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Rajasthan Board Class 12 (RBSE) Inorganic Chemistry — practice questions

Explore 4 free Inorganic Chemistry questions in Rajasthan Board Class 12 (RBSE) Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Inorganic Chemistry questions and exercises with answers

Showing questions 1–4 of 4.

Among the 3d-series transition metals, the element that exhibits the largest number of oxidation states is:Read the answer and worked solution →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 →Highest oxidation states of transition metals are typically observed only in their:Read the answer and worked solution →

Worked practice questions

Among the 3d-series transition metals, the element that exhibits the largest number of oxidation states is:

  1. Scandium (Sc)
  2. Iron (Fe)
  3. Manganese (Mn)
  4. Zinc (Zn)
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Correct answer: C. Manganese (Mn)

Manganese has the configuration $[\mathrm{Ar}]3d^5 4s^2$. All five 3d electrons and the two 4s electrons can participate in bonding so it can lose anywhere from one to seven electrons giving oxidation states $+2$ through $+7$. **Examples:** $\mathrm{Mn}^{2+}$ in $\mathrm{MnSO_4}$, $\mathrm{Mn}^{4+}$ in $\mathrm{MnO_2}$, $\mathrm{Mn}^{7+}$ in $\mathrm{KMnO_4}$.

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

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

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

  1. $1.73\,\text{BM}$
  2. $2.83\,\text{BM}$
  3. $4.90\,\text{BM}$
  4. $5.92\,\text{BM}$
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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.

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