Periodic Table & Periodicity

First, find the number of moles of lithium. We are given the mass of lithium as $3.5 \, \text{mg} = 3.5 \times 10^{-3} \, \text{g}$ and the molar mass as $7 \, \text{g mol}^{-1}$.

So,

$\text{Number of moles of } \mathrm{Li} = \frac{3.5 \times 10^{-3}}{7} = 5 \times 10^{-4}$

The process $\mathrm{Li(g)} \rightarrow \mathrm{Li^{2+}(g)}$ involves removing two electrons. The energy required for this is equal to the sum of first and second ionization enthalpies of lithium.

Total ionization energy per mole of lithium:

$520 + 7297 = 7817 \, \mathrm{kJ\,mol^{-1}}$

Now, calculate the energy for $5 \times 10^{-4}$ moles of lithium:

$\text{Energy required} = 5 \times 10^{-4} \times 7817 = 3.9085 \, \mathrm{kJ}$

Therefore, the total energy required is approximately $4 \, \mathrm{kJ}$.

The IUPAC name "Unununnium" is derived from a systematic element naming method based on the atomic number of elements, primarily used for elements that have not yet been discovered or were not yet officially named at the time. The name "Unununnium" is composed of Latin and Greek numerical roots and can be broken down into "Un-un-un-nium" where "un" represents one and "nnium" is the suffix used for elements. The full name "Unununnium" hence corresponds to the temporary elemental name for an element with the atomic number 111 (one-one-one).

The elements in the periodic table are arranged by their atomic number and grouped by their electron configurations, particularly the valence electrons which determine their chemical properties. Each column of the periodic table is known as a group. The atomic number 111 element, which was named "Unununnium" before its official name "Roentgenium (Rg)" was adopted, is in the same group as copper (Cu), silver (Ag), and gold (Au). This group is group 11 of the periodic table. Therefore:

$ n = 11 $

$\mathrm{N_2O_3,NO_2,Cl_2O_7,SO_2}$ are acidic in nature.

An amphoteric compound is a molecule or ion that can react with both as an acid or as a base.

BeO = Amphoteric

BaO = Basic

Be(OH)₂ = Amphoteric

Sr(OH)₂ = Basic

Both beryllium compound BeO and Be(OH)₂ are amphoteric in nature while compound BaO and Sr(OH)₂ are basic in nature, they form alkaline solution in H₂O.

Historically, the systematic (IUPAC temporary) name "Unnilunium" was used for the element with the atomic number 101, which is now officially named Mendelevium. Thus,

$ \boxed{101} $

By observing the values of different ionisation energies, I₁, I₂ and I₃ for
atomic number (n + 2), it is observed that there is very large difference between the second ionisation energy and first ionisation energy (I₂ >> I₁).

This indicates that number of valence shell electrons is 1 and atomic number (n + 2) should be an alkali metal.

Also for atomic number (n + 3), I₃ >> I₂. This indicates that it will be an alkaline earth metal which suggests that atomic number (n + 1) should be a noble gas and atomic number (n) should belong to halogen family. Since, n < 10; hence, n = 9 (F atom)

Answer: 2

Detailed Reasoning:

We are looking for elements from the given list that exhibit only one non-zero oxidation state. The elements given are:

O (Oxygen)

Cl (Chlorine)

F (Fluorine)

N (Nitrogen)

P (Phosphorus)

Sn (Tin)

Tl (Thallium)

Na (Sodium)

Ti (Titanium)

Let's analyze each:

Oxygen (O):

Oxygen typically shows multiple non-zero oxidation states, such as -2 (most compounds), -1 (peroxides), and even positive oxidation states in compounds like OF₂ (+2). Hence, it has more than one non-zero oxidation state.

Chlorine (Cl):

Chlorine exhibits a wide range of oxidation states: -1 (HCl), +1 (HOCl), +3 (HClO₂), +5 (HClO₃), and +7 (HClO₄). Clearly multiple non-zero states.

Fluorine (F):

Fluorine is the most electronegative element and almost always exhibits an oxidation state of -1 in its compounds. It does not show other stable non-zero oxidation states. Thus, fluorine has only one non-zero oxidation state: -1.

