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# The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of $C{{r}^{3+}}$ ion is:A. 2.87 B.M.B. 3.87 B.M.C. 3.47 B.M.D. 3.57 B.M.

Last updated date: 20th Jun 2024
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Hint: To solve this question you should have basic knowledge about electronic configuration. Electronic configuration, also called electronic structure, the arrangement of electrons in energy levels around an atomic nucleus. You should know electronic configuration of $C{{r}^{3+}}$ ion. The formula to find a magnetic moment is $=\sqrt{n(n+2)}$.

The Principal quantum number ‘n’ is a positive integer with value of n = 1, 2, 3……. The principal quantum number determines the size and to large extent the energy of the orbital. For hydrogen atom and hydrogen-like species $H{{e}^{+}},L{{i}^{2+}},.....etc$ energy and size of the orbital depends only on ‘n’.
Azimuthal quantum number ‘l’ is known as orbital angular momentum or subsidiary quantum number. It defines that three-dimensional shape of the orbital. For a given value of n, l can have n values ranging from 0 to n-1. It should be noted that for a given value of n, the possible value of l is: l = 0, 1, 2, ………… (n-1).
Spin quantum number ‘s’: an electron spins around its own axis, much in a similar way as earth spins around its own axis while revolving around the sun. Spin angular momentum of the electron is a vector quantity and it generally has two orientations relative to the chosen axis. These orientations are distinguished by the spin quantum numbers ms which can take the values of +1/2 or -1/2. These are called the two spin states of the electron and normally represented by two arrows, γ (spin up) and γ (spin down).
To find the magnetic moment value of any ion, firstly we have to write its electronic configuration then find the number of unpaired electrons. Then, use this formula:
$Magnetic Moment=\sqrt{n(n+2)}$B.M.
Where n is equal to the number of unpaired electrons.
As in question magnetic moment of $C{{r}^{3+}}$ ion is asked,
Electronic configuration of Cr is $[Ar]3{{d}^{5}}4{{s}^{1}}$
Hence, electronic configuration of $C{{r}^{3+}}$ ion is $[Ar]3{{d}^{3}}4{{s}^{0}}$.
Hence, the number of unpaired electrons is 3.
Hence, Magnetic moment of $C{{r}^{3+}}$ ion $=\sqrt{3(3+2)}$ =$\sqrt{15}$ = 3.87 B.M.

Hence, option B is the right answer.

Note:
Always remember electronic configuration of Cr is $[Ar]3{{d}^{5}}4{{s}^{1}}$ not $[Ar]3{{d}^{4}}4{{s}^{2}}$ as half-filled subshells are more stable.
An orbital cannot hold more than two electrons and these two electrons should have opposite spins.
Always remember the formula of magnetic moment, $Magnetic Moment=\sqrt{n(n+2)}$B.M.