D And F Block Elements

To determine which pair of transition metal ions is colourless, we need to understand the concept of colour in transition metals. The colour in transition metal ions is primarily due to the electronic transitions between the d-orbitals. Here, we will analyze each pair to identify if they're colourless:v

• Sc³⁺ and Zn²⁺:
  • Sc³⁺: Scandium ion Sc³⁺ has an empty d-subshell (d⁰ configuration) as all three electrons are removed upon forming the ion. No d-d transitions can occur, thus making it colourless.
  • Zn²⁺: Zinc ion Zn²⁺ has a completely filled d-subshell with an electronic configuration of [Ar]3d¹⁰. It does not allow d-d transitions, which implies it is also colourless.
• Ti⁴⁺ and Cu²⁺:
  • Ti⁴⁺: Similar to Sc³⁺, Ti⁴⁺ has a d⁰ configuration, which makes it colourless.
  • Cu²⁺: Copper ion Cu²⁺ has a d⁹ configuration, which supports d-d transitions, making it coloured.
• V²⁺ and Ti³⁺:
  • V²⁺: Vanadium ion V²⁺ has a d³ configuration that allows d-d transitions, implying it is coloured.
  • Ti³⁺: Titanium ion Ti³⁺ has a d¹ configuration, allowing for d-d transitions and therefore is coloured.
• Zn²⁺ and Mn²⁺:
  • Zn²⁺: As mentioned earlier, Zn²⁺ is colourless due to the fully filled d-subshell.
  • Mn²⁺: Manganese ion Mn²⁺ has a d⁵ configuration. It allows electronic transitions within partly filled d-orbitals, making it coloured.

Based on the above analysis, the pair Sc³⁺ and Zn²⁺ is the only one where both ions are colourless. This is due to the absence of d-d transitions in Sc³⁺ and the fully filled 3d-subshell of Zn²⁺.

Mn → 4s²3d⁵

Mn⁺⁴ → 3d³

n = 3

So, the spin-only magnetic moment (μ) value of B in BM is 4.

• Ce⁴⁺ acts as a good oxidizing agent because it readily accepts electrons to attain the more stable Ce³⁺ configuration.
• Tb4+ also acts as a good oxidizing agent because it gets reduced to the more stable Tb³⁺ configuration.
• Sm²⁺ and Ce²⁺ are reducing agents as they tend to lose electrons to attain stable configurations.
Thus, the good oxidizing agents are Ce⁴⁺ and Tb⁴⁺.

The electron configurations of the elements are as follows:

• 1.
• 2.
• 3.
• 4.
• 5.

Thus, the elements with half-filled -orbitals in their ground state are Eu (Europium) and Gd (Gadolinium), corresponding to option (1).

The correct option is (B):

The third ionization energy refers to the energy required to remove the third electron from an atom. Lanthanides exhibit varying ionization energies based on their electronic configurations.
For lanthanides like Eu (Europium) and Yb (Ytterbium), the third ionization energy is high because both elements tend to form stable, highly charged ions after the removal of the first two electrons.
- Europium ( ) has a stable configuration, making the removal of the third electron significantly harder.
- Ytterbium ( ) also has a stable configuration in its trivalent state, making the third ionization energy high.
Thus, the pair Eu, Yb has high third-ionization energies.

The correct option is (C): N
Nessler’s Reagent

The electronic configuration of Mn is .
In , Mn is in oxidation state. Water is a weak field ligand.
Electronic configuration of is .
Since is a weak field ligand, there will be no pairing of electrons in the orbitals.
Number of Unpaired Electrons ( )

Spin-Only Magnetic Moment ( )


Nearest Integer:
The nearest integer is:

Step 1: Lone Pairs in IF₅

Valence electrons:

• Iodine (I): 7 valence electrons
• Fluorine (F): 7 valence electrons × 5 fluorine atoms = 35
• Total valence electrons:

Bonding electrons:

• 5 I-F bonds = bonding electrons

Remaining electrons:

• electrons used to complete the octet of fluorine atoms
• electrons left on iodine, forming 1 lone pair

Step 2: Lone Pairs in IF₇

Valence electrons:

• Iodine (I): 7 valence electrons
• Fluorine (F): 7 valence electrons × 7 fluorine atoms = 49
• Total valence electrons:

Bonding electrons:

• 7 I-F bonds = bonding electrons

Remaining electrons:

• electrons used to complete the octet of fluorine atoms
• No electrons left on iodine, so there are 0 lone pairs on iodine

Step 3: Total Lone Pairs on the Central Atom

The sum of lone pairs on iodine in IF₅ and IF₇ is:

Final Answer:

The total number of lone pairs on the central atoms in IF₅ and IF₇ is: 1.

