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Hi

Well I didn't understand the 1st question because I feel the CuCN has to react with something to give those products. Anyway once you get the balanced equation we solve the problem. Because we know the the no of moles of CuCn and water I feel and we know eqiuvalents of water.

Thus reciprocal of ratio between no of moles of water and CuCN is equal to ratio between their eqivalents. Thus we can find the n factor of CuCN.

I am not sure with your 2nd question but I feel it is because of its structure.It is like in Cr2O7.

Please let me know if my method for the 1st question is right.      

Hi

I got your 1st answer. It is kind of simple.

Now Cu is oxidised from +1 to +2 state.

C is oxidised to +4 from 0.

N is oxidised -1 to +5.

Thus total change is 1+4+6=11.

This total change is defined as n-factor.

(1.) n-factor is defined as the total change in oxidation number per mole of the substance .

Here , Cu⁺ is converted to Cu^{2+ _.}

There is no change in oxidation number for C ( it remains as +4 )

N is converted from -5 to +5 state .

So the net change in oxidation number , n-factor = (2-1) + [5-(-5)]  

                                                                       =11 .

So Answer is (A.)

(2.) CrO_{5 ,} is actually  Cr(O₂)₂O .

In CrO_{5 ,} there are 2 peroxo linkages , that is , -O-O- bonds .

There is only one Cr=O bond .

In the peroxo linkages , each O has a charge of -1 .

The O that is double-bonded has a charge of -2 .

So the net charge on the 5 O atoms = 2*(-2) + (-2)

                                                    = -6

So the charge on Cr is +6 .

So Answer is (C.)

 i don,t know answer of i st qestion but answer of second qestion is due to the per oxide linkage of oxygen with cr 4 O atoms join with cr in peroxide  way  and each per oxide will have -1 oxd. no. so making a total of -4 the next oxygen will join in a double bond in normal way so the total oxd.no. becomes -6 and Cr will have oxd.no +6 which is its maximum 

CN cynide has charge -1.........

structure of CN is   'C#N   thus,

N has -3 and C has +3-1=+2

Cu has +1

after reaction=> C= +4 , N=+5 ,Cu=2+

 

change in charge = copper => +2-+1=1

= nitrogen => +5-(-3)= 8

=carbon => +4-(+2) = 2

total change = 1+2+8=11

 

justin george and Suvarthi Datta have got it wrong

The oxidation state of CrO₅ is 6 because oxidation state is also decided by Roman Group no. which in case of Cr is 6!!!