The correct order of increasing energy of orbitals in a multi electron atom is
The correct order of increasing energy of orbitals in a multi electron atom is
Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mn>4</mn> <mi>p</mi> <mo><</mo> <mn>4</mn> <mi>s</mi> <mo><</mo> <mn>3</mn> <mi>d</mi> <mo><</mo> <mn>2</mn> <mi>p</mi> </math> </span></p>
Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mn>2</mn> <mi>p</mi> <mo><</mo> <mn>3</mn> <mi>d</mi> <mo><</mo> <mn>4</mn> <mi>p</mi> <mo><</mo> <mn>4</mn> <mi>s</mi> </math> </span></p>
Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mn>2</mn> <mi>p</mi> <mo><</mo> <mn>3</mn> <mi>d</mi> <mo><</mo> <mn>4</mn> <mi>s</mi> <mo><</mo> <mn>4</mn> <mi>p</mi> </math> </span></p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mn>2</mn> <mi>p</mi> <mo><</mo> <mn>4</mn> <mi>s</mi> <mo><</mo> <mn>3</mn> <mi>d</mi> <mo><</mo> <mn>4</mn> <mi>p</mi> </math> </span></p>
2 Views|Posted 4 months ago
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4 months ago
Correct Option - 4
Detailed Solution:
For a multi electron system, the energy of orbitals increase with increase in (n + l) value.
For same n + l value, the orbital with higher value of n has higher energy.
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CH3COOH + NaOH → CH3COONa + H2O
ΔH = –50.6 kJ/mol
NaOH + SA [HCl] → NaCl + H2O
ΔH = –55.9 kJ/mol
the value of ΔH for ionisation of CH3COOH
⇒ ΔH = +55.9 – 50.6
5.3 kJ/mol
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