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lastzealot新蟲 (著名寫手)
【亂世的奸雄】
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[求助]
關(guān)于單重態(tài)和三重態(tài)
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看了一個資料里面講道 把乙烯分子的C=C鍵旋轉(zhuǎn)90度后,C=C鍵之間的pi鍵消失,只剩下sigma鍵。這一點我理解。 然后文中說旋轉(zhuǎn)后變成了“open-shell singlet”,也就是開殼層單重態(tài) 我就不明白了,為什么不是三重態(tài)呢?畢竟有兩個單電子的呀? 那個資料中的相關(guān)文字如下 After your calculations, you should always check that the electronic states that you have obtained are reasonable. In this exercise we want you to take a closer look on the electronic state of your ethene singlet rotated 90o. What is the spin state (Mulliken spin population) on the two carbon atoms? Does this correspond to an open or closed shell singlet? To be able to rotate the double bond of ethene, one of the bonds has to be broken (the pi overlap must vanish). Where do you expect the two electrons formerly associated with the bond to be in the staggered conformation (90orotation)? A reasonable assumption is that they are located in different orbitals, approximately located on the two carbons. The spins should still be antiparallel in the singlet. This state is referred to as an open-shell singlet. Finding the right solution for an open-shell singlet is not straightforward. 謝謝。 |
新蟲 (著名寫手)
【亂世的奸雄】
木蟲 (著名寫手)
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他沒說體系一定在單重態(tài)啊,他只是說有這樣一個態(tài),也沒說是基態(tài)。這里也沒有講任何具體的過程,也不可能知道會在哪個態(tài)。他只想介紹這個概念,講的是這個構(gòu)型下單重態(tài)會是開竅層的而已。 發(fā)自小木蟲IOS客戶端 |
新蟲 (初入文壇)
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你好,我想問問為什么開殼層會是自旋單態(tài)呢?按我得理解,開殼層是有未成對電子,設(shè)未成對電子數(shù)為s(s≠0),自旋多重度S=2s+1≠1,不等于1為什么會有開殼層單態(tài)。新手概念可能理解錯了,希望高手能指正出來。 發(fā)自小木蟲Android客戶端 |
木蟲 (著名寫手)
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首先他想說的是這個結(jié)構(gòu)上單態(tài)是開竅層,而不是開竅層是單態(tài)。 自旋多重度是2s+1但s是總自旋量子數(shù),和未成對電子數(shù)沒有直接的確定關(guān)系。如果未成對電子數(shù)是n,那么最大可能的s就是n/2,但如果n是偶數(shù)s最小可以是0,如果是奇數(shù)最小是1/2。角動量耦合是即使實驗化學家都應(yīng)該熟練掌握的基本功,如果這么基本的東西還沒學扎實會遺禍無窮的,還是好好復(fù)習一下吧 發(fā)自小木蟲IOS客戶端 |
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