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Q.(a) Why Zn, Cd and Hg are not regarded as transition metals?

(1)
(b) What is the Action of heat on K2Cr2O7?
(1)
(c) Why transition-metals form alloys?
(1)
(d) Why La(OH)3 is more basic than Lu(OH)3?
(1)
(e) Why it is difficult to separate lanthanoid elements in pure state? (1)
Himachal HpboseHPBOSE Plus Two Board 2025Subjective· 5mImportance★★★★★
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Zn/Cd/Hg have no partially-filled d-subshell so fail the defining test for transition metals; K2Cr2O7 needs strong heat to release O2; alloy formation and hydroxide basicity/separability of lanthanoids all trace back to atomic/ionic size effects (lanthanoid contraction).

(a) Why Zn, Cd, Hg are not regarded as transition metals:

The defining criterion for a transition element is having a partially filled d-subshell either in the ground state or in any of its commonly observed oxidation states. Zn, Cd, and Hg all have the configuration (n−1)d10ns2(n-1)d^{10}ns^2 in the ground state, and in their common oxidation state (+2), they lose the ns2ns^2 electrons to give (n−1)d10(n-1)d^{10} — a completely filled d-subshell in every case. Since the d-subshell is never partially filled, they do not satisfy the transition-element criterion and are excluded (though they are still d-block elements).

(b) Action of heat on K2Cr2O7K_2Cr_2O_7:

Potassium dichromate is fairly stable and does not decompose on mild heating, but on strong/prolonged heating (above its melting point, ~670 K) it decomposes, releasing oxygen gas:

4K2Cr2O7→Δ4K2CrO4+2Cr2O3+3O2↑4K_2Cr_2O_7 \xrightarrow{\Delta} 4K_2CrO_4 + 2Cr_2O_3 + 3O_2\uparrow

(Potassium chromate and chromium(III) oxide are formed, along with evolution of oxygen.)

(c) Why transition metals form alloys:

Transition metals have very similar atomic sizes/radii to one another (since successive elements across a transition series show only a small, gradual decrease in radius). This similarity allows the atoms of one transition metal to readily replace/substitute for atoms of another in the crystal lattice without significant lattice strain, forming homogeneous solid solutions — i.e., alloys.

(d) Why La(OH)3La(OH)_3 is more basic than Lu(OH)3Lu(OH)_3:

Across the lanthanoid series, ionic radius steadily decreases from La³⁺ to Lu³⁺ due to lanthanoid contraction. La3+La^{3+} is the largest lanthanoid ion, so the La−OHLa-OH bond is more ionic and readily releases OH−OH^-, making La(OH)3La(OH)_3 the most basic hydroxide in the series. Lu3+Lu^{3+}, being the smallest, forms a more covalent bond with OH−OH^- (higher charge density polarizes the O-H bond more), reducing its basicity — so La(OH)3La(OH)_3 is more basic than Lu(OH)3Lu(OH)_3.

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