Periodic Table & Periodicity

Metallic character increases down a group and decreases from left to right across a period (NCERT trend).

Given elements:

• Period 2: $N,\ O,\ F$

• Period 3: $P,\ S,\ Cl$

1) Most metallic element

In period 3, metallic character is higher than in period 2, and within period 3 it decreases left to right:

$P > S > Cl$

So, $P$ is the most metallic among the given elements.

Valence electrons in $P$ (Group 15) $= 5$.

2) Least metallic element

Least metallic means most non-metallic, which is highest towards the top-right of the periodic table. Among the given, $F$ is the most non-metallic.

Valence electrons in $F$ (Group 17) $= 7$.

Final Answer

Number of valence electrons (most metallic, least metallic) $= (5,\ 7)$.

Option C: 5 and 7

In period 4 (from $\mathrm{K}$ to $\mathrm{Kr}$), the atomic radius decreases as we move from left to right because:

• Nuclear charge increases ($Z$ increases),

• Electrons are added in the same principal shell ($n = 4$),

• So the effective nuclear charge on valence electrons increases and pulls them closer.

Hence:

• The largest atomic radius is for the first element of the period: $\boxed{\mathrm{K}}$.

• The smallest atomic radius (as per NCERT trend, minimum at halogen) is for the halogen in that period: $\boxed{\mathrm{Br}}$.

So, the elements with highest and lowest atomic radii respectively are:

$\boxed{\mathrm{K} \ \& \ \mathrm{Br}}$

Correct option: C

To compare second ionization potential configuration of mono-cation is observed

$ \begin{array}{|l|l|l|l|} \hline \mathrm{C}^{+} & \mathrm{N}^{+} & \mathrm{O}^{+} & \mathrm{F}^{+} \\ \hline[\mathrm{He}] & {[\mathrm{He}]} & {[\mathrm{He}] 2 \mathrm{~s}^2 2 \mathrm{p}^3} & {[\mathrm{He}] 2 \mathrm{~s}^2 2 \mathrm{p}^4} \\ 2 \mathrm{~s}^2 \mathrm{sp}^1 & 2 \mathrm{~s}^2 2 \mathrm{p}^2 & \begin{array}{l} \text { Half-filled } \\ \text { stable. } \end{array} & \\ \hline \end{array} $

$ \begin{aligned} &2^{\text {nd }} \text { IE order }\\ &\mathrm{O}>\mathrm{F}>\mathrm{N}>\mathrm{C} \end{aligned} $

Metallic character means the tendency of an element to lose electrons and form a positive ion. In general (as per NCERT trends), metallic character is more in s-block elements than in p-block elements.

So, elements like $\mathrm{K}$ and $\mathrm{Mg}$ (s-block) show more metallic character than $\mathrm{Al}$ and $\mathrm{B}$ (p-block).

Now, compare atomic radius and ionic radius: if an atom forms a cation, its ionic radius becomes smaller than the atomic radius. If an atom forms an anion, its ionic radius becomes larger than the atomic radius.

Therefore, ionic radius is not always smaller than atomic radius. Specifically, anionic radius is greater than atomic radius, but cationic radius is always less than atomic radius for any element.

Statement I:

Na: $1s^2\,2s^2\,2p^6\,3s^1$

After first ionisation, $\mathrm{Na}^+$ becomes $1s^2\,2s^2\,2p^6$ (noble gas configuration).

So, the second ionisation enthalpy of Na means removing an electron from a stable noble gas core, which needs very high energy.

Mg: $1s^2\,2s^2\,2p^6\,3s^2$

After first ionisation, $\mathrm{Mg}^+$ becomes $1s^2\,2s^2\,2p^6\,3s^1$.

So, the second ionisation enthalpy of Mg removes the remaining valence $3s$ electron, which needs less energy than removing from a noble gas core.

Hence, $IE_2(\mathrm{Na}) > IE_2(\mathrm{Mg})$.

So, Statement I is true.

Statement II:

$\mathrm{O}^{2-}$ has $8 + 2 = 10$ electrons, and $\mathrm{F}^-$ has $9 + 1 = 10$ electrons.

So, $\mathrm{O}^{2-}$ and $\mathrm{F}^-$ are isoelectronic (same number of electrons).

In an isoelectronic series, ionic radius decreases as nuclear charge ($Z$) increases, because higher $Z$ pulls electrons more strongly.

Here, $Z(\mathrm{O}) = 8$ and $Z(\mathrm{F}) = 9$.

So, $\mathrm{F}^-$ (higher $Z$) will be smaller than $\mathrm{O}^{2-}$.

Therefore, $r(\mathrm{O}^{2-}) > r(\mathrm{F}^-)$.

So, Statement II is true.

Correct option: Option C

Both Statement I and Statement II are true.

As per the NCERT textbook, electronegativity is the ability of an atom to attract a shared pair of electrons. For Group 15, the electronegativity decreases as we go down the group.

Let's look at their approximate electronegativity values (on the Pauling scale):

• $EN(N) = 3.0$

• $EN(P) = 2.1$

• $EN(As) = 2.0$

• $EN(Sb) = 1.9$

• $EN(Bi) = 1.9$

Notice a very important point: The drop in electronegativity from Nitrogen (N) to Phosphorus (P) is very large ($3.0 - 2.1 = 0.9$). However, for the elements below Phosphorus, the values decrease very slowly.

The question states: "The difference between the electronegativity values of 'X' and phosphorus is higher than that of the difference between phosphorus and 'Y'."

Let's write this mathematically. Let $EN$ stand for electronegativity. The absolute difference is what matters.

$ |EN(X) - EN(P)| > |EN(P) - EN(Y)| $

This means we are looking for an element 'X' whose electronegativity is very different from Phosphorus, and an element 'Y' whose electronegativity is very close to Phosphorus.

Now, let's check each option using the electronegativity values.

• Option A: X = As & Y = Bi

  • Difference 1: $|EN(X) - EN(P)| = |EN(As) - EN(P)| = |2.0 - 2.1| = 0.1$

  • Difference 2: $|EN(P) - EN(Y)| = |EN(P) - EN(Bi)| = |2.1 - 1.9| = 0.2$

  • Is $0.1 > 0.2$? No. So, this option is incorrect.

