Q24 of 27 Page 1

A metal complex having composition Cr(NH3)4Cl2Br has been isolated in two forms A and B. form A reacts with AgNO3 to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia.

(i) Write the formulae of isomers A and B.


(ii) State the hybridization of chromium in each of them.


(iii) Calculate the magnetic moment (spin only value) of the isomer.

(i) The isomer A = [Cr(NH3)4BrCl]Cl


The isomer B = [Cr(NH3)4Cl2]Br


(ii) Hybridization of Cr in both is d2sp3


(iii) 3. 87 BM


Explanation:


(i) Since, [Cr(NH3)4BrCl]Cl + AgNO3AgCl


WE know that AgCl is a white precipitate which is readily soluble in dilute aqueous ammonia we can tell that it is isomer A


And


Since, [Cr(NH3)4Cl2]Br + AgNO3AgBr


We know that AgBr gives a pale yellow precipitate which is soluble in concentrated ammonia we can tell that isomer B is the right answer.


(ii) Cr3+ is in 3d3 system; we can tell that its hybridization is d2sp3.


[NOTE:read about the different systems to know what falls in which system and its hybridization to learn this concept easily]


(iii) To find the magnetic moment:


Formula :


Where, n = the number of unpaired electrons


In Cr3 + we have d3 outermost shell configuration =


1s2 2s2 2[6 3s2 3p6 3d5 4s1 [Ar]3d5 4s1 [Ar]3d3.


Moreover, therefore it has three unpaired electrons


Substitute 3 in the formula




= 3. 87 BM


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OR


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(a) Identify A – D


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(i) Aniline and Benzylamine.


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OR


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(i) Ethanoic acid into methenamine.


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(a) A cell is prepared by dipping a zinc rod in 1M zinc sulfate solution and a silver electrode in 1M silver nitrate solution. The standard electrode potential is given:
E0Zn2+/Zn = -0.76 V,
 E0Ag2+/Ag = +0.80 V,

What is the effect of an increase in the concentration of Zn2 + on the Ecell?


(b) Write the products of electrolysis of aqueous solution of NaCl with platinum electrodes.


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OR


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