A) \[[Cr\text{ }{{(CO)}_{6}}]\]
B) \[[Fe\text{ }{{(CO)}_{5}}]\]
C) \[{{[Fe\text{ }{{(CN)}_{6}}]}^{4\,-}}\]
D) \[{{[Cr{{(N{{H}_{3}})}_{6}}]}^{3+}}\]
Correct Answer: D
Solution :
| [d] 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 |
| \[\ln \,[Cr{{(CO)}_{6}}](O.\,N.\,of\,Cr=0)\] |
| \[\underset{(Ground\,state)}{\mathop{{{\,}_{24}}Cr}}\,=1{{s}^{2}},\,2{{s}^{2}}2{{p}^{6}},\,3{{s}^{2}}3{{p}^{6}}3{{d}^{5}},\,4{{s}^{1}}\] |
 |
| Number of unpaired electron = 0 |
| \[\ln \,[Fe{{(CO)}_{5}}]\,\,(ON\,of\,Fe=0)\] |
| \[\,\underset{(Ground\,state)}{\mathop{_{26}Fe}}\,=1{{s}^{2}},\,2{{s}^{2}}\,2{{p}^{6}},\,3{{s}^{2}}\,3{{p}^{6}}\,3{{d}^{6}},4{{s}^{2}}\] |
 |
| 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. |
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