In the above example there will be no JTD because Cu (II) belongs to +2 oxidation state with a high spin filling which gives us a symmetrical electronic arrangement in eg(half filled) and t2g (full filled) by eight electrons present in d orbital of Cu (II).
Cu has an exception.. Here there will be 10 electrons in 3d and 1 electron in 4s(since half filled and full filled orbitals are more stable). Thus in cu2+, the electronic conf will be 3d94s0
Excellent explanation..it has Cleared all my doubts. Thanks sir
thank you so much for explaining so well..
👍
Fabulous explanation sir thanks
Very nicely explained.
Thank you so much sir.😊👍👍👍
Nice explanation
Thank u sooooooo much for explaing this topic😀😁😂😂😂😂
Thank you sir, it's very understandable and please improve your sound quality, its disturb plzzzz
Seriously.. tq sir..
Thanks sir nice explain by you thank you
Wow
what if only one of eg or t2g has unsymmetrical arrangement? will only one distortion occur? In Cu(H2O)6]+2 .?.
In the above example there will be no JTD because Cu (II) belongs to +2 oxidation state with a high spin filling which gives us a symmetrical electronic arrangement in eg(half filled) and t2g (full filled) by eight electrons present in d orbital of Cu (II).
Sir,could you please explain why in z in and z out distortion the t2g orbitals splits in that specific way?
thank-you sir
Thank u sir
Thank you much sir
Sir you do not say that when z out or z in will happen & why this elongation and compression occur.
Sir pls give reply
10:41
thanks bhai time bacha diya. best of luck kal ke liye
How water shows J-T distortion, could you please explain?
it is 4s2 3d9 electronic configuration of cu and 2 electrons will get removed from 4s orbital and not from 3d
Cu has an exception.. Here there will be 10 electrons in 3d and 1 electron in 4s(since half filled and full filled orbitals are more stable). Thus in cu2+, the electronic conf will be 3d94s0
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