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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