Periodic Table

Step 1: Understanding the electronic configuration.
The given electronic configuration is [Kr]4d 5s , which means the element is in the 4d block and belongs to the 5th period. The 4d configuration indicates that the element has filled 4d orbitals with 10 electrons, and the 5s configuration indicates two electrons in the 5s orbital.
Step 2: Identifying the group.
In the long-form periodic table, elements with the electronic configuration ending in 4d 5s belong to Group 11. This is because the d-block elements in this configuration are in the transition metal series and belong to the coinage metals group.
Step 3: Conclusion.
The correct position of the element is in Group 11, Period 5, so the correct answer is (D).

Step 1: Electronic configuration of La and Ce.
Lanthanum (Z = 57) has electronic configuration .
Cerium (Z = 58) has electronic configuration .

Step 2: Period determination.
Both La and Ce have their outermost electrons in the 6th shell, hence both belong to Period-6.

Step 3: Group determination.
Both elements are placed in Group-3 as they mark the beginning of the lanthanide series in the long form of the periodic table.

Step 4: Conclusion.
Therefore, the correct position is La = Group-3, Period-6 and Ce = Group-3, Period-6.

Step 1: Understanding the concept of softness in elements.
Softness or hardness in elements is often associated with their position in the periodic table and their atomic structure. Zinc (Zn) is a relatively soft metal compared to cobalt (Co), tungsten (W), and molybdenum (Mo), which are much harder.

Step 2: Analyzing the options.
(A) Co: Cobalt is a transition metal but is harder compared to Zn.
(B) Zn: Correct. Zinc is a softer metal in comparison to the other elements listed.
(C) W: Tungsten is a very hard metal with a high melting point.
(D) Mo: Molybdenum is also a relatively hard metal.

Step 3: Conclusion.
The correct answer is (B) Zn, as it is the softest element among the given options.

Step 1: Understanding Density and Weight.
Scandium is the lightest among the elements listed. It has the lowest density of the metals mentioned. Scandium's density is lower compared to cobalt, copper, and iron.

Step 2: Analyzing the options.
(A) Scandium: Correct. Scandium is the lightest and has the lowest density.
(B) Cobalt: Incorrect. Cobalt is denser than scandium.
(C) Copper: Incorrect. Copper has a higher density than scandium.
(D) Iron: Incorrect. Iron is heavier and has a higher density than scandium.

Step 3: Conclusion.
Scandium is the lightest and has the lowest density, corresponding to option (A).

Step 1: Understanding chemical twins.
Chemical twins are pairs of elements where one element is a member of a transition series and the other is its homolog in the subsequent period. Zr and Rf are not chemical twins because they belong to different groups and periods.

Step 2: Analyzing the options.
(A) Nb - Ta: These elements are chemical twins as they belong to the same group in the periodic table and have similar chemical properties.
(B) Zr - Rf: Incorrect — Zr and Rf do not form a chemical twin pair. Zr is a transition metal, and Rf is a synthetic element.
(C) Mo - W: These are chemical twins, as they belong to the same group and have similar properties.
(D) Tc - Re: These are also chemical twins, belonging to the same group and exhibiting similar chemical behaviors.

Step 3: Conclusion.
The correct answer is (B) Zr - Rf, as they do not represent chemical twins.

Step 1: Understanding entropy at absolute zero.
At , the entropy of a solid can be greater than zero due to disorder that remains in the system, even at absolute zero. This is known as residual entropy.

Step 2: Analyzing the options.
(A) Standard entropy: Standard entropy refers to the entropy at a standard temperature (usually 298 K).
(B) Residual entropy: Correct — Residual entropy refers to the entropy remaining at due to residual disorder in the solid.
(C) Formal entropy: This is not a standard term in thermodynamics.
(D) Absolute entropy: This is the entropy at any temperature, not specifically at .

Step 3: Conclusion.
The correct answer is (B) Residual entropy.

Step 1: Understanding the Van Arkel method.
The Van Arkel method is a process used for refining certain metals, including zirconium. It involves heating the impure metal in a vacuum with iodine to form a volatile compound, which is then decomposed to give pure metal. This method is effective for refining metals with high boiling points.

Step 2: Analyzing the options.
(A) Liquation: Liquation is used to separate metals with different melting points, not specifically for zirconium.
(B) Zone refining: Zone refining is used for semiconductors like silicon, not for zirconium.
(C) Polling: Polling is a method used to remove impurities from molten metals but is not applicable for zirconium refining.
(D) Van Arkel method: Correct — The Van Arkel method is used for the refining of zirconium.

Step 3: Conclusion.
The correct answer is (D) Van Arkel method.

Step 1: Understanding the element.
The third element of group-2 in the periodic table is magnesium (Mg). The atomic number of magnesium is 12, so its electronic configuration is: This is equivalent to .
Step 2: Analyzing the options.
(A) [Ne] 3S¹: Incorrect, this is the electronic configuration of sodium (Na), not magnesium.
(B) [Ne] 3S²: Incorrect, this is the configuration of magnesium, but the question asks for the third element of group-2.
(C) [Ar] 4S¹: Incorrect, the correct configuration for magnesium should be .
(D) [Ar] 4S²: Correct — This is the correct electronic configuration for magnesium.
Step 3: Conclusion.
The correct answer is (D) [Ar] 4S².

Step 1: Apply IUPAC temporary naming rules.
For atomic number 116, the digits are 1–1–6. According to IUPAC: 1 = un, 1 = un, 6 = hex
Step 2: Form the temporary name.
The name becomes un-un-hex-ium, abbreviated as Uuh.
Step 3: Conclusion.
Hence, the correct symbol is Uuh.

Concept:
Alkali metal carbonates generally decompose on heating to form metal oxide and carbon dioxide. However, is less thermally stable compared to other alkali metal carbonates due to the small size and high polarizing power of . Step 1: Understand thermal decomposition. On heating:
Step 2: Apply to lithium carbonate.
Step 3: Reason behind decomposition.

• has high charge density
• It polarizes the carbonate ion strongly
• This weakens the bonds and promotes decomposition

Step 4: Eliminate incorrect options.

• (B) LiO does not exist as a stable compound
• (C) LiC is not a valid product
• (D) forms in oxidation reactions, not here

Step 5: Conclusion.