Multiple Choice QuestionsThe electronic configuration of Cu is
[Ne] 3s2, 3p6, 3d9, 4s2
[Ne] 3s2, 3p6, 3d10, 4s1
[Ne] 3s2, 3p6, 3d3, 3d9, 4s2, 4p6
[Ne] 3s2, 3p6, 3d5, 4s2, 4p4
Assertion: Cuprous ion (Cu+) is colorless whereas cupric ion (Cu2+) is blue in the aqueous solution.
Reason: Cuprous ion (Cu+) has unpaired electrons while cupric ion (Cu2+) does not.
If both assertion and reason are true and reason is a correct explanation of the assertion.
If both the assertion and reason are true but the reason is not a correct explanation of the assertion.
If the assertion is true but the reason is false.
If both the assertion and reason are false.
The stable bivalency of Pb and trivalencey of Bi is:
due to d contraction in Pb and Bi
due to relativistic contraction of the 6s orbitals of Pb and Bi, leading to inert pair effect
due to screening effect
due to the attainment of noble liquid configuration
A magnetic moment of 1.73 BM will be shown by one among the following
[Cu(NH3)4]2+
[Ni(CN)4]2-
TiCl4
[CoCl6]4-
A.
[Cu(NH3)4]2+
Magnetic moment, is related with number of unpaired electrons as
On solving n =1
Thus, the complex/compound having one unpaired electron exhibit a magnetic moment of 1.73 BM.
(a) In [Cu(NH3)4]2+
Cu2+ = [Ar] 3d9
(Although in the presence of strong field ligand NH3, the unpaired electron gets excited to higher energy level but it still remains unpaired).
(b) In [Ni(CN)4]2-
Ni2+= [Ar]3d8
But CN- being strong field ligand pair up the unpaired electrons and hence in this complex, number of unpaired electrons = 0.
(c) In [TiCl4]
Ti4+= [Ar] 3d7
(d) In [CoCl6]4-
Co2+ = [Ar] 3d7
It contains three unpaired electrons.
Thus, [Cu(NH3)4]2+ is the complex that exhibits a magnetic moment 1.73 BM.
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