Abstract

The results of investigation of the ${\mathrm{Cr}}_{x}\mathrm{Nb}{\mathrm{Se}}_{2}$ chalcogenides $(x=0.33,0.5)$ by the nuclear magnetic resonance (NMR) method on $^{53}\mathrm{Cr}$ and $^{93}\mathrm{Nb}$ nuclei in the magnetically ordered state at zero external magnetic field have been presented. Ab initio calculations have been performed to theoretically estimate NMR parameters and interpret the experimental data. It has been shown that the intercalation of Cr atoms into $\mathrm{Nb}{\mathrm{Se}}_{2}$ results in spin and charge redistributions. The lower magnetic moment of chromium nuclei, $2.2\phantom{\rule{0.16em}{0ex}}{\ensuremath{\mu}}_{\mathrm{B}}$, compared to the theoretical value $\ensuremath{\mu}=3\phantom{\rule{0.16em}{0ex}}{\ensuremath{\mu}}_{\mathrm{B}}$ for ${\mathrm{Cr}}^{3+}$ is attributed to the high degree of hybridization of the ${a}_{1g}$ and ${e}_{g}$ orbitals of $3d$ Cr electrons with $4{d}_{{z}^{2}}$ and $5s$ niobium orbitals. Such a hybridization also results in the presence of a high local magnetic field in the niobium nuclei location in ${\mathrm{Cr}}_{x}\mathrm{Nb}{\mathrm{Se}}_{2}$.

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