Periodic Trends In Properties Of Elements

To determine the correct answer, let's analyze the given statements:

1. Statement I: Gallium is used in the manufacturing of thermometers.
   • Gallium is indeed used in the manufacturing of thermometers. This is because gallium has a low melting point of about 29.76°C, which is close to room temperature. It can exist as a liquid at slightly above room temperature, making it useful in thermometers.
2. Statement II: A thermometer containing gallium is useful for measuring the freezing point (256 K) of brine solution.
   • The freezing point of brine solution is approximately -18°C, which corresponds to 255 K (or 256 K when rounded). Gallium, with a melting point of approximately 29.76 °C (or around 303 K), would be solid at this temperature. Therefore, a gallium thermometer cannot measure temperatures as low as 256 K since gallium would be in a solid state. Therefore, this statement is incorrect.

Based on the analysis above, the correct interpretation is:

• Statement I is true but Statement II is false.

Thus, the answer is: Statement I is true but Statement II is false.

To determine the correct orders among the given chemical properties of the elements B, Al, Ga, In, and Tl, let's analyze each property step-by-step:

1. Atomic Radius:
   The atomic radius generally increases down a group in the periodic table due to the addition of electron shells. Thus, the expected order should be: . This order is correct.
2. Electronegativity:
   Electronegativity typically decreases as you move down a group because the distance between the nucleus and valence electrons increases. The order given is: . For main group elements, Boron being the uppermost element should have the highest electronegativity among these. Thus, this order is incorrect.
3. Density:
   Density generally increases down the group for metals due to increasing atomic mass. However, given that there are exceptions due to the unique property of each element, the provided order: conflicts with expected trends and known data, making it incorrect.
4. 1st Ionisation Energy:
   Ionisation energy generally decreases down a group due to increasing atomic size. However, within the same group, exceptions can occur. The order given is: . Boron has the highest ionisation energy as expected. This order is correct.

Based on the given options, the correct sequences are: B and D Only

The question is about determining which element has the lowest first ionization enthalpy from the given atomic numbers. Ionization enthalpy is the energy required to remove an electron from a gaseous atom or ion. Typically, elements with a lower ionization enthalpy can lose electrons more easily.

Step-by-step Explanation:

1. The atomic numbers given are: 87, 19, 32, 35.
2. To find the corresponding elements, we refer to the periodic table:
   • Atomic number 87: Francium (Fr)
   • Atomic number 19: Potassium (K)
   • Atomic number 32: Germanium (Ge)
   • Atomic number 35: Bromine (Br)
3. Ionization enthalpy tends to decrease as we move down a group in the periodic table because the outermost electrons are further away from the nucleus and are more shielded by inner electrons.
4. Among these elements:
   • Francium (Fr) is an alkali metal and belongs to Group 1, which is known for having low ionization energies.
   • Potassium (K) is also in Group 1 but is higher up in the column compared to Francium.
   • Germanium (Ge) and Bromine (Br) are further right in the periodic table and have higher ionization enthalpy compared to Group 1 elements.
5. Thus, Francium, being lower in the periodic table specifically in Group 1, has the lowest ionization enthalpy among the options provided.

Conclusion: The element with the lowest first ionization enthalpy among the options is Francium (Atomic number 87).

To answer the question, we first need to analyze each statement separately in the context of chemistry, specifically the properties of hydrides within the chalcogen group.

1. Statement I: is more acidic than

The acidity of hydrides in the group 16 elements (chalcogen group) increases as we move down the group. Therefore, the correct trend in acidity for hydrides (where is a chalcogen) is: . Hence, is more acidic than due to an increase in atomic size and decrease in bond strength, which facilitates the release of protons.

Therefore, Statement I is false.

1. Statement II: has a higher bond enthalpy for dissociation than

Bond enthalpy typically decreases as we move down the group due to the increasing size of the atoms, which results in weaker bonds. Therefore, , which is higher in the group than , will indeed have a higher bond enthalpy (i.e., stronger bonds) than .

Thus, Statement II is true.

Considering the above analysis, the correct option is that Statement I is false but Statement II is true.

The large difference between the melting and boiling points of oxygen (O₂) and sulfur (S₈) is best explained by their atomicity.

1. Atomicity: Atomicity refers to the number of atoms in a molecule. Oxygen exists as a diatomic molecule (O₂), meaning it consists of two atoms. In contrast, sulfur commonly exists as an octatomic molecule (S₈), consisting of eight atoms.
2. Intermolecular Forces: The structure of sulfur as S₈ makes it larger and more complex, which results in stronger van der Waals forces of attraction between its molecules compared to O₂. These stronger forces require more energy to overcome, leading to higher melting and boiling points for sulfur compared to oxygen.
3. Explanation of Options:
   • Atomic Size: While size affects melting and boiling points, the difference in atomicity is more significant in this context.
   • Electronegativity: Electronegativity does not directly affect the melting or boiling points in this case.
   • Electron Gain Enthalpy: This property is related to the energy change when an electron is added to an atom and is irrelevant to the discussion of melting and boiling points.
4. Conclusion: The correct explanation for the large difference between the melting and boiling points of oxygen and sulfur is their atomicity. The molecular forms (O₂ vs. S₈) account for the difference in the strength of intermolecular forces, thus affecting their melting and boiling points.

