Consider the following ions and calculate $(X + Y)$ :
Pb²⁺, Sn²⁺, Cd²⁺, Sb³⁺, Fe³⁺, Zn²⁺, Mg²⁺, Hg²⁺, Cr³⁺.
Here,
X = number of ions that give a coloured precipitate with dilute HCl + H₂S.
Y = number of ions that give a precipitate with dilute HCl.

Question

Consider the following ions and calculate $(X + Y)$ :

Pb²⁺, Sn²⁺, Cd²⁺, Sb³⁺, Fe³⁺, Zn²⁺, Mg²⁺, Hg²⁺, Cr³⁺.

Here,

X = number of ions that give a coloured precipitate with dilute HCl + H₂S.

Y = number of ions that give a precipitate with dilute HCl.

TestHub · Accepted Answer

X = ions forming colored precipitates with dilute HCl + H₂S (Group II cations): Pb²⁺ (black PbS), Sn²⁺ (brown SnS), Cd²⁺ (yellow CdS), Sb³⁺ (orange Sb₂S₃), Hg²⁺ (black HgS). So, X = 5.

Y = ions forming precipitates with dilute HCl (Group I cations): Pb²⁺ (white PbCl₂). So, Y = 1.

Therefore, X + Y = 5 + 1 = 6.

Chemistry - Salt Analysis Question with Solution | TestHub

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JEE Advanced (2028)