The pair of compounds that can exist together is

A \(FeCl_3, SnCl_2\)
B \(HgCl_2, SnCl_2\)
C \(FeCl_2, SnCl_2\)
D \(FeCl_3, KI\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(FeCl_2, SnCl_2\)

Step-by-step Solution

Detailed explanation

The compounds with lower oxidation number and which cannot reduced by one another can exist together. Thus, \(F e C l_{2}\) and \(S n C l_{2}\) can exist together as \(F e^{2+}\) cannot be reduced by \(S n^{2+}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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The compound \(A\) on treatment with \(Na\) gives \(B\), and with \(\mathrm{PCl}_{5}\) gives \(\mathrm{C} . \mathrm{B}\) and \(\mathrm{C}\) react together to give diethyl ether. \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are in the order

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Match List I with List II :

List I (Order of reaction)	List II (Unit of rate constant)
(A) Zero order	(I) \(mol ^{-1} Ls ^{-1}\)
(B) First order	(II) \(mol ^{-2} L^2 s^{-1}\)
(C) Second order	(III) \(s ^{-1}\)
(D) Third order	(IV) \(mol L ^{-1} s^{-1}\)

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A. ERV (Expiratory Reserve Volume)	I. \(2500-3000 mL\)
B. RV (Residual Volume)	II. 500 mL
C. IRV (Inspiratory Reserve Volume)	III. \(1000-1100 mL\)
D. TV (Tidal Volume)	IV. \(1100-1200 mL\)

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The standard heat of formation, in \(kcal / mol ^{\text {of }} Ba ^{2+}\) is :
[Given : standard heat of formation of \(SO _4^{2-}\) ion \(( aq )=-216 kcal / mol\),
Standard heat of crystallisation of
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