<p>My question concerns this homework problem I have: </p>
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<p>"How many unpaired electrons are there in the following transition metal ions. Write your answer as a sequence of numbers, for example, 23542141. </p>
<p>A) Mn5+ B) Fe3+ C) V5+ D) Fe4+ E) Mn4+ F) Ni4+ G) Cu4+ H) Sc2+ I)Fe2+ " </p>
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<p>My text wasn't really helpful. So, wouldn't these be the answers for e/one? 25043651 </p>
<p>For example, for Mn5+ I picked 2 (2 unpaired e-'s) b/c the config. is 4s2 3d5. so that is 5 + 2 = 7 unpaired electrons, minus 5 (Mn5+) = 2. Is this the right way to think about it? </p>
<p>And for instance, for V5+ I said..... since it's s2 d3, that = 5 unpaired electrons, and so 5-5 = 0. So am thinking it through the right away? And what about Cu4+, since it is one of the exceptions having a config of s1 d10? </p>
<p>Thanks in advance for any help</p>
<p>No, the two s electrons are paired with each other. The d electrons are unpaired if there are 5 or fewer.</p>
<p>also note that if the cation has the electron configuration of Cr, then it would have 6 unpaired electrons because the electron configuration is [Ar]4s^13d^5, where the d shell borrowed an electron from the s shell to be more stable.</p>
<p>i think the answer is 030216516. someone check if thats right.</p>
<p>btw, i've never heard of Fe4+, is it only formed via forced 4th ionization?</p>
<p>Electrons in each orbital don't pair unless they have to - so the second s electron has to pair with the first, but the second p electron doesn't pair with the first. </p>
<p>I get 030214514...so stanfordream we disagree on Ni4+ and Fe2+, both of which have 24e-. 1s2 2s2 2p6 3s2 3p6 4s2 3d4. Did you fill 3d before 4s? 4s is lower energy than 3d.</p>
<p>thanks, so do i fill 4s first than 3d?</p>
<p>you know there are some exceptions, right? like the configuration of Cr is [Ar] 4s1 3d5.....how do you explain this?</p>
<p>Oh I forgot about that. You're right.</p>
<p>Transition metals...the fourth row become d5 and the ninth d10 (accepted configuration, rather than the 4s2 3d4 predicted configuration). They're more stable that way.</p>
<p>thanks guys</p>
<p>i've another question: at 19.0 degree C a liquid has a 40mmHg vapor pressure; its normal boiling oint is 78.3. what is its enthalpy of vaporization in kJ(mol)^-1</p>
<p>the answer is 42 kJ(mol)^-1</p>
<p>but i have no idea how to get to the correct answer, please help. thank you</p>