Question: The sums of first and second ionization energies and those of third and fourth ionization energies (...

Question

The sums of first and second ionization energies and those of third and fourth ionization energies (in MJ/mol) of nickel and platinum are:

Metal	${{(IE)}_{1}}+{{(IE)}_{2}}$	${{(IE)}_{3}}+{{(IE)}_{4}}$
Nickel	2.49	8.80
Platinum	2.66	6.70

Based on this information, the most common oxidation states of Ni and Pt are respectively:
(A) +2, +2
(B) +4, +2
(C) +2, +4
(D) +4, +4

Answer 1

Solution

According to the modern periodic table, the elements are arranged based on their order of increasing atomic number and periodic properties of elements. The periodic properties are in terms of ionization energy, electron affinity, atomic radius, electronegativity, and ionic radius. To predict the element's physical, chemical, and atomic properties by using periodic law and table formation.

Complete step by step solution:
The minimum energy required to remove one electron from a neutral gaseous atom in its ground state to form gaseous ion is called ionization energy and ionization potential.
$X(g)\to {{X}^{+}}(g)+{{e}^{-}}$
The energy required for removing one electron is called first ionization potential ${{(IE)}_{1}}$ and the smaller ionization energy which is easier to convert an atom into its positive ion.
The ionization energies to remove the first, second, third, and fourth electron from an isolated gaseous atom is called successive ionization energies.
The order of successive ionization energies as follows,
${{(IE)}_{1}}<{{(IE)}_{2}}<{{(IE)}_{3}}<{{(IE)}_{4}}$
Given, the sums of first and second ionization energies and those of third and fourth ionization energies (in MJ/mol) of nickel and platinum are:

Metal	${{(IE)}_{1}}+{{(IE)}_{2}}$	${{(IE)}_{3}}+{{(IE)}_{4}}$
Nickel	2.49	8.80
Platinum	2.66	6.70

From the above data,
For Nickel ${{(IE)}_{1}}+{{(IE)}_{2}}<<{{(IE)}_{3}}+{{(IE)}_{4}}$ , the difference is very large. So the stable oxidation state of Ni is +2.
For Pt, ${{(IE)}_{1}}+{{(IE)}_{2}}<{{(IE)}_{3}}+{{(IE)}_{4}}$ , the difference is small. So, the stable oxidation state of Pt is +4.

So, the correct answer is option C.

Note: Variation of ionization energy in the periodic table decreases in the group when the atomic number increases. In periods, IE increases with increasing the atomic number from left to right. When the atoms are half-filled or filled, they have more ionization energy.