Question

Magnetic moment of $\left[Ti\right(H_{2}O{)}_6{]}^{4+},\left[Mn\right(H_{2}O{)}_6{]}^{2+}$ and $\left[Cr\right(H_{2}O{)}_6{]}^{3+}$ can be represented as $X,Y$ and $Z$. They are in order of:

• A. $X<Z<Y$
• B. $Z<Y<X$
• C. $X<Y<Z$
• D. $Z<Y>X$

Answer 1

The correct option is A $X<Z<Y$

In $\left[Ti\right(H_{2}O{)}_6{]}^{4+}$,

oxidation state of Ti = +4

electronic configuration = $d^0$

number of unpaierd electron = 0

In $\left[Mn\right(H_{2}O{)}_6{]}^{2+}$

oxidation state of Mn = +2

electronic configuration = $d^5$

number of unpaired electron = 5

In $\left[Cr\right(H_{2}O{)}_6{]}^{3+}$

oxidation state of Cr = +3

electronic configuration = $d^3$

number of unpaierd electron = 3

Magnetic moment is directly proportional to the number of unpaired electrons.

Hence, the of magnetic moment becomes $X<Z<Y$