Question

The complex for which the calculation of crystal field splitting can be most easily done, by knowing its absorption spectrum, will be :

• A. ${\left[{\mathrm{CoF}}_6\right]}^{-3}$
• B. ${\left[{\mathrm{TiCl}}_6\right]}^{2-}$
• C. $[\mathrm{Ti}{\left(\mathrm{CN}\right)}_6{]}^{3-}$
• D. $[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{]}^{2+}$

Answer 1

The correct option is C

$[\mathrm{Ti}{\left(\mathrm{CN}\right)}_6{]}^{3-}$

The explanation for the correct answer:-

Option (C) $[\mathrm{Ti}{\left(\mathrm{CN}\right)}_6{]}^{3-}$

• This, $\mathrm{Ti}{}^{+3}$ ion has one electron in a 3d subshell.
• Since its configuration is $3{\mathrm{d}}^{2}4{\mathrm{s}}^2$.
• Due to charge, it will lose three electrons from the valence shell.
• Thus it is one electron complex.

The explanation for incorrect answer:-

Option (A) ${\left[{\mathrm{CoF}}_6\right]}^{-3}$ .

• In this complex, cobalt will have 4 unpaired electrons so, the calculation will be difficult.
• Thus it is incorrect.

Option (B) ${\left[{\mathrm{TiCl}}_6\right]}^{2-}$

• In this, Titanium will have no electron in the 3d shell, so there will be no absorption spectrum.
• Thus it is incorrect.

Option (D) $[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{]}^{2+}$

• In this also ferrous has 4 unpaired electrons, the calculation will be difficult.
• Thus it is incorrect.

Therefore, the correct option is (c) $[\mathrm{Ti}{\left(\mathrm{CN}\right)}_6{]}^{3-}$.