Question

$\left( a \right)$ Which metal in the first transition series (${3^{rd}}$ series) exhibits $ + 1$ oxidation state most frequently and why?
$\left( b \right).$ Which of the following cations are colored in aqueous solution and why? $S{c^{3 + }},{V^{3 + }},T{i^{4 + }}$ and $M{n^{2 + }}$
$\left( {Sc = 21,V = 23,T = 22,Mn = 25} \right)$

Answer 1

Hint: There are $10$ elements in the first transition series ($3d$ series) namely, Scandium, Vanadium, Titanium, chromium, Manganese, Iron, Cobalt, Nickel, Copper and Zinc. They show variable oxidation state due to availability of vacant $d - $ orbital.

Complete step by step answer:
$\left( a \right).$ Copper $\left( {Cu} \right)$ metal in the first transition series (${3^{rd}}$ series) which is a last element in the ${3^{rd}}$ series. $Cu$ exhibit $ + 1$ oxidation state most frequently. This can be explained by electronic configuration. The electronic configuration of $Cu$ is, $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}4{s^1}$ . After losing one electron, it acquires the stable $3{d^{10}}4{s^0}$ configuration which is a fully filled electronic configuration, so we can say that $Cu$ metal exhibits $ + 1$ oxidation state most frequently.
$\left( b \right).$ The color of the cations depends on the number of unpaired electrons present in $d - $ orbital. The cations which have unpaired electrons show color because they emit energy when higher energy to lower energy in the visible region. The color of the compound can be explained on the basis of electronic configuration.
Now, The Electronic configuration of the given cation as follow;
 $Sc\left( {Z = 21} \right) = \left[ {Ar} \right]3{d^1}4{s^2}$and $S{c^{3 + }} = \left[ {Ar} \right]3{d^0}4{s^0}$ .No unpaired electron present in $d - $ orbital. So it doesn’t show any colour.
 $V\left( {Z = 23} \right) = \left[ {Ar} \right]3{d^3}4{s^2}$ and ${V^{3 + }} = \left[ {Ar} \right]3{d^2}4{s^0}$ . It has two unpaired electrons in $d - $ orbital. So it shows colour in aqueous solutions.
$Ti\left( {Z = 22} \right) = \left[ {Ar} \right]3{d^2}4{s^2}$ and $T{i^{4 + }} = 3{d^0}4{s^0}$ .It has no unpaired electron in $d - $ orbital. So it doesn’t show any colour.
$Mn\left( {Z = 25} \right) = \left[ {Ar} \right]3{d^5}4{s^2}$ and $M{n^{2 + }} = 3{d^5}4{s^0}$ .It has five unpaired electron in $d - $ orbital. So, it shows colour in aqueous solutions.
Hence, There are two cations which show colour in aqueous solution namely, ${V^{3 + }}$ and $M{n^{2 + }}$ .

Note:
The electronic configuration of elements should be done using Aufbau's principle. It states that the first electron occupied lower energy to higher energy atomic orbitals. Mainly transition elements show different applications in the colour industry (use in inorganic pigment).