Chemistry Class-11: Chapter –8. Redox Reactions Part – 4

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Question: 23

Calculate the oxidation number of each sulphur atom in the following compounds:

(a)

(b)

(c)

(d)

Answer:

(a)

(b)

(c)

(d)

Justification:

Write Lewis structure of each ion then assign electron pair shared between atoms of different Electronegativity to more electronegative atom and distribute the electron pair shared between atoms of same element equally. Now count the number of electrons possessed by each atom. Find out the difference in number of electrons possessed by neutral atom and that possessed by atom in the compound. This difference is the oxidation number. If atom present in the compound possesses more electrons than the neutral atom, the oxidation number is negative. If it possesses less electrons then oxidation number is positive.

(i) Lewis structure of can be written as follows

Lewis structure of S 2 O 4^(2-)

Lewis Structure

Electron pair shared between sulphur and oxygen is assigned to oxygen atoms because of more Electronegativity of oxygen. Thus each sulphur atom is deficient of two electrons with respect to neutral sulphur atom hence, each sulphur atom is in oxidation state. Each oxygen atom gets two excess electrons hence, it is in oxidation state. Lewis structure of can be written as follows:

Lewis structure S 4 O 6^(2-)

Lewis Structure

To find out oxidation state of each atom we distribute electrons of electron pair shared between two sulphur atoms equally (i.e. one electron is assigned to each sulphur atom). Both the electrons of electron pair shared between sulphur and oxygen atom are assigned to oxygen as oxygen is more electronegative. Thus we find that each of the central sulphur atoms obtain six electrons. This number is same as that in the outer shell of neutral sulphur atom hence oxidation state of central sulphur atoms is zero. Each of the sulphur atoms attached to oxygen atoms obtain only one electron as its share. This number is less by five electrons in comparison to the neutral sulphur atom. So, outer sulphur atoms are in oxidation state. Therefore average oxidation state of sulphur atoms is:

By using the formula we obtain average oxidation state of the particular type of atoms. Real oxidation state can be obtained only by writing the complete structural formula. Similarly we can see that each oxygen atom is in oxidation state.

In the same way one can find out the oxidation state of each atom in and ions. Oxidation state of metal atoms will be as these will lose one electron in each case.

Question: 24

Balance the following equations by the oxidation number method.

(i)

(ii)

(iii)

(iv)

Answer:

Question: 25

(i)

(ii)

(iii)

(iv)

(v)

Answer:

Question: 26

Balance the following ionic equations

(i)

(ii)

(iii)

(iv)

Answer:

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