• Option B: X = Bi & Y = N

  • Difference 1: $|EN(X) - EN(P)| = |EN(Bi) - EN(P)| = |1.9 - 2.1| = 0.2$

  • Difference 2: $|EN(P) - EN(Y)| = |EN(P) - EN(N)| = |2.1 - 3.0| = 0.9$

  • Is $0.2 > 0.9$? No. So, this option is incorrect.

• Option C: X = N & Y = As

  • Difference 1: $|EN(X) - EN(P)| = |EN(N) - EN(P)| = |3.0 - 2.1| = 0.9$

  • Difference 2: $|EN(P) - EN(Y)| = |EN(P) - EN(As)| = |2.1 - 2.0| = 0.1$

  • Is $0.9 > 0.1$? Yes. This matches the condition given in the question.

• Option D: X = As & Y = Sb

  • Difference 1: $|EN(X) - EN(P)| = |EN(As) - EN(P)| = |2.0 - 2.1| = 0.1$

  • Difference 2: $|EN(P) - EN(Y)| = |EN(P) - EN(Sb)| = |2.1 - 1.9| = 0.2$

  • Is $0.1 > 0.2$? No. So, this option is incorrect.

Therefore, X is Nitrogen (N) and Y is Arsenic (As).

In general on moving from left to right in a period ionization energy increases as $\mathbf{Z}_{\text {eff }}$ increases.

Statement I (First ionization enthalpy)

Across a period, first ionization enthalpy generally increases due to increasing nuclear charge and decreasing atomic size.

But there is an important NCERT exception:

• Nitrogen ($\mathrm{N}$) has a half-filled configuration $2p^3$, which is extra stable.

• Oxygen ($\mathrm{O}$) has configuration $2p^4$, where one $p$-orbital contains a pair of electrons, causing extra electron–electron repulsion, so it is easier to remove one electron.

So,

$\mathrm{IE}_1(\mathrm{O}) < \mathrm{IE}_1(\mathrm{N})$

Also, in general:

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

✅ Statement I is true.

Statement II (Magnitude of electron gain enthalpy)

Electron gain enthalpy usually becomes less negative down a group due to increasing size (incoming electron feels less attraction).

But for group 16, NCERT notes:

• Oxygen is very small, so the incoming electron experiences strong repulsion in the compact $2p$ subshell.

• Hence oxygen has less negative electron gain enthalpy than sulphur.

• Sulphur has the most negative (highest magnitude) electron gain enthalpy in this group.

Therefore, the magnitude order is:

$\mathrm{S} > \mathrm{Se} > \mathrm{Te} > \mathrm{Po} > \mathrm{O}$

✅ Statement II is true.

Correct option: D

✅ Both Statement I and Statement II are true.

Statement I:

The order given is:

Al > Mg > Mg²⁺ > Al³⁺

We are comparing atomic and ionic radii.

• Mg (Z = 12) → configuration: [Ne]3s²

• Al (Z = 13) → configuration: [Ne]3s²3p¹

Across a period (from Mg to Al), the effective nuclear charge increases, so atomic radius decreases.

Therefore, Mg > Al (not Al > Mg).

For the ions:

• Mg²⁺ and Al³⁺ are isoelectronic (both have 10 electrons, same as Ne).

  Among isoelectronic species, the greater the nuclear charge, the smaller the radius.

  Hence, Al³⁺ < Mg²⁺.

So the correct order should be:

Mg > Al > Mg²⁺ > Al³⁺

Therefore, Statement I is false.

Statement II:

Electron gain enthalpy (EGE) refers to the energy change when an atom gains an electron.

The more negative the EGE, the greater the electron affinity.

The general trend for electron affinity (magnitude):

Cl > Br > S > O

Reasoning:

• Across a period and down a group, EGE usually becomes less negative (magnitude decreases down group 17).

• Halogens have high (negative) EGEs.

• Among main-group nonmetals, oxygen and sulfur have less negative EGE due to smaller size and electron–electron repulsion in the compact 2p orbital of oxygen.

Thus, Cl > Br > S > O is the correct order in magnitude.

So Statement II is true.

✅ Final Answer:

Option C: Statement I is false but Statement II is true.

Ionisation Energy (Option A)

• Across a period (left ➜ right) atoms get smaller and the nucleus pulls more strongly, so the first ionisation energy (IE) goes up.
• Down a group (top ➜ bottom) atoms get bigger, so the IE goes down.

$ \mathrm{F}>\mathrm{P}>\mathrm{S}>\mathrm{B} \text { (IE order) } $

Metallic Character (Option C)

• Metallic character means how easily an atom loses electrons.
• Across a period it falls because atoms hold their electrons more tightly.
• Down a group it rises because outer electrons are farther from the nucleus.

$ \mathrm{K}>\mathrm{Mg}>\mathrm{Al}>\mathrm{B} \text { (Metallic character order) } $

Basic Nature of Oxides (Option D)

• Metal oxides are basic; non-metal oxides are acidic.
• As metallic character increases down a group, basic strength of the oxides also increases.

$ \mathrm{K}_2 \mathrm{O}>\mathrm{Na}_2 \mathrm{O}>\mathrm{MgO}>\mathrm{Al}_2 \mathrm{O}_3 $

Electron Affinity (Option B)

• Electron affinity (EA) usually becomes more negative from left to right because atoms want to gain electrons.
• Among the main-group elements, EA follows Group 17 > Group 16 > Group 15.

$ \mathrm{Cl}>\mathrm{F}>\mathrm{S}>\mathrm{P} $

(A) The electronic configurations are:

For carbon (C): $1s^2\\ 2s^2\\ 2p^2$

For $\mathrm{C}^{+}$: $1s^2\\ 2s^2\\ 2p^1$

For boron (B): $1s^2\\ 2s^2\\ 2p^1$

For $\mathrm{B}^{+}$: $1s^2\\ 2s^2$

The second ionisation energy ($\\mathrm{IE}_2$) means the energy required to remove one electron from a singly charged positive ion. Since in $\mathrm{B}^{+}$, the next electron is removed from a completely filled 2s orbital, which is more stable, more energy is needed. Hence, $\\mathrm{IE}_2$ of B is greater than $\\mathrm{IE}_2$ of C. This statement is correct.

(B) The ions $\mathrm{Al}^{3+}$, $\mathrm{Mg}^{2+}$, and $\mathrm{Na}^{+}$ all have the same number of electrons (10), so they are called isoelectronic species.

For isoelectronic ions, the ionic size depends on the number of protons; more protons mean greater attraction, leading to a smaller size.

