Question

Identify the complexes which are expected to be colored.
(i) [Ti(NO${}_3$)${}_4$]
(ii) [Cu(NCCH${}_3$)]${}^{+}$BF${}_4$
(iii) [Cr(NH${}_3$)${}_6$]${}^{3+}$3Cl${}^{-}$
(iv) K${}_3$[VF${}_6$]

• A. (i)
• B. (ii)
• C. (iii)
• D. (iv)

Answer 1

Answer: (C), (D)

The correct options are
C (iv)
D (iii)

$\left[Ti\right(N{O}_3{)}_4]$
The oxidation state of Ti is +4.
The atomic number of titanium is 22.
$Ti=\left[Ar\right]3d^{2}4s^2$
$T{i}^{4+}=\left[Ar\right]$

It acquires electronic configuration of stable Ar and d-d transition is not possible here , so it is colorless.

$\left[Cu\right(NCC{H}_3){]}^{+}B{F}_4^{-}$
The oxidation state of Cu is +1.
The atomic number of copper is 29.
$Cu=\left[Ar\right]3d^{10}4s^1$
$C{u}^{+}=\left[Ar\right]3d^{10}$

All d-orbitals are fullfilled so d-d transition won't possible, and hence colorless.

$\left[Cr\right(N{H}_3{)}_6{]}^{+3}3C{l}^{-}$
The oxidation state of Cr is +3.
The atomic number of chromium is 24.
$Cr=\left[Ar\right]3d^{4}4s^2$
$C{r}^{3^{+}}=\left[Ar\right]3d^3$
Empty d-orbitals are there, that means d-d transition is possible, so it is colored.

$K_3\left[V{F}_6\right]$
The oxidation state of V is +3.
The atomic number of vanadium is +3.
$V=\left[Ar\right]3d^{3}4s^2$
$V^{3+}=\left[Ar\right]3d^2$

F is weak field ligand
Incomplete d-orbitals are there, so it is colored.

Hence options C & D are correct.