
Which is more paramagnetic \[\;F{e^{ + 2}}\] or $F{e^{ + 3}}$?
Answer
160.8k+ views
Hint: Iron (Fe) is a d block element having $26$atomic number and $56$ mass number. Iron has an electronic configuration [Ar] \[3{d^6}4{s^2}\] written in accordance with the Aufbau Principle. The d configuration of \[\;F{e^{ + 2}}\]and $F{e^{ + 3}}$ is \[{d^6}\] and \[{d^5}\] respectively.
Complete Step by Step Solution:
Para refers to unpaired electrons and magnetic refers to magnetic properties of an electron. Thus, the substance is said to have paramagnetic properties if it has an unpaired number of electrons. More the no. of unpaired electrons more will be the paramagnetic nature. In other words, Para magnetism is directly proportional to the unpaired number of electrons.
The no. of unpaired electron in \[\;F{e^{ + 2}}\] is \[4\] unpaired electrons and in $F{e^{ + 3}}$ there are \[5\]unpaired electrons. As the no. of unpaired electron is more in case of $F{e^{ + 3}}$ than \[\;F{e^{ + 2}}\] hence $F{e^{ + 3}}$is more paramagnetic than\[\;F{e^{ + 2}}\].
Note: The filling of electrons in d orbitals must follow Hund's Rule of maximum multiplicity which states that first electrons are singly filled then only pairing takes place. If Hund's rule is not followed then the no. of unpaired electrons in \[\;F{e^{ + 2}}\] is zero and that in $F{e^{ + 3}}$ is \[1\] which is wrong. The concern is only for d electrons as rest electrons present in other subshells are already paired.
Complete Step by Step Solution:
Para refers to unpaired electrons and magnetic refers to magnetic properties of an electron. Thus, the substance is said to have paramagnetic properties if it has an unpaired number of electrons. More the no. of unpaired electrons more will be the paramagnetic nature. In other words, Para magnetism is directly proportional to the unpaired number of electrons.
The no. of unpaired electron in \[\;F{e^{ + 2}}\] is \[4\] unpaired electrons and in $F{e^{ + 3}}$ there are \[5\]unpaired electrons. As the no. of unpaired electron is more in case of $F{e^{ + 3}}$ than \[\;F{e^{ + 2}}\] hence $F{e^{ + 3}}$is more paramagnetic than\[\;F{e^{ + 2}}\].
Note: The filling of electrons in d orbitals must follow Hund's Rule of maximum multiplicity which states that first electrons are singly filled then only pairing takes place. If Hund's rule is not followed then the no. of unpaired electrons in \[\;F{e^{ + 2}}\] is zero and that in $F{e^{ + 3}}$ is \[1\] which is wrong. The concern is only for d electrons as rest electrons present in other subshells are already paired.
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