For 2nd row and 3rd-row transition metal ions the magnetic moment in Bohr magnetons is calculated by the formula:

\sqrt{4 s (s + 1) + l (l + 1)}

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\sqrt{4 s (s + 1)}

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\sqrt{n (n + 1)}

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all of the above

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Solution

The correct option is B $\sqrt{4 s (s + 1) + l (l + 1)}$
For 2nd row and 3rd row transition metal ions the magnetic moment in Bohr magnetons is calculated by the formula $\sqrt{4 s (s + 1) + l (l + 1)}$ .
It has contributions from spin angular momentum and orbital angular momentum.
Orbital motion of the electron generates orbital magnetic moment and spin motion of the electron generates spin magnetic moment.

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For first-row transition metal ions the magnetic moment in Bohr magnetons is calculated by the formula.

3.87

Find the correct statement.

i) Magnetic moment $M n^{2 +}$ ion is $\sqrt{35}$

ii) $T i^{4 +}$ is exhibit purple color

iii) 1 Bohr Magneton= $9.273 \times 10^{- 24}$ Tesla $^{- 1}$

iv) Secondary valency of transition metal ion is equal to its co-ordination number

Δ = ∣ ∣ ∣ ∣ \begin{matrix} 5 & 3 & 8 2 & 8 & 1 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣,

3^{rd}

2^{nd}

Magnetic moment of diamagnetic substance in Bohr Magnetons is:

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