JEE MAIN - Chemistry (2022 - 25th July Morning Shift - No. 16)
Consider the following metal complexes :
$$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$$
$$\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}$$
$$\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}$$
$$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{3+}$$
The spin-only magnetic moment value of the complex that absorbes light with shortest wavelength is _____________ B. M. (Nearest integer)
Answer
0
Explanation
In all complexes, Co is present in $+3$ oxidation state and all complexes are low spin or inner orbital complex.
The stronger the ligand, the higher the crystal field splitting.
So, the order of crystal field splitting is
$\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}>\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}>\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{3+}>$ $\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}$
Shortest wavelength is shown by complex having maximum crystal field splitting.
_25th_July_Morning_Shift_en_16_1.png)
Spin only magnetic moment $=\sqrt{0(0+2)}=0$ B.M
The stronger the ligand, the higher the crystal field splitting.
So, the order of crystal field splitting is
$\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}>\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}>\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{3+}>$ $\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}$
Shortest wavelength is shown by complex having maximum crystal field splitting.
Spin only magnetic moment $=\sqrt{0(0+2)}=0$ B.M
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