Q. 22 A5.0( 2 Votes )

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Answer :

(i) For scandium (Sc3+)

Theatomic number of Scandium (Sc) is Z = 21.

The electronic configuration of 21Sc= [Ar] 3d1 4s2

And, the electronic configuration of Sc3+= [Ar]3d0, i.e., there are no electrons in 3d subshell.

For chromium (Cr2+)

Theatomic number of chromium (Cr) is Z = 24.

The electronic configuration of 24Cr= [Ar] 3d5 4s1

And, the electronic configuration of Cr2+= [Ar]3d4, i.e., there are 4 unpaired electrons.

For nickel (Ni2+)

Theatomic number of nickel (Ni) is Z = 28.

The electronic configuration of 28Ni= [Ar] 4s2 3d8

And, the electronic configuration of Ni2+= [Ar]3d8, i.e., there are 2 unpaired electrons.

For titanium (Ti3+)

Theatomic number of titanium (Ti) is Z = 22.

The electronic configuration of 22Ti= [Ar] 3d2 4s2

And, the electronic configuration of Ti3+= [Ar]3d1, i.e., there is one unpaired electron.

Thus, Cr3+ having greater no. of unpaired electrons i.e., 4

(ii) As we know that those with fully-filled or empty d-orbitals are colourless. Thus, Sc3+ forms colourless compounds because it has empty d-orbital.

Note: Only those ions will be coloured which have incompletely filled d-orbitals.

(iii) Sc3+ exhibits the most stable +3 oxidation state. Because it has an empty d-orbital. As Ti3+ also exists in +3 oxidation state but is not stable due to the presence of an unpaired electron in the d-orbital.

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