
Is $V^{3+}$ paramagnetic or diamagnetic?
Answer
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Hint: Periodic table consists of different atoms arranged in the increasing order of atomic numbers. These atoms turned into ions by losing or gaining electrons. The presence of unpaired electrons makes the ion to be paramagnetic. The absence of unpaired electrons makes the ion to be diamagnetic.
Complete answer:
Periodic table is the representation of elements arranged in vertical columns and horizontal rows. The vertical columns are called as groups and the horizontal rows are called as periods. There are $ 7 $ periods and $ 18 $ groups. There are a total of $ 118 $ elements in the periodic table.
The periodic table is divided into four blocks namely s, p, d and f. The d-block consists of transition metals and is divided into three series namely $ 3d $ , $ 4d $ and $ 5d $ series.
Vanadium is the element with atomic number $ 23 $ and has five valence electrons and vanadium loses three electrons and becomes $ {V^{ + 3}} $ ion.
Thus, vanadium ion $ \left( {{V^{ + 3}}} \right) $ has two unpaired electrons and leads to the paramagnetic nature.
If an ion has unpaired electrons, it can be called a paramagnetic substance.
If an ion does not have any unpaired electrons, it can be called as diamagnetic substances.
Thus, due to the presence of two unpaired electrons in vanadium ion, it is paramagnetic.
Note:
The paramagnetic substances align in the direction of the applied magnetic field due to the possessing of magnetic properties, caused by the presence of unpaired electrons. Diamagnetic substances do not align in the direction of the applied magnetic field due to the absence of unpaired electrons.
Complete answer:
Periodic table is the representation of elements arranged in vertical columns and horizontal rows. The vertical columns are called as groups and the horizontal rows are called as periods. There are $ 7 $ periods and $ 18 $ groups. There are a total of $ 118 $ elements in the periodic table.
The periodic table is divided into four blocks namely s, p, d and f. The d-block consists of transition metals and is divided into three series namely $ 3d $ , $ 4d $ and $ 5d $ series.
Vanadium is the element with atomic number $ 23 $ and has five valence electrons and vanadium loses three electrons and becomes $ {V^{ + 3}} $ ion.
Thus, vanadium ion $ \left( {{V^{ + 3}}} \right) $ has two unpaired electrons and leads to the paramagnetic nature.
If an ion has unpaired electrons, it can be called a paramagnetic substance.
If an ion does not have any unpaired electrons, it can be called as diamagnetic substances.
Thus, due to the presence of two unpaired electrons in vanadium ion, it is paramagnetic.
Note:
The paramagnetic substances align in the direction of the applied magnetic field due to the possessing of magnetic properties, caused by the presence of unpaired electrons. Diamagnetic substances do not align in the direction of the applied magnetic field due to the absence of unpaired electrons.
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