The value of the magnetic moment of a particular ion is 2.83 Bohr magneton. The ion is :

F e^{2 +}

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N i^{2 +}

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M n^{2 +}

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C o^{3 +}

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Solution

The correct option is B $N i^{2 +}$
The value of $μ =$ 2.83 BM ( $= \sqrt{n (n + 2)}$ ) corresponds to the presence of two unpaired electrons.

So, the ion is $N i^{2 +} (3 d^{8})$ since it has two unpaired electrons in d-subshell.

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Similar questions

Magnetic moment of which ion is 2.88 BM ?

M^{2 +}

Z n^{2 +} > S c^{2 +} = C u^{2 +} > T i^{2 +} = N i^{2 +} > V^{2 +} = C o^{2 +} > C r^{2 +} = F e^{2 +} > M n^{2 +}

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

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

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