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 \[=\sqrt{n(n+2)}\] \[=\sqrt{2(2+2)}\] \[=2.83\,B.M\]