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Q.(a) Why do Zr and Hf have similar properties ?

(b) What happens when K2Cr2O7 is treated with H2SO4 and NaCl solution ?
(c) Why Transition metals form alloys easily ? OR
(a) What happens when K2Cr2O7 is heated strongly ?
(b) Transition elements form complexes easily ? Justify.
(c) Why 5d transition series elements have higher ionization enthalpies than 4d transition series elements ?
Punjab PsebPSEB Punjab Class 12 Board 2018Subjective· 6mImportance★★★★★
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(a) The lanthanide contraction is why Zr and Hf are almost twins in size and chemistry. (b) K₂Cr₂O₇ + NaCl + H₂SO₄ produces the diagnostic orange-red chromyl chloride fumes. (c) Similar atomic radii let transition metals substitute freely for one another, forming alloys.

(a) Zr and Hf have similar properties:

Zirconium (period 5, 4d4d series) and hafnium (period 6, 5d5d series) lie in the same group. Normally, atomic/ionic radius should increase noticeably on descending a group. However, hafnium comes right after the lanthanides (elements 58–71) in the periodic table. Across the lanthanide series, there is a steady, cumulative decrease in atomic/ionic radius called the lanthanide contraction, caused by the poor shielding of one 4f electron by another (4f orbitals are diffuse and shield the nuclear charge inefficiently), so the effective nuclear charge experienced by the outer electrons increases steadily across the series.

This contraction is large enough that by the time the periodic table reaches hafnium, the expected size increase (from being one period below Zr) is almost exactly cancelled out. As a result, ZrZr and HfHf end up with nearly identical atomic and ionic radii, and consequently very similar chemical properties (making them notoriously difficult to separate from each other).

(b) K2Cr2O7K_2Cr_2O_7 + H2SO4H_2SO_4 + NaClNaCl:

This combination produces orange-red, pungent vapours of chromyl chloride, CrO2Cl2CrO_2Cl_2 — this is the basis of the chromyl chloride test, used to confirm the presence of a chloride ion (as distinct from bromide/iodide, which do not give this test):

K2Cr2O7+4NaCl+6H2SO4→2CrO2Cl2+2KHSO4+4NaHSO4+3H2OK_2Cr_2O_7 + 4NaCl + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 + 2KHSO_4 + 4NaHSO_4 + 3H_2O

(c) Transition metals form alloys easily:

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