Periodic Table & Periodicity

$ \mathrm{Kr}, \mathrm{Xe}, \mathrm{Rn} $ has positive electron gain enthalpies. All of above gases are noble gases. All the occupied subshell of noble gases are completely filled. Therefore, the additional electron has to be placed in an orbital of next higher shell. We can also say that energy has to be supplied to add an additional electron and hence, electron gain enthalpy of all noble gases are positive.

A is not correct but R is correct.

The correct form of A is $ \mathrm{Na}^{+} $has smaller size than $ \mathrm{F}^{-} $because after loosing an electron Na will has less electron cloud and more effective nuclear charge whereas F after gaining an electron will have less effective nuclear charge than the parent atom.

Hence, $ \mathrm{F}^{-} $size increases.

$ \mathrm{Na}^{+} $has smaller size than $ \mathrm{F}^{-} $. It is because after losing an electron Na will have more effective nuclear charge whereas $ F $ after gaining an electron will have less effective nuclear charge than the parent atom so the size increases.

In the given species the ratio of $s$-electrons and $p$-electrons are same in $\mathrm{K}^{+}, \mathrm{Cr}^{3+}$.

$\mathrm{K}^{+}$(18 electrons) : $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 \quad \ldots \text { (i) }$

$ \begin{gathered} s \text {-electrons }=6 ; p \text {-electron }=12 \\\\ s: p=6: 12 \text { or } 1: 2 \quad \ldots \text { (ii) } \end{gathered} $

Similarly,

$\mathrm{Cr}^{3+}$ (21 electrons) :

$ \begin{gathered} 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^3 \quad \ldots \text { (iii) } \\\\ s \text {-electrons }=6 ; p \text {-electron }=12 \\\\ s: p=6: 12 \text { or } 1: 2 \quad \ldots \text { (iv) } \end{gathered} $

Option D is correct:

A - II: Ra (Radium) belongs to the s-block as it is an alkaline earth metal.

B - I: Uuq (Ununquadium) is now known as Fl (Flerovium) and belongs to the p-block, as it is a part of the post-transition metals.

C - IV: Ds (Darmstadtium) belongs to the d-block, as it is a transition metal.

D - III: Fm (Fermium) belongs to the f-block, as it is an actinide element.

These elements are classified based on their electron configurations, which determine their position in the periodic table's blocks.

Electronic configuration of $P$

$ \Rightarrow \quad \begin{aligned} & 1 s^2 2 s^2 3 p^6 3 s^2 3 p^3 \\\\ & s_{\text {electron }}=6, p_{\text {electron }}=9 \\\\ & s: p \Rightarrow 6: 9 \text { or } 2: 3 \quad \ldots \text { (i) } \end{aligned} $

Electronic configuration of Ca

$ \begin{aligned} & 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 \\\\ & \Rightarrow \quad s_{\text {electron }}=8, p_{\text {electron }}=12 \\\\ & \quad s: p \Rightarrow 8: 12 \text { or } 2: 3 \quad \ldots \text { (ii) } \end{aligned} $

Hence, P and Ca have same ratio of $S_{\text {electron }}$ and $p_{\text {electron }}$.

Electron gain enthalpy generally becomes more negative across a period as we move from left to right. Within a group, electron gain enthalpy becomes less negative down the group. However adding an electron to the $2 p$-orbital leads to greater repulsion than adding an electron to the larger $3 p$-orbital. Hence, most electronegative enthalpy is chlorine and least electron gain enthalpy is oxygen.

Here’s how the elements line up by group number:

Moscovium (Mc, Z = 115) is in group 15 ⇒ III

Livermorium (Lv, Z = 116) is in group 16 ⇒ I

Flerovium (Fl, Z = 114) is in group 14 ⇒ IV

Tennessine (Ts, Z = 117) is in group 17 ⇒ II

That corresponds to Option C.

Statement I: "Nitrogen has more ionisation enthalpy and electronegativity than beryllium."

