Identify the geometries

• For Ni complex: ideal square pyramidal geometry.
• For Fe complex: ideal trigonal bipyramidal geometry.

We must count all \angle L\!-M\!-L angles equal to $90^\circ$, $120^\circ$, and $180^\circ$ in both complexes, then add them.

Square pyramidal geometry $(ML_5)$

Structure:

• 4 ligands in a square plane
• 1 ligand at the apex

Let the four basal ligands be $A,B,C,D$ and apex be $E$.

Total number of ligand pairs is: $\binom{5}{2}=10$

Now count the required angles:

(a) $90^\circ$ angles

• Adjacent ligands in the square plane: $4$
• Apex with each basal ligand: $4$

So total $90^\circ$ angles: $4+4=8$

(b) $180^\circ$ angles

• Opposite ligands in the square plane: $2$

So total $180^\circ$ angles: $2$

(c) $120^\circ$ angles

• None in square pyramidal geometry.

So for square pyramidal: $N_{90}=8,\quad N_{120}=0,\quad N_{180}=2$ Total: $8+0+2=10$

Trigonal bipyramidal geometry $(ML_5)$

Structure:

• 3 equatorial ligands
• 2 axial ligands

Total ligand pairs: $\binom{5}{2}=10$

Now count the required angles:

(a) $120^\circ$ angles

• Between equatorial ligands: each pair gives $120^\circ$
• Number of such pairs among 3 equatorial ligands: $\binom{3}{2}=3$

So: $N_{120}=3$

(b) $90^\circ$ angles

• Each axial ligand makes $90^\circ$ with each equatorial ligand
• Number: $2\times 3=6$

So: $N_{90}=6$

(c) $180^\circ$ angles

• Between the two axial ligands: $1$

So: $N_{180}=1$

Total for trigonal bipyramidal: $6+3+1=10$

Sum for both complexes

$10+10=20$

Final Answer

The required sum of the $90^\circ$, $120^\circ$, and $180^\circ$ angles in the two complexes is: $\boxed{20}$

Comparison with stored correct answer

Stored correct answer = $20$

Our derived answer = $20$

Hence, they agree.