Question

paramagnetic species among the following is na+ , mg2+,f-,p?​

Answer 1

Answer:

That's all I searched

Explanation:

Na = it has one unpaired electron as its electronic configuration is [Ne]3s

1

, so it is paramagnetic in nature.

Mg= Mg has all electrons paired so it is diamagnetic in nature.

Ca

2+

= after losing 2 electrons it behaves as inert gas so diamagnetic in nature.

Cl

2−

anion =after accepting 2 electrons it behaves as alkali metal hence, paramagnetic in nature.

Fe, Co, Ni= these are ferromagnetic in nature as they show permanent magnetism even once they are in a magnetic field.

Answer 2

To find:

Paramagnetic Species among the following:

• Na+
• Mg+2
• F-
• P

Solution:

Paramagnetic Species are those atoms, ions or radicals which have unpaired electrons.

$\therefore {Na}^{ + } \bigg|_{10} \implies \: 1 {s}^{2} 2 {s}^{2} 2{p}^{6}$

So, Na+ is diamagnetic because all its electrons are paired up.

$\therefore {Mg}^{ +2 } \bigg|_{10} \implies \: 1 {s}^{2} 2 {s}^{2} 2{p}^{6}$

So, Mg+2 is also diamagnetic because all its electrons are paired up.

$\therefore {F}^{ - } \bigg|_{10} \implies \: 1 {s}^{2} 2 {s}^{2} 2{p}^{6}$

So, F- is also diamagnetic because all its electrons are paired up.

$\therefore P\bigg|_{15} \implies \: 1 {s}^{2} 2 {s}^{2} 2{p}^{6} 3{s}^{2} 3 {p}^{3}$

So, P is paramagnetic as it contains 3p³ orbital with 3 unpaired electrons.

$\star$ Hope It Helps.