Question

Colourless ion among the following is

• A. $ Zn^{2+} $
• B. $ Cr^{3+} $
• C. $ V^{2+} $
• D. $ Ti^{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) $Zn ^{2+}$
$ Zn (30): 1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{10}, 4 s^{2}$
$Zn ^{2+}: 1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{10}$
There is no unpaired electron in $Zn ^{2+}$. So, it is colourless.
(b) $Cr ^{3+}$
$Cr(24): 1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{5}, 4 s^{1}$
$Cr ^{3+} : 1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{3}$
There are three unpaired electron in $Cr ^{3+}$ hence, it is deep green in colour.
(c) $V ^{2+}$
$V (23): 1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{3}, 4 s^{2}$ $V ^{2+}: 1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{3}$
Coloured due to presence of three unpaired electrons.
(d) $Ti ^{3+}$
$ Ti (22): 1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{2}, 4 s^{2} $
$ Ti ^{3+}: 1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{1}$
Coloured due to presence of one unpaired electron.