Periodic Table & Periodicity

In option (D) electron is removed from 4p subshell for 1st ionization enthalpy, but in all other options electron is removed from 3p subshell.

As 3p subshell is closer to nucleus than 4p, then attraction to the electrons in 3p subshell is more compared to the electrons in 4p subhell.

That is why removal of electrons from 4p subshell is easy compared to 3p subshell.

So among all the options, option (D) will have least ionization enthalpy.

Among (A), (B), (C) options, (C) has half filled 3p subshell so it is more stable compare to the others and you have to provide more energy to remove electrons from stable 3p subshell.

That is why electron configuration [Ne] 3s2 3p3 has highest ionization enthalpy.

The third ionization enthalpy refers to the energy required to remove the third electron from a di-positive ion ($M^{2+}$) to form a tri-positive ion ($M^{3+}$).

In the case of transition metals, the energies involved in removing the third electron are typically higher than for the first and second electrons due to an increasing effective nuclear charge and the electrons being removed from closer energy levels to the nucleus or from a half-filled or full-filled d-subshell which is relatively more stable.

Let us look at the electronic configurations of the ions of each metal listed to determine which would require the highest third ionisation enthalpy.

Since we are talking about $3^{\text{rd}}$ ionization, we would be considering the removal of three electrons, ending up with a $+3$ oxidation state for each metal.

Manganese ($\mathrm{Mn}$): [Ar] $3d^5 4s^2$ Iron ($\mathrm{Fe}$): [Ar] $3d^6 4s^2$ Chromium ($\mathrm{Cr}$): [Ar] $3d^5 4s^1$ Vanadium ($V$): [Ar] $3d^3 4s^2$ After removing two electrons (assuming they come from the 4s orbital and the next available d orbital), the configurations would be: $\mathrm{Mn^{2+}}$: [Ar] $3d^5$ $\mathrm{Fe^{2+}}$: [Ar] $3d^6$ $\mathrm{Cr^{2+}}$: [Ar] $3d^4$ $V^{2+}$: [Ar] $3d^3$ Now we look at the energy required to remove the third electron from the $M^{2+}$ ions: For $\mathrm{Mn^{2+}}$, removing the third electron would break the half-filled $3d^5$ configuration and require significant energy.

For $\mathrm{Fe^{2+}}$, removing the third electron would form half-filled $3d^5$ configuration, which forms stable configuration.

For $\mathrm{Cr^{2+}}$, the third electron removal will disrupt the $3d^4$ configuration, which is not particularly stable.

For $V^{2+}$, the third electron removal will disrupt the $3d^3$ configuration.

In terms of stability, the half-filled $d^5$ subshell in manganese ($\mathrm{Mn^{2+}}$) provide extra stability.

So, the $3^{\text{rd}}$ ionisation enthalpy of $\mathrm{Mn^{2+}}$ is the highest because it involves breaking a half-filled configuration which is more stable.

Thus, out of the given options, manganese ($\mathrm{Mn}$) would have the highest third ionization enthalpy.

Therefore, the correct answer is: Option A : $\mathrm{Mn}$

We can determine the number of valence electrons by looking for a large jump in successive ionization energies.

Here's how we can analyze the data: The ionization energies given are:

Notice that: The first three ionization energies are relatively low.

There is a significant jump between the third and fourth ionization energy.

What does the gap indicate?

The large jump from

to

means that after removing the first three electrons (which are the valence electrons), the next electron to be removed (a core electron) is much more strongly bound.

This implies that the element has 3 valence electrons.

Which group does an element with 3 valence electrons belong to?

Elements with 3 valence electrons are found in Group 13 of the periodic table.

Thus, the element most likely belongs to Group 13.

Correct Answer: Option C (Group 13).

Solid hydrogencarbonates are not formed by lithium and magnesium. – 6Li + N2 $\to$ 2Li3N – 3Mg + N2 $\to$ Mg3N2

Statement I: "In group 13, the stability of +1 oxidation state increases down the group."

Analysis: Group 13 Elements: Boron (B) Aluminum (Al) Gallium (Ga) Indium (In) Thallium (Tl) Common Oxidation States: The typical oxidation state for group 13 elements is +3.

However, due to the inert pair effect, the +1 oxidation state becomes more stable as we move down the group.

Inert Pair Effect: The reluctance of the s-electrons (ns²) in the valence shell to participate in bonding.

This effect becomes more significant in heavier elements due to poor shielding by inner d and f orbitals.

