Chemistry Coordination Compounds

[Ni (H₂O)₆]²⁺ Þ Ni²⁺ is octahedral and [Ni (CN)₄]^2– is square planar.

In [Ni (H₂O)₆]²⁺ Þ Ni²⁺ has two unpaired electrons and in [Ni (CN)₄]^2– Þ Ni²⁺ has no unpaired electrons. [Ni (H₂O)₆]²⁺ is coloured as it absorbs red light due to suitable d-d transition and complementary light emitted is green.

[Ni (CN)₄]^2– has strong field ligand so the electrons of Ni²⁺ pair up and it is colourless as it cannot absorb light from visible region.

(a) [Cr (H₂O)₆]⁺³ → Cr⁺³ → $t_{2g}^{3}eg°$  

(b) [Fe (H₂O)₆]⁺³ → Fe³⁺ → $t_{2g}^{3}e{g}^2$  

(c) [Ni (H₂O)₆]⁺² → Ni²⁺ → $t_{2g}^{6}e{g}^2$  

(d) [V (H₂O)₆]⁺³ → V³⁺ → $t_{2g}^{2}eg°$  

Ni²⁺ : 4s⁰3d⁸ (No pairing with Cl^–)

[Ni (CO)₄] : 4s⁰3d¹⁰ (diamagnetic)

Correct order of increasing field strength of ligands is : I^– < OH^– < NCS^–

EAN = Atomic No. – Oxidation state + 2 * CN

= 26 – 2 + 2 * 6

=36

[Co (NH₃)₆]³⁺, NH3 becomes strong ligand due to +3 oxidation state of cobalt ion. Hence electronic configuration of

$\left. [Co{\left(N{H}_3\right)}_{5}Cl\right]C{l}_2+AgN{O}_3\left(excess\right)\to \left. [Co{\left(N{H}_3\right)}_{5}Cl\right]{\left(N{O}_3\right)}_2+2AgCl\downarrow$

Secondary valence of $Co$ $$ in $\left. [Co{\left(N{H}_3\right)}_{5}Cl\right]C{l}_2$ $\left. [{\left({}_{}\right)}_{}\right]{}_{}$ is six.

Energy of the complex $\frac{}{}{}_{}$ $=-\frac{2}{5}*4{?}_0+P$

For low spin $d^4$ ${}^{}$ cation

$=1.6{?}_0+P$

  ${\left[Co{\left(N{H}_3\right)}_6\right]}^{3+},N{H}_3$ becomes strong ligand due to +3 oxidation state of cobalt ion. Hence electronic configuration of CO³⁺ is $t_{2g}^6{e}_g^0$  

Correct order of increasing field strength of ligands is : I– < OH– < NCS–