A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M KMnO₄ solution to reach the end point. Number of moles of Fe²⁺ present in 250 mL solution is x $$\times$$ 10^$$-$$2 (consider complete dissolution of FeCl₂). The amount of iron present in the sample is y% by weight.

(Assume : KMnO₄ reacts only with Fe²⁺ in the solution

Use : Molar mass of iron as 56 g mol^$$-$$1)

The value of y is ______.

The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by s-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below :

Correct match of the C-H bonds (shown in bold) in Column J with their BDE in Column K is

Column J Molecule	Column K BDE (kcal $$mo{l^{ - 1}}$$)
(P) H-CH($$C{H_3}$$)$$_2$$	(i) 132
(Q) H-CH$$_2$$Ph	(ii) 110
(R) H-CH=CH$$_2$$	(iii) 95
(S) H-C $$ \equiv $$ CH	(iv) 88

The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by s-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below :



Cl $$-$$ Cl (g) $$\to$$ Cl$$_{(g)}^ \bullet $$ + Cl$$_{(g)}^ \bullet $$ $$\Delta$$H$$^\circ$$ = 58 kcal mol^$$-$$1

H₃C $$-$$ Cl (g) $$\to$$ H₃C$$_{(g)}^ \bullet $$ + Cl$$_{(g)}^ \bullet $$ $$\Delta$$H$$^\circ$$ = 85 kcal mol^$$-$$1

H $$-$$ Cl (g) $$\to$$ H$$_{(g)}^ \bullet $$ + Cl$$_{(g)}^ \bullet $$ $$\Delta$$H$$^\circ$$ = 103 kcal mol^$$-$$1

For the following reaction

CH₄(g) + Cl₂(g) $$\buildrel {light} \over \longrightarrow $$ CH₃Cl(g) + HCl (g)

the correct statement is

The reaction of K₃[Fe(CN)₆] with freshly prepared FeSO₄ solution produces a dark blue precipitate called Turnbull's blue. Reaction of K₄[Fe(CN)₆] with the FeSO₄ solution in complete absence of air produces a white precipitate X, which turns blue in air. Mixing the FeSO₄ solution with NaNO₃, followed by a slow addition of concentrated H₂SO₄ through the side of the test tube produces a brown ring.

Precipitate X is