hybridisation for dative bond compounds
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What would be the hybridization in CO, Al2Cl6 and NH4 +?
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Originally posted by hoay:
What would be the hybridization in CO, Al2Cl6 and NH4 +?
CO : C is sp, O is sp.Al2Cl6 : Al is sp3, Cl is sp3.
NH4+ : N is sp3, H is s.
That the bond is dative (or not) is a separate matter and hence irrelevant to the orbital hybridization.
Further note : due to resonance and other factors, in some cases the orbital hybridization may not involve exact integer values, but that's for Uni level ; for A level purposes, choose the most accurate hybridization with integer values, based on the major resonance contributor.
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Thank you for ur answer.
In CO2 and SO2 is it sp2? any explanation how to conclude which hybridization is there?
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Originally posted by hoay:
Thank you for ur answer.
In CO2 and SO2 is it sp2? any explanation how to conclude which hybridization is there?
CO2 : C is sp, O is sp2.SO2 : S is sp2, O is sp2.
By observing the electron geometry.
For instance, the N atom in the amide group is sp2 hybridized, with a trigonal planar electron geometry.
The lone pair resides in an unhybridized p orbital, which allows for delocalization of the lone pair by resonance (which explains why amides are less nucleophilic and Bronsted-Lowry basic compared to amines and even imines), which is thermodynamically advantageous as it confers resonance stabilization energy by strengthening bonds in the resonance hybrid, resulting in a more exothermic enthalpy of formation, and hence a more negative Gibbs free energy of formation.
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Flask X contains 5 dm3 of helium at 12 kPa pressure and flask Y contains 10 dm3 of neon at 6 kPa
pressure.
If the flasks are connected at constant temperature, what is the final pressure?Ans. V1 = 5 dm3 ; P1 = 12kPa
V2 = 10dm3 ; P2 = 6kPa
Do u have to use P1V1 = P2V2 or PV=nRT ?
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Originally posted by hoay:
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Flask X contains 5 dm3 of helium at 12 kPa pressure and flask Y contains 10 dm3 of neon at 6 kPa
pressure.
If the flasks are connected at constant temperature, what is the final pressure?Ans. V1 = 5 dm3 ; P1 = 12kPa
V2 = 10dm3 ; P2 = 6kPa
Do u have to use P1V1 = P2V2 or PV=nRT ?
Use PV=nRT twice, to find the moles of gas (in terms of T) in each flask before connecting together.After connecting together, add up to find the total moles of gas. Plug the total moles of gas, together with the combined volume, into PV=nRT for the 3rd time, to solve for the final pressure.
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I got the answer 8 dm3 and that is correct.
Many thank to you UltimaOnline.
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Originally posted by hoay:
I got the answer 8 dm3 and that is correct.
Many thank to you UltimaOnline.
No prob, Hoay :) -
All alkane are sp3 hybridized....What about cycloalkanes? are they sp3 hybridized too?
as there is no pi bond so sp3 would be favoured.
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Originally posted by hoay:
All alkane are sp3 hybridized....What about cycloalkanes? are they sp3 hybridized too?
as there is no pi bond so sp3 would be favoured.
Yes. -
thank you.
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In 1-butene CH3CH2CH=CH2 the hybridization is
CH3 .....sp3
CH2......sp3
CH......sp2
CH2 (double bond) .....sp2
is it correct ?
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Originally posted by hoay:
In 1-butene CH3CH2CH=CH2 the hybridization is
CH3 .....sp3
CH2......sp3
CH......sp2
CH2 (double bond) .....sp2
is it correct ?
Yes, correct. -
Hi
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In the electrophilic addition reaction of propene in the carbocation the positively charged carbona atom is sp3 hybridized (in propene it is sp2). But the electron geometry is tripgonal planar so it must be sp2 why it is sp3?
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Originally posted by hoay:
In the electrophilic addition reaction of propene in the carbocation the positively charged carbona atom is sp3 hybridized (in propene it is sp2). But the electron geometry is tripgonal planar so it must be sp2 why it is sp3?
