IMO Practice Test: The d- and f-Block Elements | Class 12 Chemistry

Question 1 of 14

A 3d-series $M^{2+}$ ion has a spin-only moment of 3.87 BM. The number of unpaired electrons and the ion are:

2, $\text{Ni}^{2+}$

3, $\text{Cr}^{3+}$-like ($d^3$)

4, $\text{Fe}^{2+}$

5, $\text{Mn}^{2+}$

Explanation: $\sqrt{n(n+2)}=3.87\Rightarrow n=3$; a $d^3$ divalent ion is $\text{V}^{2+}$ (same $d^3$, 3 unpaired electrons).

Question 2 of 14

Among $\text{Mn}^{2+}$, $\text{Fe}^{2+}$, $\text{Co}^{2+}$ and $\text{Ni}^{2+}$, the one with the highest spin-only magnetic moment is:

$\text{Mn}^{2+}$

$\text{Fe}^{2+}$

$\text{Co}^{2+}$

$\text{Ni}^{2+}$

Explanation: $\text{Mn}^{2+}$ ($d^5$) has 5 unpaired electrons, the maximum, giving 5.92 BM.

Question 3 of 14

Why does the $\text{IE}_3$ of Mn exceed that of Fe?

Mn has lower nuclear charge

Removing the 3rd electron from $\text{Mn}^{2+}$ breaks the stable $d^5$ configuration

Fe is a non-metal

Mn has more protons than Fe

Explanation: $\text{Mn}^{2+}$ is $d^5$ (extra-stable), so removing a further electron is difficult.

Question 4 of 14

In the reaction of $\text{KMnO}_4$ with oxalic acid, the ratio of moles of $\text{MnO}_4^-$ to $\text{C}_2\text{O}_4^{2-}$ is:

1 : 1

2 : 5

5 : 2

1 : 2

Explanation: $2\text{MnO}_4^-+5\text{C}_2\text{O}_4^{2-}+16\text{H}^+\rightarrow2\text{Mn}^{2+}+10\text{CO}_2+8\text{H}_2\text{O}$.

Question 5 of 14

The number of moles of $\text{Fe}^{2+}$ oxidised by 1 mole of $\text{Cr}_2\text{O}_7^{2-}$ in acid is:

3

5

6

2

Explanation: Dichromate accepts 6 electrons; each $\text{Fe}^{2+}\rightarrow\text{Fe}^{3+}$ gives 1 electron, so 6 moles are oxidised.

Question 6 of 14

Which 3d ion would be both colourless AND diamagnetic?

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

$\text{Zn}^{2+}$

$\text{Cr}^{3+}$

$\text{Mn}^{2+}$

Explanation: $\text{Zn}^{2+}$ is $d^{10}$: no d-d transition (colourless) and all electrons paired (diamagnetic).

Question 7 of 14

Lanthanoid contraction makes the radius of Hf nearly equal to that of:

Ti

Zr

Ta

Y

Explanation: Zr and Hf, of the same group, have almost identical radii due to the contraction.

Question 8 of 14

Which lanthanoid ion attains stability by reaching a half-filled f subshell in the +2 state?

$\text{Ce}^{2+}$

$\text{Eu}^{2+}$

$\text{Lu}^{2+}$

$\text{La}^{2+}$

Explanation: $\text{Eu}^{2+}$ is $4f^7$, a stable half-filled configuration.

Question 9 of 14

The colour of acidified $\text{KMnO}_4$ is discharged on adding excess oxalic acid because:

$\text{MnO}_4^-$ is converted to coloured $\text{MnO}_2$

$\text{MnO}_4^-$ is reduced to nearly colourless $\text{Mn}^{2+}$

oxalic acid is purple

permanganate evaporates

Explanation: Purple $\text{MnO}_4^-$ is reduced to the pale pink/colourless $\text{Mn}^{2+}$ ion.

Question 10 of 14

Why do second- and third-row transition metals show higher melting points than the first row?

weaker metallic bonds

stronger metal-metal bonding from greater d-orbital overlap

they are non-metals

they have fewer electrons

Explanation: Larger, more diffuse 4d/5d orbitals overlap more strongly, giving stronger metallic bonding.

Question 11 of 14

The basic strength of the hydroxides $\text{La(OH)}_3$ to $\text{Lu(OH)}_3$:

increases

decreases

stays constant

first increases then decreases

Explanation: As cation size falls (lanthanoid contraction), the hydroxides become less basic from La to Lu.

Question 12 of 14

For a $d^4$ ion in an octahedral field with weak ligands, the number of unpaired electrons (high-spin) is:

0

2

4

1

Explanation: High-spin $d^4$: $t_{2g}^3 e_g^1$, all four electrons unpaired.

Question 13 of 14

Among $\text{Sc}$, $\text{V}$, $\text{Cr}$ and $\text{Mn}$, the metal expected to have the highest melting point is:

Sc

V

Cr

Mn

Explanation: Cr lies near mid-series with the maximum number of unpaired d electrons available for bonding (Mn dips due to $d^5$).

Question 14 of 14

The actinoid contraction is greater than the lanthanoid contraction mainly because:

$5f$ electrons shield even more poorly than $4f$

$5f$ electrons shield better

actinoids have fewer protons

actinoids are smaller atoms

Explanation: The poorer shielding by the $5f$ electrons produces a slightly larger contraction.

IMO Practice Test — The d- and f-Block Elements