question_answer

How many unpaired electrons are present in\[[Cr{{(N{{H}_{3}})}_{5}}]B{{r}_{3}}\]?

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\[[Cr(N{{H}_{3}})B{{r}_{3}}]\].