Number of protons:

${Al}^{3+}$ → 13

${Mg}^{2+}$ → 12

${Na}^{+}$ → 11

Therefore, the size order is: ${Al}^{3+} < {Mg}^{2+} < {Na}^{+}$. This statement is correct.

(C) Density values are:

Potassium (K): $0.86\\ {g/cm^3}$

Sodium (Na): $0.97\\ {g/cm^3}$

Since the density of Na is greater than that of K, this statement is incorrect as per the given periodic trend.

(D) Bond energies:

H–H bond energy = $435.88\\ {kJ/mol}$

F–F bond energy = $155\\ {kJ/mol}$

The F–F bond is weaker than the H–H bond because the small size of fluorine atoms leads to strong repulsion between lone pairs of electrons. Hence, this statement is incorrect.

Ionization energy is the energy needed to remove the most loosely held (outermost) electron from a neutral atom.

Ionization energy generally increases when the effective nuclear charge increases. This happens because the nucleus attracts the valence electron more strongly, so more energy is required to remove it.

Also, some electronic configurations are extra stable. Electrons in stable configurations are harder to remove, so their ionization energy is higher.

Fully filled orbitals have greater ionisation energy due to stable electronic configuration.

Order of I.E:

Here $n=2$, so we are comparing configurations of elements in the 2nd period. As we move from $s$-block to $p$-block in the same period, atomic size decreases and effective nuclear charge increases, so ionisation energy generally increases.

Using stability ideas: the configuration $n s^2 n p^6$ (noble gas, fully filled) has the maximum ionisation energy, and $n s^2 n p^1$ has comparatively lower ionisation energy.

So the decreasing order is: $D>C>A>B$

Ans. (A)

The element on the far left of a period belongs to Group 1. These elements have only one electron in their outermost shell, so they lose this electron easily. Therefore, they have the lowest first ionisation enthalpy (IE).

The element on the far right of a period, except noble gases, belongs to Group 17 (halogens). These atoms need just one electron to complete their outermost shell, so they strongly attract an extra electron. Hence, they have the most negative value of electron gain enthalpy ($\Delta \mathrm{H}_{\mathrm{eg}}$).

Evaluating Statement I:

We need to determine the chemical nature of each pair of oxides:

1. Pair 1: $[\mathrm{Al}_2 \mathrm{O}_3, \mathrm{Cr}_2 \mathrm{O}_3]$

• $\mathrm{Al}_2 \mathrm{O}_3$ (Aluminium oxide) is a standard example of an amphoteric oxide as it reacts with both acids and bases.

• $\mathrm{Cr}_2 \mathrm{O}_3$ (Chromium(III) oxide) has chromium in the $+3$ oxidation state. Transition metal oxides in intermediate oxidation states are generally amphoteric.

Both oxides are amphoteric, so they have the same nature.

2. Pair 2: $[\mathrm{Cl}_2 \mathrm{O}_7, \mathrm{Mn}_2 \mathrm{O}_7]$

• $\mathrm{Cl}_2 \mathrm{O}_7$ is a non-metal oxide in a very high oxidation state ($+7$), which makes it strongly acidic.

• $\mathrm{Mn}_2 \mathrm{O}_7$ contains manganese in its highest oxidation state ($+7$). Transition metals in their highest oxidation states form highly acidic oxides.

Both oxides are acidic, so they have the same nature.

3. Pair 3: $[\mathrm{Na}_2 \mathrm{O}, \mathrm{V}_2 \mathrm{O}_3]$

• $\mathrm{Na}_2 \mathrm{O}$ is an alkali metal oxide and is strongly basic.

• $\mathrm{V}_2 \mathrm{O}_3$ contains vanadium in a lower oxidation state ($+3$). Lower oxidation state oxides of d-block elements are typically basic in nature.

Both oxides are basic, so they have the same nature.

4. Pair 4: $[\mathrm{CO}, \mathrm{N}_2 \mathrm{O}]$

• Both $\mathrm{CO}$ (carbon monoxide) and $\mathrm{N}_2 \mathrm{O}$ (nitrous oxide) are classic examples of non-metal oxides that are completely neutral.

Both oxides are neutral, so they have the same nature.

Since all 4 pairs contain oxides of the same chemical nature, Statement I is true.

Evaluating Statement II:

We are given four oxides: $\mathrm{Na}_2 \mathrm{O}$, $\mathrm{Al}_2 \mathrm{O}_3$, $\mathrm{CO}$, and $\mathrm{Cl}_2 \mathrm{O}_7$. Let us list their nature:

• $\mathrm{Na}_2 \mathrm{O}$ is a basic oxide.

• $\mathrm{Al}_2 \mathrm{O}_3$ is an amphoteric oxide.

• $\mathrm{CO}$ is a neutral oxide.

• $\mathrm{Cl}_2 \mathrm{O}_7$ is an acidic oxide.

From this list, it is clear that $\mathrm{Na}_2 \mathrm{O}$ is the most basic because it is an oxide of a highly electropositive Group 1 metal. Similarly, $\mathrm{Cl}_2 \mathrm{O}_7$ is the most acidic because it is an oxide of a highly electronegative halogen in its highest oxidation state. Thus, Statement II is true.

Conclusion:

Since both Statement I and Statement II are factually correct, the correct answer is Option A.

Option A

Statement I:

Electronegativity generally increases across a period from left to right.

For the elements nitrogen, oxygen, and fluorine in the second period, the correct order is

$ \mathrm{F} > \mathrm{O} > \mathrm{N} $

So, Statement I is true.

Statement II:

Let us check the oxidation state of oxygen in both compounds.

1. In $\mathrm{OF}_2$

Fluorine is the most electronegative element, so its oxidation state is always $-1$.

Let the oxidation state of oxygen be $x$.

Then,

$ x + 2(-1) = 0 $

$ x - 2 = 0 $

$ x = +2 $

So, in $\mathrm{OF}_2$, oxygen has oxidation state $+2$.

2. In $\mathrm{Na}_2\mathrm{O}$

Sodium has oxidation state $+1$.

Let the oxidation state of oxygen be $x$.

Then,

$ 2(+1) + x = 0 $

$ 2 + x = 0 $

$ x = -2 $

So, in $\mathrm{Na}_2\mathrm{O}$, oxygen has oxidation state $-2$.

Hence, Statement II is also true.

