A) 1
B) 2
C) 3
D) 4
Correct Answer: C
Solution :
\[[Cr{{(N{{H}_{3}})}_{5}}]B{{r}_{3}}\] Suppose oxidation state of Cr is \[x,\]then. \[x+(5\times 0)+(-1\,\times \,3)=0\] \[x-3=0\] \[x=+\,3\] \[{{\,}_{24}}Cr=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{5}},4{{s}^{1}}\] \[C{{r}^{3+}}=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{3}}\] So, three unpaired electrons are present in \[[Cr(N{{H}_{3}})]B{{r}_{3}}.\]You need to login to perform this action.
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