During the intermediate, it is sp2. Only upon nucleophilic addition (the 2nd step of the 2-step electrophilic addition mechanism) does it become sp3. -
Actually i was referring to Q.37 of Nov 2009 Singapore A-level. It asks about the change of hybridization from sp3 to sp2 in the reaction internediate.
three reaction were given:
1 CH3-CO-CH3 + CN-
2 (CH3)HC=CH(CH3) + H+
3 C6H6 + +NO2
1 is correct since Carbon joined to Oxygen has 4 eelctron pairs
about 2 the marking scheme says its true also
3 is correct also as in Wheland Intermediate one of carbon is sp3
1 and 3 are correct...2 must be correct also ?
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Originally posted by hoay:
Actually i was referring to Q.37 of Nov 2009 Singapore A-level. It asks about the change of hybridization from sp3 to sp2 in the reaction internediate.
three reaction were given:
1 CH3-CO-CH3 + CN-
2 (CH3)HC=CH(CH3) + H+
3 C6H6 + +NO2
1 is correct since Carbon joined to Oxygen has 4 eelctron pairs
about 2 the marking scheme says its true also
3 is correct also as in Wheland Intermediate one of carbon is sp3
1 and 3 are correct...2 must be correct also ?
All 3 options are correct.In your earlier post, you asked abt the +ve formal charged C atom of the intermediate, that remains sp2 during the intermediate.
Instead Cambridge is saying, if *either* of the reactant's C atoms transform from sp2 to sp3, then the option is to be considered correct.
For option 2, of the 2 sp2 C atoms of the alkene undergoing protonation, 1 C atom (whose pi bond becomes a new sigma bond) immediately becomes sp3, while the C atom which then bears the +ve formal charge remains sp2, until nucleophilic addition of the Bronsted-Lowry acid's couter anion conjugate base (eg. Cl- ion if HCl was employed as the Bronsted-Lowry acid) occurs, then it becomes sp3 as well in the product.
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A gardener fertilizer is said to have a phosphorous content of 30.0% soluble in water.
what is the % by mass of the phosphorous in the fertilizer?
A 6.55 % B 13.1 % C 26.2 % D 30.0 %
Ans ... the balanced reaction is P2O5 + 2H2O............... 2H3PO4
Now 30.0 % of P2O5 (Mr = 142) is 42.6. so mass of phosphorous will be
142g..........62g in H3PO4
42.6..........x g in H3PO4
x = 18.6 g of Phosphorous
since only 30.0% is soluble so out of total mass of H3PO4 (196) only 58.8 is available
18.6/58.8 x 100 = 13.8 %
Is the working correct?
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Originally posted by hoay:
A gardener fertilizer is said to have a phosphorous content of 30.0% soluble in water.
what is the % by mass of the phosphorous in the fertilizer?
A 6.55 % B 13.1 % C 26.2 % D 30.0 %
Ans ... the balanced reaction is P2O5 + 2H2O............... 2H3PO4
Now 30.0 % of P2O5 (Mr = 142) is 42.6. so mass of phosphorous will be
142g..........62g in H3PO4
42.6..........x g in H3PO4
x = 18.6 g of Phosphorous
since only 30.0% is soluble so out of total mass of H3PO4 (196) only 58.8 is available
18.6/58.8 x 100 = 13.8 %
Is the working correct?
The original Cambridge question is different from the one you paraphrased.For the original question, P2O5 takes up 30% (by mass) of the fertilizer.
Calculate % (by mass) of P in P2O5, ie. (2 x Mr of P) / (Mr of P2O5).
Final answer = (% P in P2O5) x 30%
The "solubility" is just a red-herring, irrelevant to the qn, so just ignore it.
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I got 13.1 % which is the correct answer.
This was from Nov 2003 / CIE / P1. I copied the question as i have it here. Anyway thank you for your prompt answer.
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Originally posted by hoay:
I got 13.1 % which is the correct answer.
This was from Nov 2003 / CIE / P1. I copied the question as i have it here. Anyway thank you for your prompt answer.
Your original post left out "P2O5" in your question, which changes the question entirely. In future, for all CIE questions, please link directly to the question online, whenever available.Working is (as I explained in my previous post)
( 30/100 ) x ( ( (2 x 31) / ( (2 x 31) + (5 x 16) ) ) = 0.131 or 13.1%
No prob, Hoay :)
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okay thank you for the link