JEE MAIN - Chemistry (2016 - 9th April Morning Slot - No. 11)
Identify the correct trend given below : (Atomic No.=Ti : 22, Cr : 24 and Mo : 42)
$$\Delta $$o of [Cr(H2O)6]2+ >
[Mo(H2O)6]2+ and
$$\Delta $$o of [Ti(H2O)6]3+ > [Ti(H2O)6]2+
[Mo(H2O)6]2+ and
$$\Delta $$o of [Ti(H2O)6]3+ > [Ti(H2O)6]2+
$$\Delta $$o of [Cr(H2O)6]2+ >
[Mo(H2O)6]2+ and
$$\Delta $$o of [Ti(H2O)6]3+ < [Ti(H2O)6]2+
[Mo(H2O)6]2+ and
$$\Delta $$o of [Ti(H2O)6]3+ < [Ti(H2O)6]2+
$$\Delta $$o of [Cr(H2O)6]2+
< [Mo(H2O)6]2+ and
$$\Delta $$o of [Ti(H2O)6]3+ > [Ti(H2O)6]2+
< [Mo(H2O)6]2+ and
$$\Delta $$o of [Ti(H2O)6]3+ > [Ti(H2O)6]2+
$$\Delta $$o of [Cr(H2O)6]2+
< [Mo(H2O)6]2+ and
$$\Delta $$o of [Ti(H2O)6]3+ < [Ti(H2O)6]2+
< [Mo(H2O)6]2+ and
$$\Delta $$o of [Ti(H2O)6]3+ < [Ti(H2O)6]2+
Explanation
$$\Delta$$0 of complex [Mo(H2O)6]2+ is greater than that of [Cr(H2O)6]2+. It is due to the fact that Mo2+ belongs to second row transition element while Cr2+ is first row transition element.
Magnitude of crystal field splitting energy ($$\Delta$$0) depends on the charge on the metal ion. Greater the charge, greater is the crystal field splitting energy. Therefore, [Ti(H2O)6]3+ > [Ti(H2O)6]2+.
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