Answer the question on the basis of passage given below:
All transition elements display typical metallic properties like high melting point, boiling point, tensile strength, high malleability and ductility, lustre, etc. Their metallic character increases down the group and elements of 4d and 5d series exhibit higher metallic characters. They show variable oxidation states and form a number of coloured compounds due to the presence of unpaired electrons which show d-d transitions and make them paramagnetic.
Which out of the following coordination compounds are colourless in nature?
A. \([Cr(NH_3)_6]Cl_3\)
B. \([Co(NH_3)_6]Br_3\)
C. \(K_3[Fe(CN)_6]\)
D. \([MnCl_2(en)_2]^+\)
E. \([Ni(CO)_4]\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → B, E
The passage links colour to unpaired d electrons and d-d transitions. A complex with no unpaired electrons (a completely filled d10 set, or a d6 ion forced into the low-spin t2g6 arrangement by a strong-field ligand) is diamagnetic, and on the passage’s basis it is colourless.
A. [Cr(NH3)6]Cl3 — Incorrect. Cr (Z = 24) is [Ar]3d54s1; \( \ce{Cr^3+} \) is 3d3 (t2g3), with 3 unpaired electrons, so it is coloured.
B. [Co(NH3)6]Br3 — Correct. Co (Z = 27) is [Ar]3d74s2; \( \ce{Co^3+} \) is 3d6. \( \ce{NH3} \) pairs these electrons (inner orbital, d2sp3 complex), giving t2g6eg0: 0 unpaired, diamagnetic.
C. K3[Fe(CN)6] — Incorrect. Fe (Z = 26) is [Ar]3d64s2; \( \ce{Fe^3+} \) is 3d5. Even with strong-field \( \ce{CN-} \) (t2g5), one electron stays unpaired, so it is coloured.
D. [MnCl2(en)2]+ — Incorrect. The charge gives Mn +3 (x − 2 = +1). Mn (Z = 25) is [Ar]3d54s2, so \( \ce{Mn^3+} \) is 3d4: it has unpaired electrons (2 or 4, whatever the spin state) and is coloured.
E. [Ni(CO)4] — Correct. CO is neutral, so Ni is in the 0 state. Ni (Z = 28) is [Ar]3d84s2; CO pairs the electrons into 3d10, so there are 0 unpaired electrons and no d-d transition is possible.
NCERT (2026-27 edition):
Valence Bond Theory – “… In the diamagnetic octahedral complex, [Co(NH3)6]3+, the cobalt ion is in +3 oxidation state and has the electronic configuration 3d6. The hybridisation scheme is as shown in diagram. Six pairs of electrons, one from each NH3 molecule, occupy the six hybrid orbitals. Thus, the complex has octahedral geometry and is diamagnetic because of the absence of unpaired electron. In the formation of this complex, since the inner d orbital (3d) is used in hybridisation, the complex, [Co(NH3)6]3+ is called an inner orbital or low spin or spin paired complex.”