Nitrogen (N):

Nitrogen shows many oxidation states: from -3 (ammonia, NH₃) to +5 (in nitrates, NO₃⁻), passing through several intermediate states (+1, +2, +3, +4). Many non-zero oxidation states.

Phosphorus (P):

Phosphorus can exist in -3 (PH₃), +3 (PCl₃), and +5 (PCl₅) states. Multiple non-zero oxidation states.

Tin (Sn):

Tin commonly shows +2 and +4 oxidation states. Hence more than one non-zero oxidation state.

Thallium (Tl):

Thallium typically shows +1 and +3 oxidation states. Again, multiple non-zero oxidation states.

Sodium (Na):

Sodium, in stable compounds, is almost always present as Na⁺ (oxidation state +1). It does not commonly display any other non-zero oxidation state. So sodium has only one non-zero oxidation state.

Titanium (Ti):

Titanium can exhibit +2, +3, and +4 oxidation states, among others. Multiple non-zero oxidation states.

Conclusion:

Only fluorine (F) and sodium (Na) have exactly one non-zero oxidation state.

Number of such elements = 2

When comparing the basic character of oxides, we must consider several factors, including the position of the metal in the periodic table and the nature of the oxide (whether it is ionic or covalent). Basic character generally increases as you move down a group in the periodic table due to the decrease in electronegativity and the increase in metallic character. Basic character also tends to be higher in oxides of metals as compared to those of nonmetals.

We have the following compounds to consider: MgO, SrO, K₂O, NiO, and Cs₂O. Oxides of alkali metals (K₂O, Cs₂O) and alkaline earth metals (MgO, SrO) tend to be more basic than oxides of transition metals (NiO). Within each group (alkali and alkaline earth metals), the basic character increases as we move down the group. This is due to the larger atomic size and reduced ionic character, making the oxide ion ($ O^{2-} $) more able to accept a proton.

Among the given oxides:

• Alkali metal oxides (K₂O and Cs₂O) are generally more basic than alkaline earth metal oxides.
• In the alkaline earth group, beryllium and magnesium oxides behave more amphoteric and less basic compared to oxides of calcium, strontium, and barium. So, SrO is more basic than MgO.
• For alkali metals, cesium oxide (Cs₂O) is more basic than potassium oxide (K₂O) due to being further down the group in the periodic table.
• Nickel oxide (NiO), being a transition metal oxide, is generally less basic compared to oxides from either the alkali or alkaline earth metals.

Therefore, the increasing order of basic strength for these oxides is:

$ \text{NiO} < \text{MgO} < \text{SrO} < \text{K}_2\text{O} < \text{Cs}_2\text{O} $

This order reflects moving from transition metal oxides to more typical alkaline earth metal oxides, and then to more basic alkali metal oxides, with the basicity increasing significantly as we move from lighter to heavier elements within each metal group.

To arrange the ions Li⁺, Mg²⁺, K⁺, and Al³⁺ in order of their increasing radii, we need to consider several factors, primarily the atomic number (i.e., the number of protons in the nucleus), the number of electrons, and the effective nuclear charge experienced by the outer electrons.

Here is a brief overview of each ion:

• Li⁺: This is a lithium ion with one electron removed, leaving it with 2 electrons. The original lithium atom has an atomic number of 3. The loss of one electron increases the effective nuclear charge experienced by the remaining electrons, pulling them closer and reducing the size of the ion compared to the neutral atom.
• Mg²⁺: This is a magnesium ion with two electrons removed, leaving it with 10 electrons. Its atomic number is 12. This ionization results in a smaller ionic radius, as the loss of two electrons greatly increases the effective nuclear charge, causing the electrons to be held more tightly and closer to the nucleus.
• K⁺: This is a potassium ion with one electron removed, leaving it with 18 electrons. It has an atomic number of 19. Like Li⁺, losing one electron increases its effective nuclear charge, but the effect is less pronounced compared to Li⁺ and Mg²⁺ because potassium starts with more electron shells.
• Al³⁺: This is an aluminum ion with three electrons removed, leaving it with 10 electrons. Its atomic number is 13. Removing three electrons substantially increases effective nuclear charge, thereby significantly decreasing the ionic radius.

