question_answer

The "spin-only" magnetic moment \[[in\text{ }units\text{ }of\text{ }Bohr\text{ }magneton,\,({{\mu }_{B}})]\]of \[N{{i}^{2+}}\] in aqueous solution would be: \[(At.\text{ }no.\text{ }Ni=28)\]

A) 0                     

B) 1.73   

C) 2.84                 

D) 4.90

Correct Answer: B

Solution :

 [c] 2.84 Electronic configuration of \[N{{i}^{2+}}\]is It has two unpaired electrons.             Spin only magnetic moment \[=\sqrt{n(n+2)}\] \[=\sqrt{2(2+2)}\] \[=2.83\,B.M\]