Question

1.) Identify the most likely transition metal, M and provide a suitable explanation:
a.) A complex ion with the formula [M(H₂O)₆]³⁺, where M is a second row transition metal, that has a LFSE (CFSE) of -2.4 _{$\Delta _{o}$} (ignoring pairing and exchange energies). Hint: only the low spin configuration is observed in second row transition metals.

b.) A tetrahedral complex ion with the formula [MCl₄]^-, where M is a first row transtion metal, that has five unpaired electrons.

Answer 1

1.a) Its given that its a second row transition metal and has low spin. The CFSE is given as $-2.4\Delta_0$ . Its an octahedral complex. The formula for CFSE

$CFSE=n0.6\Delta_o-m0.4\Delta_o$

n- Number of electrons in $e_g$

m- Number of electrons in $t_{2g}$

So for the CFSE to be the given value n should be zero and m six.

So the outer electronic configuration will be

$M^{+3}$ - $4d^6$

$M-4d^75s^2$

The metal with this outer electronic configuration is Rhodium.

b.) Its a tetrahedral complex and is high spin as there are 5 unpaired electrons. The oxidation state of M is +3. The outer electronic configuration is

$M^{+3}-3d^5$

$M-3d^64s^2$

The metal with this outer configuration is Iron (Fe).