Indian Navy INET Inorganic Chemistry — practice questions

Explore 2 free Inorganic Chemistry questions in Indian Navy INET Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

Practise Inorganic Chemistry with Quill →

Inorganic Chemistry questions and exercises with answers

Showing questions 1–2 of 2.

The 'spin only' magnetic moment $\mu$ (in Bohr magnetons) of a transition metal ion with $n$ unpaired d-electrons is given by:Read the answer and worked solution →The spin-only magnetic moment of $\mathrm{Fe}^{3+}$ (ground state) is approximately:Read the answer and worked solution →

Worked practice questions

The 'spin only' magnetic moment $\mu$ (in Bohr magnetons) of a transition metal ion with $n$ unpaired d-electrons is given by:

$\mu = n$
$\mu = \sqrt{n}$
$\mu = \sqrt{n(n+2)}$
$\mu = n(n+1)$

Show answer and solution

Correct answer: C. $\mu = \sqrt{n(n+2)}$

The spin-only formula is $\mu = \sqrt{n(n + 2)}\,\text{BM}$, where $n$ is the number of unpaired electrons. It comes from the spin angular momentum quantum number $S = n/2$ and the relation $\mu = 2\sqrt{S(S+1)}\,\text{BM} = \sqrt{n(n+2)}$. The actual magnetic moment can differ slightly due to orbital contribution this is small for first-row transition ions in octahedral fields.

Open this question →

The spin-only magnetic moment of $\mathrm{Fe}^{3+}$ (ground state) is approximately:

$1.73\,\text{BM}$
$2.83\,\text{BM}$
$4.90\,\text{BM}$
$5.92\,\text{BM}$

Show answer and solution

Correct answer: D. $5.92\,\text{BM}$

$\mathrm{Fe}^{3+}: [\mathrm{Ar}]3d^5$, so $n = 5$ unpaired electrons (high-spin, weak-field assumption). $\mu = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92\,\text{BM}$. The other options correspond to $n = 1, 2$ and $4$ unpaired electrons.