Question

Which has highest paramagnetism?

• A. ${\left[Cr{\left(H_{2}O\right)}_6\right]}^{3+}$
• B. ${\left[Fe{\left(H_{2}O\right)}_6\right]}^{2+}$
• C. ${\left[Cu{\left(H_{2}O\right)}_6\right]}^{2+}$
• D. ${\left[Zn{\left(H_{2}O\right)}_6\right]}^{2+}$

Answer 1

Answer: (B)

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

Paramagnetism is directly proportional to the number of unpaired electrons. The complex having the highest number of unpaired electrons will show the highest paramagnetism.

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

Oxidation state of $Cr$ is $+3$ (i.e. $x-0=+3$), hence $C{r}^{3+}$ have $3d^3$ electron in outermost orbit, so it have three unpaired electrons $(n=3)$.

$\left(B\right)$ $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$ oxidation state of $Fe$ is $+2$ (i.e. $x-0=+2$), hence $F{e}^{2+}$ have $3d^6$ electron in outermost orbit, so it have four unpaired electron $(n=4)$.

$\left(C\right)$ $\left[Cu\right(H_{2}O{)}_6{]}^{2+}$ oxidation state of $Cu$ is $+2$ (i.e. $x-0=+2$), hence $C{u}^{2+}$ have $3d^9$ electron in outermost orbit, so it have one unpaired electron $(n=1)$.

$\left(D\right)$ $\left[Zn\right(H_{2}O{)}_6{]}^{2+}$ oxidation state of $Zn$ is $+2$ (i.e. $x-0=+2$), hence $Z{n}^{2+}$ have $3d^{10}$ electron in outermost orbit, so all electrons are paired $(n=0)$.

So as explained above, the highest number of the unpaired electron is $(n=4)$ for $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$, hence it shows the highest paramagnetism.