Question

Identify the complexes which is/are expected to be coloured.

• A. $Ti(N{O}_3{)}_4$
• B. $Cu\left(NCC{H}_3\right){]}^{+}B{F}_4^{-}$
• C. $\left[Cr\right(N{H}_3{)}_6]C{l}_3$
• D. $K_3\left[V{F}_6\right]$

Answer 1

Answer: (C), (D)

$\left[Cr\right(N{H}_3{)}_6]C{l}_3$
$K_3\left[V{F}_6\right]$

Option (C), (D) are correct. $\left[Cr\right(N{H}_3{)}_6]C{l}_3$ and $K_3\left[V{F}_6\right]$ are coloured.

For the colouration of complexes, the complex must have unpaired electrons present in outer available $d$-orbitals.

(A) : $\left[Ti\right(N{O}_3{)}_4]$, titanium is in $T{i}^{4+}$ : $\left[Ar\right]3d^{0}4s^0$ and has no unpaired electrons.

( B) : $\left[Cu\right(NCC{H}_3{)}_4{]}^{+}B{F}_4^{-}$; copper is in $C{u}^{+}$ : $\left[Ar\right]3d^{10}4s^0$ and has no unpaired electrons.

(C) : $\left[Cr\right(N{H}_3{)}_6]C{l}_3$; chromium is in $C{r}^{3+}$ : $\left[Ar\right]3d^{3}4s^0$ has three unpaired electrons and thus it is coloured complex.

(D) : $K_3\left[V{F}_6\right]$; vanadium is in $V^{3+}$ : $\left[Ar\right]3d^{2}4s^0$ has two unpaired electrons and thus it is coloured complex.