Question

Question No. 2
Which species among the
following is coloured?
Answer
A. Cu (+1)
B. Zn (2+)
C. Ti (3+)
D. Ti (4+)
M​

Answer 1

Answer:

ti (4+)

Explanation:

NaVO

4

is colourless species due to absence of unpaired electrons. d-d transitions are not

possible. V has +5 oxidation state and [Ar]3d

0

electronic configuration.

Answer 2

The correct option is C.

• The ions which have half-filled, full-filled or vacant d orbitals can't undergo d-d transition and remain colourless.
• The electronic configuration of $Cu^{+1}$ is
• ${1s}^{2} {2s}^{2} {2p}^{6} {3s}^{2} {3p}^{6} {3d}^{10}$

Thus it contains filled d orbital and remains colourless.

• The electronic configuration of $Zn^{+2}$ is
• ${1s}^{2} {2s}^{2} {2p}^{6} {3s}^{2} {3p}^{6} {3d}^{10}$
• Thus it contains filled d orbital and remains colourless.
• The electronic configuration of $Ti^{+3}$ is
• ${1s}^{2} {2s}^{2} {2p}^{6} {3s}^{2} {3p}^{6} {3d}^{1}$
• Thus it contains one electron in d orbital and undergoes a d-d transition and give colour.
• The electronic configuration of $Ti^{+4}$ is
• ${1s}^{2} {2s}^{2} {2p}^{6} {3s}^{2} {3p}^{6}$
• Thus it contains vacant d orbital and remains colourless.