Question

The spin only magnetic moment value of $\mathop {Cr(CO)}\nolimits_6 $
A.$2.84B.M.$
B.$4.90B.M.$
C.\[2.84B.M.\]
D.$0B.M.$

Answer 1

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 the electronic configuration of $\mathop {Cr(CO)}\nolimits_6 $ ion. The formula to find a magnetic moment is $ = \sqrt {n(n + 2)} $

Complete answer:
-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 $\mathop {He}\nolimits^ + ,\mathop {Li}\nolimits^{2 + } $ etc energy and size of the orbital depends only on ‘n’.
-The Azimuthal quantum number ‘l’ is known as orbital angular momentum. 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. The 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 states quantum numbers ms which can take the values of $ + \dfrac{1}{2}$ or $ - \dfrac{1}{2}$.
-These are called the two spin states of the electron and normally represented by two arrows, Y(spin up) and Y(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
$MagneticMoment = \sqrt {n(n + 2)} B.M.$
$\mathop {Cr(CO)}\nolimits_6 \Rightarrow \mathop {Cr}\nolimits^0 \Rightarrow [Ar]\mathop {4s}\nolimits^1 \mathop {3d}\nolimits^5 $
$CO = $ Strong field ligand
In this, all the electrons are paired due to the strong field ligand
‘n’ is the number of unpaired electrons
So, the no. of unpaired electrons $ = 0$
Hence, the magnetic moment of $\mathop {Cr(CO)}\nolimits_6 $ = $\sqrt {0(0 + 2)} B.M. = 0B.M.$
So, our correct answer is option D.

Note:
We should always remember the electronic configuration of $\mathop {Cr(CO)}\nolimits_6 $ is $\mathop {Cr(CO)}\nolimits_6 \Rightarrow \mathop {Cr}\nolimits^0 \Rightarrow [Ar]\mathop {4s}\nolimits^1 \mathop {3d}\nolimits^5 $ as half-filled subshells are more stable. An orbital cannot hold more than two electrons and these electrons should have opposite spins.
Always remember the formula of magnetic moment.
$MagneticMoment = \sqrt {n(n + 2)} B.M.$