The question requires us to identify the element from Group 14 of the periodic table that possesses the lowest melting point, based on the given atomic numbers. The atomic numbers listed correspond to specific elements as follows:

• Carbon (C) - Atomic number 6
• Silicon (Si) - Atomic number 14
• Tin (Sn) - Atomic number 50
• Lead (Pb) - Atomic number 82

Group 14 elements include Carbon, Silicon, Germanium, Tin, and Lead. One distinctive property of these elements is their varying melting points. Generally, upon moving down the group, the melting points tend to decrease with some local exceptions due to structural differences.

Let's compare the melting points based on typical properties of the elements:

• Carbon (C) has a very high melting point as it exists as graphite or diamond.
• Silicon (Si) has a high melting point due to its covalent bonding structure.
• Tin (Sn) exists in several allotropes, one of which, called grey tin, has a low melting point.
• Lead (Pb) has a relatively low melting point but still higher than grey tin.

Among the given elements, Tin has the allotrope with the lowest melting point (especially grey tin, which is metallic and has a lower melting point compared to other elements mentioned).

Thus, the element with the lowest melting point among the given atomic numbers is Tin (Atomic number 50).

Correct Answer: 50

To determine the elements of Group 13 with the highest and lowest first ionization enthalpies, we need to understand the concept of ionization enthalpy and the trends in the periodic table.

Ionization Enthalpy: Ionization enthalpy is the energy required to remove the outermost electron from an isolated gaseous atom to form a cation. In general, the ionization enthalpy depends on factors such as the atomic size, nuclear charge, and electron shielding effect.

Trends in Group 13: As we move down the group in the periodic table, the atomic size increases due to the addition of new electron shells, which leads to an increase in distance between the nucleus and the outermost electron. This generally results in a decrease in ionization enthalpy down the group. However, due to the poor shielding effect of d and f orbitals (present in heavier elements like Ga and Tl), there are some exceptions to this trend.

Let's analyze the elements of Group 13:

• Boron (B): Being the first element of Group 13, Boron has the smallest atomic size and highest nuclear charge among its group, which leads to the highest ionization enthalpy.
• Gallium (Ga): Although it's lower in the group than Aluminum, the poor shielding provided by the d-electrons results in a slightly higher ionization enthalpy than expected.
• Indium (In) and Thallium (Tl): Both these elements have lower ionization enthalpies compared to B and Ga due to the increased size and effective nuclear charge.

From this analysis, we can conclude that:

• The element with the highest ionization enthalpy in Group 13 is Boron (B).
• The element with the lowest ionization enthalpy, among the provided options, is Gallium (Ga) due to the d-block contraction effect.

Therefore, the elements of Group 13 with the highest and lowest first ionization enthalpies are Boron (B) and Gallium (Ga), respectively.

The question involves two statements about the properties of elements in Group 13 and Group 14 of the periodic table. Let's analyze each statement individually to determine their correctness and to substantiate the given answer.

1. Statement I: The first ionisation enthalpy of group 14 elements is higher than the corresponding elements of group 13.
   • Ionisation enthalpy is the energy required to remove the outermost electron from an atom in its gaseous state. It is influenced by atomic size, nuclear charge, and electronic configuration.
   • Group 14 elements (e.g., Carbon, Silicon) have more protons and a greater effective nuclear charge compared to Group 13 elements (e.g., Boron, Aluminium) for the same period, leading to a higher effective hold on their valence electrons. This results in higher ionisation enthalpy.
   • Therefore, the first ionisation enthalpy of Group 14 elements is indeed higher than that of Group 13 elements. Statement I is correct.
2. Statement II: Melting points and boiling points of group 13 elements are in general much higher than those of the corresponding elements of group 14.
   • The melting and boiling points are determined by the type of bonding and structure of the elements.
   • Group 14 elements typically have a covalent network structure, which usually leads to higher melting and boiling points (as seen in Carbon (diamond) and Silicon) compared to metallic bonding in Group 13 elements.
   • For example, Carbon in the form of diamond has one of the highest melting points, much higher than Boron or Aluminium.
   • Thus, generally, the melting and boiling points of Group 14 elements are higher compared to those in Group 13. Statement II is incorrect.

Based on our analysis, we conclude: Statement I is correct but Statement II is incorrect. Hence, the correct option is:

Statement I is correct but Statement II is incorrect