Therefore, the correct answer is:

$ \boxed{\text{Option A}} $

Both Statement I and Statement II are true.

For magnesium, the electronic configuration is

$ \mathrm{Mg} = 1s^2\,2s^2\,2p^6\,3s^2 $

The first ionization enthalpy corresponds to:

$ \mathrm{Mg(g)} \rightarrow \mathrm{Mg^+(g)} + e^- $

This is given as

$ \Delta_i H_1 = +737 \,\mathrm{kJ\,mol^{-1}} $

Now the second ionization enthalpy corresponds to removing one more electron from $\mathrm{Mg^+}$:

$ \mathrm{Mg^+(g)} \rightarrow \mathrm{Mg^{2+}(g)} + e^- $

After losing one electron, $\mathrm{Mg^+}$ has configuration

$ 1s^2\,2s^2\,2p^6\,3s^1 $

The second electron is still being removed from the same outermost shell, but now from a positively charged ion. So this electron is held more strongly than in neutral Mg.

Hence:

• second ionization enthalpy must be greater than first ionization enthalpy,

• ionization enthalpy is always positive.

So options with negative values are impossible, and $+590 \,\mathrm{kJ\,mol^{-1}}$ is less than the first ionization enthalpy, so it is also not possible.

Therefore, the most probable value is

$ \boxed{+1450 \,\mathrm{kJ\,mol^{-1}}} $

So, the correct option is C.

To find the correct set, we need one oxide of each type:

• Basic oxide

• Acidic oxide

• Amphoteric oxide

• Neutral oxide

Now let us check the given oxides.

Important classifications

• Basic oxides: Usually metallic oxides like $Na_2O$, $K_2O$

• Acidic oxides: Usually non-metallic oxides like $Cl_2O_7$

• Amphoteric oxides: Oxides that react with both acids and bases, like $Al_2O_3$, $As_2O_3$

• Neutral oxides: Neither acidic nor basic, like $CO$, $NO$, $N_2O$

Option A

$Na_2O,\ K_2O,\ Al_2O_3,\ As_2O_3$

• $Na_2O$ = basic

• $K_2O$ = basic

• $Al_2O_3$ = amphoteric

• $As_2O_3$ = amphoteric

This set does not contain acidic and neutral oxides.

So, Option A is incorrect.

Option B

$Al_2O_3,\ As_2O_3,\ CO,\ NO$

• $Al_2O_3$ = amphoteric

• $As_2O_3$ = amphoteric

• $CO$ = neutral

• $NO$ = neutral

This set does not contain basic and acidic oxides.

So, Option B is incorrect.

Option C

$K_2O,\ Cl_2O_7,\ As_2O_3,\ NO$

• $K_2O$ = basic

• $Cl_2O_7$ = acidic

• $As_2O_3$ = amphoteric

• $NO$ = neutral

This set contains all four kinds of oxides.

So, Option C is correct.

Option D

$Na_2O,\ N_2O,\ Al_2O_3,\ CO$

• $Na_2O$ = basic

• $N_2O$ = neutral

• $Al_2O_3$ = amphoteric

• $CO$ = neutral

This set does not contain any acidic oxide.

So, Option D is incorrect.

Final Answer

$\boxed{\text{Option C}}$

For Statement (I), let us check the trend of first ionisation enthalpy.

Across a period, first ionisation enthalpy generally increases from left to right because effective nuclear charge increases.

For the elements given:

• Na

• Mg

• Cl

• Ar

the correct order of first ionisation enthalpy is

$ \text{Na} < \text{Mg} < \text{Cl} < \text{Ar} $

But the statement says

$ \text{Na} > \text{Mg} > \text{Cl} > \text{Ar} $

This is clearly wrong.

So, Statement (I) is false.

For Statement (II), we compare the third ionisation enthalpy of Ca, Al, Fe and B.

Third ionisation enthalpy means removal of the third electron, that is, from the species $M^{2+}$.

Now consider calcium:

• Ca has electronic configuration

  $ [\text{Ar}]\,4s^2 $

• After losing two electrons,

  $ \text{Ca}^{2+} = [\text{Ar}] $

  which is a noble gas configuration.

Removing the third electron from $\text{Ca}^{2+}$ means breaking a very stable noble gas configuration, so the third ionisation enthalpy is very high.

Hence, Statement (II) is true.

Therefore:

• Statement (I) is false

• Statement (II) is true

So, the correct answer is:

$ \boxed{\text{Option D}} $

To determine which element has the lowest first ionization enthalpy among the given options, we look at the atomic numbers:

Atomic number 32 corresponds to Germanium (Ge).

Atomic number 35 corresponds to Bromine (Br).

Atomic number 87 corresponds to Francium (Fr).

Atomic number 19 corresponds to Potassium (K).

The element with the lowest first ionization energy is Francium (Fr) with atomic number 87. This can be represented by its electron configuration: $\text{Fr} - [\text{Rn}] 7s^1$. Francium, being a part of the alkali metal group, has the most loosely bound outer electron compared to the other elements listed, resulting in a lower first ionization enthalpy.

To determine the relative acidity and bond enthalpy of H₂Se and H₂Te, let's examine both:

Acidic Strength:

The acidity of hydrides increases as we move down the group in the periodic table. This is because the bond strength between hydrogen and the central atom decreases, making it easier for the hydrogen ions to dissociate. Therefore, H₂Te, being further down the group than H₂Se, is more acidic:

$ \mathrm{H}_2\mathrm{Se} < \mathrm{H}_2\mathrm{Te} $

Bond Enthalpy:

Bond enthalpy refers to the energy required to dissociate a bond. The bond enthalpy decreases down the group due to weaker bonds forming as the atomic size increases. Hence, H₂Se has a higher bond enthalpy compared to H₂Te:

$ \Delta_{\text{dis}} \mathrm{H}: \mathrm{H}_2\mathrm{Se} > \mathrm{H}_2\mathrm{Te} $

$ \text{H}_2\mathrm{Se} \, [276 \, \text{KJ/mol}] > \text{H}_2\mathrm{Te} \, [238 \, \text{KJ/mol}] $

Thus, H₂Se is indeed less acidic than H₂Te, but it has a higher bond enthalpy for dissociation.

In a period (row on the periodic table), the atomic size generally decreases as you move from left to right. This is because additional electrons are being added to the same outer shell while the number of protons in the nucleus increases, pulling the electron cloud closer to the nucleus and reducing the atomic radii.

