Question

the magnetic moment of cobalt in hg[co(scn4)]

Answer 1

Coordination entity is [Co(SCN)4]^--

Let oxidation no of Co be x, therefore,

x+4(-1)=-2==»x=2

Co^2+==[Ar]3d7,i.e, 3 unpaired electrons.

Magnetic moment={3(3+2)}sqrt=sqrt15

Answer 2

Answer: 3.87 BM

Explanation: The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

The given coordination sphere is, $[Co(SCN)_4]^{-2}$

Let the oxidation state of Co be, 'x'

$x+4(-1)=-2\\\\x=+2$

Hence, the oxidation state of Co is, (+2)

The electronic configuration of Co is, $1s^22s^22p^63s^23p^64s^23d^7$

$Co^{2+}:1s^22s^22p^63s^23p^63d^7$

The number of unpaired electrons in Co= 3Formula used for magnetic moment :

$\mu=\sqrt{n(n+2)}$

where,

$\mu$ = magnetic moment

n = number of unpaired electrons

Now put all the given values in the above formula, we get the magnetic moment.

$\mu=\sqrt{3(3+2)}=\sqrt{15}=3.87BM$

Therefore, the magnetic moment is 3.87 BM.