Trends In Periodic Table

Let's evaluate each option:

• (A) The lowest oxidation state shown by them is +2: This is incorrect. The lowest oxidation state of elements like Scandium (Sc) is +3, not +2.
• (B) 4s orbital is completely filled in the ground state: This is incorrect. The 4s orbital is not completely filled in all elements from Sc to Cu. For example, Ti has only one electron in the 4s orbital in the ground state.
• (C) 3d orbital is not completely filled in the ground state: This is incorrect. The 3d orbital is not completely filled in many elements from Sc to Cu in their ground state.
• (D) The ions in +2 oxidation states are paramagnetic: This is correct. Most of the transition metals from Sc to Cu, when in the +2 oxidation state, have unpaired electrons in their d orbitals, making them paramagnetic.

Therefore, the correct answer is (D) : The ions in +2 oxidation states are paramagnetic.

Let's break down each option one by one to identify the one that does not represent the correct property.

Option (A) – Ionic size:
The ionic size generally increases with the increasing atomic number when ions are in the same oxidation state.
- has the smallest ionic radius, followed by , and then .
- This order is correct for ionic sizes, so option (A) is correct.

Option (B) – Number of oxidation states:
- Titanium (Ti) typically has oxidation states of +2 and +4.
- Vanadium (V) can exhibit +2, +3, +4, and +5 oxidation states.
- Manganese (Mn) shows a broader range of oxidation states from +2 to +7.
- The increasing number of oxidation states from Ti to Mn is correct. Therefore, option (B) is correct.

Option (C) – Paramagnetic behavior:
- has 4 unpaired electrons, has 5 unpaired electrons, and has 4 unpaired electrons.
- This order is consistent with paramagnetic behavior, where more unpaired electrons correspond to stronger paramagnetism.
- Therefore, option (C) is correct.

Option (D) – Density:
- The density of Scandium (Sc) is about 2.985 g/cm³, Chromium (Cr) is about 7.19 g/cm³, and Iron (Fe) is about 7.87 g/cm³.
- This shows that the density of Fe is higher than Cr, and Cr is higher than Sc, which makes the order incorrect.
- Therefore, option (D) is incorrect.

Conclusion:
The correct answer is (D), as it does not represent the correct order of density for the given elements.

1. Analyze the Periodic Trends

The given sequence represents hydrides of Group 15 elements (Nitrogen family): , , , , .

- Thermal Stability: As we move down the group, the size of the central atom increases. This leads to a longer and weaker M-H bond (where M is the Group 15 element). As the bond weakens, it becomes easier to break the bond, and hence the thermal stability decreases.

- Reducing Property: As the M-H bond weakens down the group, it becomes easier for the hydride to lose hydrogen atoms and act as a reducing agent. Therefore, the reducing property increases down the group.

- Bond Angle: As the electronegativity difference between the central atom and hydrogen decreases down the group, the bond pair electrons are less attracted to the central atom, and the bond angle decreases. This also affects the contribution of 's' character to bonding.

- Acidic Character: The acidic character increases down the group because the M-H bond strength decreases, making it easier to release ions.

2. Compare the trends with the given sequence
The given sequence shows a decreasing trend.


Now, let's check which property matches this trend:

- (A) Reducing property: The reducing property increases down the group, which is opposite to the given trend.

- (B) Thermal stability: The thermal stability decreases down the group, which matches the given trend.

- (C) Bond angle: The bond angle decreases down the group, which matches the given trend.

- (D) Acidic character: The acidic character increases down the group, which is opposite to the given trend.

Final Answer:
The correct answer is both (B) : Thermal stability and (C) : Bond angle

1. Understand Ionization Enthalpy
Ionization enthalpy is the energy required to remove an electron from an isolated gaseous atom or ion. Successive ionization enthalpies (first, second, third, etc.) refer to the removal of subsequent electrons. The third ionization enthalpy is the energy required to remove the third electron.

2. Consider the Electronic Configurations

Alkali Metals (Group 1): General electronic configuration is ns¹. After removing one electron (first ionization), they attain a noble gas configuration. Removing a second electron requires significantly more energy (very high second ionization enthalpy). Removing a third electron would be even harder, but the third ionization energy is still extremely high.

Alkaline Earth Metals (Group 2): General electronic configuration is ns². After removing two electrons (second ionization), they attain a noble gas configuration. Removing a third electron would require breaking this stable noble gas configuration, requiring a very high amount of energy.

