The reaction in which hybridization of the underlined atom is affected is:

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <msub> <mrow> <mrow> <mover accent="true"> <mrow> <mi mathvariant="normal">X</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">F</mi> </mrow> <mo>¯</mo> </mover> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mrow> <mi mathvariant="normal">S</mi> <mi mathvariant="normal">b</mi> <mi mathvariant="normal">F</mi> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> </msub> <mo>→</mo> </math> </span></p>
Option 2 - <p><span class="mathml" contenteditable="false"> <math> <msub> <mrow> <mrow> <mi mathvariant="normal">H</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi mathvariant="normal">S</mi> <mi mathvariant="normal">O</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msub> <mo>+</mo> <mi mathvariant="normal">N</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">l</mi> <mover> <mrow> <mrow> <mo>⟶</mo> </mrow> </mrow> <mrow> <mrow> <mn>420</mn> <mtext> </mtext> <mi mathvariant="normal">K</mi> </mrow> </mrow> </mover> </math> </span></p>
Option 3 - <p><span class="mathml" contenteditable="false"> <math> <msub> <mrow> <mrow> <mi mathvariant="normal">H</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi mathvariant="normal">P</mi> <mi mathvariant="normal">O</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> <mover> <mrow> <mrow> <mo>⟶</mo> </mrow> </mrow> <mrow> <mrow> <mtext>&nbsp;</mtext> <mtext>Disproportionation</mtext> <mtext>&nbsp;</mtext> </mrow> </mrow> </mover> </math> </span></p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <msub> <mrow> <mrow> <mi mathvariant="normal">N</mi> <mi mathvariant="normal">H</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> <mover> <mrow> <mrow> <mo>⟶</mo> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi mathvariant="normal">H</mi> </mrow> </mrow> <mrow> <mrow> <mo>+</mo> </mrow> </mrow> </msup> </mrow> </mrow> </mover> </math> </span>&nbsp;(Bonding)</p>
2 Views|Posted 6 months ago
Asked by Shiksha User
1 Answer
A
6 months ago
Correct Option - 1
Detailed Solution:

Sol. 4Li+O2? 2Li2O oxide

2Na+O2? Na2O2 peroxide

K+O2? KO2 superoxide

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ΔG° = –RT * 2.303 log K

–nFE° = +RT * 2.303 log K

2 * 96500 * 0.295 = 8.314 * 298 * 2.303 log10 K

10 = log10 K = 1010

It has chiral centre and differently di substituted double bonded carbon atoms.

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