IMO Practice Test: Coordination Compounds | Class 12 Chemistry

Question 1 of 14

A 3d $M^{3+}$ complex is octahedral, low-spin and diamagnetic. If $M^{3+}$ is $d^6$, the number of unpaired electrons is:

4

2

0

6

Explanation: Low-spin $d^6$ = $t_{2g}^6 e_g^0$, so 0 unpaired electrons (diamagnetic).

Question 2 of 14

If $\mu = 5.92$ BM for a complex, the number of unpaired electrons is:

3

4

5

2

Explanation: $\sqrt{n(n+2)}=5.92\Rightarrow n=5$.

Question 3 of 14

For [Cr(NH₃)₆]³⁺ (Cr³⁺ = $d^3$), the spin-only moment is:

1.73 BM

2.83 BM

3.87 BM

4.90 BM

Explanation: $d^3$ has 3 unpaired e⁻: $\mu=\sqrt{3(3+2)}=3.87$ BM.

Question 4 of 14

The total number of ions produced in solution by 1 formula unit of [Co(NH₃)₆]Cl₃ is:

2

3

4

5

Explanation: One complex cation + three Cl⁻ = 4 ions.

Question 5 of 14

How many geometrical isomers exist for octahedral [Ma₃b₃]?

1

2

3

4

Explanation: The fac and mer arrangements give 2 geometrical isomers.

Question 6 of 14

If $\Delta_o = 21600\ \text{cm}^{-1}$, then $\Delta_t$ for the same system is:

$4800\ \text{cm}^{-1}$

$9600\ \text{cm}^{-1}$

$14400\ \text{cm}^{-1}$

$48600\ \text{cm}^{-1}$

Explanation: $\Delta_t=\frac{4}{9}\times21600=9600\ \text{cm}^{-1}$.

Question 7 of 14

The CFSE of a high-spin octahedral $d^4$ ion (ignore pairing) is:

$-1.2\,\Delta_o$

$-0.6\,\Delta_o$

$-0.4\,\Delta_o$

$0$

Explanation: $t_{2g}^3 e_g^1$: $(-0.4\times3 + 0.6\times1)\Delta_o=-0.6\,\Delta_o$.

Question 8 of 14

The oxidation number of iron in [Fe(H₂O)₅NO]SO₄ (brown ring, NO⁺) is:

+1

+2

+3

0

Explanation: With NO as NO⁺ and outer SO₄²⁻: $x+(+1)=+2\Rightarrow x=+1$.

Question 9 of 14

Among [Fe(CN)₆]³⁻ ($d^5$ low-spin) the number of unpaired electrons is:

5

3

1

0

Explanation: Low-spin $d^5$ = $t_{2g}^5 e_g^0$, so 1 unpaired electron.

Question 10 of 14

Which arrangement of $\Delta_o$ (increasing) is correct for Co³⁺?

CN⁻ < H₂O < F⁻

F⁻ < H₂O < CN⁻

H₂O < F⁻ < CN⁻

CN⁻ < F⁻ < H₂O

Explanation: Spectrochemical order: F⁻ < H₂O < CN⁻.

Question 11 of 14

The number of unpaired electrons in tetrahedral [NiCl₄]²⁻ ($d^8$) is:

0

1

2

4

Explanation: High-spin tetrahedral $d^8$ ($e^4 t_2^4$) has 2 unpaired electrons.

Question 12 of 14

For the octahedral series, which $d^n$ high-spin ion has CFSE = 0?

$d^3$

$d^5$

$d^6$

$d^8$

Explanation: High-spin $d^5$ ($t_{2g}^3 e_g^2$): $(-1.2 + 1.2)\Delta_o = 0$.

Question 13 of 14

The hybridisation and unpaired electrons of [Mn(CN)₆]³⁻ (Mn³⁺=$d^4$, strong field) are:

$sp^3d^2$, 4

$d^2sp^3$, 2

$dsp^2$, 0

$sp^3$, 4

Explanation: Strong-field CN⁻: low-spin $t_{2g}^4$ → $d^2sp^3$, 2 unpaired electrons.

Question 14 of 14

One mole of [Pt(NH₃)₆]Cl₄ gives how many ions in water?

2

3

4

5

Explanation: One complex cation + four Cl⁻ = 5 ions.

IMO Practice Test — Coordination Compounds