Ionisation Enthalpy: Nitrogen, being a non-metal and located in group 15, has a higher ionisation energy than beryllium, a metal in group 2. For example, the first ionisation energy of nitrogen is around $1402 \, \text{kJ/mol}$, while for beryllium it is around $900 \, \text{kJ/mol}$.

Electronegativity: Nitrogen is significantly more electronegative than beryllium. This reflects in their chemical behavior, as non-metals tend to attract electrons more strongly than metals.

Conclusion for Statement I: This statement is correct.

Statement II: "$\mathrm{CrO}_3, \mathrm{~B}_2 \mathrm{O}_3$ are acidic oxide."

Chromium Trioxide ($\mathrm{CrO}_3$): This oxide is acidic. It reacts with water to form chromic acid:

$\mathrm{CrO}_3 + \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2\mathrm{CrO}_4.$

Diboron Trioxide ($\mathrm{B}_2 \mathrm{O}_3$): This oxide is also acidic. It reacts with water to yield boric acid:

$\mathrm{B}_2 \mathrm{O}_3 + 3 \, \mathrm{H}_2 \mathrm{O} \rightarrow 2 \, \mathrm{H}_3 \mathrm{BO}_3.$

Conclusion for Statement II: This statement is also correct.

Since both statements are correct, the answer is:

Option A

Both Statement I and Statement II are correct.

In option D, the elements are incorrectly arranged according to their metallic nature.

Metallic Character: This property increases as you move down a group in the periodic table and decreases as you move from left to right across a period.

Therefore, the correct order reflecting metallic nature is: $ \text{K} > \text{Ba} > \text{Na} $.

The general trend for atomic radius in the periodic table is as follows:

As we move down a group, the atomic radius increases. This happens because additional electron shells are added, which make the atoms larger.

As we move from left to right across a period, the atomic radius decreases. This decrease is due to the increased electrostatic force of attraction between the electrons and the nucleus, which pulls the electrons closer to the nucleus and reduces the size of the atom.

Based on these trends, the correct order of atomic radii for $\mathrm{N}$, $\mathrm{F}$, $\mathrm{Al}$, and $\mathrm{Si}$ is:

$ \mathrm{Al} > \mathrm{Si} > \mathrm{N} > \mathrm{F} $

Let's analyze the atomic radii trends for the elements involved:

In the same period (period 2), atomic radii decrease from left to right due to increasing nuclear charge.

Beryllium (Be) is to the left of Boron (B), but since Boron has an extra proton with a similar shielding effect, its electrons are pulled in tighter.

Therefore, in period 2: $B < Be.$

Comparing elements in different periods:

Magnesium (Mg) is in period 3, meaning it has an additional electron shell compared to Be and B.

More electron shells result in a larger atomic radius; hence, $Mg$ is larger than both $Be$ and $B.$

Combining these observations, the order from smallest to largest is:

$B < Be < Mg.$

Thus, the correct option is Option A.

Atomic radii of group 13 elements are lower than those of alkaline earth metal of group 2 due to greater nuclear charge of group 13 element.

On moving down the group, atomic radius of Ga is slightly lower than of Al . This is due to the presence of $d$-electrons in Ga which does not sheild the nucleus effectively.

Hence, the correct order of atomic radius is

$ \mathrm{B}<\mathrm{Ga}<\mathrm{Al}<\mathrm{In}<\mathrm{Tl} $

The correct match is A-III, B-I, C-IV, D-II.

(A) Electron gain enthalpy generally increases as we move left to right in the period. As we move from left to right the size of element

decreases and effective nuclear charge increase.

There is exception of Be and N .

$ \mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N} . $

(B) Metallic character decreases as We move from left to right in the period

$ \mathrm{P}<\mathrm{Si}<\mathrm{Mg}<\mathrm{Na} $

(C) Generally electron gain enthalpy decreases as we go from top to bottom, but there is exception of d in group 17 element.

$ \mathrm{I}<\mathrm{Br}<\mathrm{F}<\mathrm{Cl} $

(D) Electronegativity increases as we move from left to right and from top to bottom.

