CSIR NET 2024 | Chemical Bonding Chemistry | Revision Series | Questions Session 02 | By Nadeem Sir

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[@Vanshika_075]

Thank you so muchh sir❤

[@sumitrasaini8333]

H.w question answer c. Sicl4. Td. , trans sncl4 (py) 2

[@Prince-vv4ge]

But according to GT any molicule contain horizontal plane they are non polar molicule. ??????? but Td doesn't posses horizontal plane???

[@jheelugadwal625]

Option c answer of hw question

[@chemistrymagnets]

Option 3

[@harshadatamtharekar496]

C

[@hempushpasahu9377]

Option c is correct

[@Sohitrao234]

Option c

[@sujjuuraoyadav]

Option C..sicl4 perfect td so dipole moment 0...and trans-sncl4py2 plane present hoga..nonpolar so 0

[@pawanlipne8624]

C option

[@monikadhull1195]

Hw- option c

[@brajendrakumar3706]

3

[@biswajitbjm]

ans-3

[@Akash_neon11]

HW ans - opt c

[@ashakandpal9260]

Option c sicl4 td dm zero and trans sncl4py2 dm zero

[@hempushpasahu9377]

Sir master class join nahi hota h server error btata hai 2 days se try kr rhi join hi nahi ho pa raha

[@kk-kk6yr]

So option is C

[@monayadav614]

Option C correct..sicl4 perfect td &trans sicl4py2 have d4h point grp so possess horizontal plane then that is also non polar

[@kk-kk6yr]

Home work - SiCl4 perfect td and trans mai resultant dipole moment xero hota hai sir

[@vaidyarajprajapati1452]

H. W. Answer Option 3 SiCl4 is td dipole moment is 0 trans-[ Sn Cl4 ( py) 2] Oh dipole moment is Zero Thank you so much sir जी🙏❤

[@sumaylasumayla1599]

Hw option c correct. Sicl4 is perfect td. So dipole moment is zero and trans [Sncl4 py2]. Point grp is D4H and net dipole moment is zero Thank you so much sir

[@manishakumawat3354]

H.W. Option C ✅ SiCl4 and trans [SnCl4(py)2] has zero dipole moment

[@poojabajad5560]

Option c is correct sicl4 is non polar mol (u=o) td mol and sncl4(py)2 is (u=0) d4h non polar mol thank you sir

[@Prince-vv4ge]

But according to GT any molicule contain horizontal plane they are non polar molicule. ???????

[@Amit4UU2]

Hw ans) correct option 3)✅ a) SiCl4 -- perfect TD - Show hight symmetry point group.. Net dipole moment is 0. b) [Sncl4(py)2] ----- square bipy- D4h point group.... Dipole moment 0. C) P(O) Cl3 and, NF3 has pyramidal like shape... Point group C3V.... And polar molecule Tq sir🙏🙏🙏🙏

[@manishamishra3993]

Homework Ans :- With zero dipole moment : Correct option -> (c) SiCl4 and trans-[SnCl4Py2] SiCl4 is perfect td , so dipole moment is zero. And trans-[SnCl4Py2] has D4h point group, since horizontal plane present, so it has zero dipole moment. Thank you so much sir 🙏

[@sunitasaini1456]

Option c