
Magnetic moment \[2.84BM\] is given by
Answer
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Hint:Magnetic moment is the strength and orientation of a magnetic field . The SI unit of magnetic moment is $J/Tesla$. The magnetic moment is a vector quantity which is defined as $\mu = iA$. Whose direction is perpendicular to $A$ and is determined by right hand rule.
Complete step by step answer:
Magnetic moment is calculated by $\mu = \sqrt {n(n + 2)} BM$
Where
\[n\] is numbered of unpaired electrons
$\mu $ which is given $2.84BM$
Now putting the values
$ \Rightarrow $$2.84 = \sqrt {n(n + 2)} BM$
$ \Rightarrow $${(2.84)^2} = n(n + 2)$
$ \Rightarrow $$8 = {n^2} + 2n$
$ \Rightarrow $${n^2} + 2n - 8 = 0$
$ \Rightarrow $${n^2} + 4n - 2n - 8 = 0$
$ \Rightarrow $$n(n + 4) - 2(n + 4) = 0$
$ \Rightarrow $$n = 2$
Now electronic configuration of each elements given are
\[N{i^{2 + }} = \left[ {Ar} \right]3{d^8}4{s^0}\] It has two unpaired electrons
$T{i^{3 + }} = \left[ {Ar} \right]3{d^1}4{s^0}$ It has one unpaired electron
$C{r^{3 + }} = \left[ {Ar} \right]3{d^3}$ It has three unpaired electrons
$C{o^{2 + }} = \left[ {Ar} \right]3{d^7}4{s^0}$ it has three unpaired electrons
So, only $N{i^{2 + }}$ has two unpaired electrons.
Additional information:
Magnetic moment is sometimes referred to as a magnetic dipole moment. The magnetic south and north poles are separated by a small distance. We can say that the magnetic dipole moment of an object experienced torque with larger magnetic moments. The strength of the torque depends upon the magnitude of the magnetic moment but also in its orientation relative to direction of the magnetic field. Magnetic moment is a vector quantity. Any system having magnetic dipole produces a dipolar magnetic field.
Note:
We can find the number of unpaired electrons present by writing the electronic configuration. Similarly when we wrote the electronic configuration of nickel, we saw that in the d shell two electrons were unpaired when we fill the orbitals according to the hund's rule . hund's rule said that every orbitals must be singly occupied first before it is doubly occupied . The singly occupied electrons should have the same spin.
Complete step by step answer:
Magnetic moment is calculated by $\mu = \sqrt {n(n + 2)} BM$
Where
\[n\] is numbered of unpaired electrons
$\mu $ which is given $2.84BM$
Now putting the values
$ \Rightarrow $$2.84 = \sqrt {n(n + 2)} BM$
$ \Rightarrow $${(2.84)^2} = n(n + 2)$
$ \Rightarrow $$8 = {n^2} + 2n$
$ \Rightarrow $${n^2} + 2n - 8 = 0$
$ \Rightarrow $${n^2} + 4n - 2n - 8 = 0$
$ \Rightarrow $$n(n + 4) - 2(n + 4) = 0$
$ \Rightarrow $$n = 2$
Now electronic configuration of each elements given are
\[N{i^{2 + }} = \left[ {Ar} \right]3{d^8}4{s^0}\] It has two unpaired electrons
$T{i^{3 + }} = \left[ {Ar} \right]3{d^1}4{s^0}$ It has one unpaired electron
$C{r^{3 + }} = \left[ {Ar} \right]3{d^3}$ It has three unpaired electrons
$C{o^{2 + }} = \left[ {Ar} \right]3{d^7}4{s^0}$ it has three unpaired electrons
So, only $N{i^{2 + }}$ has two unpaired electrons.
Additional information:
Magnetic moment is sometimes referred to as a magnetic dipole moment. The magnetic south and north poles are separated by a small distance. We can say that the magnetic dipole moment of an object experienced torque with larger magnetic moments. The strength of the torque depends upon the magnitude of the magnetic moment but also in its orientation relative to direction of the magnetic field. Magnetic moment is a vector quantity. Any system having magnetic dipole produces a dipolar magnetic field.
Note:
We can find the number of unpaired electrons present by writing the electronic configuration. Similarly when we wrote the electronic configuration of nickel, we saw that in the d shell two electrons were unpaired when we fill the orbitals according to the hund's rule . hund's rule said that every orbitals must be singly occupied first before it is doubly occupied . The singly occupied electrons should have the same spin.
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