$ \mathrm{I}<\mathrm{N}<\mathrm{O}<\mathrm{F} $

Let's break down the reasoning:

The first ionisation energy is the energy needed to remove the outermost electron from an atom in its gaseous state.

The given values are:

Lithium (Li): $520 \text{ kJ/mol}$

Beryllium (Be): $899 \text{ kJ/mol}$

Carbon (C): $1086 \text{ kJ/mol}$

Moving across a period in the periodic table, ionisation energy generally increases. However, there is an anomaly between Be and B:

Beryllium (Be) has a $2s^2$ configuration, which is a stable, fully-filled subshell.

Boron (B), on the other hand, has a $2s^2 2p^1$ configuration. The electron in the $2p$ orbital is less strongly held (more shielded and farther from the nucleus) than the electrons in the $2s$ orbital, which leads to a lower ionisation energy than one might expect from a simple trend.

Experimentally, the first ionisation energy of boron is about $801 \text{ kJ/mol}$.

Therefore, the correct answer is:

Option C: $801\text{ kJ/mol}$

The correct match is

A-II, B-IV, C-I, D-III.

Electron gain enthalpy is defined as the amount of energy released when an electron is added to an isolated gaseous atom. It increases from left to right in period. As we go down electronegativity becomes less negative.

$\mathrm{F} \rightarrow-328 \mathrm{~kJ} / \mathrm{mol}$

$\mathrm{Cl} \rightarrow-349 \mathrm{~kJ} / \mathrm{mol}$

$\mathrm{O} \rightarrow-141 \mathrm{~kJ} / \mathrm{mol}$

$\mathrm{S} \rightarrow-200 \mathrm{~kJ} / \mathrm{mol}$

We'll analyze each oxide based on its chemical behavior regarding acid–base properties:

$\mathrm{C}_2\mathrm{O}_3$

Compounds of carbon with oxygen are generally acidic. In this case, although $\mathrm{C}_2\mathrm{O}_3$ is not commonly encountered, nonmetal oxides tend to be acidic rather than basic.

$\mathrm{CrO}_3$

This is a well-known oxide of chromium in a high oxidation state. It is acidic in nature (in fact, it reacts with water to eventually form chromic acid).

$\mathrm{V}_2\mathrm{O}_5$

Vanadium(V) oxide is also in a high oxidation state. Like many high oxidation state transition metal oxides, it behaves as an acidic oxide (or sometimes amphoteric but predominantly leaning towards acidic behavior).

$\mathrm{V}_2\mathrm{O}_3$

Vanadium(III) oxide, with vanadium in a lower oxidation state, tends to be a basic oxide. In transition metals, as the oxidation state decreases, the oxide becomes more basic because the metal acts more like a typical metal ion.

Thus, the basic oxide among the options is:

$\boxed{\mathrm{V}_2\mathrm{O}_3}$

In summary:

Nonmetal oxides like $\mathrm{C}_2\mathrm{O}_3$ are acidic.

High oxidation state metal oxides such as $\mathrm{CrO}_3$ and $\mathrm{V}_2\mathrm{O}_5$ are acidic.

The lower oxidation state metal oxide, $\mathrm{V}_2\mathrm{O}_3$, exhibits basic properties.

So, the correct answer is Option D: $\mathrm{V}_2\mathrm{O}_3$.

When $ \mathrm{O}^{-} $changes into $ \mathrm{O}^{2-} $, energy is absorbed i.e the change is endothermic because $ \mathrm{O}^{-} $repels the incoming electron due to similar charge.

Hence, it needs energy to add (accept) the electron.

So, electron affinity $ \left(\mathrm{EA}_{2}\right) $ is positive.

$ \mathrm{N}^{3-}, \mathrm{O}^{2-} $ and $ \mathrm{F}^{-} $are isoelectronics ( 10 electrons) species. Their ionic radii will decrease with increase in magnitude of the nuclear charge from $ \mathrm{N}(\mathrm{Z}=7) $ to O $ (Z=8) $ and $ F(Z=9) $.

i.e.

