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# In which of the following will the ionic radii of chromium be the smallest

$(a)\;K_2Cr_2O_4\qquad(b)\;CrCl_3\qquad(c)\;CrO_2\qquad(d)\;CrF_2$

Oxidation number of Cr in options a, b, c and d are +6, + 4, + 3, + 3 respectively. In given options $K_2Cr_2O_4$ has high oxidation number therefore its radii will be low. Atomic radii decreases with increase in oxidation no.
edited Aug 2, 2014