When arranging ions in order of increasing ionic radius, it's also critical to consider periodic trends:

1. Within a period, as the nuclear charge increases and the number of electrons remains similar (across cations), the radius decreases.
2. Across periods, ions with more electron shells tend to be larger if the number of missing electrons (from being cations) is the same or similar.

Hence, considering that Li⁺, Mg²⁺, and Al³⁺ are found in Period 2, and K⁺ is in Period 4, we can ascertain that K⁺ will generally be larger than those in Period 2. Within Period 2, the increasing removal of electrons (with Al³⁺ losing most) implies that Al³⁺ should be the smallest, followed by Mg²⁺ and then Li⁺.

Thus, the ions in order of increasing radii are:

$ \text{Al}^{3+} < \text{Mg}^{2+} < \text{Li}^+ < \text{K}^+ $

This arrangement follows both the periodic trend of decreasing radius with increasing nuclear charge (and ionic charge) in a period and the general increase in size with additional electron shells across periods.

To arrange ions in order of increasing ionic size, we need to consider two main factors: the nuclear charge of the ions and the number of electrons each ion possesses. A general rule is that within an isoelectronic series (ions having the same number of electrons), an ion with a higher nuclear charge will be smaller because of the stronger attraction between the nucleus and the electron cloud.

First, identify the total number of electrons in each species for comparison:

• N^3- would have 7 (number of protons in N) + 3 = 10 electrons.
• Na⁺ would have 11 (number of protons in Na) - 1 = 10 electrons.
• F^- would have 9 (number of protons in F) + 1 = 10 electrons.
• O^2- would have 8 (number of protons in O) + 2 = 10 electrons.
• Mg²⁺ would have 12 (number of protons in Mg) - 2 = 10 electrons.

All these ions are isoelectronic, having the same number of electrons, but the nuclear charge varies, affecting their sizes. The ionic radius decreases typically with an increase in nuclear charge due to the increased pull on the electrons:

• N^3- (10 protons, most negative, least nuclear pull relative to its electron cloud)
• O^2- (8 protons)
• F^- (9 protons)
• Na⁺ (11 protons)
• Mg²⁺ (12 protons, most positive, strongest nuclear pull)

Thus, in increasing order of ionic size, the sequence is:

$\text{Mg}^{2+} < \text{Na}^+ < \text{F}^- < \text{O}^{2-} < \text{N}^{3-}$

This arrangement is due to the increasing effect of the nuclear charge in pulling the electron cloud closer as the charge becomes more positive, and the increasing number of electrons or increased negative charge would generally cause an increase in the ionic radius.

To arrange the ions Cl^-, S^2-, Ca²⁺, and the neutral atom Ar in order of increasing size, we need to consider several factors including nuclear charge, number of electrons, and electron-electron interactions in the electron shells.

1. Atomic and Ionic Sizes: In general, neutral atoms and ions with more electrons and similar nuclear charge have larger sizes. This is due to increased electron-electron repulsion within the outer shells.

2. Isoelectronic Species: Cl^-, Ca²⁺, and Ar are isoelectronic, meaning they each have the same number of electrons (18 electrons), but different nuclear charges. A higher nuclear charge (more protons in the nucleus) pulls the electron cloud closer, decreasing the size of the ion or atom. In this case, Ar (18 protons) has a higher nuclear charge than Cl^- (17 protons) and Ca²⁺ (20 protons) has the highest.

3. Comparing a Non-Isoelectronic Species: S^2- has 18 electrons, the same as Cl^-, Ca²⁺, and Ar, but sulfur typically has fewer protons (16 protons) than Argon. However, the addition of two electrons to form S^2- significantly increases its radius compared to the neutral sulfur atom due to increased electron-electron repulsion outweighing the nuclear charge.

Ordered by Size: Smaller ions/atoms have higher nuclear charge relative to the number of electrons. Here’s the order:

• Ca²⁺ (smallest): higher nuclear charge tightly holds the electrons.


• Ar: similar electron count but slightly less nuclear charge than Ca²⁺.


• Cl^-: one fewer proton than Ar, leading to less pull on the same number of electrons, resulting in a slightly larger size.


• S^2- (largest): despite having an equal number of electrons as the others, the significantly lower nuclear charge (16 protons) allows more spread out electron cloud.

