question_answer

                Atomic number of Cr and Fe are respectively 24 and 26, which of the following is paramagnetic with the spin of electron:                                                                                                                                                       

A)                 \[[Cr{{(CO)}_{6}}]\]

B)                 \[[Fe{{(CO)}_{5}}]\]

C)                 \[{{[Fe{{(CN)}_{6}}]}^{4-}}\]

D)                 \[{{[Cr{{(N{{H}_{3}})}_{6}}]}^{3+}}\]

Correct Answer: D

Solution :

                Atoms, ions or molecules having unpaired electrons are paramagnetic. In \[[Cr(N{{H}_{3}})_{6}^{3+}\]Cr is present as Cr (III) or \[C{{r}^{3+}}\]                 So electronic configuration is             \[\underset{Ground\,state}{\mathop{_{24}Cr+}}\,1{{s}^{2}},\,2{{s}^{2}}\,2{{p}^{6}},\,3{{s}^{2}}\,3{{p}^{6}}\,3{{d}^{5}},\,4{{s}^{1}}\]                 \[C{{r}^{3+}}=1{{s}^{2}},\,2{{s}^{2}}2{{p}^{6}},\,3{{s}^{2}}3{{p}^{6}}3{{d}^{3}}\]                                 Number of unpaired electrons = 3                 \[In\,[Cr{{(CO)}_{6}}](O.\,N.\,of\,Cr=0)\]   \[\underset{(Ground\,state)}{\mathop{{{\,}_{24}}Cr}}\,=1{{s}^{2}},\,2{{s}^{2}}2{{p}^{6}},\,3{{s}^{2}}3{{p}^{6}}3{{d}^{5}},\,4{{s}^{1}}\]                                 Number of unpaired electron = 0                 \[In\,[Fe{{(CO)}_{5}}]\,\,(ON\,of\,Fe=0)\]   \[\,\underset{(Ground\,state)}{\mathop{_{26}Fe}}\,=1{{s}^{2}},\,2{{s}^{2}}\,2{{p}^{6}},\,3{{s}^{2}}\,3{{p}^{6}}\,3{{d}^{6}},4{{s}^{2}}\]                                 Number of unpaired electron = 0                 In \[{{[Fe{{(CN)}_{6}}]}^{4-}}\,(O\,No\,of\,Fe=+2)\] \[\underset{(Ground\,state)}{\mathop{F{{e}^{2+}}}}\,=1{{s}^{2}},\,2{{s}^{2}}\,2{{p}^{6}},\,3{{s}^{2}}\,3{{p}^{6}}\,3{{d}^{6}}\]                                 Number of unpaired electron = 0                 Hence, in above complex ion paramagnetic character is in \[{{[Cr{{(N{{H}_{3}})}_{6}}]}^{3+}}\] as it contains three unpaired electrons.