To determine the number of valence electrons in the metal with the lowest enthalpy of atomization among Chromium (Cr), Cobalt (Co), Iron (Fe), and Nickel (Ni), we need to examine their respective electron configurations.

Chromium has the electron configuration:

$ \text{Cr} = [\text{Ar}] 3d^5 4s^1 $

Based on this configuration, Chromium has a total of six valence electrons (1 from the 4s orbital and 5 from the 3d orbital).

Chromium is noted for having the lowest enthalpy of atomization among the given metals. Thus, the number of valence electrons in Chromium, the metal with the lowest enthalpy of atomization, is 6.

Given the ionization energy values for elements $A$ and $B$ in group 14, where $A$ has an ionization energy of 708 kJ/mol and $B$ has 715 kJ/mol, these values are the lowest among their group. This helps identify the elements as tin (Sn) and lead (Pb), respectively. In this context:

The $\mathrm{A}^{2+}$ ion corresponds to $\mathrm{Sn}^{2+}$, which acts as a reducing agent.

The $\mathrm{B}^{4+}$ ion corresponds to $\mathrm{Pb}^{4+}$, which acts as an oxidizing agent.

Therefore, Sn in the +2 oxidation state tends to donate electrons (reduce), while Pb in the +4 oxidation state tends to accept electrons (oxidize), underscoring their respective roles as reducing and oxidizing agents.

$\mathrm{Mn}^{2+}:[\mathrm{Ar}] 3 \mathrm{~d}^5$

Half filled stability

More 1E than $\mathrm{Fe}^{2+}$

$\mathrm{Fe}^{2+}:[\mathrm{Ar}] 3 \mathrm{~d}^6$

$\begin{aligned} &\text { IE order }\\ &\mathrm{B}>\mathrm{T} \ell>\mathrm{Ga}>\mathrm{Al}>\mathrm{In} \end{aligned}$

$\begin{array}{lc} \text { Radius Order } & \mathrm{T} \ell>\mathrm{In}>\mathrm{A} \ell>\mathrm{Ga}>\mathrm{B} \\ \text { EN Order } & \mathrm{B}>\mathrm{T} \ell>\mathrm{In}>\mathrm{Ga}>\mathrm{Al} \\ \text { Density Order } & \mathrm{T} \ell>\mathrm{In}>\mathrm{Ga}>\mathrm{A} \ell>\mathrm{B} \\ \mathrm{IE}_1 \text { Order } & \mathrm{B}>\mathrm{T} \ell>\mathrm{Ga}>\mathrm{A} \ell>\mathrm{In} \end{array}$

To correctly match the elements with their respective families and symbols, consider the following associations:

Pnictogen is matched with Moscovium (Mc), which has the atomic number 115.

Chalcogen corresponds to Livermorium (Lv), with the atomic number 116.

Halogen is associated with Tennessine (Ts), with atomic number 117.

Noble gas correlates with Oganesson (Og), which has an atomic number of 118.

This alignment clarifies the classification of elements based on their family groups and chemical symbols.

(A) The process of adding an $\mathrm{e}^{-}$to a neutral gaseous atom is not always exothermic it may be exothermic or endothermic.

$\begin{array}{cc}\text { (C) } \mathrm{Be} & \mathrm{B} \\ 1 \mathrm{~s}^2 2 s^2 & 1 s^2 2 s^2 2 p^1\end{array}$

In Be 2s subshell in fully filled

So, need high energy to remove $\mathrm{e}^{-}$as compared to B .

due to partially positive charge $z_{\text {eff }} \uparrow, \mathrm{EN} \uparrow$

So, EN of $\mathrm{C} \Rightarrow \mathrm{CCl}_4>\mathrm{CH}_4$

(E) Cs is most electropositive.

The electronic configurations of four elements, A, B, C, and D, are presented as follows:

(A) $1s^2 2s^2 2p^3$

(B) $1s^2 2s^2 2p^4$

(C) $1s^2 2s^2 2p^5$

(D) $1s^2 2s^2 2p^2$

To determine the correct order of increasing electronegativity according to Pauling's scale, analyze the elements as follows:

Nitrogen ($1s^2 2s^2 2p^3$): Electronegativity = 3.0

Oxygen ($1s^2 2s^2 2p^4$): Electronegativity = 3.5

Fluorine ($1s^2 2s^2 2p^5$): Electronegativity = 4.0

Carbon ($1s^2 2s^2 2p^2$): Electronegativity = 2.55

The correct order of increasing electronegativity is:

Carbon (D) < Nitrogen (A) < Oxygen (B) < Fluorine (C)

$\Rightarrow$ The metallic radius order of $\mathrm{Al} \& \mathrm{Ga}$ is

(due to poor shielding of d-subshell electrons)

$\Rightarrow$ The ionic radius order of $\mathrm{Al}^{+3} \& \mathrm{Ga}^{+3}$ is

$\mathrm{B}^{+3}<\mathrm{Al}^{+3}<\mathrm{Ga}^{+3}<\mathrm{In}^{+3}<\mathrm{T} \ell^{+3}$

The order of first four elements of group-17 are as follows.

$\mathrm{F}<\mathrm{Cl}<\mathrm{Br}<\mathrm{I} \text { (Covalent radius) }$

$\mathrm{Cl}>\mathrm{F}>\mathrm{Br}>\mathrm{I}$ (Electron affinity)

$\mathrm{F}<\mathrm{Cl}^{-}<\mathrm{Br}^{-}<\mathrm{I}$ (Ionic radius)

$\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$ ( $\mathrm{I}^{\mathrm{st}}$ ionization energy)

Electron affinity order is irregular.

The elements given are

Li, Na, Be, Mg, B and Al

2$^{nd}$ period elements $\to$ Li, Be, B

Order of atomic radii $\to$ Li > Be > B

3$^{rd}$ period elements $\to$ Na, Mg, Al

Order of atomic radii $\to$ Na > Mg > Al

In a period, atomic radii decreases on moving from left to right.

So, the element with least atomic radii in 2$_{nd}$ period (Li, Be, B) is B. The element with least atomic radii 3$_{rd}$ period (Na, Mg, Al) is Al.

In B and Al $\Rightarrow$ B and Al are same group elements. Down the group, radii increases. So, B has least radii.

B is the element with least atomic radius to form the oxide $A_2O_3$.

