Properties Of D Block Elements

- The density of the elements typically increases with increasing atomic mass as you go down a group in the periodic table.
- For the alkaline earth metals (Group 2), this general trend applies except for some anomalies due to atomic packing and atomic weight.
- Strontium (Sr) has the highest density among the given elements, followed by beryllium (Be), magnesium (Mg), and calcium (Ca).

Step 1: Use the freezing point depression formula where: - is the depression in freezing point, - is the cryoscopic constant, - is the molality of the solute.
Step 2: Calculate the new freezing point where: - is the normal freezing point of water (273 K), - is given as 0.052 K. Step 3: Compute the final answer Thus, the correct answer is .

The root-mean-square (RMS) velocity of an ideal gas is given by the equation:

Here, is the universal gas constant, is the absolute temperature, and is the molar mass of the gas. To find the relationship between RMS velocity and temperature, we square both sides of the equation:

Rearranging, we get:

This resembles the equation of a straight line , where is on the y-axis and is on the x-axis. Comparing both equations:

• (since no additional terms are present)

In the provided graph, since is plotted against , the slope of the straight line represents:

Therefore, the solution states that the slope cannot be as given in some options. To find the correct correspondence, the correct slope should reflect:

Thus, the option showing is indeed the correct answer, representing the direct relationship of temperature with RMS velocity squared.

The primary structure of a protein refers to the unique sequence of amino acids in a polypeptide chain. This sequence dictates the protein's higher-level structures (secondary, tertiary, and quaternary).
Step 1: The primary structure of a protein is the most fundamental level of organization, representing its constitution in terms of amino acid sequence.
Step 2: Secondary structure refers to local folded structures that form within a polypeptide chain, such as alpha helices and beta sheets. Tertiary structure refers to the overall three-dimensional shape of the protein, while quaternary structure refers to the arrangement of multiple polypeptide chains. Thus, the correct answer is Primary structure, as it represents the constitution of the protein.

The element Fe (Iron) has an atomic number of 26. Therefore, its electronic configuration is: . The number of electrons in Fe is 6 from the 3d subshell.

(i) Total number of -electrons in Mg:
Mg (Magnesium) has an atomic number of 12, with the electronic configuration: . The total number of -electrons is .

(ii) Total number of -electrons in Cl:
Cl (Chlorine) has an atomic number of 17, with the electronic configuration: . The total number of -electrons is .

(iii) Total number of -electrons in Ne:
Ne (Neon) has an atomic number of 10, with the electronic configuration: . The total number of -electrons is 6.

Matching the number of electrons in Fe (6) with the options:
(i) s-electrons in Mg: 6 - Match
(ii) p-electrons in Cl: 11 - No Match
(iii) p-electrons in Ne: 6 - Match

Thus, the correct answer is: i, iii only.

Step 1: Analyze the graph showing the variation of probability density with distance . The graph shows a single peak, followed by a monotonic decrease, and then no further peaks. This is characteristic of a 2s orbital, where the probability density function first increases, reaches a peak, and then decreases after crossing a node (zero probability at some distance from the nucleus).
Step 2: In contrast, a 1s orbital would have only a single peak and no node, while a 3s orbital would have multiple peaks and nodes. Therefore, the graph represents a 2s orbital.

Anti-Markovnikov addition occurs when HBr adds to an alkene such that the hydrogen atom attaches to the carbon atom with the fewest alkyl groups, resulting in the formation of the less stable carbocation. This addition typically occurs with peroxides (as in the case of the anti-Markovnikov mechanism).
Step 1: Propene (C H ) undergoes anti-Markovnikov addition of HBr in the presence of peroxides. In this reaction, the hydrogen adds to the carbon with fewer alkyl groups, and the bromine adds to the carbon with more alkyl groups.
Step 2: 1-Butene (C H ) also undergoes anti-Markovnikov addition of HBr in the presence of peroxides, similar to propene.
Step 3: 2-Butene (C H ) undergoes Markovnikov addition of HBr, where the bromine atom attaches to the carbon with more alkyl groups, and the hydrogen attaches to the carbon with fewer alkyl groups. Therefore, 2-Butene does not undergo anti-Markovnikov addition.
Step 4: 3-Methyl-2-pentene (C H ) undergoes Markovnikov addition and does not follow the anti-Markovnikov addition rule.
Thus, the correct answer is 2-Butene, which does not undergo anti-Markovnikov addition.

