Question

1.625 g of an interhalogen compound ICI(x), was
subjected to electric discharge which split it into
constituents and brought to STP. It occupied
224 ml. The interhalogen compound is
1.ICl 2.ICl3
3.ICl5 4.ICl7​

Answer 1

Answer:

IHC-Inter Halogen Compounds.An interhalogen compound is a molecule which contains two or more different halogen atoms (fluorine, chlorine, bromine, iodine, or astatine) and no atoms of elements from any other group.

Most interhalogen compounds known are binary (composed of only two distinct elements). Their formulae are generally XYn, where n = 1, 3, 5 or 7, and X is the less electronegative of the two halogens. They are all prone to hydrolysis, and ionize to give rise to polyhalogen ions. Those formed with astatine have a very short half-life due to astatine being intensely radioactive.

No interhalogen compounds containing three or more different halogens are definitely known,[1] although a few books claim that IFCl

2 and IF

2Cl have been obtained,[2][3][4][5] and theoretical studies seem to indicate that some compounds in the series BrClF

n are barely stable

Answer 2

Answer:

lodine is solid at STP, therefore CI must have occupied 224 mL.

Now, 1 mole of a gas at STP corresponds to 22.4 L or 22400 mL Therefore, 224 mL corresponds to 224 0.01 mole of Cl 22400

Hence, 0.01 mole of chlorine is produced at STP.

1 mole of chlorine = 35.5 g

So, 0.01 mole of chlorine 0.01 x 35.5 = 0.355 g

=

ICIx → I₂ + xCl₂ Let us suppose that 1 mole of I₂ was produced in the above reaction.

Atomic number of I = 127g

Therefore, amount of iodine produced = 127 a

Amount of Chlorine produced I = 35.5ax

Now, amount of compound d = 1.625g Therefpre. 127a + 35.5ax = 1.625

and, 35.5ax = 0.355g

Hence, 127a + 35.5g = 1.62g

r, a = 0.01n of iodine

Now, 35.5ax = 0.355

35.5 * 0.02x = 0.355

or, x = 1

Therefore, the given compound is ICI and the value of x is 1.