$ \begin{array}{c} \mathrm{N}^{3-} > \mathrm{O}^{2-} > \mathrm{F}^{-} \\ 1.71 \mathring{A} \quad 1.40 \mathring{A} \quad 1.36 \mathring{A} \end{array} $

Hence, option (a) is correct.

The ionization energy of an atom or molecule describes the minimum amount of energy required to remove an electron from the atom or molecule in the gaseous state. It generally increases across a period (from left to right) on the periodic table because the number of protons is increasing, making the nucleus more positively charged and therefore more strongly attracting the negatively charged electrons.

However, there are exceptions to this trend when half-filled or fully-filled sub-shells are present, which are particularly stable configurations. The increased stability of these configurations makes it harder to remove an electron, increasing the ionization energy.

The order of first ionization enthalpy for these elements, considering the above factors, should be:

Li < B < Be < C < O < N < F

This is because:

• Lithium (Li) has the lowest nuclear charge in this group and thus the lowest ionization energy.


• Beryllium (Be) has a fully filled 2s subshell, which increases its ionization energy above that of boron (B).


• Carbon (C) has more protons than boron, increasing its ionization energy.


• Nitrogen (N) has a half-filled 2p subshell, which increases its ionization energy above that of oxygen (O).


• Fluorine (F) has the highest nuclear charge in this group and thus the highest ionization energy.

So the correct answer is: $\mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}<\mathrm{F}$

The correct answer is Option C: Both A and R are correct and R is the correct explanation of A.

• Assertion A: The energy required to form Mg²⁺ from Mg is much higher than that required to produce Mg⁺.

This is true because it takes more energy to remove two electrons from an atom than it does to remove one electron. The second electron is more tightly bound to the nucleus than the first electron, so it requires more energy to remove it.

• Reason R: Mg²⁺ is a small ion and carry more charge than Mg⁺.

This is also true. Mg²⁺ has a smaller radius than Mg⁺ because it has lost two electrons. The smaller radius means that the protons in the nucleus are more tightly packed together, which makes the charge of the nucleus more effective at attracting the electrons. This makes it more difficult to remove an electron from Mg²⁺ than it is to remove an electron from Mg⁺.

The reason is the correct explanation of the assertion because it explains why it takes more energy to form Mg²⁺ than it does to form Mg⁺. The smaller radius of Mg2+ and the higher charge of the nucleus make it more difficult to remove electrons from Mg²⁺, which requires more energy.

Electronegativity is a measure of an atom's ability to attract shared electrons to itself. On the periodic table, electronegativity generally increases as we move from left to right across a period due to increase in effective nuclear charge.

Here are the Pauling electronegativities for the elements:

• Carbon (C): 2.55


• Phosphorus (P): 2.19


• Astatine (At): 2.2


• Bromine (Br): 2.96

Based on these values, the correct order of electronegativity from highest to lowest is Br > C > At > P.

Therefore, Option B is the correct answer:

Br > C > At > P

The melting point of gallium is very low (about 29.76°C or 85.57°F). Therefore, gallium can melt just from the heat of one's hand. The boiling point of gallium, on the other hand, is quite high (2400°C or 4352°F).

On the contrary, water boils at 100°C (212°F) under normal atmospheric pressure. Therefore, inside boiling water, gallium would melt into a liquid but would not boil or evaporate, unlike the other elements listed. Lithium (Li), Bromine (Br), and Cesium (Cs) have boiling points below 100°C, so they would turn into a gas in boiling water.

The correct order of bond strength is

$\mathrm{H_2O > H_2S > H_2Se > H_2Te}$

Statement I is correct as the decrease in first ionization enthalpy from B to Al is indeed much larger than from Al to Ga. This is due to the shielding effect of the electron in the outermost shell and the increased effective nuclear charge as one moves from B to Al and from Al to Ga.

Statement II is also correct, as the d orbitals in Ga are indeed completely filled. This results in the shielding effect being maximum for Ga and thus, the decrease in first ionization enthalpy from Al to Ga is smaller compared to that from B to Al.

