NEET 2024, Chemistry Q31: Colour, Magnetism And Stability Of Co-ordination Compounds

Count unpaired electrons in each $d$ configuration: Ti${}^{3+}$ ($d^1$, 1), Cr${}^{2+}$ ($d^4$, 4), Mn${}^{2+}$ ($d^5$, 5), Fe${}^{2+}$ ($d^6$, 4), Sc${}^{3+}$ ($d^0$, 0).

$\mu =\sqrt{n(n+2)}$ BM, so Cr${}^{2+}$ and Fe${}^{2+}$ both have 4 unpaired electrons and hence the same spin-only moment.