Chemistry, 20.04.2020 20:30 Anshuman2002

In which of the following species does sulfur have the lowest oxidation number? A) S8 B) SO(3)2- C)SO(4)2- D)S(2)O(3)2- E)S2-

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Answer from: Heartlesskiidkii

The option S2- has the lowest oxidation number of sulphur.

The first option which is S8 has an oxidation number of zero because it is in elemental form and free state.

The oxidation number of sulphur in SO(3)2- is 4 .

The charge on the atom is -2, therefore the sum of the oxidation numbers of all the elements should be equal to -2.

x+3(-2)= -2

x = 4

The oxidation number of S in SO(4)2- is 6. The charge on the atom is -2, therefore the sum of the oxidation numbers of all the elements should be equal to -2.

x +4(-2) = -2

x = +6

The oxidation number of S in S(2)O(3)2- is 2.

The charge on the atom is -2, so sum of the oxidation numbers of all the elements should be equal to -2.

2x + 3(-2) = -2

2x = 4

x = 2

The oxidation number of S in S2- is -2 which is the lowest.

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Answer from: lupitaca888

$S^{2-}$ is having lowest oxidation number of sulphur in teh options given. The oxidation number is -2.

The correct option is E.

Explanation:

$S^{2-}$ have oxidation number of -2, which is lowest in all the options given.

In the given examples $S_{8}$ will have oxidation number zero because it is in elemental form. In the rules for oxidation number it is given that any element in free state has its oxidation number zero.

All other options are in compound form as:

S $O_{3} ^{2-}$ = oxidation number is +4. It is calculated as oxygen having O.N of -2,

The charge on the atom is -2, so sum of the oxidation numbers of all the elements should be equal to -2.

x+3(-2)= -2

x = 4

S $O_{4} ^{2-}$ = The charge on the atom is -2, so sum of the oxidation numbers of all the elements should be equal to -2.

x +4(-2) = -2

x = +6

$S_{2}O_{3^{2-} }$

The charge on the atom is -2, so sum of the oxidation numbers of all the elements should be equal to -2.

2x + 3(-2) = -2

2x = 4

x = 2

$S^{2-}$ = oxidation number is -2 as the charge on ion in equal to its oxidation number.

Answer from: Quest

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

Answer from: Quest

D)can be increased in a closed system

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