Question

Colourless ion among the following is

• A. $Z{n}^{2+}$
• B. $C{r}^{3+}$
• C. $V^{2+}$
• D. $T{i}^{3+}$

Answer 1

Answer: (A)

The colour of the ion depends on the presence of unpaired electrons. In the absence of upaired electrons ion becomes colourless.
(a) $Z{n}^{2+}$
$Zn(30):1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^{10},4s^2$
$Z{n}^{2+}:1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^{10}$
There is no unpaired electron in $Z{n}^{2+}$. So, it is colourless.
(b) $C{r}^{3+}$
$Cr(24):1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^5,4s^1$
$C{r}^{3+}:1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^3$
There are three unpaired electron in $C{r}^{3+}$ hence, it is deep green in colour.
(c) $V^{2+}$
$V(23):1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^3,4s^2$ $V^{2+}:1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^3$
Coloured due to presence of three unpaired electrons.
(d) $T{i}^{3+}$
$Ti(22):1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^2,4s^2$
$T{i}^{3+}:1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^1$
Coloured due to presence of one unpaired electron.