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Central Atom and Ligands

Which NH4Cl...

Question

Which $N H_{4} C l$ is not used together with $N H_{4} O H$ in group-III reagent which of the following cation will not be precipitated?

A

F e^{2 +}

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B

C r^{3 +}

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C

Z n^{2 +}

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D

B a^{2 +}

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Solution

The correct option is D $B a^{2 +}$
As $B a (O H)_{2}, K s p = 5.0 \times 10^{- 3}$ has high solubility product so it will not form ppt.

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Similar questions

Q. In the third group of qualitative analysis, the precipitating reagent is

N H_{4} C l + N H_{4} O H

. The function of

N H_{4} C l

Q. In III group precipitation,

N H_{4} C l

is added before adding

N H_{4} O H

List I	List II	List III
A. $Z n^{2 +}, N i^{2 +}$	1. $H_{2} S / H^{+}$	P. I
B. $A g^{+}, H g_{2}^{2 +}$	2. $(N H_{4})_{2} C O_{3} / N H_{4} O H$	Q. II
C. $C a^{2 +}, B a^{2 +}$	3. $H_{2} S / N H_{4} O H_{,} N H_{4} C l$	R. III
D. $C u^{2 +}, C d^{2 +}$	4.dil. HCl	S. IV
E. $A l^{3 +}, C r 3 +$	5. $N H_{4} C l / N H_{4} O H$	T. V

Pick the WRONG pairing from the following
1. A3S
2. B4P
3. C2Q
4. E5R

Q. The filtrate of group II has been given to you for analysis of group III ( which may contain

F e^{2 +}, C r^{3 +} a n d A l^{3 +}

). mention systematic steps you would follow

Step I : Boil off

H_{2} S

Step II : Add a drop of

H N O_{3}

F e^{2 +}

F e^{3 +}

N H_{4} C l

boil, cool and add

N H_{4} O H

Which of following is correct option?

Q. Which of the following cations (i.e., basic radicals) forms colored (not white) precipitate with aqueous solution of potassium iodide but the precipitate does not dissolve in excess of reagent ?

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