Coordination Compounds — JEE Main practice

Statement I

Compare the number of unpaired electrons in each complex.

• $[\mathrm{Cu(NH_3)_4}]^{2+}$:

  $$\mathrm{Cu}^{2+};(3d^9)$$

  It has 1 unpaired electron.

• $[\mathrm{Ni(en)_3}]^{2+}$:

  $$\mathrm{Ni}^{2+};(3d^8)$$

  Octahedral complex with 2 unpaired electrons.

• $[\mathrm{Ni(NH_3)_6}]^{2+}$:

  $$\mathrm{Ni}^{2+};(3d^8)$$

  Octahedral complex with 2 unpaired electrons.

• $[\mathrm{Mn(H_2O)_6}]^{2+}$:

  $$\mathrm{Mn}^{2+};(3d^5)$$

  Since $\mathrm{H_2O}$ is a weak-field ligand, the complex is high spin and has 5 unpaired electrons.

Therefore, $[\mathrm{Mn(H_2O)_6}]^{2+}$ has the maximum number of unpaired electrons.

Hence, Statement I is true.

Statement II

Determine whether both species in each pair are diamagnetic.

• ${[\mathrm{NiCl_4}]^{2-},[\mathrm{Ni(CO)_4}]}$

  • $[\mathrm{NiCl_4}]^{2-}$ is paramagnetic.
  • $[\mathrm{Ni(CO)_4}]$ is diamagnetic.
  • Not a pair containing only diamagnetic species.

• ${[\mathrm{NiCl_4}]^{2-},[\mathrm{Ni(CN)_4}]^{2-}}$

  • $[\mathrm{NiCl_4}]^{2-}$ is paramagnetic.
  • $[\mathrm{Ni(CN)_4}]^{2-}$ is diamagnetic.
  • Not a pair containing only diamagnetic species.

• ${[\mathrm{Ni(CO)_4}],[\mathrm{Ni(CN)_4}]^{2-}}$

  • Both complexes are diamagnetic.
  • This is one pair containing only diamagnetic species.

Thus, the number of such pairs is 1, not 2.

Hence, Statement II is false.

Since Statement I is true and Statement II is false, the correct answer is Option B.

Statement I

Determine the electronic configuration of each metal ion in an octahedral field with $\mathrm{H_2O}$, which is a weak-field ligand. Therefore, both complexes are high-spin.

• For $[\mathrm{Cr(H_2O)*6}]^{2+}$:

  $$\mathrm{Cr}^{2+}: 3d^4$$

  Electronic configuration:

  $$t*{2g}^{3}e_g^{1}$$

  CFSE:

  $$3(-0.4\Delta_o)+1(+0.6\Delta_o) =-0.6\Delta_o$$

• For $[\mathrm{Mn(H_2O)*6}]^{2+}$:

  $$\mathrm{Mn}^{2+}: 3d^5$$

  Electronic configuration:

  $$t*{2g}^{3}e_g^{2}$$

  CFSE:

  $$3(-0.4\Delta_o)+2(+0.6\Delta_o) =0$$

Since

the chromium complex has greater stabilization (more negative CFSE).

Hence, Statement I is true.

Statement II

• Potassium ferricyanide:

  $$\mathrm{K_3[Fe(CN)*6]}$$

  Here,

  $$\mathrm{Fe}^{3+}:3d^5$$

  Since $\mathrm{CN^-}$ is a strong-field ligand, the complex is low spin:

  $$t*{2g}^{5}e_g^{0}$$

  Number of unpaired electrons:

  $$n=1$$

  Spin-only magnetic moment:

  $$\mu=\sqrt{n(n+2)} =\sqrt{1(1+2)} =\sqrt{3}\ \mathrm{BM}$$

• Sodium ferrocyanide:

  $$\mathrm{Na_4[Fe(CN)*6]}$$

  Here,

  $$\mathrm{Fe}^{2+}:3d^6$$

  With strong-field $\mathrm{CN^-}$:

  $$t*{2g}^{6}e_g^{0}$$

  All electrons are paired, so

  $$n=0,\qquad \mu=0$$

Therefore, potassium ferricyanide has a greater spin-only magnetic moment than sodium ferrocyanide.

Hence, Statement II is true.

Since both Statement I and Statement II are true, the correct answer is Option C.

Determine the oxidation state of nickel.

• $\mathrm{PPh_3}$ is a neutral ligand.
• Each $\mathrm{Cl^-}$ contributes $-1$.

Therefore,

and

Since the complex is stated to be paramagnetic, it cannot be the usual square-planar $d^8$ complex (which is diamagnetic).

Therefore, it must be tetrahedral (NCERT treatment), having $sp^3$ hybridisation.

Statement A

A tetrahedral complex of the type $[\mathrm{MA_2B_2}]$ does not exhibit geometrical (cis-trans) isomerism.

Hence, Statement A is incorrect.

Statement B

The complex is not white in colour.

Since it is a coloured transition metal complex, Statement B is incorrect.

Statement C

A tetrahedral $d^8$ complex has two unpaired electrons.

The spin-only magnetic moment is

where $n=2$.

Therefore,

Hence, Statement C is correct.

Statement D

For a tetrahedral $d^8$ configuration,

The CFSE is

It is not $-0.8\Delta_o$.

Hence, Statement D is incorrect.

Statement E

$\mathrm{Ni(CO)_4}$ is also a tetrahedral complex.

Therefore, both complexes have the same geometrical arrangement.

Hence, Statement E is correct.

Thus, the incorrect statements are A, B and D.

Therefore, the correct answer is Option C.