Oxidation states of the A in oxides :

${A_0} \to + 2\,\,{A_{{0_2}}} \to + 2\,\,{A_{{2^0}}} \to + 1\,\,{A_{2_3^0}} \to + 3$

B outer configuration is s$^2$P$^1$. 3 valence electrons and form A$_2$O$_3$ type oxide.

Apalite family - A group of phosphate minerals that includes fluorapatite, chlorapatite, and hychoxyapatite. Chemical formula $C{a_{10}}{(P{O_4})_6}{(OH,F,Cl)_2}$

$\left. \matrix{ Fluorapatite \to C{a_{10}}{(P{O_4})_6}{F_2} \hfill \cr Chlorapatite \to C{a_{10}}{(P{O_4})_6}C{l_2} \hfill \cr Hydroxyapatite \to C{a_{10}}{(P{O_4})_6}{(OH)_2} \hfill \cr} \right\}\matrix{ {Main\,components:} \cr {Calcium\,and\,phosphate} \cr }$

So, the element considered is phosphorus.

The first four group 15 elements are Nitrogen (N), Phosphorus (P), Arsenic (As) and Antimony (Sb). The outer configuration of phosphorus is half filled and hence stable $({s^2}{p^3})$.

For a stable configuration, ionization enthalpy is high. Ionization enthalpy - Energy required to remove an electron from the gaseous atom.

Nitrogen has the highest first ionization enthalpy in group 15.

Down the group, ionization enthalpy decreases.

Down the group, atomic size increases, valence electrons are less attracted to the nucleus. And here, electron removal will become easier. So, ionization enthalpy is also decreases.

The highest value of ionization enthalpy 1402 KJ mol$^{-1}$ corresponds to nitrogen element.

So, the first ionization enthalpy of the element that is a main component of apatite family is option D) 1012 kJ mol$^{-1}$.

Element $E \rightarrow$ Ionisation enthalpy value $\rightarrow 374 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Element $A \rightarrow$ Electron gain enthalpy value $\rightarrow-328 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Element $C \rightarrow$ Electron gain enthalpy value $\rightarrow-325 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}$

Element $D \rightarrow$ Electron gain enthalpy value $\rightarrow-295 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Ionization enthalpy is the energy required to remove an election from an atom or ion in the gas phase. An element with lower ionization enthalpy indicate easier cation formation and this lead to higher ionic character. Electron gain enthalpy is the energy released when an election is added to an atom or ion in the gas phase. A highly negative election gain enthalpy Indicate that the atom readily gains electrons, favouring the formation of anions. This leads to stronger ionic character in compounds. High ionic character generally occurs when the difference in ionisation enthalpy and electron gain enthalpy between the two atoms is large.

For $E A$,

Difference between ionisation enthalpy and electron gain enthalpy

$\begin{aligned} & =374 \mathrm{kJmol}^{-1}-\left(-328 \mathrm{~kJ} \mathrm{~mol}^{-1}\right) \\ & =702 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$

For $E B$,

$\begin{aligned} &\text {Difference between ionisation enthalpy and elation gain enthalpy } \\ & =374 \mathrm{~kJ} \mathrm{~mol}^{-1}-\left(-349 \mathrm{~kJ~mol}^1\right) \\ & =723 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$

For $E C$,

$\begin{aligned} & \text { Difference between ionisation enthalpy and election gain enthalpy } \\ & =374 \mathrm{~kJ} \mathrm{~mol}^{-1}-\left(-325 \mathrm{~kJ} \mathrm{~mol}^{-1}\right) \\ & =699 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$

For $ED$,

$\begin{aligned} & =374 \mathrm{~kJ} \mathrm{~mol}^{-1}-\left(-295 \mathrm{~kJ} \mathrm{~mol}^{-1}\right) \\ & =669 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$

Higher difference between ionisation enthalpy and election gain enthalpy is for compound EB. So, EB has higher ionic character.

The Lower value of difference between ionisation enthalpy and election gain enthalpy is for compound ED. So, ED has lower ionic character. The increasing order of compounds for the difference between ionisation enthalpy and electron gain enthalpy $\to$

$E B< E C < E A< E B$

So, increasing. order of compounds for ionic character:

$E D< E C < E A< E B$

In the decreasing order,

$E B>E A>E C>E D$

Correct option (1)

Statement I : Correct

The radii of isoelectronic species increases in the order $\mathrm{Mg^{2+} < Na^+ < F^- < O^{2-}}$ (number of electrons = 10)

In general, radius of cation is smaller than anion $\mathrm{A^+ < A^-}$ cation with positive charge have higher effective nuclear charge.

Reason : In cations, number of electrons decreases, but number of protons remains the same. This causes a stronger pull on the remaining electrons by the nucleus, resulting in a higher effective nuclear charge.

With increase in effective nuclear charge, radii decreases so, increase in positive charge indicate more number of electron removal and hence increase in effective nuclear charge. As a result, radii decreases. So, Mg$^{2+}$ has lower radii than Na$^+$.

For anion, number of electrons are increased. The electron-electron repulsion weakens the attraction between the nucleus and the electrons, resulting in a decrease in effective nuclear charge. So, effective nuclear charge of O^{2-}$ is lower than F^-$.

So, O$^{2-}$ with lower effective nuclear charge causes longer radii of it.

So, ${F^ - } < {O^{2 - }}$

So, $M{g^{2 + }} < N{a^ + } < {F^ - } < {O^{2 - }}$

Ionic radii of isoelectronic species decreases as nuclear charge increases. The cation with greater +ve charge will have a smaller radius and the anion with the greater $-$ve charge will have a bigger radius for isoelectronic species.

Statement II : Correct

The magnitude of electron gain enthalpy of halogen decreases in the order Cl > F > Br > I.

Electron gain enthalpy is the amount of energy released when an atom accepts the electron from any neutral isolated gaseous atom to form an anion.

Generally, electron gain enthalpy decreases down the group. The exception is for F and cl. Electron gain enthalpy of F is less than Cl. It is due to the comparatively small size of F atom.

For Cl, Br and I, the electron gain enthalpy decreases as go from Cl to I.

So, both the statements are correct.

Law of octaves was arranged in the increasing order of their atomic weight.

Lothar Meyer plotted the physical properties such as atomic volume, melting point and boiling point against atomic weight.

