229.

Give reasons:
(i) The d-orbital of Cu is completely filled (3d¹⁰4s¹) still it is considered as transition metal but Zn (3d¹⁰4s²) is not.

(ii) Zinc salts are colourless at room temperature but nickel salts are coloured.

(iii) The atomic radii of europium and ytterbium do not fit into the concept of lanthanide contraction but ionic radii fit in.

(iv) The atomic radius of Cu is greater than that of Cr but ionic radius of Cr²⁺ is greater than that of Cu²⁺.

(v) Cu²⁺ is colourless while Cu²⁺ is blue in aqueous solution.

 

(i) The quantum mechanical definition of a transition element demands that there should be incomplete d-orbital either in the atom or ion of that element. In Cu atom (3d¹⁰4s¹) the 3d¹⁰ 1s completely filled but in its Cu²⁺ ion (3d⁹4s⁰) one of the 3d-orbitals is only half filled. Therefore, copper is a transition elements. On the other hand, in Zn (3d¹⁰4s²) or Zn²⁺ (3d¹⁰4s⁰) the 3d-orbital is completely filled. Therefore, zinc is not considered as a true transition element, but its electronic configuration does not fit in the quantum mechanical definition.


(ii) Zinc salts are colourless because in Zn²⁺ ion (3d⁰4s⁰) there is no unpaired electron in its d-orbital rather 3d orbital is completely filled. Therefore, the visible light is not absorbed but it is transmitted through zinc salts. On the other hand, nickel salts are coloured because in Ni²⁺ ion (3d⁸4s⁰) there are 2 unpaired electrons in 3d orbital
$(\downarrow \uparrow \downarrow \uparrow \downarrow \uparrow \downarrow ).$
Therefore, some light is absorbed for d-d transition and the complementary part is transmitted. The colour of nickel salt is because of the partly transmitted radiation.


(iii)    Ca. In europium (Eu = 4f⁷5d⁰ 6s²) the f-orbital is half filled. There are only 2 electrons in 6s-orbitals available for the formation of metallic bond in the crystal lattice of Europium. So, the crystal lattice is less compact. The atomic separation in Eu is large as compared to other metals in which 3 or more electrons can take part in the formation of metallic bonds.
In Ytterbium (Yb = 4f¹⁴5d⁰6s²) the f-orbitals is completely filled. Therefore, the crystal lattice is less compact and the atomic separation is large. Therefore, the metallic radius of Yb is larger than other members of lanthanide series.
The lanthanide contraction in ionic radii r(M³⁺) of tri-positive cations is quite smooth because three electrons are removed from the valence orbitals.


(iv)    The atomic radius of Cu is greater than that of Cr because in Cu atom (3d¹⁰4s¹) the d-d electron repulsion is large due to pairing of electrons $(\uparrow \downarrow \uparrow \downarrow \uparrow \downarrow \uparrow \downarrow \uparrow \downarrow ).$
But in Cr atom (3d³4s¹) the d-d electron repulsion is minimum as the 3d orbitals are singly occupied $(\uparrow \uparrow \uparrow \uparrow \uparrow ).$
The ionic radius of Cu²⁺ is less that of Cr⁺. Because in Cu²⁺ ion (3d⁹) the d-d electron repulsion is reduced due to unpairing of electron spins $(\uparrow \downarrow \uparrow \downarrow \uparrow \downarrow \uparrow \downarrow \uparrow )$ and the other electrons are attracted by 29 protons of the nucleus.
Therefore, the attraction of nucleus on the outer electron has become strong and d-electrons are pulled inward.
In Cr²⁺ the 3d⁴ electrons are still unpaired $(\uparrow \uparrow \uparrow \uparrow ),$ and there are attracted by only 24 protons of nucleus.


(v) The ion which has unpaired electron is coloured in aqueous solution. Thus Cu⁺(ion) [s²2s²2p⁶3s²3p⁶3d¹⁰4s⁰] does not have any unpaired electron, hence its aqueous solution is colourless.
Cu²⁺ (ion) [1s²2s²2p⁶3s²3p⁶3d⁹4s⁰] has a unpaired electron in one of the 3d orbitals, therefore its aqueous solution is coloured (blue).