713. The no. of electrons in the t2g set of orbitals for the complex [Co(NO2)6]- 4 is ?
11 Answers
71
2621) p=4 q=2 r=2 Ans:8 ?
3) Co2+ - 3d7 → 6 electrons in t2g
9) is it 0.5 ?
71For question 1, I agree with P,Q, but for R why is it 2? We have to chirality centers there, one with -Me and other with -Me and -OH .. ?
3. ) The answer given is 0
9) It's an integer type question. So obviously wrong. How did you attempt that.
13) 7
6C12 is the stable isotope of carbon having the desired np ratio of 1 for the elements in that range.
so if n/p ration is 1.33 = 4/3 then it must be C14
=> mv = hλ
=> 14 * 10-3 6.023 6 * 1023 = hλ
717 is correct. I too got it later.. hadn't noticed that it is given that it is carbon isotope.
1. I don't see any possibility of favored ring expansion. You may force one but that won't be right.
Some More Problems
For 7, I think something is missing. Even Naturally, Benzene has BP 354 K. It can't deviate so much.
For 8, I think it should be C, Since ortho effect doesn't work for Phenol.
262@vivek - 1)ring expansion will take place .
9 ) i forgot to multiply with 14 :P
3) if t2g is empty then it means configuration is 3d0 .Hence wrong!
8) i think C)
713. There is some really mind blowing reason given behind it.. It's somewhat like exception type. I thought someone might be aware of it here.
1. How is ring expansion taking place. Enlighten me a bit. If you could sketch some rough figure.
Any ideas about 7 ?
213)Vivek ans can be 0 for eg not for t2g
No2 is a strong ligand and attacks octahedrally.Due to crystal field splitting we have 1 elec. in eg and 6 elec in t2g.Now we know the energy of eg is raised and becomes comparable with the next s orbital.
Hence it prefers d2sp3[combination of similar energy orbitals] instead of dsp3d.Hence the 1 elec. in eg is shifted to orbital beyond the np orbital.
7)An interesting result is 52.25kJ/122J=428[though the formula is correct,it uses values of 2 diff. sub]
I think they have mixed up something.
71Yes *my mistake) !! By the way swaraj,, How do you/Where do you find such things.. ?
For the seven, I think that is just not write to get anyhow the options :P
13-the answer for third will be zero if the compound given has Co+3.
1-how did you find the number of sterioisomer in q?
and in r it would be 4 and p would be 4.