Assertion (A) is true but Reason (R) is false.

Nitrogen has more ionisation enthalpy than oxygen due to half-filled $p$-orbital, which is more stable.

The atomic radius of gallium is less than that of aluminium. This is due to poor shielding effect of $d$-electrons which leads to a greater attraction of the outermost electron, in turn leads to smaller size of gallium.

The correct match is A-IV, B-III, C-I, D-II.

${ }_{56} \mathrm{Ba}$ is an alkaline earth metal,

So, group number = 2

Block $=s$-block.

For, $s$-block element, period number can be calculated for its outermost electronic configuration.

$ { }_{56} \mathrm{Ba}=[\mathrm{Xe}] 6 s^2 $

Hence, period number $=6$

(ii) ${ }_{50} \mathrm{Sn}=[\mathrm{Kr}] 4 d^{10} 5 s^2 5 p^2$

Its a $p$-block element.

Hence, period number = 5

Group number $=10+4=14$

(iii) ${ }_{27} \mathrm{Co}=[\mathrm{Ar}] 3 d^7 4 s^2$

Its a $d$-block element.

Hence, period number = 4

Group number $=7+2=9$

(iv) ${ }_{58} \mathrm{Ce}=[\mathrm{Xe}] 4 f^1 5 d^1 6 s^2$

Its a $f$-block element.

Hence, period number = 6

Group number = 3

Metallic character generally increases as the ionisation enthalpy decreases. Thus, it is highest for alkali metals and lowest for halogen or we can say that it decreases on moving from left to right in the period.

Thus, the order is, $\mathrm{Sn}<\mathrm{Cd}<\mathrm{Zr}<\mathrm{Sr}$ (14th group) (12th group) (4th group) (2nd group)

The correct order of second ionisation enthalpy is oxygen $>$ fluorine > nitrogen > carbon.

Electronic configuration of elements after removal of first electron,

$ \begin{aligned} & \mathrm{O}^{+}=1 s^2, 2 s^2, 2 p^3 \\ & \mathrm{~F}^{+}=1 s^2, 2 s^2, 2 p^4 \\ & \mathrm{~N}^{+}=1 s^2, 2 s^2, 2 p^2 \\ & \mathrm{C}^{+}=1 s^2, 2 s^2, 2 p^1 \end{aligned} $

The second I.E. of oxygen is more, as it acquires a half-filled stable electronic configuration. So, require a lot of energy to remove electron from oxygen. Carbon has least I.E. as it will remove one electron to attain a stable fully-filled configuration.

The electron gain enthalpy $(E A)$ increases along a period and decreases down the group. So, elements of group? 17 i.e., Cl and F have more (-ve) ${ }^{\text {EA }}$ than 16th group element i.e., $s$ has more (-ve) EA than 15 th group element P. However, EA of F is less than Cl due to small size of fluorine. Therefore, correct order of electron affinity is $\mathrm{Cl}>\mathrm{F}>\mathrm{S}>\mathrm{P}$.

$ \begin{array}{llll} \hline \text { Elements } & \begin{array}{l} \text { Boiling } \\ \text { point }\left(T_2(\mathrm{~K})\right) \end{array} & \begin{array}{l} \text { Melting } \\ \text { point }\left(T_1(\mathrm{~K})\right) \end{array} & \begin{array}{l} T_2-T_1 \\ =T(\mathrm{~K}) \end{array} \\ \hline \mathrm{Al} & 2740 & 933.5 & 1806.5 \\ \hline \mathrm{Ga} & 2478 & 302.92 & 2175.08 \\ \hline \mathrm{In} & 2350 & 429.78 & 1920.22 \\ \hline \mathrm{~B} & 4275 & 2365 & 1910 \\ \hline \end{array} $

From above table, it clearly seen that, $T_2-T_1$ is maximum for Ga .