The given reaction can be broken down into two steps: 1. Mg(g) Mg (g) + e (IE ) 2. Mg (g) Mg (g) + e (IE ) The overall reaction is the sum of these two steps: Mg(g) Mg (g) + 2e The energy required for the overall reaction is the sum of the ionization enthalpies:
Energy = IE + IE Given IE = 178 kcal mol and IE = 348 kcal mol , we have: Energy = 178 + 348 = 526 kcal mol
Final Answer: +526 kcal mol .

To find the change in internal energy , we use the first law of thermodynamics: where is the heat added to the system and is the work done on the system.
Step 1: The heat evolved by the system is given as . Since the system loses heat, .
Step 2: The work done by the system during contraction is given by the formula: where and . Thus, We convert the work to joules using the conversion factor : Step 3: Now, using the first law of thermodynamics: Step 4: Since the magnitude of internal energy has been reduced, we also have to adjust for the provided answer options. The correct internal energy change value matches with the corrected options as:

Step 1: Determining the number of unpaired electrons
- The number of unpaired electrons in a complex depends on the oxidation state of the metal and the ligand strength.
- Cyanide ( ) is a strong field ligand, leading to low spin configurations.
- Fluoride ( ) and chloride ( ) are weak field ligands, leading to high spin configurations.
Step 2: Electron Configurations
- → High spin ( , ) → 5 unpaired electrons
- → High spin ( , ) → 5 unpaired electrons
- → Low spin ( , ) → 3 unpaired electrons
- → Low spin ( , ) → 1 unpaired electron
Step 3: Arranging in Increasing Order - Thus, the correct order is:

Step 1: Understanding the Blast Furnace Process - The blast furnace is used for extracting iron from its ore (hematite or magnetite).
- The raw materials used include iron ore, coke, and limestone, which undergo reduction reactions to produce molten iron.
- The first product obtained from the blast furnace is pig iron, which contains high amounts of carbon (about 3.5 - 4.5%), making it brittle and hard.
Step 2: Explanation of Other Types of Iron - Wrought Iron: It is nearly pure iron with very low carbon content and is obtained by further refining pig iron.
- Cast Iron: It is made by remelting pig iron and has a slightly lower carbon content (2-3%).
- Steel: It is an alloy of iron with carbon and other elements and is not directly obtained from the blast furnace.
Final Answer: The correct answer is Pig Iron (Option B).

- Electron precise hydrides are those in which the total number of valence electrons equals the sum of the bonding electrons around the central atom.
- Group 14 elements form typical electron precise hydrides like CH₄ (methane), where each hydrogen atom contributes 1 electron and carbon contributes 4, making a stable electron count of 8 around carbon.

Step 1: Use the density formula for a unit cell Where: - = number of atoms per unit cell - = molar mass of Fe = 56 g/mol - = Avogadro’s number = - pm = cm - Given density g/cm³
Step 2: Calculate the unit cell volume Step 3: Solve for Thus, , indicating a body-centered cubic (BCC) structure.

To determine the correct order regarding their covalent character for the given compounds, we need to understand the concept of covalent character in ionic compounds. The covalent character of an ionic bond is influenced by factors like the polarizing power of the cation and the polarizability of the anion.
1. Fajans’ Rules: According to Fajans' rules, increased covalent character in an ionic bond is observed with: a) Higher charge of the cation or anion. b) Smaller size of the cation. c) Larger size of the anion.
Now, let's analyze each statement:
(i)
- The size of the iodide ion ( ) is larger than the fluoride ion ( ), leading to higher polarizability. Thus, has more covalent character than .
(ii)
- Lithium ion ( ) is smaller than potassium ion ( ), leading to higher polarization in than . Thus, should actually have more covalent character than . Hence this order is incorrect.
(iii)
- has tin in a higher oxidation state compared to , increasing the polarizing power of the tin ion and enhancing covalent character. Thus, is more covalent than .
(iv)
- The ion has a higher charge density and polarizing power than , so is more covalent than .
Considering these analyses, the correct order for covalent character is in statements i, iii, and iv.

Element classifications:
- Technetium is a Transition metal.
- Fluorine is a Non-metal.
- Tellurium is a Metalloid.
- Dysprosium is a Lanthanoid.

Step 1: The Heisenberg Uncertainty Principle is given by: where is the uncertainty in position and is the uncertainty in momentum, with being the reduced Planck's constant.
Step 2: The uncertainty in velocity is related to the uncertainty in momentum by the equation: Step 3: Since (as given in the problem statement), we substitute this into the formula for velocity: Step 4: Now, using the Heisenberg Uncertainty Principle: Step 5: Substituting into the equation for : Simplifying the expression, we get: