Class 12 Chemistry – d-Block Elements

 

Class 12 Chemistry – d-Block Elements

10 Important Questions with Answers

1. Why do transition elements show variable oxidation states?
Answer:
Transition elements have comparable energies of $(n-1)d$ and $ns$ orbitals. Therefore, electrons from both orbitals can participate in bonding, resulting in variable oxidation states.

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2. Why are transition metal compounds generally coloured?
Answer:
Transition metal ions usually have partially filled $d$-orbitals. Electrons absorb specific wavelengths of visible light and undergo d–d transitions. The remaining transmitted/reflected light gives the compound its colour.

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3. Why do transition elements form complexes?
Answer:
They form complexes because they have:

• Small atomic/ionic size
• High nuclear charge
• Vacant or partially filled orbitals
• Ability to accept electron pairs from ligands

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4. Why do transition elements show catalytic activity?
Answer:
Transition metals show catalytic activity because they can:

1. Exhibit variable oxidation states.
2. Provide a suitable surface for adsorption of reactants.
3. Form intermediate compounds during reactions.

Example: $Fe$ is used as a catalyst in the Haber process.

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5. Why is $Zn$ not considered a transition element?
Answer:
The electronic configuration of Zn is:

$$Zn = [Ar]\,3d^{10}4s^2$$

and $Zn^{2+}$ has:

$$Zn^{2+} = [Ar]\,3d^{10}$$

Since both Zn and $Zn^{2+}$ have completely filled $d$-orbitals, Zn does not qualify as a transition element.

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6. Why is $Mn^{2+}$ particularly stable?
Answer:
The electronic configuration of $Mn^{2+}$ is:

$$Mn^{2+} = [Ar]\,3d^5$$

It has a half-filled $d^5$ configuration, which is particularly stable due to greater exchange energy and symmetrical distribution of electrons.

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7. Why is $Cu^+$ unstable in aqueous solution?

Answer:

$Cu^+$ undergoes disproportionation:

$$2Cu^+ \rightarrow Cu^{2+}+Cu$$

This occurs because $Cu^{2+}$ is strongly hydrated in aqueous solution, making the disproportionation energetically favourable.

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8. Calculate the oxidation state of Mn in $KMnO_4$.

Answer:

Let oxidation state of Mn = $x$.

$$(+1)+x+4(-2)=0$$ $$1+x-8=0$$ $$\boxed{x=+7}$$

Therefore, oxidation state of Mn is +7.

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9. Why is $Cr^{2+}$ a strong reducing agent while $Mn^{3+}$ is a strong oxidising agent?

Answer:

$$Cr^{2+} \rightarrow Cr^{3+}$$

$Cr^{3+}$ has a stable $3d^3$ configuration. Therefore, $Cr^{2+}$ readily loses an electron and acts as a reducing agent.

Similarly,

$$Mn^{3+} \rightarrow Mn^{2+}$$

$Mn^{2+}$ has a stable half-filled $3d^5$ configuration. Therefore, $Mn^{3+}$ readily accepts an electron and acts as an oxidising agent.

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10. Why do transition metals have high melting and boiling points?

Answer:
Transition metals generally have strong metallic bonding because both $ns$ and $(n-1)d$ electrons can participate in metallic bonding. Hence, considerable energy is required to break these bonds, giving them high melting and boiling points.