Stability Trend: Boron (B): Exhibits only the +3 oxidation state.

Aluminum (Al): Predominantly +3 oxidation state.

Gallium (Ga): +3 is more stable, but +1 oxidation state starts appearing.

Indium (In): +1 oxidation state becomes significant.

Thallium (Tl): The +1 oxidation state is more stable than the +3 oxidation state.

Conclusion for Statement I: Correct.

The stability of the +1 oxidation state increases as we move down group 13 due to the inert pair effect.

Statement II: "The atomic size of gallium is greater than that of aluminium."

Analysis: Expected Trend: Generally, atomic size increases down a group because each successive element has an additional electron shell.

Actual Atomic Radii: Aluminum (Al): Approximately 143 picometers (pm) Gallium (Ga): Approximately 135 pm Anomalous Behavior: **Gallium's atomic radius is *slightly smaller* than that of aluminum.

** This anomaly is due to the presence of 10 d-electrons in gallium's electron configuration (Ga: [Ar] 3d¹⁰ 4s² 4p¹).

Poor Shielding Effect: 3d electrons do not shield the nuclear charge effectively.

As a result, the effective nuclear charge increases, pulling the outer electrons closer to the nucleus.

This causes gallium to have a smaller atomic radius compared to aluminum.

Conclusion for Statement II: Incorrect.

The atomic size of gallium is less than that of aluminum due to poor shielding by d-electrons.

Final Answer: Statement I is correct, as the stability of the +1 oxidation state increases down group 13.

Statement II is incorrect, because gallium has a smaller atomic radius than aluminum.

Answer: Option B

P3$-$ > S2$-$ > Cl$-$ > K+ > Ca2+ (Correct order of ionic radii) all the given species are isoelectronic species.

In isoelectronic species size increases with increase of negative charge and size decreases with increase in positive charge.

The peroxide and oxides of alkali metals are colourless when pure.

Superoxides are paramagnetic while peroxides are diamagnetic.

Electronic configuration of $\mathrm{O}_{2}^{2-}$ (peroxide) $\left(\sigma 1 s^{2}\right)\left(\sigma^{*} 1 s^{2}\right)\left(\sigma 2 s^{2}\right)\left(\sigma^{*} 2 s^{2}\right)\left(\sigma 2 p_{z}^{2}\right)\left(\pi 2 p_{x}^{2} \pi 2 p_{y}^{2}\right)$ $\left(\pi^{*} 2 p_{x}^{2} \pi^{*} 2 p_{y}^{2}\right)$ Electronic configuration of $\mathrm{O}_{2}^{-}$ $\left(\sigma 1 s^{2}\right)\left(\sigma^{*} 1 s^{2}\right)\left(\sigma 2 s^{2}\right)\left(\sigma^{*} 2 s^{2}\right)\left(\sigma 2 p_{z}^{2}\right)\left(\pi 2 p_{x}^{2} \quad \pi 2 p_{y}^{2}\right)$ $\left(\pi^{*} 2 p_{x}^{2} \pi^{*} 2 p_{y}^{1}\right)$ superoxide is paramagnetic due 1 unpaired electron.

Oxygen is the second most electronegative element in comparison to fluorine.

In group - 16 family $(\mathrm{O}, \mathrm{S}, \mathrm{Se}, \mathrm{Te})$, O-atom is smallest in size.

So, electron density on O-atom is very high in group -16 During addition of a free electron to gaseous $\mathrm{O}$-atom,

We have to supply a significant amount of energy (endothermic) to overcome the electrostatic repulsion between the approaching electron and O-atom of very high electron density.

So, the net value of electron affinity (EA) or (negative) electron gain enthalpy $\left[\Delta_{\mathrm{eg}} H\right.$ or $\left.\left|\Delta_{\mathrm{eg}} H\right|\right]$ of oxygen decreases to a higher extent in comparison to other elements of group -16 who have larger size and lower electronegativity.

So, the correct order of EA or $\left|\Delta_{\mathrm{eg}} \mathrm{H}\right|$ of group $-16$ elements will be $\mathrm{S}$ $>\mathrm{Se}>\mathrm{Te}>\mathrm{O}$

Molecular hydrogen is produced as a byproduct in the industrial production of

by electrolysis of aq

solution

Cathode :

Anode :

is crystallised from the remaining part of electrolyte.

First ionisation energy increases along the period. Along the period

increases which outweighs the shielding effect.