Let's analyze the options based on the periodic trends in atomic radii. Recall two key trends:

Atomic radii decrease as you move from left to right in a period due to increasing effective nuclear charge.

Atomic radii increase as you move down a group because additional electron shells are added.

We'll review each option:

Option A: $\mathrm{Si} > \mathrm{P} > \mathrm{Cl} > \mathrm{F}$

Silicon (Si), phosphorus (P), and chlorine (Cl) are in the same period, with a gradual decrease in atomic radius from left to right.

Fluorine (F) is in an earlier period (it has only 2 electron shells) and is much smaller.

Hence, the order follows the expected trend.

Option B: $\mathrm{Mg} > \mathrm{Al} > \mathrm{C} > \mathrm{O}$

Magnesium (Mg) and aluminum (Al) are period 3 elements, whereas carbon (C) and oxygen (O) belong to period 2.

Period 3 elements generally have larger radii than period 2 counterparts.

The order is consistent: Mg and Al are larger than C and O, and within each period, radii decrease from left to right.

Option C: $\mathrm{Al} > \mathrm{B} > \mathrm{N} > \mathrm{F}$

Aluminum (Al) is in period 3 and is expected to be larger than boron (B), nitrogen (N), and fluorine (F) from period 2.

Within period 2 (B, N, F), the atomic radii decrease from left to right.

Thus, this order is correct as well.

Option D: $\mathrm{Be} > \mathrm{Mg} > \mathrm{Al} > \mathrm{Si}$

Beryllium (Be) is in period 2, while magnesium (Mg), aluminum (Al), and silicon (Si) are in period 3.

Since atomic radii increase with the number of electron shells, Be (with only 2 shells) should be smaller than Mg.

This means that stating $\mathrm{Be} > \mathrm{Mg}$ is incorrect.

Thus, the incorrect decreasing order of atomic radii is given in Option D.

Among $\mathrm{Al}, \mathrm{In}, \mathrm{Tl}, \mathrm{Ge}, \mathrm{Sn}, \mathrm{Pb}$, the metal having highest $\mathrm{IE}_1$ is Ge and lowest $\mathrm{IE}_1$ is $\mathrm{In}$.

Most stable oxidation state of Ge is +4 and In is +3 .

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:

$IE_1 = 800 \, \text{kJ/mol}$

$IE_2 = 2427 \, \text{kJ/mol}$

$IE_3 = 3658 \, \text{kJ/mol}$

$IE_4 = 25024 \, \text{kJ/mol}$

$IE_5 = 32824 \, \text{kJ/mol}$

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 $IE_3$ to $IE_4$ 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).

In group 14, the general trend is that ionization energy decreases down the group because atomic size increases.

Now check each statement carefully.

For Statement (I):

Pb and Sn belong to the same group.

Usually, ionization energy should decrease from Sn to Pb. But Pb shows slightly higher ionization energy than expected because of poor shielding by inner $d$ and $f$ electrons and relativistic effects. Due to this, the outer electrons of Pb are held more strongly.

So, the first ionization energy of Pb is slightly greater than that of Sn.

Statement (I) is true.

For Statement (II):

Ge lies below Si in group 14. As atomic size increases from Si to Ge, the outer electron becomes easier to remove.

Hence, the first ionization energy of Ge is less than that of Si.

So, Statement (II) is false.

Therefore, the correct option is:

$\boxed{\text{Option D}}$

Statement I is true but Statement II is false.

Let's analyze each statement one by one:

$\textbf{A. The properties of elements are function of atomic weights.}$

Originally, elements were arranged by atomic weights, but it was later understood that their properties depend on the atomic number (the number of protons) rather than the atomic weight.

Thus, this statement is NOT true.

$\textbf{B. The properties of elements are function of atomic numbers.}$

The periodic table is arranged by increasing atomic number, and an element’s properties directly relate to its number of protons and the corresponding electron configuration.

This statement is true.

$\textbf{C. Elements having similar outer electronic configurations are arranged in same period.}$

Elements that have similar outer electronic configurations are actually grouped in the same column (group) rather than the same row (period).

Therefore, this statement is NOT true.

$\textbf{D. An element's location reflects the quantum numbers of the last filled orbital.}$

The arrangement of elements corresponds to the electron configurations determined by quantum numbers, especially for the last electron added.

This statement is true.

$\textbf{E. The number of elements in a period is same as the number of atomic orbitals available in energy level that is being filled.}$

In an energy level, the number of atomic orbitals does not directly equal the number of elements in a period. For example, period 2 has 1s filled, and then the 2s and 2p orbitals are filled (a total of 4 orbitals), yet there are 8 elements because each orbital can accommodate 2 electrons.

Hence, this statement is NOT true.

So, the statements that are NOT true about the periodic table are A, C, and E.

Thus, the correct answer is:

Option A: A, C and E Only.

Order of M.P. of group 14: $\mathrm{C}>\mathrm{Si}>\mathrm{Ge}>\mathrm{Pb}>\mathrm{Sn}$

Let's analyze the periods for each element:

Platinum (Pt) has an atomic number of 78, which places it in period 6.

Osmium (Os) has an atomic number of 76, also in period 6.

Iridium (Ir) has an atomic number of 77, which is in period 6 as well.

Palladium (Pd) has an atomic number of 46, placing it in period 5.

Since Palladium is in period 5 while the other elements are in period 6, the element that does not belong to the same period is:

$\textbf{Option D: Palladium.}$

First, let us re-state the two statements clearly:

Statement (I): An element in the extreme left of the periodic table forms acidic oxides.

Statement (II): Acid is formed during the reaction between water and oxide of a reactive element present in the extreme right of the periodic table.

Analyzing Statement (I)

The extreme left of the periodic table corresponds to the alkali metals (Group 1) and alkaline earth metals (Group 2).

These metals typically form basic (or occasionally amphoteric, in the case of some Group 2 elements) oxides, not acidic oxides.

For example, Na₂O, K₂O, MgO, CaO, etc., all form basic solutions (e.g., Na₂O + H₂O → 2 NaOH).

Hence, Statement (I)—that an element in the extreme left forms acidic oxides—is false.

Analyzing Statement (II)

The extreme right of the periodic table corresponds to the nonmetals in Groups 15, 16, 17 (and noble gases in Group 18).

Nonmetal oxides (such as those of sulfur, phosphorus, chlorine) are generally acidic.

When these oxides dissolve in water, they typically form acids.