Permanganate Thiosulphate Reaction Conditions Analysis

The reaction of permanganate with thiosulphate depends on the pH of the medium. Permanganate can be reduced to different manganese species depending on the conditions.

In a neutral or weakly alkaline medium, permanganate is reduced to manganese(IV) oxide :

In this reaction, the oxidation state of manganese changes from +7 in to +4 in , a change of 3 units.

In an acidic medium, permanganate is reduced to manganese(II) ions :

Here, the oxidation state of manganese changes from +7 to +2, a change of 5 units.

Conclusion:

Since the question specifies a change in oxidation state of 3, the reaction must be occurring in a neutral or weakly alkaline medium.

To solve this problem, we need to match the alkali metals in List-I with their corresponding emission wavelengths given in List-II. Alkali metals are known for their characteristic colors when excited, and these colors correspond to specific wavelengths of light emitted during transitions. Let's explore each metal's emission wavelength:

1. Lithium (Li) is known to emit a red color when it is excited, with a characteristic emission wavelength of approximately 670.8 nm. This corresponds to option (III).
2. Sodium (Na) produces a bright yellow flame, which corresponds to 589.2 nm. This is option (I).
3. Rubidium (Rb) exhibits a reddish-violet flame with an emission wavelength of approximately 780.0 nm. This matches option (IV).
4. Cesium (Cs) shows a violet-blue flame, associated with an emission wavelength of 455.5 nm. This is option (II).

From the above analysis, we can match the metals with their corresponding wavelengths as follows:

• (A) Li - (III) 670.8 nm
• (B) Na - (I) 589.2 nm
• (C) Rb - (IV) 780.0 nm
• (D) Cs - (II) 455.5 nm

Therefore, the correct answer is: (A)-(III), (B)-(I), (C)-(IV), (D)-(II).

To solve this matching question, we must determine the electronic configuration for each ion in List-I and match it with the appropriate electronic distribution given in List-II. Let's analyze each species one by one:

1. Cr⁺²: Chromium (Cr) in its neutral state has the electronic configuration . When Cr loses two electrons to form Cr⁺², the electrons are removed from the 4s orbital first, followed by the 3d orbital. This results in the configuration , corresponding to option III.
2. Mn⁺: Manganese (Mn) has the electronic configuration . Upon losing one electron to form Mn⁺, the electron is removed from the 4s orbital, resulting in the configuration , corresponding to option IV.
3. Ni⁺²: Nickel (Ni) has the electronic configuration . When Ni loses two electrons to form Ni⁺², both electrons are removed from the 4s orbital, resulting in the configuration , matching option I.
4. V⁺: Vanadium (V) normally has the electronic configuration . When it loses one electron to form V⁺, the electron is removed from the 4s orbital, resulting in the configuration , which corresponds to option II.

Based on the above analysis, the correct matching is:

Hence, the correct answer is: (A)-III, (B) – IV, (C) – I, (D)-II.

To determine which compound is white in color, we need to examine the properties of each option. Let's evaluate them one by one:

1. Ammonium sulphide: This compound is typically yellow due to the presence of sulphur, reflecting its character as a sulfide compound. Thus, it is not white.
2. Lead sulphate: Lead sulphate ( ) is known to be a white solid and is insoluble in water. This makes it a promising candidate for being the white compound among the given options.
3. Lead iodide: Lead iodide ( ) typically appears as a bright yellow solid. Due to its characteristic color, it is not the white compound in question.
4. Ammonium arsinomolybdate: Beyond common compounds, this is not as widely recognized and usually involves complex formation with variable colors. Thus, without specific reference indicating otherwise, it can't be assumed to be white.

From the above analysis, we can conclude that among the listed options, lead sulphate is distinctly known for its white coloration.

Therefore, the correct answer is lead sulphate.

To determine which statements about Zn, Cd, and Hg are correct, let's evaluate each option provided:

