Question

What is the number of pairs of ion which are coloured in aqueous solution?

$\left(i\right)T{i}^{3+},V^{3+}\left(ii\right)C{u}^{+},S{c}^{3+}\left(iii\right)F{e}^{2+},F{e}^{3+}\left(iv\right)C{o}^{2+},N{i}^{2+}\left(v\right)Z{n}^{2+},A{g}^{+}$

• A. 3
• B. 4
• C. 5
• D. 2

Answer 1

The correct option is A 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 [Ar]$3d^1$. No. of unpaired electron =1
The electronic configuration of $V{}^{3+}$ is [Ar]$3d^2$
No. of unpaired electron = 2

(ii) The electronic configuration of $C{u}^{+}$ is [Ar]$3d^{10}$. No. of unpaired electron =0
The electronic configuration of $S{c}^{3+}$ is [Ar]$3d^0$. No. of unpaired electron =0

(iii) The electronic configuration of $F{e}^{2+}$ is [Ar]$3d^6$. No. of unpaired electron =4
The electronic configuration of $F{e}^{3+}$ is [Ar]$3d^5$. No. of unpaired electron =5

(iv) The electronic configuration of $C{o}^{2+}$ is [Ar]$3d^7$. No. of unpaired electron =3
The electronic configuration of $N{i}^{2+}$ is [Ar]$3d^8$. No. of unpaired electron =2

(v) The electronic configuration of $Z{n}^{2+}$ is [Ar]$3d^{10}$. No. of unpaired electron =0
The electronic configuration of $A{g}^{+}$ is [Kr]$3d^{10}$.
No. of unpaired electron =0
$A{g}^{+},Z{n}^{+2}$ show no colour due to absence of unpaired electron.
Thus 3 pairs show colour.