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Introduction to d and f Block Elements

For first-row...

Question

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

\sqrt{n (n + 1)}

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\sqrt{4 s (s + 1)}

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\sqrt{n (n + 2)}

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both

B

and

C

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Solution

The correct option is D both $B$ and $C$
For first row transition metal ions the magnetic moment in Bohr magnetons is calculated by $\sqrt{n (n + 2)}$ , where n is no. of unpaired electrons and spin magnetic moment by $\sqrt{4 s (s + 1)}$ , where s is spin quantum number.

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Similar questions

For 2nd row and 3rd-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

μ = \sqrt{n (n + 2)} B M

n =

B M

1 B M = \frac{e h}{4 π m c} = 9.27 \times 10^{- 24} {A m}^{2}

J T^{- 1}

μ = \sqrt{n (n + 2)} B M

n =

B M

1 B M = \frac{e h}{4 π m c} = 9.27 \times 10^{- 24} {A m}^{2}

J T^{- 1}

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