1. High Enthalpy of Atomization: Statement A suggests that Zn, Cd, and Hg exhibit high enthalpy of atomization because the d-subshell is full. The enthalpy of atomization is generally high when atoms form strong metallic bonds. However, as Zn, Cd, and Hg have completely filled d-subshells and are in the d-block but behave more like s-block elements (due to their filled d and s subshells), they have lower enthalpy of atomization than typical transition metals. Thus, this statement is incorrect.
2. Oxidation States: Statement B says Zn and Cd do not show variable oxidation states, while Hg shows +I and +II. Zn and Cd typically only exhibit a +II oxidation state due to the stability of their filled d and s orbitals. Mercury can have oxidation states of +I and +II due to its ability to form covalent bonds. This makes statement B correct.
3. Paramagnetic Compounds: Statement C claims that compounds of Zn, Cd, and Hg are paramagnetic. Compounds of Zn, Cd, and Hg are typically diamagnetic, as Zn²⁺, Cd²⁺, and Hg²⁺ ions have fully filled d-subshells, rendering them non-paramagnetic. Hence, this statement is incorrect.
4. Soft Metals: Statement D states that Zn, Cd, and Hg are called soft metals. Indeed, these metals are considered soft, with mercury being a liquid at room temperature. This makes the statement D correct.

Considering the evaluation:

• Statement B is correct, as it describes the oxidation characteristics of Zn, Cd, and Hg accurately.
• Statement D is also correct, categorizing these elements as soft metals.

Therefore, the correct answer is: B, D only.

The question provides two statements related to the chemistry of manganese compounds, and we need to determine the veracity of these statements based on chemical reactions and principles.

Statement I: Fusion of with and an oxidising agent gives dark green .

Explanation: When manganese dioxide ( ) is fused with potassium hydroxide ( ) in the presence of an oxidizing agent such as chlorine ( ) or potassium nitrate ( ), it forms potassium manganate ( ), which is dark green.

This reaction can be represented as:

Therefore, Statement I is true.

Statement II: Manganate ion on electrolytic oxidation in alkaline medium gives permanganate ion.

Explanation: In alkaline medium, the manganate ion ( ) can undergo further oxidation to form permanganate ion ( ), especially under electrolytic conditions. This transformation occurs due to the gain of an electron by :

This shows an increase in oxidation state from Mn (+6 in manganate) to Mn (+7 in permanganate).

Hence, Statement II is also true.

Conclusion: Based on the explanations, both Statement I and Statement II are true. Therefore, the correct answer is: Both Statement I and Statement II is true.

The reaction involving potassium dichromate, potassium chloride, and concentrated sulfuric acid typically leads to the formation of chromyl chloride ( ). In this process, let's consider the reaction:



In potassium dichromate ( ), the oxidation state of chromium is . During the formation of chromyl chloride ( ), each chromium atom maintains its oxidation state. We verify this by calculating the oxidation state in :

Let the oxidation state of be . Oxygen has an oxidation state of , and chlorine typically exhibits :



This computation confirms chromium retains an oxidation state of in the product , matching the given range (6,6). Therefore, the oxidation state of chromium in the product is correctly .

Let's analyze each statement given in the question to determine which are correct:

1. Statement A: is an oil at room temperature.
   This statement is correct. Manganese heptoxide ( ) is a volatile liquid that exists as an oily green liquid at room temperature. It is known for being a dangerous oxidizing agent and can decompose explosively.
2. Statement B: reacts with acid to give .
   This statement is incorrect. is actually a vanadium(IV) oxide, which does not react with acid to directly give ions. Typically, vanadium compounds form a different set of oxidation states related to the oxidation and reduction processes.
3. Statement C: is a basic oxide.
   This statement is correct. Chromium(II) oxide ( ) is considered a basic oxide. It can react with acids to form salts and water, which is characteristic behavior of basic oxides.
4. Statement D: does not react with acid.
   This statement is incorrect. Vanadium(V) oxide ( ) does react with acids. It's an amphoteric oxide and can react with both acids and bases, forming different salts like vanadates.

Based on the analysis, the correct statements are A and C. Thus, the option "A and C only" is the correct answer.

Problem Statement:
Ammonia (NH₃) acts as a WFL (Weak Field Ligand) with Ni²⁺, and the hybridization of the complex is .

Electronic Configuration of Ni²⁺:
The electron configuration of (with atomic number 28) is:
This implies that the 4s orbital is empty, and there are 8 electrons in the 3d orbital.

Electron Configuration Diagram:
The 3d orbitals are filled as follows, showing the electron arrangement for the 3d orbitals of Ni²⁺:

Hybridization:
The hybridization of the complex is , as it involves six ligands (NH₃), suggesting the use of six orbitals (one from each of the 3d, 4s, and 4p orbitals) to form the bonds.

Number of Unpaired Electrons:
There are 2 unpaired electrons in the 3d orbitals, which will contribute to the magnetic properties of the complex.

Magnetic Moment Calculation:
The magnetic moment is given by the formula: where is the number of unpaired electrons. Here, (since there are 2 unpaired electrons), so: Thus, the magnetic moment is

Conclusion:
The magnetic moment , which corresponds to .