Block D and F Elements

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3 weeks ago

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A
alok kumar singh

Contributor-Level 10

Cu is the only element of 3d – series whose M2+ / M value is positive because of fact that low hydration enthalpy and high sublimation & ionization enthalpies.

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4 weeks ago

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Vishal Baghel

Contributor-Level 10

Ethylene diamine, en is bidentate, chelating ligand. Chelating ligands increase stability due to higher entropy factor.

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4 weeks ago

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Vishal Baghel

Contributor-Level 10

Eutrophication occurs when the water body becomes overly enriched with nutrients. Eutrophication leads to decrease in the level of dissolved oxygen (DO) in water bodies.

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4 weeks ago

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Vishal Baghel

Contributor-Level 10

The greater stability of Cu²? (aq) rather than Cu? (aq) is due to the much more negative ΔhydH of Cu²? (aq) than Cu? , which more than compensates for the second ionisation enthalpy of Cu.

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a month ago

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Raj Pandey

Contributor-Level 9

The E° value for Ce? /Ce³? is +1.74 V, which suggests that Ce? is a strong oxidant, reverting to its common +3 oxidation state. So, Ce³? is more stable than Ce?

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a month ago

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Raj Pandey

Contributor-Level 9

The size of the Bk³? ion is less than the Np³? ion because Berkelium (Bk) lies beyond Neptunium (Np) in the actinoid series, and the size variation here is because of the actinoid contraction.

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a month ago

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alok kumar singh

Contributor-Level 10

Inversion of Sucrose:
C? H? O? (Sucrose) + H? O - (Invertase)-> C? H? O? (Glucose) + C? H? O? (Fructose)

Fermentation of Glucose:
C? H? O? (Glucose) - (Zymase)-> 2C? H? OH (Ethanol) + 2CO?

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a month ago

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alok kumar singh

Contributor-Level 10

KMnO? oxidizes alkanes that contain a tertiary hydrogen (3°H) to form tertiary alcohols.

n-alkanes, which do not contain tertiary hydrogens, are not oxidized by KMnO? under these conditions.

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a month ago

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Raj Pandey

Contributor-Level 9

Functional group isomers

 

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Vishal Baghel

Contributor-Level 10

A divalent metal ion with an atomic number (Z) of 25 is Mn²?
Mn²? (Z=25) has the electronic configuration: 3d? 4s? (n=5, five unpaired electrons).
The magnetic moment (μ) is calculated as: μ = √ [n (n+2)] = √ [5 (5+2)] = √35 = 5.92 BM.

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