inorganic cumulatively !

hi everyone ! i was of the opinion that we can handle inorganic chemistry for coming year's JEE in a different way ! all of us have observed the questions in the recent 5 years JEE papers ! so, what we can do is collect bits and pieces of information on inorganic chemistry from the sources that individuals get and put those up in this thread ! that would be of great help to everybody !!
please start contributing !

12 Answers

24
eureka123 ·

nice initiative...

but i am confused what we have to contribute ??
reactions or properties ??

1
Shreyan ·

ya...nice thread..

well i start with this:

NaNO3 ----500oC---------> NaNO2 + O2
|
--------800oC----------> Na2O + O2 + N2

(not many people know about the second reaction...)

19
Debotosh.. ·

contribute anything important and sensible for JEE 2010 !

19
Debotosh.. ·

dead thread !!!!

11
Devil ·

Can someone come up with something on the unsymmetrical and symmetrical cleavages of diborane?

1
Little Angel ·

Let's start with Qualitative Analysis : -

IMPORTANT POINTS TO BE NOTED:-

(i) Cu Cl 2, Zn Cl2, C6H5NH3 + Cl - (Anilinium chloride) , Ba Cl2 gives chromyl chloride test. But the Chromyl chloride test is not given by chlorides of Ag, Hg, Sb, Sn.

(ii)Ag Cl is not precipitated if AbHO3 is added to [Co (NH3)3 Cl 3]. A Cl is not soluble in H2O, HI and HNO2.

(iii)Ag Cl is not soluble in H2O and dil. HNO3 but soluble in NH4OH, NaCN , KCN, hypo, etc. due to the formation of complexes.

(iv) Ag Br is not soluble in HNO3 but completely soluble in hypo.

(v) Ag Cl, Ag Br , Ag I all are soluble in hypo.

(vi) Ag F is soluble in water.

(vii) Ring test is not reliable in presence of nitrite, bromide, and iodide.

24
eureka123 ·

I will keep adding compounds here....

here is first one

Mercuric oxide, HgO : It is obtained as a red solid by heating mercury in air or oxygen

for a long time at 673 K
679K
2Hg + O2 → 2 HgO ( red )

or by heating mercuric nitrate alone or in the presence of Hg
Heat
2Hg ( NO3 )2 → 2HgO + 4NO2 + O2
red

When NaOH is added to a solution of HgCI2, yellow precipitate of HgO are obtained.
Hg2 CI2 + 2NaOH → HgO ↓+ H2O + 2NaCl
(yello)

Red and yellow forms of HgO differ only in their particle size. On heating to 673 K, yellow form changes to red form.
673K
HgO → HgO
Yellow red

It is used in oil paints or as a mild antiseptic in ointments.

If anyone dislike it..plz tell me.[1]

19
Debotosh.. ·

this was posted by tushar, sometimes back :

24
eureka123 ·

Mercurous chloride, Hg2Cl2 :

It is obtained as under :

(a) Hg2 ( NO3 ) + 2NaCl → Hg2 Cl2 + 2NaNo3
white ppt.
Heat in an iron retort
(b) HgCl2 + Hg → Hg2 Cl2(condenses on cooling)

It is purified by sublimation.

Mercurous chloride is also called calomel. It is a white powder insoluble in H2O. On heating, it decomposes to give HgCl2 and Hg.

Heat
Hg2 Cl2 → Hgcl2 + Hg

It dissolves in chlorine water forming mercuric chloride.

Hg2Cl2 + Cl2 → 2HgCl2

With ammonia, it turns black due to the formation of a mixture of finely divided black Hg and mercuric amino chloride.

Hg2Cl2 + 2NH3 → Hg + NH2 HgCl + NH4Cl

(black)

It is used to prepare standard calomel electrode and as a purgative in medicine.

19
Debotosh.. ·

Bent's rule, which concerns orbital hybridisation of chemical bonds, was stated in 1961 by the American chemist Henry Bent. Originally, it was expressed as follows:
• Atomic s character concentrates in orbitals directed toward electropositive substituents
In other words, more understandable in terms of p-character of hybridised bonds:
• A central atom tends to direct hybrids of higher p character toward more electronegative substituents

courtesy:wikipedia

1
aieeee ·

Some important complex formation reactions.

Pb ( CH3COO )2 + NaHCO3 → PbCO3 . Pb (OH)2 ( basic lead carbonate )

NH3 + K2 [ HgI4 ] → HgO . Hg (NH2) I ( iodide of millon's base )

Zn2+ + K4 [ Fe (CN)6 ] → K2Zn3 [ Fe (CN)6 ]2 ( white precipitate )

Fe2+ + ( S2O3 )2- → [ Fe (S2O3) ]- ( violet solution )

Na2S2O3 + AgBr (yellow) → Na3 [ Ag (S2O3)2 ] ( soluble complex ) + NaBr

Be + 2NaOH + H2O → Na2 [ Be (OH)4 ] (sodium beryllate) + H2

B2H6 + NH3 (at 1000o C) → B2H6 . 2NH3 ( at 450K ) → B3N3H6 (Borazole or inorganic graphite ) + H2

Na2CO3 (at 1400o C ) → CO2 + Na2O ( + SiO2 ) → ( Na2SiO3 )n ( a type of glass )

CuSO4 + Na2S2O3 → CuS2O3 + Na2SO4

CuS2O3 + Na2S2O3 → Cu2S2O3 + Na2S4O6

Cu2S2O3 + Na2S2O3 → Na4 [ Cu6 ( S203 )5 ]

(these three are series of reactions shown by hypo )

19
Debotosh.. ·

someone please take care to pink these important contributions because this is not possible to accumulate on anyone's own !

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