The quantum number of four electrons are given below:
$I . n = 4, l = 2, {m}_{l} = – 2, {m}_{s} = – 1 / 2$
$II . n = 3, l = 2, {m}_{l} = 1, {m}_{s} = + 1 / 2$
$III . n = 4, l = 1, {m}_{l} = 0, {m}_{s} = + 1 / 2$
$IV . n = 3, l = 1, {m}_{l} = 1, {m}_{s} = – 1 / 2$
The correct order of their increasing energies will be:

Question

The quantum number of four electrons are given below:

I . n = 4, l = 2, {m}_{l} = – 2, {m}_{s} = – 1 / 2

II . n = 3, l = 2, {m}_{l} = 1, {m}_{s} = + 1 / 2

III . n = 4, l = 1, {m}_{l} = 0, {m}_{s} = + 1 / 2

IV . n = 3, l = 1, {m}_{l} = 1, {m}_{s} = – 1 / 2

The correct order of their increasing energies will be:

Quantrex Academy · Accepted Answer

Correct answer: B. IV < II < III < I

Higher the value of (n + l) higher will be energy. Hence, 3p _ (4) If (n + l) same, then higher the value of, higher will be energy.

The quantum number of four electrons are given below: I . n = 4, l...