Example: SO₃ + H₂O → H₂SO₄ (sulfuric acid)

Example: P₂O₅ + 3 H₂O → 2 H₃PO₄ (phosphoric acid)

Example: Cl₂O₇ + H₂O → 2 HClO₄ (perchloric acid)

Thus, Statement (II)—that acid is formed when water reacts with an oxide of a reactive element in the extreme right—is true.

Conclusion

Statement (I) is false.

Statement (II) is true.

Therefore, the correct choice is:

(D) Statement (I) is false but Statement (II) is true.

(A) $\mathrm{Al}^{3+}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{F}^{-}$

All four ions ($\mathrm{Al}^{3+},\mathrm{Mg}^{2+},\mathrm{Na}^{+},\mathrm{F}^{-}$) are isoelectronic (each has 10 electrons). For isoelectronic species, ionic radii decrease as the positive nuclear charge increases, which is why the smallest ion here is $\mathrm{Al}^{3+}$ (Z=13) and the largest is $\mathrm{F}^{-}$ (Z=9). Thus, this ordering is one of increasing ionic radius.

$ \boxed{ (A) \;\longrightarrow\; \text{(IV) Ionic radii} } $

(B) $\mathrm{B}<\mathrm{C}<\mathrm{O}<\mathrm{N}$

Check first ionization enthalpies ($\mathrm{IE}_1$):

B: $\approx 801$\,kJ/mol

C: $\approx 1086$\,kJ/mol

O: $\approx 1314$\,kJ/mol

N: $\approx 1402$\,kJ/mol

Hence, the order of increasing $\mathrm{IE}_1$ is

$ B < C < O < N. $

This matches the given sequence exactly.

$ \boxed{ (B) \;\longrightarrow\; \text{(I) Ionisation enthalpy} } $

(C) $\mathrm{B}<\mathrm{Al}<\mathrm{Mg}<\mathrm{K}$

Consider metallic character (the tendency to lose electrons easily, show metallic properties). Across a period (left to right), metallic character decreases; down a group, it increases.

B (metalloid) has the least metallic character here.

Al (group 13 metal) is more metallic than B.

Mg (group 2 metal) is typically more metallic than Al.

K (group 1 metal) is the most metallic among these.

Thus, $\mathrm{B} < \mathrm{Al} < \mathrm{Mg} < \mathrm{K}$ is an order of increasing metallic character.

$ \boxed{ (C) \;\longrightarrow\; \text{(II) Metallic character} } $

(D) $\mathrm{Si}<\mathrm{P}<\mathrm{S}<\mathrm{Cl}$

Check electronegativities:

Si: $\approx 1.90$

P: $\approx 2.19$

S: $\approx 2.58$

Cl: $\approx 3.16$

They increase in the order

$ \mathrm{Si}<\mathrm{P}<\mathrm{S}<\mathrm{Cl}, $

which matches the given sequence for increasing electronegativity.

$ \boxed{ (D) \;\longrightarrow\; \text{(III) Electronegativity} } $

Final Matching

$ (A) \to (IV),\quad (B) \to (I),\quad (C) \to (II),\quad (D) \to (III). $

Looking at the choices given:

Option A: $(A)-(IV), (B)-(I), (C)-(II), (D)-(III)$

This is exactly what we found.

Answer: Option A

Let’s compare the known electronegativities (Pauling scale) of the elements in each option:

F ~ 3.98

O ~ 3.44

N ~ 3.04

Cl ~ 3.16

C ~ 2.55

B ~ 2.04

S ~ 2.58 (sometimes listed as 2.5–2.58)

Si ~ 1.90

Al ~ 1.61

Be ~ 1.57

Mg ~ 1.31

Now, check each statement:

Option A: $\mathrm{S}<\mathrm{Cl}<\mathrm{O}<\mathrm{F}$

S (2.58) < Cl (3.16) < O (3.44) < F (3.98)

This order is correct.

Option B: $\mathrm{Al}<\mathrm{Si}<\mathrm{C}<\mathrm{N}$

Al (1.61) < Si (1.90) < C (2.55) < N (3.04)

This order is correct.

Option C: $\mathrm{Al}<\mathrm{Mg}<\mathrm{B}<\mathrm{N}$

Actual electronegativities are Al (1.61) and Mg (1.31).

The statement says “Al < Mg,” which would mean 1.61 < 1.31, which is wrong.

Therefore, Option C is incorrect.

Option D: $\mathrm{Mg}<\mathrm{Be}<\mathrm{B}<\mathrm{N}$

Mg (1.31) < Be (1.57) < B (2.04) < N (3.04)

This order is correct.

Answer: Option C is the incorrect order.

The correct order of atomic radii is given in (ii) and (iv).

The remaining are incorrect. Their correct form are $\mathrm{Li}>\mathrm{Cl}>\mathrm{F}$,

$\mathrm{Eu}>\mathrm{Sm}>\mathrm{Tm}$

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

Let's determine the block for each element provided:

(A) Cd (Cadmium)

Cadmium (atomic number 48) is located in Group 12, Period 5 of the periodic table.

Its electron configuration is [Kr] 4d¹⁰ 5s².

Elements in Groups 3 through 12, where the d-orbitals are being filled, are known as d-block elements. Cadmium fits this description.

So, (A) Cd matches with (III) d-block.

(B) Eu (Europium)

Europium (atomic number 63) is a lanthanide element.

Lanthanides and actinides are typically placed below the main body of the periodic table and comprise the f-block elements, where the f-orbitals are being filled.

Its electron configuration is [Xe] 4f⁷ 6s².

So, (B) Eu matches with (I) f-block.

(C) Se (Selenium)

Selenium (atomic number 34) is located in Group 16, Period 4 of the periodic table.

Its electron configuration is [Ar] 3d¹⁰ 4s² 4p⁴.

Elements in Groups 13 through 18, where the p-orbitals are being filled, are known as p-block elements. Selenium fits this description.

So, (C) Se matches with (IV) p-block.

(D) Ba (Barium)

Barium (atomic number 56) is located in Group 2, Period 6 of the periodic table.

Its electron configuration is [Xe] 6s².

Elements in Group 1 (alkali metals) and Group 2 (alkaline earth metals), where the s-orbitals are being filled, are known as s-block elements. Barium fits this description.

So, (D) Ba matches with (II) s-block.

Combining the matches:

A - III

B - I

C - IV

D - II