The reaction of $\mathrm{NiBr}_{2}$ with two equivalents of $\mathrm{PPh}_{3}$ in $\mathrm{CS}_{2}$ at $-78^{\circ} \mathrm{C}$ gives a red-colored diamagnetic complex. $\left[\mathrm{NiBr}_{2}\left(\mathrm{PPh}_{3}\right)_{2}\right]$. This transforms to a green-colored paramagnetic complex with the same molecular formula at $25^{\circ} \mathrm{C}$. The geometry and the number of unpaired electrons in the green-colored complex, respectively, are:
(a)square planar and 2
(b)tetrahedral and 2
(c)square planar and 4
(d)tetrahedral and 1
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
Green paramagnetic [NiBr₂(PPh₃)₂] is tetrahedral d⁸ with two unpaired electrons (the red diamagnetic isomer is square planar) (b).