8.2. What is the oxidation number of the underlined elements in each of the following and how do you rationalise your results?

(a) In Kl₃, since the oxidation number of K is +1, therefore, the average oxidation number of iodine = -1/3. But the oxidation number cannot be fractional. Therefore, we must consider its structure, K⁺ [I —I < I]^–. Here, a coordinate bond is formed between I₂ molecule and I^– ion. The oxidation number of two iodine atoms forming the I₂ molecule is zero, while that of iodine forming the coordinate bond is -1. Thus, the oxidation number of the three I atoms, atoms in Kl₃ is 0, 0 and -1, respectively.

 

(b) By conventional method O.N. of S in H₂S₄O₆is calculated as:

2 (+1) +4x + 6) (-2) = 0

Or x = +2.5

But all the four

...more

(a) In Kl₃, since the oxidation number of K is +1, therefore, the average oxidation number of iodine = -1/3. But the oxidation number cannot be fractional. Therefore, we must consider its structure, K⁺ [I —I <— I]^–. Here, a coordinate bond is formed between I₂ molecule and I^– ion. The oxidation number of two iodine atoms forming the I₂ molecule is zero, while that of iodine forming the coordinate bond is -1. Thus, the oxidation number of the three I atoms, atoms in Kl₃ is 0, 0 and -1, respectively.

 

(b) By conventional method O.N. of S in H₂S₄O₆is calculated as:

2 (+1) +4x + 6) (-2) = 0

Or x = +2.5

But all the four S atoms cannot be in the same oxidation state.

So by chemical bonding method:

The O.N. of each of the S-atoms linked with each other in the middle is zero while that of each of the remaining two S-atoms is +5.

(c) By conventional method O.N. of Fe in Fe₃O₄ is calculated as:

3x + 4 (-2) = 0 or x = 8/3

By stoichiometry O.N. of Fe in Fe₃O₄ is +2 and +3.

(d) By conventional method, O.N. of C in CH3CH2OH is calculated as:

2x + 6 (+1) +1 (-2) = 0 or x = -2

(e) By conventional method, O.N. of C in CH₃COOH is calculated as:

2x + 4 - 4 = 0 or x = 0

By chemical bonding method, C2 is attached to three H-atoms (less electronegative than carbon) and one -COOH group (more electronegative than carbon).

 

Therefore, O.N. of C₂ = 3 (+1) + x + 1 (-1) = 0 or x = -2

But C₁ is attached to one oxygen atom by a double bond, one OH (-1) and one CH3 (+1) group, therefore, O.N. of C₁ = +1 + x+ 1 (-2) + 1 (-1) = 0 or x = +2

less

<p><strong> (a)</strong> In Kl₃, since the oxidation number of K is +1, therefore, the average oxidation number of iodine = -1/3. But the oxidation number cannot be fractional. Therefore, we must consider its structure, K⁺ [I —I &lt;— I]^–. Here, a coordinate bond is formed between I_2&nbsp;molecule and I^–&nbsp;ion. The oxidation number of two iodine atoms forming the&nbsp;I₂ molecule is zero, while that of iodine forming the coordinate bond is -1. Thus, the oxidation number of the three I atoms, atoms in Kl₃ is 0, 0 and -1, respectively.</p><p>&nbsp;</p><p><strong> (b)</strong> By conventional method O.N. of S in H₂S₄O₆is calculated as:</p><p>2 (+1) +4x + 6) (-2) = 0</p><p>Or x = +2.5</p><p>But all the four S atoms cannot be in the same oxidation state.</p><p>So by chemical bonding method:</p><p>The O.N. of each of the S-atoms linked with each other in the middle is zero while that of each of the remaining two S-atoms is +5.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1741860389phpnRRuBv_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1741860389phpnRRuBv.jpeg" alt="" width="217" height="80"></picture></div></div><p><strong> (c)</strong> By conventional method O.N. of Fe in Fe₃O₄ is calculated as:</p><p>3x + 4 (-2) = 0 or x = 8/3</p><p>By stoichiometry O.N. of Fe in Fe₃O₄is +2 and +3.</p><p><strong> (d)</strong> By conventional method, O.N. of C in CH3CH2OH is calculated as:</p><p>2x + 6 (+1) +1 (-2) = 0 or x = -2</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1741860454phpvefBqt_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1741860454phpvefBqt.jpeg" alt="" width="301" height="41"></picture></div></div><p><strong> (e) </strong>By conventional method, O.N. of C in CH₃COOH is calculated as:</p><p>2x + 4 - 4 = 0 or x = 0</p><p>By chemical bonding method, C2 is attached to three H-atoms (less electronegative than carbon) and one -COOH group (more electronegative than carbon).</p><p>&nbsp;</p><p>Therefore, O.N. of C₂ = 3 (+1) + x + 1 (-1) = 0 or x = -2</p><p>But C₁ is attached to one oxygen atom by a double bond, one OH (-1) and one CH3 (+1) group, therefore, O.N. of C₁ = +1 + x+ 1 (-2) + 1 (-1) = 0 or x = +2</p>

0 Upvotes 0 Upvotes 0 Downvotes

• • Report Abuse
• Comment