Chalcogens (Group 16): General electronic configuration is ns²np⁴. They need two more electrons to attain a noble gas configuration. Removing three electrons will not result in a noble gas configuration, and the ionization enthalpies will be lower than for alkali and alkaline earth metals after they have achieved noble gas configurations.

Pnictogens (Group 15): General electronic configuration is ns²np³. They need three more electrons to attain a noble gas configuration. Removing three electrons will result in a noble gas configuration, but with higher nuclear charge than alkali and alkaline earth metals.

3. Analyze the Third Ionization Enthalpy
Alkali metals will have extremely high second and third ionization energies due to the stable noble gas configuration they attain after the first ionization. Alkaline earth metals will have high third ionization energies since after two ionizations they reach a stable configuration. However, the increase is much more prominent in alkaline earth metals because they attain the noble gas configuration only after removing 2 electrons. So the third ionization involves removing an electron from a noble gas configuration.

Final Answer:
(B) Alkaline earth metals

The correct order of first ionization enthalpy for the elements carbon (C), boron (B), beryllium (Be), and lithium (Li) is indeed:

C. Li < B < Be < C

Here's a brief explanation of why this is the case:

Lithium (Li): Being an alkali metal, it has a single electron in its outermost shell (ns configuration). This electron is relatively loosely held and is easily removed, making Li have a low first ionization enthalpy.

Boron (B): Boron is a metalloid and has two electrons in its outermost shell (ns np configuration). The first ionization involves removing an electron from the np orbital, which is more tightly bound than the ns orbital in Li, thus requiring more energy.

Beryllium (Be): Beryllium is an alkaline earth metal with two electrons in its outermost shell (ns configuration). The first ionization enthalpy of Be is higher than B because the two electrons are in the same s orbital and are more strongly attracted to the nucleus due to the increased nuclear charge compared to Li.

Carbon (C): Carbon is a non-metal with four electrons in its outermost shell (ns np configuration). The first ionization enthalpy of carbon is higher than Be because it involves breaking a half-filled np orbital, which is a more stable configuration and thus requires more energy to remove an electron.

The trend generally follows that ionization enthalpy increases across a period from left to right due to increasing nuclear charge, which holds the electrons more tightly. However, there are exceptions due to electron configurations and orbital stability, as seen with Be and B.

Understanding Basicity:

• Basicity refers to the ability of a substance to donate a lone pair of electrons.
• In the case of Group 15 hydrides (NH3, PH3, AsH3, SbH3), basicity is related to the availability of the lone pair on the central atom.

Factors Affecting Basicity:

• Electronegativity: As we move down the group, the electronegativity of the central atom decreases. This means the lone pair becomes more available for donation, which would seem to increase basicity. However, this is not the dominant factor.
• Bond Dissociation Enthalpy: As we move down the group, the size of the central atom increases, and the bond length increases. This results in a decrease in bond dissociation enthalpy, making it harder for the central atom to hold onto its lone pair.
• Size: The size of the central atom increases as we move down the group. The larger the atom, the more diffuse the lone pair becomes, and the less available it is for donation.

Trend:

• The dominant factor is the size of the central atom.
• Therefore, as we move down the group, the basicity of the hydrides decreases.

Correct Order:

• The correct decreasing order of basicity is: NH3 > PH3 > AsH3 > SbH3.

Conclusion:

The correct answer is Option 4.

We are comparing the melting points of transition elements based on their atomic structures and bonding characteristics. Generally, the melting points of transition elements are influenced by the number of unpaired electrons and the strength of metallic bonds.

Let's analyze the given options:

Option 1: V > Cr

This is incorrect. Chromium (Cr) has a higher melting point than Vanadium (V), due to the unique electron configuration of Cr, which results in a more stable metallic bonding.

Option 2: Cr > Mn

This is correct. Chromium has a higher melting point than Manganese because Cr has a more stable electronic configuration and stronger bonding.

Option 3: Mn > Fe

This is incorrect. Manganese (Mn) has a lower melting point than Iron (Fe). Iron has a stronger metallic bond compared to Manganese.

Option 4: Ti > V

This is incorrect. Titanium (Ti) has a lower melting point than Vanadium (V), as the metallic bonding in Vanadium is stronger.

Conclusion:

The correct answer is: Option 2: Cr > Mn.