Question

Which of the following arrangements does not represent the correct order of the property stated against it?
A. $\text{N}{{\text{i}}^{2+}}\text{ }\le \text{ C}{{\text{o}}^{2+}}\text{ }\le \text{ F}{{\text{e}}^{2+}}\text{ }\le \text{ M}{{\text{n}}^{2+}}$; ionic size
B. $\text{C}{{\text{o}}^{3+}}\text{ }\le \text{ F}{{\text{e}}^{3+}}\text{ }\le \text{ C}{{\text{r}}^{3+}}\text{ }\le \text{ S}{{\text{c}}^{3+}}$; stability in aqueous solution
C. $\text{Sc }\le \text{ Ti }\le \text{ Cr }\le \text{ Mn}$; a number of oxidation state
D. ${{\text{V}}^{2+}}\text{ }\le \text{ C}{{\text{r}}^{2+}}\text{ }\le \text{ M}{{\text{n}}^{2+}}\text{ }\le \text{ F}{{\text{e}}^{2+}}$; paramagnetic behaviour

Answer 1

Hint: For this problem, we have to analyse each aspect that is of ionic size, aqueous solution's stability, oxidation state and paramagnetic behaviour. Due to which we can arrange the given elements correctly.

Complete step by step answer:
- In the given question, we have to identify the order which has an incorrect arrangement of the elements according to the given aspect.
- Now, in the given arrangement firstly we will see the order of the paramagnetic behaviour of the elements.
- As we know that those elements which have unpaired electrons in the orbital are considered as the paramagnetic.

- So, ${{\text{V}}^{2+}}$ has the electronic configuration $(\text{Ar})3{{\text{d}}^{3}}$. So, we know that there are 5 orbitals among which only three orbitals are filled with one electron each.
- So, the number of unpaired electrons will be 3.

- Similarly, $\text{C}{{\text{r}}^{2+}}$ the electronic configuration will be $(\text{Ar})3{{\text{d}}^{4}}$. So, we know that there are 5 orbitals among which all the five orbitals are filled with one electron in four and two electrons in the first orbital.
- So, the number of the unpaired electrons will be 4.

- $\text{M}{{\text{n}}^{2+}}$ has the electronic configuration $(\text{Ar})3{{\text{d}}^{5}}\text{ }4{{\text{s}}^{0}}$. So, we know that there are 5 orbitals among which all the five orbitals are filled with one electron in each orbital.
- So, the number of unpaired electrons will be 5.

- Now, $\text{F}{{\text{e}}^{2+}}$ has the electronic configuration $(\text{Ar})3{{\text{d}}^{6}}$. So, we know that there are 5 orbitals among which all the four orbitals are filled with one electron in each orbital and one orbital with two electrons.
- So, the number of unpaired electrons will be 4.
- So, the correct order of the paramagnetic behaviour will be:
${{\text{V}}^{2+}}\text{ }\le \text{ C}{{\text{r}}^{2+}}\text{ }\le \text{ F}{{\text{e}}^{2+}}\text{ }\le \text{ M}{{\text{n}}^{2+}}$

- Now, the order of stability of elements in the aqueous solution is given as:
$\text{C}{{\text{o}}^{3+}}\text{ }\le \text{ F}{{\text{e}}^{3+}}\text{ }\le \text{ C}{{\text{r}}^{3+}}\text{ }\le \text{ S}{{\text{c}}^{3+}}$
- As we know that the hydration enthalpy is directly proportional to the charge density or inversely proportional to the ionic radius.

- So, here the correct order of the ionic radii will be:
$\text{Co }\le \text{ Fe }\le \text{ Cr }\le \text{ Sc}$ because as we know that from left to right in the period the ionic radius decreases.

- So, the charge density on cobalt will be maximum and minimum in the scandium. So, the correct order of the stability of the aqueous solution will be:
 $\text{C}{{\text{o}}^{3+}}\text{ }\ge \text{ F}{{\text{e}}^{3+}}\text{ }\ge \text{ C}{{\text{r}}^{3+}}\text{ }\ge \text{ S}{{\text{c}}^{3+}}$
So, the correct answer is “Option B and D”.

Note: The order of the ionic radius is correct because when we move from left to the right he ionic radius will decrease whereas the oxidation state of the given elements in the option C is also correct because when we move from left to right in the period there is an increase in the oxidation state due to the presence of a high number of unpaired electrons.