in the ground status three D six for s two. So here we're at three d five for s zero. As you can see, manganese two plus an iron three plus have the same electron configuration Now we go to copper to us. Copper in the ground state is three D 10 for s one. So we take away no one s electron and one D electron we look at can be, um, re plus in me neutral state. It would be three d one for us to, but here we have three b zero four. That's zero. Since we've taken away those three outer electrons and then lastly, we look at Itanium four plus, we will actually see that we end up with the same electron configuration three D zero for s zero.

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