Question

Which of the following is false:
A. ${\text{C}}{{\text{r}}^{2 + }}_{\left( {\text{g}} \right)}$ ion has greater magnetic moment compared to ${\text{C}}{{\text{o}}^{3 + }}_{\left( {\text{g}} \right)}$.
B. The magnitude of ionization potential of iron anion would be equal to electron gain enthalpy of iron.
C. Lanthanide contraction is the cause of lower ionization potential of ${\text{Pb}}$ and ${\text{Sn}}$.
D. If successive ionization energy are $332,738,849,4080,4958{\text{kJ}}.{\text{mol}}$, then this element can be of ${15^{{\text{th}}}}$ group.

Answer 1

Hint:Magnetic moment depends on the number of unpaired electrons.
The value of each ionization energy will increase with each removed electron, since the attractive influence of the nucleus increases and will require more energy for the removal of electrons from positive charges. Electron affinity and ionization energy are opposite.

Complete step by step answer:
We can solve this problem by considering each option.
A. Chromium has atomic number $24$. The electronic configuration of ${\text{Cr}}$ is $1{{\text{s}}^2}2{{\text{s}}^2}2{{\text{p}}^6}3{{\text{s}}^2}3{{\text{p}}^6}4{{\text{s}}^1}3{{\text{d}}^5}$.
In ${\text{C}}{{\text{r}}^{2 + }}$, two electrons are lost. Thus electronic configuration will be $1{{\text{s}}^2}2{{\text{s}}^2}2{{\text{p}}^6}3{{\text{s}}^2}3{{\text{p}}^6}4{{\text{s}}^1}3{{\text{d}}^3}$

$ \uparrow $

$4{\text{s}}$

$ \uparrow $

$ \uparrow $

$ \uparrow $

$3{\text{d}}$
Therefore the number of unpaired electrons of ${\text{C}}{{\text{r}}^{2 + }}$ is $4$.
Now we have to calculate the magnetic moment. Magnetic moment is calculated by the following formula:
$\mu = \sqrt {{\text{n}}\left( {{\text{n}} + 2} \right)} $, where ${\text{n}}$ is the number of unpaired electron.
Substituting the values, we get $\mu = \sqrt {4\left( {4 + 2} \right)} = \sqrt {4 \times 6} = \sqrt {24} = 4.89{\text{BM}}$
Thus the magnetic moment is $4.89{\text{BM}}$
${\text{C}}{{\text{o}}^{3 + }} - 1{{\text{s}}^2}2{{\text{s}}^2}2{{\text{p}}^6}3{{\text{s}}^2}3{{\text{p}}^6}4{{\text{s}}^0}3{{\text{d}}^6}$

$4{\text{s}}$

$ \uparrow \downarrow $

$ \uparrow $

$ \uparrow $

$ \uparrow $

$ \uparrow $

$3{\text{d}}$
There are three unpaired numbers of electrons. Thus magnetic moment can be calculated by
$\mu = \sqrt {4\left( {4 + 2} \right)} = \sqrt {4 \times 6} = \sqrt {24} = 4.89{\text{BM}}$
Thus both have equal magnetic moments. The statement is false.
B. The given statement is true. To form iron anion, electron gas to be gained. Thus electron gain enthalpy is equal to the ionization potential.
C. The ionization potential decreases while moving down the group. ${\text{Pb}}$ and ${\text{Sn}}$ are at ${6^{{\text{th}}}}$ and ${5^{{\text{th}}}}$ periods. Thus they have low ionization potential.
D. From the given values of ionization energy, the fourth ionization energy is greater than the third ionization energy. It indicates that there is a valency of three electrons. Therefore it comes under the ${13^{{\text{th}}}}$ group. Thus the statement is wrong.
Hence, the options A and D are false.

Note:
The changes in the values of ionization energy tells about the group number. While electron affinity values increase on moving left to right. It becomes more exothermic here. It decreases moving down the group.