A) 3 and 6
B) 2 and 4
C) 4 and 2
D) 4 and 4
Correct Answer: B
Solution :
\[[\underset{x}{\mathop{Ni}}\,\,\,{{(\underset{0}{\mathop{N{{H}_{3}}}}\,)}_{4}}]\underset{-2}{\mathop{S{{O}_{4}}}}\,\] \[x+0+(-2)=0\Rightarrow x=+\,2\]is valency and 4 is C.N. of Ni.You need to login to perform this action.
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