Let's analyze each statement one by one:

Statement (I): "The order of first ionisation enthalpy of first three elements of 3rd period is $\mathrm{Mg}>\mathrm{Al}>\mathrm{Na}.$"

The first three elements in the 3rd period are sodium (Na), magnesium (Mg), and aluminium (Al). The experimental ionisation energies are approximately:

Na: 496 kJ/mol

Mg: 738 kJ/mol

Al: 578 kJ/mol

Since magnesium has the highest ionisation energy, followed by aluminium and then sodium, the order $\mathrm{Mg}>\mathrm{Al}>\mathrm{Na}$ is correct.

Statement (II): "The element with electronegativity of 3.5 is chlorine."

On the commonly used Pauli electronegativity scale, the values are approximately:

Fluorine: 4.0

Oxygen: 3.5

Chlorine: 3.16

Hence, an electronegativity of 3.5 corresponds more nearly to oxygen, not chlorine. This statement is incorrect.

Statement (III): "The order of sizes of ions $\mathrm{Mg}^{2+}, \mathrm{Na}^{+}, \mathrm{F}^{-}, \mathrm{O}^{2-}$ is $\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{F}^{-}<\mathrm{O}^{2-}.$"

All these ions are isoelectronic (they each have 10 electrons). Their sizes depend on the nuclear charge. More protons pull the electrons closer, giving a smaller ionic radius.

$\mathrm{Mg}^{2+}$ (12 protons) is the smallest,

followed by $\mathrm{Na}^{+}$ (11 protons),

then $\mathrm{F}^{-}$ (9 protons),

and finally $\mathrm{O}^{2-}$ (8 protons) is the largest.

Thus, the ordering given is correct.

Statement (IV): "The IUPAC name of the element with atomic number 106 is bohrium."

According to IUPAC, the element with atomic number 106 is called seaborgium (Sg). Bohrium (Bh) is element 107. Therefore, this statement is incorrect.

Since only Statements (I) and (III) are correct, the answer is:

Option C: I and III only.

Statements given in II and III are correct. Whille those in I and IV are incorrect. It is because Au and Pt both are soluble in aqua regia. Also, the correct stability order of oxides of halogens are $\mathrm{I}>\mathrm{Cl}>\mathrm{Br}$.

The given statements pertain to the concepts of atomic and ionic radii in chemical bonding and periodic properties. Let's analyze each statement individually for clarity.

Statement I: In $\mathrm{Cl}_{2}$ molecule the covalent radius is double of the atomic radius of chlorine.

The atomic radius of an element like chlorine refers to the size of its atoms, typically measured when the element is in its gas phase and not bonded to anything else. The covalent radius, on the other hand, is a measure of the size of an atom that forms part of a single covalent bond - it's essentially half the distance between two atoms bonded together.

In the case of a $\mathrm{Cl}_{2}$ molecule, the covalent bond is formed between two chlorine atoms. Therefore, the distance from the nucleus of one chlorine atom to the nucleus of the other (the bond length) is essentially twice the covalent radius of chlorine. This makes the statement true only if correctly interpreted: the covalent bond length is double the covalent radius, but it tends to be conflated with the concept of the atomic radius. However, it is important to clarify that the atomic radius and covalent radius are different measures. In a diatomic molecule like $\mathrm{Cl}_2$, saying the covalent radius is "double of the atomic radius" is inaccurate. The precise statement should be that the bond length (twice the covalent radius) is not directly twice the atomic radius but rather the distance between the nuclei of the bonding atoms. Therefore, this statement as phrased is misleading or incorrect.

Statement II: Radius of anionic species is always greater than their parent atomic radius.

This statement is based on the idea that when an atom gains electrons and becomes an anion, the increased electron-electron repulsions in the electron cloud cause it to expand. This is generally true across the periodic table. For example, when chlorine gains an electron to become $\mathrm{Cl}^{-}$, its radius increases due to the addition of an extra electron which increases repulsion among electrons and expands the electron cloud. This principle holds for virtually all anions compared to their parent atoms, so Statement II is correct.

Given the analysis, the correct option is:

Option D: Statement I is incorrect but Statement II is correct.