In case of nitrogen, if M 1 represents spin multiplicity if Hund's rule is following and M 2 represents spin multiplicity if only Hund's Rule is violated (electrons are paired first) the

Question

In case of nitrogen, if M 1 ​ represents spin multiplicity if Hund's rule is following and M 2 ​ represents spin multiplicity if only Hund's Rule is violated (electrons are paired first) then the value of M 2 ​ M 1 ​ ​ will be:

TestHub · Accepted Answer

Nitrogen (N) has 3 valence electrons (2p³).

M₁ (Hund's rule followed): Electrons are unpaired, S = 3/2. Multiplicity M₁ = 2S + 1 = 2(3/2) + 1 = 4.

M₂ (Hund's rule violated): Two electrons are paired, one is unpaired, S = 1/2. Multiplicity M₂ = 2S + 1 = 2(1/2) + 1 = 2.

Ratio: $\frac{M _{1}}{M _{2}} = \frac{4}{2} = 2$ .

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