Question

What is the number of pairs of ion which are coloured in aqueous solution?
(i) $T{i}^{3+},V^{3+}\left(ii\right)C{u}^{+},S{c}^{3+}\left(iii\right)F{e}^{2+},F{e}^{3+}\left(iv\right)C{o}^{2+},N{i}^{2+}\left(v\right)Z{n}^{2+},A{g}^{+}$

• A. $4$
• B. $3$
• C. $5$
• D. $2$

Answer 1

The correct option is B $3$

Transition elements form different coloured compounds. The compound in which metal ions has an unpaired electron that compound will be coloured.

(i) The electronic configuration of $T{i}^{3+}$ is $\left[Ar\right]3d^1$. No of unpaired electron = 1
The electonic configuration of $V^{3+}$ is $\left[Ar\right]3d^2$
No of unpaired electron = 2
(ii) The electonic configuration of $C{u}^{+}$ is $\left[Ar\right]3d^{10}$. No of unpaired electron = 0
The electonic configuration of $S{c}^{3+}$ is $\left[Ar\right]3d^0$. No of unpaired electron = 0
(iii) The electonic configuration of $F{e}^{2+}$ is $\left[Ar\right]3d^6$. No of unpaired electron = 4
The electonic configuration of $F{e}^{3+}$ is $\left[Ar\right]3d^5$. No of unpaired electron = 5
(iv) The electonic configuration of $C{o}^{2+}$ is $\left[Ar\right]3d^7$. No of unpaired electron = 3
The electonic configuration of $N{i}^{2+}$ is $\left[Ar\right]3d^8$. No of unpaired electron = 2
(v) The electonic configuration of $Z{n}^{2+}$ is $\left[Ar\right]3d^{10}$. No of unpaired electron = 0
The electonic configuration of $A{g}^{+}$ is $\left[Kr\right]3d^{10}$. No of unpaired electron = 0

$A{g}^{+},Z{n}^{+2}$ show no colour due to absence of unpaired electron.
Thus 3 pairs show colour.