In conclusion, the order of increasing size will be: $ Ca^{2+} < Ar < Cl^{-} < S^{2-} $. S^2- is the largest due to its lower nuclear charge compared to its electron count, resulting in greater electron cloud expansion.

To determine the decreasing order of ionic size for the ions Mg²⁺, O^2-, Na⁺, and F^-, we need to consider several factors such as nuclear charge, electrons’ configurations, and their positions in the periodic table.

Firstly, when ions have the same electronic configuration (which we can think of as isoelectronic), the nuclear charge and its effective pull on the electrons will determine their sizes. A higher positive charge (more protons in the nucleus) results in a smaller ionic radius because the protons pull the electron cloud more tightly towards the nucleus.

The ions Mg²⁺, O^2-, Na⁺, and F^- are all approximately isoelectronic, having the same number of electrons as the noble gas Ne (10 electrons). Thus, the order of their ionic sizes will mainly depend on their nuclear charge:

• O^2- (8 protons)
• F^- (9 protons)
• Na⁺ (11 protons)
• Mg²⁺ (12 protons)

Higher proton count correlates with greater effective nuclear charge, leading to a more strongly pulled-in electron cloud, and thus a smaller ionic radius.

Therefore, the decreasing order of ionic size from largest to smallest is:

$ \text{O}^{2-} > \text{F}^{-} > \text{Na}^{+} > \text{Mg}^{2+} $

O^2-, having the fewest protons among these ions, exhibits the largest ionic size due to a lesser effective nuclear pull on its electron cloud, compared to the other ions with more protons. Conversely, Mg²⁺ is the smallest as it has the highest number of protons exerting the strongest pull on the same number of electrons.

When evaluating the acidic properties of compounds, we must consider the nature of the oxides formed by different elements. The acidity of an oxide depends on the position of the respective element in the periodic table, particularly their electronegativity and the period to which they belong.

Oxides can be broadly categorized into four types based on the nature of the elements forming them:

• Metallic oxides (usually basic)
• Non-metallic oxides (usually acidic)
• Amphoteric oxides (show both acidic and basic behavior)
• Neutral oxides (do not show acidic or basic behavior)

Let's analyze the given compounds:

• ZnO: Zinc oxide is amphoteric. It reacts with both acids and bases, but tends more towards basic than acidic properties.
• Na₂O₂: Sodium peroxide is generally a basic oxide, but it can also react with water to form hydroxide ions, indicating it is less acidic than ZnO.
• MgO: Magnesium oxide is a basic oxide. It reacts with acids to form salts and water but does not react with bases.
• P₂O₅: Diphosphorus pentoxide is strongly acidic. It reacts readily with water to form phosphoric acid.

Based on the typical behavior of their chemical classification, here is the order of increasing acidic property:

1. Na₂O₂ (mostly basic)
2. MgO (basic)
3. ZnO (amphoteric, weakly acidic)
4. P₂O₅ (strongly acidic)

The increasing acidic behavior aligns with moving from basic, through amphoteric, to strongly acidic oxides.

On the Mulliken scale, the average of ionization potential and electron affinity is known as electronegativity. In essence, the Mulliken electronegativity scale is defined by the formula:

$ \chi_{Mulliken} = \frac{I + E}{2} $

Where:

• $ \chi_{Mulliken} $ is the Mulliken electronegativity,
• $ I $ is the ionization potential (the energy required to remove an electron from an atom in the gaseous state),
• $ E $ is the electron affinity (the energy released when an electron is added to a neutral atom in the gaseous state).

This formula essentially provides a measure of how strongly an atom in a molecule attracts electrons towards itself. The concept of electronegativity is crucial in understanding chemical bonding, particularly the polarity of bonds and molecular structures. By using the average of ionization potential and electron affinity, Mulliken's approach provides a quantitative way of assessing an atom's ability to attract shared electrons in a chemical bond.

The energy released when an electron is added to a neutral gaseous atom is called the electron affinity of the atom. Electron affinity is a measure of the energy change that occurs when an electron is added to a neutral atom in the gas phase to form a negative ion. This property is important in understanding the chemical behavior of atoms. It can either release energy, indicated by a negative value, meaning the process is exothermic, or absorb energy, indicated by a positive value, indicating an endothermic process.