Question

Identify the correct trend given below :
(Atomic No. = Ti : 22, Cr : 24 and Mo : 42)

• A. \(\Delta_o\) of \([Cr(H_2O)_6]^{2+} \gt [Mo(H_2O)_6]^{2+}\) and \(\Delta_o\) of \([Ti(H_2O)_6]^{3+} \gt [Ti(H_2O)_6]^{2+}\)
• B. \(\Delta_o\) of \([Cr(H_2O)_6]^{2+} \gt [Mo(H_2O)_6]^{2+}\) and \(\Delta_o\) of \([Ti(H_2O)_6]^{3+} \lt [Ti(H_2O)_6]^{2+}\)
• C. \(\Delta_o\) of \([Cr(H_2O)_6]^{2+} \lt [Mo(H_2O)_6]^{2+}\) and \(\Delta_o\) of \([Ti(H_2O)_6]^{3+} \gt [Ti(H_2O)_6]^{2+}\)
• D. \(\Delta_o\) of \([Cr(H_2O)_6]^{2+} \lt [Mo(H_2O)_6]^{2+}\) and \(\Delta_o\) of \([Ti(H_2O)_6]^{3+} \lt [Ti(H_2O)_6]^{2+}\)

Hint:

Crystal field splitting increases down a group (3d < 4d < 5d) and with higher oxidation state of the metal.

Answer 1

The Correct Option is C

Both Cr and Mo are in group 6 and both complexes have the metal in the +2 state, so the only difference is the principal quantum number of the d orbitals.
The 4d orbitals of Mo are larger and overlap better with the ligands than 3d orbitals of Cr, so \(\Delta_o\) of \([Mo(H_2O)_6]^{2+}\) is greater: \(\Delta_o(Cr^{2+}) \lt \Delta_o(Mo^{2+})\).
For the same metal and ligand, a higher oxidation state gives a larger charge/size ratio and stronger attraction of the ligands, so \(\Delta_o\) of \([Ti(H_2O)_6]^{3+}\) is greater than that of \([Ti(H_2O)_6]^{2+}\).
Hence the correct trend is option (3).
Answer: (3) \(\Delta_o\) of \([Cr(H_2O)_6]^{2+} \lt [Mo(H_2O)_6]^{2+}\) and \(\Delta_o\) of \([Ti(H_2O)_6]^{3+} \gt [Ti(H_2O)_6]^{2+}\)