Wrapping up d block.... at last..

As per J.D LEE

Zn Cl₂ and Cd Cl₂ are ionic, but Hg Cl ₂ is covalent (Pg no. 842)

The halides and alkyls of lithium are far more covalent than the corresponding sodium compounds, and bcoz of this covalency, they r soluble in organic solvents. (Pg no. 310)

q2) b
Ag+ has completely filled d so cannot reduce and the d orbital cannot be used for hybridisation hence sp2

q5) 5 ions this is a double salt

q6) Zn(ll) is in group 6
Fe (lll) is in group 3 n Cu(ll) in group 2

by adding the group reagent of group 3 i.e. NH4OH in NH4Cl we will get reddish brown ppt. of Fe(OH)₃ which is filtered off to separate Fe(lll) from the solution.

q7. C Vanadium(V) 1 electron in d and blue coloured as given in NCERT

q8) A, D by fajans rules

@tush....i am getting confused becoz JD lee says something else and wiki something else

@rick
ans2,6,8 is wrong
ans5,7 are rite....

can u explain ans 5 a bit ??

Ans 9) (a) , (b) ?????????????[135]

9) B,D ??

Q.1) c. fr sure.

Q.2) It should be d).

Ag - 4d¹⁰ 5s¹ , Ag⁺ - 4d¹⁰

and in [Ag(NH3)2]⁺ , [Ag(CN)2]^- , [Ag(S2O3)2]^3- , its sp hybridised.

and it acts as a lewis acid.

so, the only statement right is d) all are linear.

Q.3) Actually , all the carbides, borides , arsenides etc. of transition metals are strong and hard compounds. so, i think with reference to that it might be due to covalent bonding.
i also remember sea borg's model of atomwhich said metallic bonding to be weak.

Q.6) put some sodium acetate in the mixture.

CH₃COONa + Fe³⁺ → [ Fe₃(OH)₂ (CH₃COO)₆]⁺ (a blood red colour solution)

thus , the other two ions remain undissolved.

Also , ammonium sulphide can be used as it gives black precipitate with Fe³⁺.

Q.7 ) Certainly V⁴⁺ is coloured , with one unpaired electron left.

Q.8) I feel it should be a) , b) , d).

as ionisation of HgCl₂ is not so high. this is the reason for which it can't give the chromyl chloride test , while others can.

Q.9) It should be a) , b) , c) , d).

as b) , c) , d) give direct precipitate with the reagent while , a) Fe₂₊ gives precipitate using this reagent as an intermediate.

Fe²⁺ + K₃[Fe(CN)₆] → Fe³⁺ + K₄[Fe(CN)₆]

then these products form Fe₄ [ Fe(CN)₆]₃ ( turnbull's blue )

so, practically, it should be all , but depends a) has been considered or not.

For ques 8, it should be (a) , (d)

ans. 6 is also right.

ans.8 is still doubtful.

ans.9 is surely wrong. as Cd²⁺ gives bluish white precipitate with the reagent.

well... I gave ans of q8) as a,d n u told it's wrong....[2]

for 6) we can use many reagents... the one i told is the one which
we have used in our school practicals (see qualitative analysis)

in q 9) Fe+3 gives prussian blue
Cu+2 gives reddish brown ppt

@arshad n aieee..

Fe+2 gives turnbull's blue with [Fe(CN)6]+3 not +4
for the Cd+2 test CH3COOH must be added

in Q5)

we see that it is not a complex as no"box" is there
it is just a double salt

so when we add water all ions are dissociated and show their independent tests

thx guys.....
All questions done now
but dbts still remain in Q8[2]

in q8)

the anion is same in all i.e. Cl
by fajans rule smaller the cation, more polarising power hence more covalent

now Li+ and Hg+2 are pretty small

so ans a,d

The question asks whether LiCl is ionic or not which it surely is.
This can be gauged from the fact that LiCl is used for electrolysis
in extraction of Lithium metal.
NO doubts about its covalent character as well.
Hope this clarifies the doubts. :)

QUOTING j.d.lee :" the halides and alkyls of lithium are far more covalent than the corresponding sodium compounds and because of this covalency they are soluble in organic solvents"

also, the first element in each of the main groups (Li, Be, B ,C ,N,O,F) differs from the rest of the group,this is partly because the first element is much smaller than the subsequent elements and consequently it is more likely to form covalent compounds (fajan's rule) and complexes !

@ uttara and arshad

both are wrong

Q1to 7 are single choice correct

Q3) b

Nops[2]

kk..sorry ...thx...it was a stupid dbt

9) B is one ans

I think C is also right