Question

c) What is the predicted magnetic moment of Cu+
and Cu2+ ions in B Unit (Atomic number
of Cu 29)?
7. a) Given that the spacing of the lines in the microwave spectrum of 24Al1H is constant at
12.604 cm1
. Calculate the moment of inertia and bond length of the molecule.

Answer 1

Answer on Question #76270 – Chemistry – Inorganic Chemistry

c) What is the predicted magnetic moment of $\mathrm{Cu^{+}}$ and $\mathrm{Cu^{2+}}$ ions in BM Unit (Atomic number of Cu 29)?

7. a) Given that the spacing of the lines in the microwave spectrum of $^{27}\mathrm{Al}^1\mathrm{H}$ is constant at $12.604~\mathrm{cm^{-1}}$. Calculate the moment of inertia and bond length of the molecule.

Solution:

c) Copper (Cu) atomic number = 29

The ground state electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10}$ or [Ar] $4s^{1}3d^{10}$

For $\mathrm{Cu^{+}}$ one electron is removed from 4s orbital of Cu, the electronic configuration becomes [Ar] $4s^{0}3d^{10}$. So, it has no unpaired electrons.

The formula to find the magnetic moment is $\sqrt{(n)(n + 2)}$, $n =$ unpaired electrons

$$\mu = \sqrt{(n)(n + 2)} = \sqrt{(0)(0 + 2)} = 0 \text{ BM}$$

For $\mathrm{Cu^{2+}}$ the electronic configuration becomes [Ar] $4s^{0}3d^{9}$, here in 3d orbital there is one unpaired electron. So, it has definitely some magnetic moment.

$$\mu = \sqrt{(n)(n + 2)} = \sqrt{(1)(1 + 2)} = \sqrt{3} \text{ BM}$$

7. a) If the spacing of lines is constant, the effects of centrifugal distortion are negligible. Hence we may use for the wavenumbers of the transitions

$$F(J) - F(J - 1) = 2BJ$$

Since $J = 1, 2, 3, \ldots$, the spacing of the lines is $2B$

$$12.604~\mathrm{cm^{-1}} = 2B,$$ $$B = 6.302~\mathrm{cm^{-1}} = 6.302 \times 10^{2}~\mathrm{m^{-1}}$$ $$I = \frac{\eta}{4\pi c B} = m_{\text{eff}} R^{2}$$ $$\frac{\eta}{4\pi c B} = \frac{1.0546 \times 10^{-34} J \cdot s}{(4\pi) \times (2.9979 \times 10^{8} m \cdot s^{-1})} = 2.7993 \times 10^{-44} kg \cdot m$$ $$I = \frac{2.7993 \times 10^{-44} kg \cdot m}{6.302 \times 10^{2} m^{-1}} = 4.442 \times 10^{-47} kg \cdot m^{2}$$ $$m_{\text{eff}} = \frac{m_{Al} m_{H}}{m_{Al} + m_{H}} = \left(\frac{(26.98) \times (1.008)}{(26.98) + (1.008)}\right) u \times \left(1.6605 \times 10^{-27} kg \cdot u^{-1}\right) = 1.6136 \times 10^{-27} kg$$ $$R = \left(\frac{I}{m_{\text{eff}}}\right)^{1/2} = \left(\frac{4.442 \times 10^{-47} kg \cdot m^{2}}{1.6136 \times 10^{-27} kg}\right)^{1/2} = 1.659 \times 10^{-10} m = 165.9~pm$$

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