Problem Solving Strategies

TrY iS CheM

3:37 PM

Hello

Can we ask questions here

user41111

4:52 PM

Can anyone help

Yashas Samaga

5:35 PM

There is no current through the capacitor.

Remove the branch which has the capacitor from the node X.

Now solve the problem.

After solving, u get x.

Now find the V across the capacitor such that no current flows in that branch.

You get V and u kno C.

You can find q

@user41111

16 micro coulomb is the answer.

Koolman

5:58 PM

3

Q: Confusion in hybridisation of coordination compound

I want to know the configuration and hybridisation of the compound, hexaaquavanadium (III) ion that is $\ce{[V(H2O)6]^{+3}}$. In this Vanadium is in $+3$ oxidation state its electronic configuration is $\mathrm{[Ar]4s^0 3d^2}$. So there are $3$ more empty $\mathrm{d}$ orbitals. Hence the hybri...

@TrYiSCheM nice question

Yashas Samaga

You should stop using VBT.

Use CFT.

CFT is more accurate and gaurntees a correct answer for JEE level questions.

The only problem you will have is with cobalt.

Cobalt has too many exceptions.

0

A: RMS Values in AC Circuits

The voltage drop and the current flowing through a circuit element are changing with time. For an AC source, there is no current through the circuit element at one point of time and at another time, the maximum current flows through it. Therefore, multiplying the peak values of voltage drop and t...

A detailed explaination of RMS values of current/voltage in AC circuits.

There will be at least one Q from AC circuits.

And AC circuits is a very simple chapter.

Koolman

but in that in which shell the lone pairs will be donated

Yashas Samaga

4 marks for free

@Koolman ?

Koolman

@YashasSamaga why

Yashas Samaga

that msg was for AC circuits

I did not understand what you asked... long pairs being donated to which shell?

Koolman

6:03 PM

@YashasSamaga the lone pair wil donate electrons to d sub shel or s subshell

Yashas Samaga

what lone pair?

I have no idea in what context you are asking that question from

Koolman

6 mins ago, by Koolman

3

Q: Confusion in hybridisation of coordination compound

I want to know the configuration and hybridisation of the compound, hexaaquavanadium (III) ion that is $\ce{[V(H2O)6]^{+3}}$. In this Vanadium is in $+3$ oxidation state its electronic configuration is $\mathrm{[Ar]4s^0 3d^2}$. So there are $3$ more empty $\mathrm{d}$ orbitals. Hence the hybri...

inorganic-chemistry coordination-compounds

The oxygen has lone pairs which will donate

Yashas Samaga

d2sp3?

The person who answered the question also stated what I stated

don't use VBT :)

if you use VBT, then you will have one electron in the d orbital + donated electrons

Koolman

ohk

Yashas Samaga

as there are 2 or more empty d orbitals, the donated electrons will use two of the orbitals

two of the d2 orbitals

the rest are 4s and 4p

Koolman

6:09 PM

i see

Yashas Samaga

0

Q: How do you determine the hybridisation state of a coordinate complex?

If an exercise asks for the hybridisation state of a coordinate complex (ion in solution) consisting of a central cobalt atom surrounded by $\ce{NH3}$ ligands, and neither coordination number or complex ion is given, how do you decide which configuration will occur? The answer is that in a solut...

coordination-compounds transition-metals hybridization

read the first line in that answer

So start practicing using CFT to describe coordination compounds.

VBT is unreliable

If you were preparing for olympiads, you might have learned about LFT but that is not required for JEE.

Koolman

oh

have you learned that

Yashas Samaga

ye

the basics

but I prefer to use CFT

they are nearly the same

in LFT, you consider covalent character of the bonds

in CFT, you treat ligands as point charges

and consider electrostatic attraction only

Ligand field theory

Ligand field theory (LFT) describes the bonding, orbital arrangement, and other characteristics of coordination complexes. It represents an application of molecular orbital theory to transition metal complexes. A transition metal ion has nine valence atomic orbitals - consisting of five nd, three (n+1)p, and one (n+1)s orbitals. These orbitals are of appropriate energy to form bonding interaction with ligands. The LFT analysis is highly dependent on the geometry of the complex, but most explanations begin by describing octahedral complexes, where six ligands coordinate to the metal. Othe...

Koolman

anyway I don't like chemistry

Yashas Samaga

I don't like chemistry too :D

Koolman

6:15 PM

@YashasSamaga any idea about that

Yashas Samaga

The first reaction gives you N-substituted amide

In the second rection, LAH reduces it to amine

What's Mel?

Koolman

CH3I

Yashas Samaga

I can guess that it is CH3-Ph lol

Koolman

methyl iodide

Yashas Samaga

aw ok not that bad guess

OMG

It was MeI

I was like what is Mel

Koolman

6:18 PM

lol

@YashasSamaga you mean both carbon group will leave nitrogen

Yashas Samaga

the C=O will get reduced to alcohol

Koolman

yes

Yashas Samaga

the first reaction will convert the acid chloride to amide

N substituted amide

LAH reduces the C=O to OH

MeI replaces OH through SN2

Koolman

What will MeI do

I am confused

Yashas Samaga

I am confused too

this question is tricky

the dimethyl amine reaction produces two products

Koolman

6:22 PM

Then will Ag2O again add OH

Yashas Samaga

u get CH3CH2NHCl as a product too

nvm

The reaction sequence indicates that there cud be Hoffman elimination

but there are other stuff in that compound too

I need to check the mechanism :/

it's probably Hoffman elimination

+ SN2

let me try to get the final answer

Koolman

oh god

Yashas Samaga

omg it is Hoffman's elimination

I made a mistake in my diagram

I had retained a H in my amine

for nitrogen

the H isn't there

Koolman

@YashasSamaga Yeah

Yashas Samaga

now you have a 3* amine with a beta hydrogen

now it is straightforward then

Me from MeI attacks the amine

Koolman

6:28 PM

Hofmann elimination

Hofmann elimination, also known as exhaustive methylation, is a process where a quaternary amine is reacted to create a Tertiary amine and an alkene by treatment with excess methyl iodide followed by treatment with silver oxide, water, and heat. After the first step, a quaternary ammonium iodide salt is created. After replacement of iodine by an hydroxyl anion, an elimination reaction takes place to the alkene. With asymmetrical amines, the major alkene product is the least substituted and generally the least stable, an observation known as the Hofmann rule. This is in direct contrast to normal...

Yashas Samaga

you get a (+) charged nitrogen intermediate

AgI will precipate

Koolman