d- and f-Block Elements

Question 1

Given below are two statements: In light of the above statements, choose the most appropriate answer from the options given below: Assertion A: Generally, 3d transition metals have high melting points. Reason R: Involvement of 3d-electrons in addition to 4s-electrons in the interatomic metallic bonding.

Accepted Answer

Both A and R are correct and R is the correct explanation of A. Transition metals melt at high temperatures precisely because their (n−1)d electrons join the ns electrons in the metallic bond. More delocalised electrons mean a stronger lattice, so the reason is not merely true alongside the assertion — it is the explanation for it.

Answer

Both A and R are correct and R is NOT the correct explanation of A

Answer

A is correct but R is not correct

Answer

A is not correct but R is correct

Question 2

The lanthanide ion having four unpaired electrons is (Given: Atomic numbers of Ce = 58, Nd = 60, Tb = 65 and Ho = 67)

Accepted Answer

Ho³⁺. Lanthanide tripositive ions have the configuration 4fⁿ. Ho³⁺ is 4f¹⁰: seven orbitals take ten electrons as three pairs plus four singles, so four are unpaired. For comparison Ce³⁺ is 4f¹ with one, Nd³⁺ is 4f³ with three, and Tb³⁺ is 4f⁸ with six.

Answer

Nd³⁺

Answer

Ce³⁺

Answer

Tb³⁺

Question 3

The green paramagnetic species formed by heating KMnO₄ at 513 K is

Accepted Answer

K₂MnO₄. Heating permanganate above about 500 K decomposes it: 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂. The manganate ion MnO₄²⁻ is green and, with manganese in the +6 state, has one unpaired electron, so it is paramagnetic.

Answer

Mn₃O₄

Answer

MnO

Answer

KO₂

Q1.Given below are two statements: In light of the above statements, choose the most appropriate answer from the options given below:

Q2.The lanthanide ion having four unpaired electrons is (Given: Atomic numbers of Ce = 58, Nd = 60, Tb = 65 and Ho = 67)

Q3.The green paramagnetic species formed by heating KMnO₄ at 513 K is