Answer to Question #276692 in General Chemistry for Hutchii

Question

Provide the electron configuration for each element then determine the magnetic property of each.

1. Cl
2. Cu2+
3. Na+
4. O2-
5. Gd

Answer 1

Cl $[Ne]\>3s^2\>3p^5$

$1s^2\>2s^2\>2p^6\>3s^2\>3p^5$

There is one unpaired electron in p orbital.

Cl atoms are paramagnetic although weak.

Cu²⁺

$1s^2\>2s^2\>2p^6\>3s^2\>3p^6\>3d^9$

There is one unpaired electron in its outermost energy level.

So Cu²⁺ compounds are paramagnetic in nature.

$Na^+$

$1s^2\>2s^2\>2p^6\>$

$Na^+$ is diamagnetic, it has 10 electrons which are paired.

$O^{2-}$

$1s^2\>2s^2\>2p^6$

It is diamagnetic. All electrons are paired.

$Gd$

$[Xe]\>4f^7\>5d^1\>6s^2$

It has 8 unpaired electrons (4f⁷ and 5d¹ )

And its sum of spin $=8×\frac{1}{2}=4$

Therefore it is ferromagnetic material.