Bond angle of SO2 H2S SO3

Explanation:

In SO2 , the bond angle is 119^∘ whereas in SO3 , the bond angle is 120^∘ .H2S = 92.1°

In SO2 , the bond angle is 119^∘ whereas in SO3 , the bond angle is 120^∘ .

Explanation:

S atom in both SO2 and SO3 is sp^2-hybridized.

There are 3 bonding sets of electrons and one non-bonding pair on the sulphur. So the shape is based on a tetrahedral structure - but the extra repulsion of the non-bonding pair will decrease the O-S-O angles by a few degrees from the 'perfect' tetrahedral angle.

The bonding in SO3 can be described as a combination of 3 resonance forms of 1 double and 2 single bonds. The SO2 molecule has a dipole moment, The starting point of 109.5 degrees is only applicable for 4 electron regions round the central atom. In SO2, we have 2 double bonds and one lone pair ; 3 electron regions so the default angle is 120 degrees. The extra repulsion of the lp vs double bonds accounts for reduction to 119. S atom in both SO2 and SO3 is sp2-hybridized. But it is related to dipole moment and assertion is the explanation from bent angle so here reason is unable to give explanation of assertion.

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years