K2Pt+4Cl6 is well known compound and corresponding Ni4+ Salt it unknown ? Whereas Ni+2 is more stable than Pt+2.
Give Proper Explanation .

13 Answers

sakshi pandey pandey ·

bhaiyya is it bcoz in d block heavier species favour high coordination number.........& Ni being less heavier than Pt doesnt favour +4 oxidation state......?

mentor_Layak Singh ·

nopes !
try to solve carefully.
apply some periodicity concepts ...will be easy to explain.
try again !

mentor_Layak Singh ·

No one is there to solve this problem correctly !!!!!!!!!

eureka123 ·

concept to be used is ionization energies...

mentor_Layak Singh ·

the concept i have already told !!!

mentor_Layak Singh ·

The stability of the compounds depend upon sum of ionization enthalpies :
IE1 + IE2 < IE1 + IE2
in Ni in Pt

Ni2+ is stable than Pt+2.
IE1 + IE2 + IE3 + IE4 < IE1 + IE2 + IE3 + IE4
in Pt4+ in Ni4+
Pt4+ is stable,
K2PtCl6 is well known compound.

eureka123 ·

but we dont have to remember the values how to solve this ques othrewise??[7]

mentor_Layak Singh ·

Need not to be remember the values .
but one thing you should remember the trend of Ionization enthalpy in excited states.

Debotosh.. ·

great explanation mentor sahab ! wow !

Rajiv Agarwal ·

The electronic configuration of Ni : 1s2 2s2p6 3s2p6d8 4s2

And for Pt: 1s2 2s2p6 3s2p6d10 4s2p6d10f14 5s2p6d9 6s1

Now lets try to see what happens when we are trying to form the +2 states for both these elements:

The config of Ni+2 is: 1s2 2s2p6 3s2p6d8
While the config of Pt2 is: 1s2 2s2p6 3s2p6d10 4s2p6d10f14 5s2p6d8

Now config for Ni+4 is: 1s2 2s2p6 3s2p6d6
While the config for Pt+4 is: 1s2 2s2p6 3s2p6d10 4s2p6d10f14 5s2p6d6

Let me give a hint here so that you guys can try to find the logic for the values given by Layak (I agree with both Debotosh and Layak need not remember the values but the trend one may find out and keep in mind). Anyway, the hint goes: Screening Effect :) ...Now try to give an explanation.

Also see the weird electronic config of Platinum... its 5d96s1 ... unlike Ni which is 3d84s2 ...if you can explain this..the first one gets automatically answered

$ourav @@@ -- WILL Never give ·

giving a try...,first to form Ni+2 and Pt+2 ,...Pt is near to full filled d orbital than,more energy is required for Pt to remove 2 electrons
but while forming +4 cations,screening effect of 3d electrons is inadequate,so,dey are bound more closely and tightly to the nucleous.....hense to form Ni+4,more energy required to pull out the electrons....
is it correct rajiv sir???

Rajiv Agarwal ·

@Sourav ... your explanation for the +2 state is not correct....

i lll post the correct answre in the evening once more ppl have tried this... a related question can be from inert pair effect.

Akash Raj ·

Its because pt has 4f subshell which is less effective in shielding(lanthnoid contraction) and hence the 5s omsubshell contracts closer to nucleus.. accordingly it requires more energy to remove an electron from 5s subshell..hence the sum of 1st to I.E is higher than Nickel which lack f subshell..

Now taking about the sum of first 4 I.E.

Once the electron is removed from 5s orbital it is rendered empty..and the next is removed from d subshell which is not shielded much by 4f orbital..hand extraction electron is relatively pt is large molecule and its effective nuclear charge decreases ...and Correspondingly sum of first 4 IE decreases..

Hope it hepled..

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